Instructor's Solutions Manual for Engineering Mechanics Statics, SI Edition, 14E [14E in SI ed.] 1292089261, 9781292089263

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1–1. Evaluate each of the following and express with an appropriate prefix: (a) (430 kg)2, (b) (0.002 mg)2, and (c) (230 m)3.

Solution a) (430 kg)2 = 0.185(106) kg2 = 0.185 Mg2 Ans. b) (0.002 mg)2 = [2(10 - 6) g]2 = 4 mg2 Ans. c) (230 m)3 = [0.23(103) m]3 = 0.0122 km3

Ans.

Ans: 0.185 Mg2 4 mg2 0.0122 km3 1

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1–2. Represent each of the following combinations of units in the correct SI form: (a) Mg>ms, (b) N>mm, (c) mN>(kg # ms).

Solution (a)

Mg ms

=

103 kg 10-3 s

= 106 kg>s = Gg>s

Ans.

(b)

N 1N = = 103 N>m = kN>m mm 10-3 m

Ans.

(c)

mN 10-3 N = = kN>(kg # s) (kg # ms) 10-6 kg # s

Ans.

Ans: Gg>s kN>m kN>(kg # s) 2

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1–3. What is the weight in newtons of an object that has a mass of (a) 8 kg, (b) 0.04 kg, and (c) 760 Mg?

Solution Ans.

(a)

W = 9.81(8) = 78.5 N

(b)

W = 9.81(0.04) ( 10 - 3 ) = 3.92 ( 10 - 4 ) N = 0.392 mN

Ans.

(c)

W = 9.81(760) ( 10

Ans.

3

) = 7.46 ( 10 ) N = 7.46 MN 6

Ans: W = 78.5 N W = 0.392 mN W = 7.46 MN 3

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*1–4. Represent each of the following combinations of units in the correct SI form: (a) KN >ms, (b) Mg >mN, and (c) MN>(kg # ms).

Solution (a)

kN>ms = 103N> ( 10 - 6 ) s = GN>s

(b)

Mg>mN = 106g > (10- 3)N = Gg>N

(c)

MN>(kg # ms)

6

= 10 N>kg(10

-3

)s =

Ans. Ans. GN>(kg # s)

Ans.

Ans: GN>s Gg>N GN>(kg # s) 4

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1–5.

Represent each of the following quantities in the correct SI form using an appropriate prefix: (a) 0.000 431 kg, (b) 35.3(103) N, and (c) 0.005 32 km.

SOLUTION a) 0.000 431 kg = 0.000 431 A 103 B g = 0.431 g

Ans.

b) 35.3 A 103 B N = 35.3 kN

Ans.

c) 0.005 32 km = 0.005 32 A 103 B m = 5.32 m

Ans.

Ans: 0.431 g 35.3 kN 5.32 m 5

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1–6.

Represent each of the following combinations of units in the correct SI form using an appropriate prefix: (a) m>ms, (b) mkm, (c) ks>mg, and (d) km # mN.

SOLUTION a) m>ms = ¢

11023 m m = ≤ ¢ ≤ = km>s s 1102-3 s

Ans.

b) mkm = 1102-611023 m = 1102-3 m = mm 3

c) ks>mg =

1102 s 1102

d) km # mN =

-6

kg

3

m

10

Ans.

9

= 10

1102 s kg -6

= Gs>kg

N = 10

-3

Ans.

m # N = mm # N

Ans.

Ans: km>s mm Gs>kg mm # N 6

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1–7. Represent each of the following as a number between 0.1 and 1000 using an appropriate prefix: (a) 45 320 kN, (b) 568(105) mm, and (c) 0.00563 mg.

Solution (a)

45 320 kN = 45.3 MN

Ans.

(b)

568 ( 10

Ans.

(c)

0.00563 mg = 5.63 mg

5

) mm = 56.8 km

Ans.

Ans: 45.3 MN 56.8 km 5.63 mg 7

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*1–8. Represent each of the following combinations of units in the correct SI form: (a) GN # mm, (b) kg>mm, (c) N>ks2, and  (d) KN>ms.

Solution (a)

GN # mm = 109 ( 10-6 ) N # m = kN # m 3

-6

Ans. Ans.

(b)

kg>mm = 10 g>10

(c)

N>ks2 = N>106 s2 = 10-6 N>s2 = mN>s2

Ans.

-6

Ans.

(d)

3

m = Gg>m

9

kN>ms = 10 N>10 s = 10 N>s = GN>s

Ans: kN # m Gg>m mN>s2 GN>s 8

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1–9. Represent each of the following combinations of units in the correct SI form using an appropriate prefix: (a) Mg/mm, (b) mN>ms, (c) mm # Mg.

SOLUTION a) Mg>mm = b) mN>ms =

103 kg -3

10 m 10 - 3 N 10 - 6 s

=

106 kg = Gg>m m

Ans.

=

103 N = kN>s s

Ans.

c) mm # Mg = C 10-6 m D

# C 103 kg D

= (10)-3 m # kg

= mm # kg

Ans.

Ans: Gg>m kN>s mm # kg 9

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1–10. Represent each of the following with SI units having an appropriate prefix: (a) 8653 ms, (b) 8368 N, (c) 0.893 kg.

SOLUTION a) 8653 ms = 8.653(10)3(10-3) s = 8.653 s

Ans.

b) 8368 N = 8.368 kN

Ans.

c) 0.893 kg = 893(10-3)(103) g = 893 g

Ans.

Ans: 8.653 s 8.368 kN 893 g 10

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1–11.

Using the SI system of units, show that Eq. 1–2 is a dimensionally homogeneous equation which gives F in newtons. Determine to three significant figures the gravitational force acting between two spheres that are touching each other. The mass of each sphere is 200 kg and the radius is 300 mm.

SOLUTION Using Eq. 1–2, F = G N = a

m 1 m2 r2

kg # kg kg # m m3 2b a 2 b = # kg s m s2

F = G

(Q.E.D.)

m 1 m2 r2

= 66.73 A 10 - 12 B c

200(200) 0.62

d

= 7.41 A 10 - 6 B N = 7.41 mN

Ans.

Ans: 7.41 mN 11

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*1–12. Round off the following numbers to three significant figures: (a) 58 342 m, (b) 68.534 s, (c) 2553 N, and (d) 7555 kg.

SOLUTION a) 58.3 km

b) 68.5 s

c) 2.55 kN

Ans.

d) 7.56 Mg

Ans: 58.3 km 68.5 s 2.55 kN 7.56 Mg 12

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1–13. A rocket has a mass of 3.529(106) kg on earth. Specify (a) its mass in SI units and (b) its weight in SI units. If the rocket is on the moon, where the acceleration due to gravity is gm = 1.61 m/s2,determine to three significant figures (c) its weight in SI units and (d) its mass in SI units.

Solution a) 3.529(106) kg = 3.53 Gg Ans.

3 3.529 1 106 2 kg 4 1 9.81 m>s2 2 34.619 1 106 2 kg # m>s2

b) We = mg = =

= 34.6 MN

Ans.

3 3.529 1 106 2 kg 4 1 1.61 m>s2 2 5.682 1 106 2 N = 5.68 MN Ans.

c) Wm = mgm = = Or

Wm = We a

1.61 m>s2 gm b = (34.619 MN) a b = 5.68 MN g 9.81 m>s2

d) Since the mass is independent of tis location, then mm = me = 3.53 Gg Ans.

Ans: 3.53 Gg 34.6 MN 5.68 MN mm = me = 3.53 Gg 13

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1–14.

Evaluate each of the following to three significant figures and express each answer in Sl units using an appropriate prefix: (a) 354 mg(45 km) > (0.0356 kN), (b) (0.004 53 Mg) (201 ms), and (c) 435 MN> 23.2 mm.

SOLUTION a) (354 mg)(45 km)>(0.0356 kN) =

=

C 354 A 10-3 B g D C 45 A 103 B m D 0.0356 A 103 B N

0.447 A 103 B g # m N

= 0.447 kg # m>N

Ans.

b) (0.00453 Mg)(201 ms) = C 4.53 A 10-3 B A 103 B kg D C 201 A 10-3 B s D = 0.911 kg # s c) 435 MN>23.2 mm =

435 A 106 B N 23.2 A 10-3 B m

=

Ans. 18.75 A 109 B N m

= 18.8 GN>m

Ans.

Ans: 0.447 kg # m>N 0.911 kg # s 18.8 GN>m 14

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1–15. Evaluate each of the following to three significant figures and express each answer in SI units using an appropriate prefix: (a) (212 mN)2, (b) (52 800 ms)2, and (c) [548(106)]1>2 ms.

Solution 3 212(10)-3 N 4 2

(a)

(212 mN)2 =

(b)

2

(52 800 ms) =

(c)

3 548(10)6 4

1 2

= 0.0449 N2 = 44.9(10)-3 N2

3 52 800(10) 4

s = 2788 s = 2.79 ( 10

-3 2 2

2

3

)s

2

ms = (23 409)(10)-3 s = 23.4(10)3(10)-3 s = 23.4 s

Ans. Ans. Ans.

Ans: 44.9(10)-3 N2 2.79 ( 103 ) s2 23.4 s 15

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*1–16. Evaluate each of the following to three significant figures and express each answer in SI units using an appropriate prefix: (a) 1684 mm2>143 ms2, (b) 128 ms210.0458 Mm2>1348 mg2, (c) (2.68 mm)(426 Mg).

SOLUTION a) (684 mm)>43 ms =

684(10 -6) m -3

43(10 ) s = 15.9 mm>s

=

15.9(10-3) m s Ans.

b) (28 ms)(0.0458 Mm)>(348 mg) = =

C 28(10-3) s D C 45.8(10-3)(10)6 m D 348(10-3)(10-3) kg

3.69(106) m # s = 3.69 Mm # s>kg kg

Ans.

c) (2.68 mm)(426 Mg) = C 2.68 A 10-3 B m D C 426 A 103 B kg D = 1.14 A 103 B m # kg = 1.14 km # kg

Ans.

Ans: 15.9 mm>s 3.69 Mm # s>kg 1.14 km # kg 16

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1–17. A concrete column has a diameter of 350 mm and a length of 2 m. If the density 1mass>volume2 of concrete is 2.45 Mg>m3, determine the weight of the column.

SOLUTION 2 3 V = pr2h = p A 0.35 2 m B (2 m) = 0.1924 m

m = rV = ¢

2.45(103)kg m3

≤ A 0.1924 m3 B = 471.44 kg

W = mg = (471.44 kg) A 9.81 m>s2 B = 4.6248 A 103 B N = 4.63 kN

Ans.

Ans: 4.63 kN 17

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1–18. Determine the mass of an object that has a weight of (a) 20 mN, (b) 150 kN, (c) 60 MN. Express the answer to three significant figures.

Solution a) m = b) m =

c) m =

20 1 10 - 3 2 kg # m>s2 W = = 2.04 g Ans. g 9.81 m>s2 150 1 103 2 kg # m>s2 W = = 15.3 Mg Ans. g 9.81 m>s2

60 1 106 2 kg # m>s2 W = = 6.12 Gg Ans. g 9.81 m>s2

Ans: 2.04 g 15.3 Mg 6.12 Gg 18

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1–19. If a man weighs 690 newtons on earth, specify (a) his mass in kilograms. If the man is on the moon, where the acceleration due to gravity is gm = 1.61 m>s2, determine (b) his weight in newtons, and (c) his mass in kilograms.

Solution 690 = 70.3 kg Ans. 9.81 1.61 (b) W = 690 c d = 113 N Ans. 9.81 690 (c) m = = 70.3 kg Ans. 9.81 (a) m =

Ans: 70.3 kg 113 N 70.3 kg 19

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*1–20. Evaluate each of the following to three significant figures and express each answer in SI units using an appropriate prefix: (a) (200 kN)2, (b) (0.005 mm)2, and (c) (400 m)3.

Solution (a) (200 kN)2 = 40 000(106) N2 = 0.04(1012) N2 = 0.04 MN2 Ans. (b) (0.005 mm)2 = 25(1012) m2 = 25 mm2 Ans. (c) (400 m)3 = 0.064(109) m3 = 0.064 km3 Ans.

Ans: 0.04 MN2 25 mm2 0.064 km3 20

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1–21. Two particles have a mass of 8 kg and 12 kg, respectively. If they are 800 mm apart, determine the force of gravity acting between them. Compare this result with the weight of each particle.

SOLUTION F = G

m1 m2 r2

Where G = 66.73 A 10-12 B m3>(kg # s2) F = 66.73 A 10 - 12 B B

8(12) (0.8)2

R = 10.0 A 10 - 9 B N = 10.0 nN

Ans.

W1 = 8(9.81) = 78.5 N

Ans.

W2 = 12(9.81) = 118 N

Ans.

Ans: F = 10.0 nN W1 = 78.5 N W2 = 118 N 21

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2–1. v

Determine the magnitude of the resultant force FR = F1 + F2 and its direction, measured clockwise from the positive u axis.

30 75

F1  4 kN 30 u F2  6 kN

Solution Parallelogram Law. The parallelogram law of addition is shown in Fig. a, Trigonometry. Applying Law of cosines by referring to Fig. b,

FR = 242 + 62 - 2(4)(6) cos 105° = 8.026 kN = 8.03 kN

Ans.

Using this result to apply Law of sines, Fig. b, sin u sin 105° = ; 6 8.026

u = 46.22°

Thus, the direction f of FR measured clockwise from the positive u axis is

Ans.

f = 46.22° - 45° = 1.22°

Ans: f = 1.22° 22

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2–2. v

Resolve the force F1 into components acting along the u and v axes and determine the magnitudes of the components. 30 75

F1  4 kN 30 u F2  6 kN

Solution Parallelogram Law. The parallelogram law of addition is shown in Fig. a, Trigonometry. Applying the sines law by referring to Fig. b.

(F1)v sin 45° (F1)u sin 30°

=

4 ; sin 105°

(F1)v = 2.928 kN = 2.93 kN

Ans.

=

4 ; sin 105°

(F1)u = 2.071 kN = 2.07 kN

Ans.

Ans: (F1)v = 2.93 kN (F1)u = 2.07 kN 23

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2–3. v

Resolve the force F2 into components acting along the u and v axes and determine the magnitudes of the components. 30 75

F1  4 kN 30 u F2  6 kN

Solution Parallelogram Law. The parallelogram law of addition is shown in Fig. a, Trigonometry. Applying the sines law of referring to Fig. b,

(F2)u sin 75° (F2)v sin 30°

=

6 ; sin 75°

(F2)u = 6.00 kN

Ans.

=

6 ; sin 75°

(F2)v = 3.106 kN = 3.11 kN

Ans.

Ans: (F2)u = 6.00 kN (F2)v = 3.11 kN 24

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*2–4. If 60° and 450 N, determine the magnitude of the resultant force and its direction, measured counterclockwise from the positive x axis.

y F

15

x

700 N

SOLUTION

The parallelogram law of addition and the triangular rule are shown in Figs. a and b, respectively. Applying the law of consines to Fig. b, 7002 497.01 N

4502

2(700)(450) cos 45° Ans.

497 N

This yields sin 700

sin 45° 497.01

Thus, the direction of angle positive axis, is 60°

of F 95.19°

95.19° measured counterclockwise from the 60°

Ans.

155°

Ans: FR = 497 N f = 155° 25

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2–5. y

If the magnitude of the resultant force is to be 500 N, directed along the positive y axis, determine the magnitude of force F and its direction u.

F

u 15

x

700 N

SOLUTION

The parallelogram law of addition and the triangular rule are shown in Figs. a and b, respectively. Applying the law of cosines to Fig. b, F = 25002 + 7002 - 2(500)(700) cos 105° Ans.

= 959.78 N = 960 N Applying the law of sines to Fig. b, and using this result, yields sin (90° + u) sin 105° = 700 959.78

Ans.

u = 45.2°

Ans: F = 960 N u = 45.2° 26

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2–6. If FB = 2 kN and the resultant force acts along the positive u axis, determine the magnitude of the resultant force and the angle u.

y

FA ⫽ 3 kN u 30⬚

B

x

A

u FB

Solution The parallelogram law of addition and the triangular rule are shown in Figs. a and b, respectively. Applying the law of sines to Fig. b, yields sin f sin 30° = 3 3

f = 48.59°

Thus, Ans.

u = 30° + f = 30° + 48.59° = 78.59° = 78.6° With the result u = 78.59°, applying the law of sines to Fig. b again, yields FR 2 = sin(180° - 78.59°) sin 30°

Ans.

FR = 3.92 kN

Ans: 78.6° FR = 3.92 kN 27

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2–7. If the resultant force is required to act along the positive u axis and have a magnitude of 5 kN, determine the required magnitude of FB and its direction u.

y

FA ⫽ 3 kN u 30⬚

B

x

A

u FB

Solution The parallelogram law of addition and the triangular rule are shown in Figs. a and b, respectively. Applying the law of cosines to Fig. b, FB = 232 + 52 - 2(3)(5) cos 30°

Ans.

= 2.832 kN = 2.83 kN

Using this result and realizing that sin(180° - u) = sin u, the application of the sine law to Fig. b, yields

sin u sin 30° = 5 2.832

Ans.

u = 62.0°

Ans: 2.83 kN u = 62.0 28

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*2–8. Determine the magnitude of the resultant force FR = F1 + F2 and its direction, measured clockwise from the positive x axis.

y F1 = 600 N

F2 = 800 N 60° 45°

x F3 = 450 N

75°

Solution FR = 2(600)2 + (800)2 - 2(600)(800)cos 75° = 866.91 = 867 N

Ans.

866.91 800 = sin 75° sin u u = 63.05°

Ans.

f = 63.05° + 45° = 108°

Ans: u = 108° 29

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2–9. Resolve F1 into components along the u and v axes and determine the magnitudes of these components.

v F1

F2

SOLUTION

150 N

30

250 N

u

30

Sine law:

105

F1v 250 = sin 30° sin 105°

F1v = 129 N

Ans.

F1u 250 = sin 45° sin 105°

F1u = 183 N

Ans.

Ans: F1v = 129 N F1u = 183 N 30

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2–10. Resolve F2 into components along the u and v axes and determine the magnitudes of these components.

v F1

F2

SOLUTION

150 N

30

250 N

u

30

Sine law:

105

F2v 150 = sin 30° sin 75°

F2v = 77.6 N

Ans.

F2u 150 = sin 75° sin 75°

F2u = 150 N

Ans.

Ans: F2v = 77.6 N F2u = 150 N 31

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2–11. The device is used for surgical replacement of the knee joint. If the force acting along the leg is 360 N, determine its components along the x and y ¿ axes.

y¿

y 10⬚

x¿x¿ xx 60⬚

Solution 360 N

-Fx 360 = ; Fx = -125 N sin 20° sin 100° Fy′ sin 60°

=

Ans.

360 ; Fy′ = 317 N sin 100°

Ans.

Ans: Fx = - 125 N Fy = 317 N 32

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*2–12. The device is used for surgical replacement of the knee joint. If the force acting along the leg is 360 N, determine its components along the x ¿ and y axes.

y¿

y 10⬚

x¿ x 60⬚

Solution 360 N

-Fx′ 360 = ; Fx′ = - 183 N sin 30° sin 80° Fy sin 70°

=

Ans.

360 ; Fy = 344 N sin 80°

Ans.

Ans: Fx′ = - 183 N Fy = 344 N 33

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2-13. If the tension in the cable is 400 N, determine the magnitude and direction of the resultant force acting on the pulley. This angle defines the same angle T of line AB on the tailboard block. y

400 N 30

u B

x 400 N

A

Solution FR

F1  F 1  2F 1 F 1 cos  90   T 1 2

T1

30 

FR

400 N

2

sin  T 1

sin 90   T FR

F1

§ FR

90  asin ¨

T

© F1

T

60

·

sin  T 1 ¸

¹ Ans.

Ans: u = 60 34

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2–14. The truck is to be towed using two ropes. Determine the magnitude of forces FA and FB acting on each rope in order to develop a resultant force of 950 N directed along the positive x axis. Set u = 50°.

y

A

B

FA 20°

x

θ FB

Solution Parallelogram Law: The parallelogram law of addition is shown in Fig. (a). Trigonometry: Using law of sines [Fig. (b)], we have FA 950 = sin 50° sin 110° Ans.

FA = 774 N FB 950 = sin 20° sin 110°

Ans.

FB = 346 N

Ans: FA = 774 N FB = 346 N 35

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2–15. The plate is subjected to the two forces at A and B as shown. If u = 60°, determine the magnitude of the resultant of these two forces and its direction measured clockwise from the horizontal.

FA u

8 kN

A

SOLUTION 40

Parallelogram Law: The parallelogram law of addition is shown in Fig. a.

B

Trigonometry: Using law of cosines (Fig. b), we have 2

FB

6 kN

2

FR = 28 + 6 - 2(8)(6) cos 100° Ans.

= 10.80 kN = 10.8 kN The angle u can be determined using law of sines (Fig. b). sin 100° sin u = 6 10.80 sin u = 0.5470 u = 33.16° Thus, the direction f of FR measured from the x axis is

Ans.

f = 33.16° - 30° = 3.16°

Ans: FR = 10.8 kN f = 3.16° 36

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*2–16. Determine the angle of u for connecting member A to the plate so that the resultant force of FA and FB is directed horizontally to the right. Also, what is the magnitude of the resultant force?

FA u

SOLUTION

8 kN

A

40

Parallelogram Law: The parallelogram law of addition is shown in Fig. a.

B

Trigonometry: Using law of sines (Fig .b), we have

FB

6 kN

sin (90° - u) sin 50° = 6 8 sin (90° - u) = 0.5745 Ans.

u = 54.93° = 54.9°

From the triangle, f = 180° - (90° - 54.93°) - 50° = 94.93°. Thus, using law of cosines, the magnitude of FR is FR = 282 + 62 - 2(8)(6) cos 94.93° Ans.

= 10.4 kN

Ans: u = 54.9° FR = 10.4 kN 37

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2–17. Two forces act on the screw eye. If F1 = 400 N and F2 = 600 N, determine the angle u(0° … u … 180°) between them, so that the resultant force has a magnitude of FR = 800 N.

F1

u

F2

SOLUTION The parallelogram law of addition and triangular rule are shown in Figs. a and b, respectively. Applying law of cosines to Fig. b, 800 = 24002 + 6002 - 2(400)(600) cos (180° - u°) 8002 = 4002 + 6002 - 480000 cos (180° - u) cos (180° - u) = - 0.25 180° - u = 104.48 Ans.

u = 75.52° = 75.5°

Ans: u = 75.5° 38

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2–18. Two forces F1 and F2 act on the screw eye. If their lines of action are at an angle u apart and the magnitude of each force is F1 = F2 = F, determine the magnitude of the resultant force FR and the angle between FR and F1.

F1

u

SOLUTION

F2

F F = sin f sin (u - f) sin (u - f) = sin f u - f = f f =

u 2

Ans.

FR = 2(F)2 + (F)2 - 2(F)(F) cos (180° - u) Since cos (180° - u) = -cos u FR = F A 22 B 21 + cos u u 1 + cos u Since cos a b = 2 A 2 Then u FR = 2F cosa b 2

Ans.

Ans: f =

u 2

u FR = 2F cos a b 2 39

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2–19. y

Determine the magnitude and direction of the resultant force, FR measured counterclockwise from the  positive x axis. Solve the problem by first finding the resultant F′ = F1 + F2 and then forming FR = F′ + F3.

F1  400 N 90º

F2  200 N

150º

x

F3  300 N

Solution Parallelogram Law. The parallelogram law of addition for F1 and F2 and then their resultant F′ and F3 are shown in Figs. a and b, respectively. Trigonometry. Referring to Fig. c,

F ′ = 22002 + 4002 = 447.21 N

Thus f′ = 90° - 30° - 26.57° = 33.43°

u ′ = tan-1 a

200 b = 26.57° 400

Using these results to apply the law of cosines by referring to Fig. d,   FR = 23002 + 447.212 - 2(300)(447.21) cos 33.43° = 257.05 N = 257 kN Ans. Then, apply the law of sines,

sin u sin 33.43° = ; 300 257.05

u = 40.02°

Thus, the direction f of FR measured counterclockwise from the positive x axis is

Ans.

f = 90° + 33.43° + 40.02° = 163.45° = 163°

Ans: FR = 257 N  f = 163° 40

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*2–20. Determine the magnitude and direction of the resultant force, FR measured counterclockwise from the positive x axis. Solve the problem by first finding the resultant F = F2 + F3 and then forming FR = F + F1.

y F1  400 N 90º

F2  200 N

150º

x

F3  300 N

Solution Parallelogram Law. The parallelogram law of addition for F2 and F3 and then their resultant F′ and F1 are shown in Figs. a and b, respectively. Trigonometry. Applying the law of cosines by referring to Fig. c,

F ′ = 22002 + 3002 - 2(200)(300) cos 30° = 161.48 N

Ans.

Using this result to apply the sines law, Fig. c, sin u′ sin 30° = ; 200 161.48

u′ = 38.26°

Using the results of F′ and u′ to apply the law of cosines by referring to Fig. d,   FR = 2161.482 + 4002 - 2(161.48)(400) cos 21.74° = 257.05 N = 257 N Ans. Then, apply the sines law,

sin u sin 21.74° = ; 161.48 257.05

u = 13.45°

Thus, the direction f of FR measured counterclockwise from the positive x axis is

Ans.

f = 90° + 60° + 13.45° = 163.45° = 163°

Ans: f = 163° FR = 257 N 41

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2–21. The log is being towed by two tractors A and B. Determine the magnitude of the two towing forces FA and FB if it is required that the resultant force have a magnitude FR = 10 kN and be directed along the x axis. Set u = 15°.

y

FA 30° FB

A x

θ

B

Solution Parallelogram Law: The parallelogram law of addition is shown in Fig. (a). Trigonometry: Using law of sines [Fig. (b)], we have FA 10 = sin 15° sin 135° Ans.

FA = 3.66 kN FB 10 = sin 30° sin 135°

Ans.

FB = 7.07 kN

Ans: FA = 3.66 kN FB = 7.07 kN 42

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2–22. If the resultant FR of the two forces acting on the log is to be directed along the positive x axis and have a magnitude of 10 kN, determine the angle u of the cable, attached to B such that the force FB in this cable is minimum. What is the magnitude of the force in each cable for this situation?

y

FA 30° FB

A x

θ

B

Solution Parallelogram Law: In order to produce a minimum force, FB, FB has to act perpendicular to FA. The parallelogram law of addition is shown in Fig. (a). Trigonometry: Fig.(b). FB = 10 sin 30° = 5.00 kN

Ans.

FA = 10 cos 30° = 8.66 kN

Ans.

The angle u is Ans.

u = 90° - 30° = 60.0°

Ans: FB = 5.00 kN FA = 8.66 kN u = 60.0 43

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2–23. Determine the magnitude and direction of the resultant FR = F1 + F2 + F3 of the three forces by first finding the resultant F¿ = F1 + F2 and then forming FR = F¿ + F3.

y F1

30 N 3

5

F3

4

50 N

20

SOLUTION

F2

20 N

F¿ = 2(20)2 + (30)2 - 2(20)(30) cos 73.13° = 30.85 N 30 30.85 = ; sin 73.13° sin (70° - u¿)

u¿ = 1.47°

FR = 2(30.85)2 + (50)2 - 2(30.85)(50) cos 1.47° = 19.18 = 19.2 N

Ans.

30.85 19.18 = ; sin 1.47° sin u

Ans.

u = 2.37°

Ans: FR = 19.2 N u = 2.37° c 44

x

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*2–24. Determine the magnitude and direction of the resultant FR = F1 + F2 + F3 of the three forces by first finding the resultant F¿ = F2 + F3 and then forming FR = F¿ + F1.

y F1

30 N 3

5

F3

4

50 N

20

SOLUTION ¿

2

F2

20 N

2

F = 2(20) + (50) - 2(20)(50) cos 70° = 47.07 N 20 sin u

¿

=

47.07 ; sin 70°

u¿ = 23.53°

FR = 2(47.07)2 + (30)2 - 2(47.07)(30) cos 13.34° = 19.18 = 19.2 N 30 19.18 = ; sin 13.34° sin f

Ans.

f = 21.15° Ans.

u = 23.53° - 21.15° = 2.37°

Ans: FR = 19.2 N u = 2.37° c 45

x

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2–25. If the resultant force of the two tugboats is 3 kN, directed along the positive x axis, determine the required magnitude of force FB and its direction u.

y

FA

A

2 kN 30

x

u C

SOLUTION The parallelogram law of addition and the triangular rule are shown in Figs. a and b, respectively.

FB

B

Applying the law of cosines to Fig. b, FB = 222 + 32 - 2(2)(3)cos 30° Ans.

= 1.615kN = 1.61 kN Using this result and applying the law of sines to Fig. b, yields sin u sin 30° = 2 1.615

Ans.

u = 38.3°

Ans: FB = 1.61 kN u = 38.3° 46

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2–26. If FB = 3 kN and u = 45°, determine the magnitude of the resultant force of the two tugboats and its direction measured clockwise from the positive x axis.

y

FA

A

2 kN 30

x

u C

SOLUTION The parallelogram law of addition and the triangular rule are shown in Figs. a and b, respectively.

FB

B

Applying the law of cosines to Fig. b, FR = 222 + 32 - 2(2)(3) cos 105° Ans.

= 4.013 kN = 4.01 kN Using this result and applying the law of sines to Fig. b, yields sin 105° sin a = 3 4.013

a = 46.22°

Thus, the direction angle f of FR, measured clockwise from the positive x axis, is Ans.

f = a - 30° = 46.22° - 30° = 16.2°

Ans: FR = 4.01 kN f = 16.2° 47

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2–27. If the resultant force of the two tugboats is required to be directed towards the positive x axis, and FB is to be a minimum, determine the magnitude of FR and FB and the angle u.

y

FA

A

2 kN 30

x

u

SOLUTION

C

FB

For FB to be minimum, it has to be directed perpendicular to FR. Thus, Ans.

u = 90°

B

The parallelogram law of addition and triangular rule are shown in Figs. a and b, respectively. By applying simple trigonometry to Fig. b, FB = 2 sin 30° = 1 kN

Ans.

FR = 2 cos 30° = 1.73 kN

Ans.

Ans: u = 90° FB = 1 kN FR = 1.73 kN 48

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*2–28. Determine the magnitude of force F so that the resultant FR of the three forces is as small as possible. What is the minimum magnitude of FR?.

8 kN

F 30 6 kN

Solution Parallelogram Law. The parallelogram laws of addition for 6 kN and 8 kN and then their resultant F′ and F are shown in Figs. a and b, respectively. In order for FR to be minimum, it must act perpendicular to F. Trigonometry. Referring to Fig. b, F′ = 262 + 82 = 10.0 kN Referring to Figs. c and d,

8 u = tan - 1 a b = 53.13°. 6



FR = 10.0 sin 83.13° = 9.928 kN = 9.93 kN

Ans.



F = 10.0 cos 83.13° = 1.196 kN = 1.20 kN

Ans.

Ans: FR = 9.93 kN F = 1.20 kN 49

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2–29. y

Determine the magnitude and direction u of FA so that the resultant force is directed along the positive x axis and has a magnitude of 1250 N.

FA

θ O

A

x

30° B

SOLUTION

FB = 800 N

+ FR = ΣFx; S x

FRx = FA sin u + 800 cos 30° = 1250

+ c FRy = ΣFy;

FRy = FA cos u - 800 sin 30° = 0 u = 54.3°

Ans.

FA = 686 N

Ans.

Ans: u = 54.3° FA = 686 N 50

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2–30. y

Determine the magnitude and direction, measured counterclockwise from the positive x axis, of the resultant force acting on the ring at O, if FA = 750 N and u = 45°.

FA

θ O

A

x

30° B

SOLUTION

FB = 800 N

Scalar Notation: Suming the force components algebraically, we have + FR = ΣFx; S x

FRx = 750 sin 45° + 800 cos 30° = 1223.15 N S

+ c FRy = ΣFy;

FRy = 750 cos 45° - 800 sin 30° = 130.33 N c

The magnitude of the resultant force FR is FR = 3F 2Rx + F 2Ry = 21223.152 + 130.332 = 1230 N = 1.23 kN

Ans.

The directional angle u measured counterclockwise from positive x axis is u = tan-1

FRy FRx

= tan-1 a

130.33 b = 6.08° 1223.15

Ans.

Ans: FR = 1.23 kN u = 6.08° 51

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2–31. Two forces act on the screw eye. If F = 600 N, determine the magnitude of the resultant force and the angle u if the resultant force is directed vertically upward.

y F 500 N

u

30⬚

x

SOLUTION The parallelogram law of addition and triangular rule are shown in Figs. a and b respectively. Applying law of sines to Fig. b, sin 30° sin u = ; sin u = 0.6 600 500

u = 36.87° = 36.9°

Ans.

Using the result of u, f = 180° - 30° - 36.87° = 113.13° Again, applying law of sines using the result of f, FR 500 = ; sin 113.13° sin 30°

FR = 919.61 N = 920 N

Ans.

Ans: u = 36.9 u = 920 N 52

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*2–32. Resolve each force acting on the gusset plate into its x and y components, and express each force as a Cartesian vector.

y 650 N

F3 3

5

F2

4

750 N

45 F1

x 900 N

SOLUTION F1 = {900( +i)} = {900i} N

Ans.

F2 = {750 cos 45°(+i) + 750 sin 45°(+ j)} N = {530i + 530j} N

Ans.

F3 = e 650a

3 4 b (+i) + 650 a b(-j) f N 5 5

= {520 i - 390j)} N

Ans.

53

Ans: F1 = 5 900i6 N F2 = 5 530i + 530j 6 N F3 = 5 520i - 390j 6 N

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2–33. Determine the magnitude of the resultant force acting on the plate and its direction, measured counterclockwise from the positive x axis.

y 650 N

F3 3

5

F2

4

750 N

45 F1

x 900 N

SOLUTION Rectangular Components: By referring to Fig. a, the x and y components of F1, F2, and F3 can be written as (F1)x = 900 N

(F1)y = 0

(F2)x = 750 cos 45° = 530.33 N 4 (F3)x = 650 a b = 520 N 5

(F2)y = 750 sin 45° = 530.33 N 3 (F3)y = 650a b = 390 N 5

Resultant Force: Summing the force components algebraically along the x and y axes, we have + : ©(FR)x = ©Fx;

(FR)x = 900 + 530.33 + 520 = 1950.33 N :

+ c ©(FR)y = ©Fy;

(FR)y = 530.33 - 390 = 140.33 N c

The magnitude of the resultant force FR is FR = 2(FR)x2 + (FR)y2 = 21950.332 + 140.332 = 1955 N = 1.96 kN Ans. The direction angle u of FR, measured clockwise from the positive x axis, is u = tan-1 c

(FR)y 140.33 b = 4.12° d = tan-1 a (FR)x 1950.33

Ans.

Ans: FR = 1.96 kN u = 4.12° 54

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2–34. y 60

Resolve F1 and F2 into their x and y components.

30 400 N

F1

45

SOLUTION

F2

x

250 N

F1 = {400 sin 30°(+ i) + 400 cos 30°( +j)} N = {200i +346j} N

Ans.

F2 = {250 cos 45°(+ i) +250 sin 45°( -j)} N Ans.

= {177i -177j} N

55

Ans: F1 = 5200i + 346j 6 N F2 = 5177i - 177j 6 N

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2–35. Determine the magnitude of the resultant force and its direction measured counterclockwise from the positive x axis.

y 60 30 400 N

F1

SOLUTION

(F1)x = 400 sin 30° = 200 N

(F1)y = 400 cos 30° = 346.41 N

(F2)x = 250 cos 45° = 176.78 N

(F2)y = 250 sin 45° = 176.78 N

x

45

Rectangular Components: By referring to Fig. a, the x and y components of F1 and F2 can be written as F2

250 N

Resultant Force: Summing the force components algebraically along the x and y axes, we have + : ©(FR)x = ©Fx;

(FR)x = 200 + 176.78 = 376.78 N

+ c ©(FR)y = ©Fy;

(FR)y = 346.41 - 176.78 = 169.63 N c

Ans.

The magnitude of the resultant force FR is FR = 2(FR)x2 + (FR)y2 = 2376.782 + 169.632 = 413 N

Ans.

The direction angle u of FR, Fig. b, measured counterclockwise from the positive axis, is u = tan-1 c

(FR)y 169.63 d = tan-1 a b = 24.2° (FR)x 376.78

Ans.

Ans: FR = 413 N u = 24.2° 56

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*2–36. y

Determine the magnitude of the resultant force and its direction, measured counterclockwise from the positive x axis.

F1  200 N 45 x 30

SOLUTION

F2  150 N

+ R FRx = ©Fx;

FRx = - 150 cos 30° + 200 sin 45° = 11.518 N

Q+ FRy = ©Fy;

FRy = 150 sin 30° + 200 cos 45° = 216.421 N

FR = 2 (11.518)2 + (216.421)2 = 217 N

Ans.

u = tan - 1 ¢

Ans.

216.421 ≤ = 87.0° 11.518

Ans: FR = 217 N u = 87.0° 57

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2–37. y

Determine the magnitude of the resultant force and its direction, measured clockwise from the positive x axis.

400 N B 30 x 45 800 N

Solution Scalar Notation. Summing the force components along x and y axes by referring to Fig. a, + (F ) = ΣF ;  (F ) = 400 cos 30° + 800 sin 45° = 912.10 N S S R x

x

R x

+ c (FR)y = ΣFy;  (FR)y = 400 sin 30° - 800 cos 45° = - 365.69 N = 365.69 NT Referring to Fig. b, the magnitude of the resultant force is

FR = 2(FR)2x + (FR)2y = 2912.102 + 365.692 = 982.67 N = 983 N

Ans.

And its directional angle u measured clockwise from the positive x axis is

u = tan - 1 c

(FR)y (FR)x

d = tan - 1a

365.69 b = 21.84° = 21.8° 912.10

Ans.

Ans: FR = 983 N u = 21.8° 58

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2-38. Determine the magnitude of the resultant force and its direction, measured clockwise from the positive x axis.

y

50 N 30 70 N

45

x

65 N

Solution Given: F1

70 N

F2

50 N

F3

65 N

T

30 

I

45

 o

F Rx = 6Fx;

F RX

F 1  F 2 cos  T  F 3 cos  I



F Ry = 6F y;

F RY

F2 sin  T  F 3 sin  I

n

FR

T FR

T

2

FRX  FRY

2

§ FRY · ¸ © FRX ¹

atan ¨

97.8 N 46.5

Ans. Ans.

Ans: FR = 97.8 N u = 46.5 59

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2–39. y

Express F1, F2, and F3 as Cartesian vectors. F3  750 N

45

3

5

x

4

SOLUTION F1 =

30

F1  850 N

F2  625 N

4 3 (850) i - (850) j 5 5

= {680 i - 510 j} N

Ans.

F2 = - 625 sin 30° i - 625 cos 30° j = { - 312 i - 541 j} N

Ans.

F3 = - 750 sin 45° i + 750 cos 45° j { -530 i + 530 j} N

Ans.

Ans: F1 = {680i - 510j} N F2 = { - 312i - 541j} N F3 = { - 530i + 530j} N 60

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*2–40. Determine the magnitude of the resultant force and its direction measured counterclockwise from the positive x axis.

y F3  750 N

45

3

5

x

4

SOLUTION

30 F2  625 N

+ F = ©F ; : Rx x

FRx =

+ c FRy = ©Fy ;

3 FRy = - (850) - 625 cos 30° + 750 cos 45° = -520.94 N 5

4 (850) - 625 sin 30° - 750 sin 45° = - 162.83 N 5

FR = 2(- 162.83)2 + (- 520.94)2 = 546 N f = tan-1 a

F1  850 N

Ans.

520.94 b = 72.64° 162.83

u = 180° + 72.64° = 253°

Ans.

Ans: FR = 546 N u = 253° 61

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2–41. Determine the magnitude and direction u of the resultant force FR. Express the result in terms of the magnitudes of the components F1 and F2 and the angle f.

SOLUTION

F1

F 2R

=

F 21

+

F 22

FR

- 2F1F2 cos (180° - f)

Since cos (180° - f) = - cos f,

f u

FR = 2F 21 + F 22 + 2F1F2 cos f

Ans.

F2

From the figure, tan u =

F1 sin f F2 + F1 cos f

u = tan –1 ¢

F1 sin f ≤ F2 + F1 cos f

Ans.

Ans: FR = 2F12 + F22 + 2F1F2 cos f F1 sin f b u = tan-1a F2 + F1 cos f 62

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2–42. y

Determine the magnitude of the resultant force and its direction, measured counterclockwise from the positive x axis.

F3  8 kN F2  5 kN

60 45

F1  4 kN

x

Solution Scalar Notation. Summing the force components along x and y axes algebraically by referring to Fig. a, + (F ) = ΣF ;  (F ) = 4 + 5 cos 45° - 8 sin 15° = 5.465 kN S S R x

x

R x

+ c (FR)y = ΣFy;  (FR)y = 5 sin 45° + 8 cos 15° = 11.263 kN c By referring to Fig. b, the magnitude of the resultant force FR is

FR = 2(FR)2x + (FR)2y = 25.4652 + 11.2632 = 12.52 kN = 12.5 kN

Ans.

And the directional angle u of FR measured counterclockwise from the positive x axis is

u = tan - 1 c

(FR)y (FR)x

d = tan - 1 a

11.263 b = 64.12° = 64.1° 5.465

Ans.

Ans: FR = 12.5 kN u = 64.1° 63

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2–43. y

Determine the x and y components of F1 and F2.

45 F1  200 N

30

SOLUTION F1x = 200 sin 45° = 141 N

Ans.

F1y = 200 cos 45° = 141 N

Ans.

F2x = - 150 cos 30° = - 130 N

Ans.

F2y = 150 sin 30° = 75 N

Ans.

F2  150 N x

Ans: F1x = F1y = F2x = F2y = 64

141 N 141 N -130 N 75 N

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*2–44. y

Determine the magnitude of the resultant force and its direction, measured counterclockwise from the positive x axis. 45⬚

F1 ⫽ 200 N

30⬚

SOLUTION +R FRx = ©Fx;

FRx = -150 cos 30° + 200 sin 45° = 11.518 N

Q+FRy = ©Fy;

FRy = 150 sin 30° + 200 cos 45° = 216.421 N

F2 ⫽ 150 N x

FR = 2 (11.518)2 + (216.421)2 = 217 N

Ans.

216.421 ≤ = 87.0° 11.518

Ans.

u = tan - 1 ¢

Ans: FR = 217 N u = 87.0° 65

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2–45. Express each of the three forces acting on the support in Cartesian vector form and determine the magnitude of the resultant force and its direction, measured clockwise from positive x axis.

y

F1  50 N 5

4

4 3

x F3  30 N

15

F2  80 N

Solution Cartesian Notation. Referring to Fig. a,

3 4 F1 = (F1)x i + (F1)y j = 50 a b i + 50 a b j = {30 i + 40 j} N 5 5

Ans.

F2 = - (F2)x i - (F2)y j = -80 sin 15° i - 80 cos 15° j = { -20.71 i - 77.27 j} N = { -20.7 i - 77.3 j} N



Ans.

F3 = (F3)x i = {30 i}



Ans.

Thus, the resultant force is FR = ΣF ;

FR = F1 + F2 + F3 = (30i + 40 j) + ( - 20.71i - 77.27j) + 30i = {39.29 i - 37.27 j} N

Referring to Fig. b, the magnitude of FR is

FR = 239.292 + 37.272 = 54.16 N = 54.2 N

Ans.

And its directional angle u measured clockwise from the positive x axis is

u = tan - 1 a

37.27 b = 43.49° = 43.5° 39.29

Ans.

Ans:  F1 =  F2 =  F3 = FR = u = 66

{30i + 40j} N { -20.7 i - 77.3 j} N {30 i} 54.2 N 43.5°

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2–46. y

Determine the x and y components of each force acting on the gusset plate of a bridge truss. Show that the resultant force is zero.

F2  6 kN F1  8 kN 4

5

3

5

4 3

F3  4 kN

F4  6 kN

x

Solution Scalar Notation. Referring to Fig. a, the x and y components of each forces are

4 (F1)x = 8 a b = 6.40 kN S  5

Ans.

3 (F1)y = 8 a b = 4.80 kN T  5

Ans.

3 (F2)x = 6 a b = 3.60 kN S  5

Ans.

4 (F2)y = 6 a b = 4.80 kN c  5

Ans.



(F3)x = 4 kN d 

Ans.



(F3)y = 0

Ans.



(F4)x = 6 kN d 

Ans.



(F4)y = 0

Ans.

Summing these force components along x and y axes algebraically, + (F ) = ΣF ;   (F ) = 6.40 + 3.60 - 4 - 6 = 0 S R x x R x + c (FR)y = ΣFy ;     (FR)y = 4.80 - 4.80 = 0 Thus, FR = 2(FR)2x + (FR)2y = 2 0 2 + 0 2 = 0

(Q.E.D)

Ans: (F1)x (F1)y (F2)x (F2)y (F3)x (F3)y (F4)x (F4)y 67

= = = = = = = =

6.40 kN S 4.80 kN T 3.60 kN S 4.80 kN c 4 kN d 0 6 kN d 0

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2–47. Express F1, F2, and F3 as Cartesian vectors.

y

40

F2  26 kN 5

F1  15 kN

13 12

Solution

x 30

F1 = {15 sin 40°i + 15 cos 40°j} kN = {9.6i + 11.49j} kN

Ans.

F3  36 kN

F2 = -26(12>13)i + 26(5>13)j = { - 24i + 10j} kN

Ans.

F3 = 36 cos 30°i - 36 sin 30°j = {31.2i + 18j} kN

Ans.

Ans: F1 = {9.64i + 11.5j} kN F2 = { - 24i + 10j} kN F3 = {31.2i - 18j} kN 68

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*2–48. Determine the magnitude of the resultant force and its direction measured counterclockwise from the positive x axis.

y

F2  26 kN 5

40

F1  15 kN

13 12

SOLUTION

x 30

+ F = ©F ; : Rx x

FRx = 15 sin 40° -

12 (26) + 36 cos 30° = 16.82 kN 13

+ c FRy = ©Fy ;

FRy = 15 cos 40° +

5 (26) - 36 sin 30° = 3.491 kN 13

F3  36 kN

FR = 2(16.82)2 + (3.491)2 = 17.2 kN

Ans.

3.491 b = 11.7° 16.82

Ans.

u = tan-1 a Also,

F1 = {15 sin 40° i + 15 cos 40° j} kN = {9.64i + 11.5j} kN F2 = b -

12 5 (26)i + (26)j r kN = {- 24i + 10j} kN 13 13

F3 = {36 cos 30°i - 36 sin 30°j} kN = {31.2i - 18j} kN FR = F1 + F2 + F3 = {9.64i + 11.5j} + { -24i + 10j} + {31.2i - 18j}

Ans: FR = 17.2 kN, u = 11.7° 69

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2–49. If F1 = 300 N and u = 10°, determine the magnitude and direction, measured counterclockwise from the positive x¿ axis, of the resultant force acting on the bracket.

y F2  200 N

x¿ F3  180 N

SOLUTION

13

60

u

F1

5 12

+ F = ©Fx ; : Rx

FRx

+ c FRy = ©Fy ;

FRy = 300 cos 70° + 200 +

12 = 300 sin 70° (180) = 115.8 N 13

x

5 (180) = 371.8 N 13

FR = 2(115.8)2 + (371.8)2 = 389 N f = tan - 1 B

371.8 R = 72.71° 115.8

Ans.

au Ans.

f¿ = 72.71° - 30° = 42.7°

Ans: FR = 389 N f′ = 42.7° 70

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2–50. y

Three forces act on the bracket. Determine the magnitude and direction u of F1 so that the resultant force is directed along the positive x¿ axis and has a magnitude of 800 N.

F2  200 N

x¿ F3  180 N

SOLUTION

13

60

u

F1

5 12

+ F = ©F ; : Rx x

12 (180) 800 sin 60° = F1 sin(60° + u) 13

+ c FRy = ©Fy ;

800 cos 60° = F1 cos(60° + u) + 200 +

x

5 (180) 13

60° + u = 81.34° u = 21.3°

Ans.

F1 = 869 N

Ans.

Ans: u = 21.3° F1 = 869 N 71

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2–51. y

Determine the magnitude and orientation u of FB so that the resultant force is directed along the positive y axis and has a magnitude of 1500 N.

FB

B

FA  700 N 30

A

u x

SOLUTION Scalar Notation: Suming the force components algebraically, we have + F = ΣF ; S Rz x

0 = 700 sin 30° - FB cos u (1)

FB cos u = 350 + c FRy = ΣFy;

1500 = 700 cos 30° + FB sin u (2)

FB sin u = 893.8 Solving Eq. (1) and (2) yields u = 68.6°

Ans.

FB = 960 N

Ans: u = 68.6° FB = 960 N 72

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*2–52. y

Determine the magnitude and orientation, measured counterclockwise from the positive y axis, of the resultant force acting on the bracket, if FB = 600 N and u = 20°.

FB

B

FA  700 N 30

A

u x

SOLUTION Scalar Notation: Suming the force components algebraically, we have + F = ΣF ; S Rx x

FRx = 700 sin 30° - 600 cos 20° = - 213.8 N = 213.8 N d

+ c FRy = ΣFy;

FRy = 700 cos 30° + 600 sin 20° = 811.4 N c

The magnitude of the resultant force FR is FR = 2F 2Rx + F 2Ry = 2213.82 + 811.42 = 839 N

Ans.

The directional angle u measured counterclockwise from positive y axis is u = tan-1

FRx FRy

= tan-1 a

213.8 b = 14.8° 811.4

Ans.

Ans: FR = 839 N u = 14.8° 73

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2–53. y

Three forces act on the bracket. Determine the magnitude and direction u of F so that the resultant force is directed along the positive x′ axis and has a magnitude of 8 kN.

4 kN

F 15

u

x'

30 x 6 kN

Solution Scalar Notation. Equating the force components along the x and y axes algebraically by referring to Fig. a, + (F ) = ΣF ;  8 cos 30° = F sin u + 6 - 4 sin 15° S R x

x



F sin u = 1.9635(1) + c (FR)y = ΣFy ;  8 sin 30° = F cos u + 4 cos 15°



F cos u = 0.1363(2)

Divide Eq (1) by (2)

tan u = 14.406

u = 86.03° = 86.0°

Ans.

Substitute this result into Eq (1) F sin 86.03° = 1.9635

Ans.

F = 1.968 kN = 1.97 kN

Ans:  u = 86.0° F = 1.97 kN 74

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2–54. y

If F = 5 kN and u = 30°, determine the magnitude of the resultant force and its direction, measured counterclockwise from the positive x axis.

4 kN

F 15

u

x'

30 x 6 kN

Solution Scalar Notation. Summing the force components along x and y axes algebraically by referring to Fig. a, + (F ) = ΣF ; (F ) = 5 sin 30° + 6 - 4 sin 15° = 7.465 kN S S R x x R x + c (FR)y = ΣFy; (FR)y = 4 cos 15° + 5 cos 30° = 8.194 kN c

By referring to Fig. b, the magnitude of the resultant force is FR = 2(FR)2x + (FR)2y = 27.4652 + 8.1942 = 11.08 kN = 11.1 kN Ans.

And its directional angle u measured counterclockwise from the positive x axis is

u = tan - 1 c

(FR)y (FR)x

d = tan - 1 a

8.194 b = 47.67° = 47.7° 7.465

Ans.

Ans: FR = 11.1 kN u = 47.7° 75

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2–55. y

If the magnitude of the resultant force acting on the bracket is to be 450 N directed along the positive u axis, determine the magnitude of F1 and its direction f.

u

F1

f 30 F2

SOLUTION

(F1)x = F1 sin f

(F1)y = F1 cos f

(F2)x = 200 N

( F2)y = 0

(F3)x = 260 ¢

13

12

Rectangular Components: By referring to Fig. a, the x and y components of F1, F2, F3, and FR can be written as

5 ≤ = 100 N 13

(F3)y = 260 ¢

(FR)x = 450 cos 30° = 389.71 N

x 200 N

5

F3

260 N

12 ≤ = 240 N 13

(FR)y = 450 sin 30° = 225 N

Resultant Force: Summing the force components algebraically along the x and y axes, + ©(F ) = ©F ; : R x x

389.71 = F1 sin f + 200 + 100 F1 sin f = 89.71

+ c ©(FR)y = ©Fy;

(1)

225 = F1 cos f - 240 F1 cos f = 465

(2)

Solving Eqs. (1) and (2), yields f = 10.9°

F1 = 474 N

Ans.

Ans: f = 10.9 F1 = 474 N 76

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*2–56. y

If the resultant force acting on the bracket is required to be a minimum, determine the magnitudes of F1 and the resultant force. Set f = 30°.

u

F1

f 30

SOLUTION

F2

Rectangular Components: By referring to Fig. a, the x and y components of F1, F2, and F3 can be written as (F1)x = F1 sin 30° = 0.5F1

(F1)y = F1 cos 30° = 0.8660F1

(F2)x = 200 N

(F2)y = 0

(F3)x = 260a

5 b = 100 N 13

(F3)y = 260a

x 200 N

13

12 5

F3

260 N

12 b = 240 N 13

Resultant Force: Summing the force components algebraically along the x and y axes, + ©(F ) = ©F ; : R x x

(FR)x = 0.5F1 + 200 + 100 = 0.5F1 + 300

+ c ©(FR)y = ©Fy;

(FR)y = 0.8660F1 - 240

The magnitude of the resultant force FR is FR = 2(FR)x2 + (FR)y2 = 2(0.5F1 + 300)2 + (0.8660F1 - 240)2 = 2F 21 - 115.69F1 + 147 600

(1)

Thus, FR2 = F 21 - 115.69F1 + 147 600

(2)

The first derivative of Eq. (2) is 2FR For FR to be minimum,

dFR = 2F1 - 115.69 dF1

(3)

dFR = 0. Thus, from Eq. (3) dF1

2FR

dFR = 2F1 - 115.69 = 0 dF1

F1 = 57.846 N = 57.8 N

Ans.

from Eq. (1), FR = 2(57.846)2 - 115.69(57.846) + 147 600 = 380 N

Ans.

Ans: F1 = 57.8 N FR = 380 N 77

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2–57. Determine the magnitude of force F so that the resultant force of the three forces is as small as possible. What is the magnitude of the resultant force?

F

14 kN 30

45 8 kN

SOLUTION + : FRx = ©Fx ;

FRz = 8 - F cos 45° - 14 cos 30° = - 4.1244 - F cos 45°

+ c FRy = ©Fy ;

FRy = - F sin 45° + 14 sin 30° = 7 - F sin 45° FR2 = ( -4.1244 - F cos 45°)2 + (7 - F sin 45°)2 2FR

From Eq. (1);

(1)

dFR = 2( -4.1244 - F cos 45°)(- cos 45°) + 2(7 - F sin 45°)( -sin 45°) = 0 dF F = 2.03 kN

Ans.

FR = 7.87 kN

Ans.

Also, from the figure require (FR)x¿ = 0 = ©Fx¿;

F + 14 sin 15° - 8 cos 45° = 0 Ans.

F = 2.03 kN (FR)y¿ = ©Fy¿;

FR = 14 cos 15° - 8 sin 45° Ans.

FR = 7.87 kN

Ans: F = 2.03 kN FR = 7.87 kN 78

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2–58. Express F1 and F2 as Cartesian vectors.

y

F2 = 26 kN 13

12 5

x

SOLUTION F1 = -30 sin 30° i - 30 cos 30° j = 5-15.0 i - 26.0 j6 kN F2 = =

30°

Ans.

12 5 1262 i + 1262 j 13 13

F1 = 30 kN

-10.0 i + 24.0 j kN

Ans.

Ans: F1 = { - 15.0i - 26.0j} kN F2 = { - 10.0i + 24.0j} kN 79

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2–59. Determine the magnitude of the resultant force and its direction measured counterclockwise from the positive x axis.

y

F2 = 26 kN 13

12 5

x

SOLUTION + F = ©F ; : Rx x + c FRy = ©Fy ;

FRx = - 30 sin 30° FRy

5 1262 = - 25 kN 13

30°

12 = - 30 cos 30° + 1262 = -1.981 kN 13

F1 = 30 kN

FR = 21 - 2522 + 1- 1.98122 = 25.1 kN f = tan-1 a

Ans.

1.981 b = 4.53° 25

u = 180° + 4.53° = 185°

Ans.

Ans: FR = 25.1 kN u = 185° 80

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z

*2–60. Determine the coordinate angleg for F2 and then express each force acting on the bracket as a Cartesian vector.

F1

450 N

45 30 45

y

60

x F2

Solution

600 N

Rectangular Components: Since cos2 a2 + cos2 b2 + cos2 g2 = 1, then 2 2 cos g2z = { 21 - cos 45° - cos 60° = {0.5. However, it is required g2 7 90°, thus, g2 = cos - 1 1 -0.52 = 120°. By resolving F1 and F2 into their x, y, and z components, as shown in Figs. a and b, respectively F1 and F2 can be expressed in Cartesian vector form as F1 = 450 cos 45° sin 30°( - i) + 450 cos 45° cos 30°( +j) + 450 sin 45°( +k) = 5 - 159i + 276j + 318k6 N

Ans.

= 5 424i + 300j - 300k6 N

Ans.

F2 = 600 cos 45°i + 600 cos 60°j + 600 cos 120°k

Ans: F1 = 5 - 159i + 276j + 318k6 N 81

F2 = 5424i + 300j - 300k6 N

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z

2–61. Determine the magnitude and coordinate direction angles of the resultant force acting on the bracket.

F1

450 N

45 30 45

y

60

x

Solution

F2 2

2

600 N

2

Rectangular Components: Since cos a2 + cos b2 + cos g2 = 1, then cos g2z = { 21 - cos 2 45° - cos 2 60° = {0.5. However, it is required a2 7 90°, thus, g2 = cos - 1 1 - 0.52 = 120°. By resolving F1 and F2 into their x, y, and z components, as shown in Figs. a and b, respectively F1 and F2 can be expressed in Cartesian vector form, as F1 = 450 cos 45° sin 30°( - i) + 450 cos 45° cos 30°( +j) + 450 sin 45°( + k) Ans.

= { - 159.10i + 275.57j + 318.20k6 N F2 = 600 cos 45°i + 600 cos 60°j + 600 cos 120°k

Ans.

= 5 424i + 300j - 300k} N

Resultant Force: By adding F1 and F2 vectorally, we obtain FR. FR = F1 + F2 = ( -159.10i + 275.57j + 318.20k) + (424.26i + 300j - 300k) = 5 265.16i + 575.57j + 18.20k6 N

The magnitude of FR is

FR = 2(FR)2x + (FR)2y + (FR)2z

= 2265.162 + 575.572 + 18.202 = 633.97 N = 634 N

Ans.

The coordinate direction angles of FR are a = cos - 1 c b = cos - 1 c g = cos - 1 c

(FR)x FR (FR)y FR (FR)z FR

d = cos - 1a d = cos - 1a d = cos - 1a

265.16 b = 65.3° 633.97

Ans.

575.57 b = 24.8° 633.97

Ans.

18.20 b = 88.4° 633.97

Ans.

Ans: F1 = { -159.10i + 275.57j + 318.20k} N F2 = {424i + 300j - 300k} N FR = 634 N a = 65.3 b = 24.8 g = 88.4 82

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2–62. z

Determine the magnitude and coordinate direction angles of the force F acting on the support. The component of F in the x9y plane is 7 kN. F

y

30 40 7 kN x

Solution Coordinate Direction Angles. The unit vector of F is uF = cos 30° cos 40°i - cos 30° sin 40°j + sin 30° k = {0.6634i - 0.5567j + 0.5 k} Thus,

cos a = 0.6634;

a = 48.44° = 48.4°

Ans.



cos b = - 0.5567; b = 123.83° = 124°

Ans.



cos g = 0.5;

Ans.

g = 60°

The magnitude of F can be determined from

F cos 30° = 7;

Ans.

F = 8.083 kN = 8.08 kN

Ans: a = 48.4° b = 124° g = 60° F = 8.08 kN 83

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2–63. The bolt is subjected to the force F, which has components acting along the x, y, z axes as shown. If the magnitude of F is 80 N, and a = 60° and g = 45°, determine the magnitudes of its components.

z

Fz g

F Fy

b y

a Fx

SOLUTION

x

cosb = 21 - cos2 a - cos2g = 21 - cos2 60° - cos2 45° b = 120° Fx = |80 cos 60°| = 40 N

Ans.

Fy = |80 cos 120°| = 40 N

Ans.

Fz = |80 cos 45°| = 56.6 N

Ans.

Ans: Fx = 40 N Fy = 40 N Fz = 56.6 N 84

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*2–64. The stock mounted on the lathe is subjected to a force of 60 N. Determine the coordinate direction angle b and express the force as a Cartesian vector.

z 60 N

45⬚

SOLUTION

b

1 = 2cos a + cos b + cos g 2

2

2

60⬚

1 = cos 60° + cos b + cos 45° 2

2

2

y x

cos b = ; 0.5 b = 60°, 120° Use b = 120°

Ans.

F = 60 N(cos 60°i + cos 120°j + cos 45°k) = {30i - 30j + 42.4k} N

Ans.

Ans: b = 120° F = {30i - 30j + 42.4k} N 85

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2–65. Specify the magnitude F3 and directions a3, b3, and g3 of F3 so that the resultant force of the three forces is FR = {9j} kN. z 10 kN

F2

F3 g3

5

13

b3

12

a3

y 30 F1

12 kN

x

SOLUTION Initial Guesses:

F 3x

Given

§ F3x · § 0 · ¨ ¸ ¨ ¸ ¨ F3y ¸  F1¨ cos  T ¸  ¨F ¸ © sin  T ¹ © 3z ¹

FR

F1

12 kN

c

5

F2

10 kN

d

12

T

1 kN

F 3y

1 kN

F 3z

1 kN

§ d · ¨0¸ ¸ 2 2¨ c d © c ¹ F2

30

FR

§0· ¨ 9 ¸ kN ¨ ¸ ©0¹

§ F3x · ¨ ¸ ¨ F3y ¸ ¨F ¸ © 3z ¹ § D3 · ¨ ¸ ¨ E3 ¸ ¨J ¸ © 3¹

Find  F3x  F 3y  F3z

§ F3 · acos ¨ ¸ © F3 ¹

F3

§ D3 · ¨ ¸ ¨ E3 ¸ ¨J ¸ © 3¹

§ F3x · ¨ ¸ ¨ F3y ¸ ¨F ¸ © 3z ¹

F3

§ 15.5 · ¨ 98.4 ¸ deg ¨ ¸ © 77.0 ¹

§ 9.2 · ¨ 1.4 ¸ kN ¨ ¸ © 2.2 ¹

F3

9.6 kN

Ans.

Ans.

Ans: F3 = a3 = b3 = g3 = 86

9.6 kN 15.5 98.4 77.0

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2–66. Determine the magnitude and coordinate direction angles of the resultant force, and sketch this vector on the coordinate system.

z F2  125 N 3

5 4

20

x

y 60 45

60 F1  400 N

Solution Cartesian Vector Notation. For F1 and F2, F1 = 400 (cos 45°i + cos 60°j - cos 60°k) = {282.84i + 200j - 200k} N 4 4 3 F2 = 125 c (cos 20°)i - (sin 20°)j + k d = {93.97i - 34.20j + 75.0k} 5 5 5

Resultant Force.

FR = F1 + F2 = {282.84i + 200j - 200k} + {93.97i - 34.20j + 75.0k} = {376.81i + 165.80j - 125.00k} N The magnitude of the resultant force is



FR = 2(FR)2x + (FR)2y + (FR)2z = 2376.812 + 165.802 + ( -125.00)2 = 430.23 N = 430 N

Ans.

The coordinate direction angles are

cos a =



cos b =



cos g =

(FR)x FR (FR)y FR (FR)z FR

=

376.81 ; 430.23

a = 28.86° = 28.9°

Ans.

=

165.80 ; 430.23

b = 67.33° = 67.3°

Ans.

=

- 125.00 ; 430.23

g = 106.89° = 107°

Ans.

Ans: FR =  a =  b =  g = 87

430 N 28.9° 67.3° 107°

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2–67. z

Determine the magnitude and coordinate direction angles of the resultant force, and sketch this vector on the coordinate system. F2  525 N

60

120

45 x

y 5

4 3

F1  450 N

Solution Cartesian Vector Notation. For F1 and F2, 3 4 F1 = 450 a j - kb = {270j - 360k} N 5 5

F2 = 525 (cos 45°i + cos 120°j + cos 60°k) = {371.23i - 262.5j + 262.5k} N

Resultant Force. FR = F1 + F2 = {270j - 360k} + {371.23i - 262.5j + 262.5k} = {371.23i + 7.50j - 97.5k} N The magnitude of the resultant force is



FR = 2(FR)2x + (FR)2y + (FR)2z = 2371.232 + 7.502 + ( - 97.5)2 = 383.89 N = 384 N

Ans.

The coordinate direction angles are

cos a =



cos b =



cos g =

(FR)x FR (FR)y FR (FR)z FR

=

371.23 ; 383.89

a = 14.76° = 14.8°

Ans.

=

7.50 ; 383.89

b = 88.88° = 88.9°

Ans.

=

- 97.5 ; 383.89

g = 104.71° = 105°

Ans.

Ans: FR = 384 N 371.23 cos a = ; a = 14.8° 383.89 7.50 cos b = ;  b = 88.9° 383.89 -97.5 cos g = ; g = 105° 383.89 88

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*2–68. Determine the magnitude and coordinate direction angles of F3 so that the resultant of the three forces acts along the positive y axis and has a magnitude of 600 N.

z F3

F1  180 N

SOLUTION

40 x

Given: F

600 N

F1

180 N

F2

300 N

D1

30 

D2

40 

y

30

F2  300 N

Initial guesses:

D

40

J

E

50 

F3

50  45 N

Given F Rx = 6F x;

0

F1  F2 cos  D 1 sin  D 2  F 3 cos  D

F Ry = 6F y;

F

F2 cos  D 1 cos  D 2  F3 cos  E

F Rz = 6F z;

0

F2 sin  D 1  F3 cos  J

cos  D  cos  E  cos  J 2

§ F3 · ¨ ¸ ¨D¸ ¨E ¸ ¨ ¸ ©J ¹

Find  F 3  D  E  J

2

F3

2

1

428 N

§¨ D ·¸ ¨E ¸ ¨J ¸ © ¹

§ 88.3 · ¨ 20.6 ¸ deg ¨ ¸ © 69.5 ¹

Ans: F3 = 428 N a = 88.3° b = 20.6° g = 69.5° 89

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2–69. Determine the magnitude and coordinate direction angles of F3 so that the resultant of the three forces is zero.

z F3

F1  180 N

SOLUTION

40 30

x

Given: F1

180 N

D1

30

F2

300 N

D2

40

Initial guesses:

y

F2  300 N

D

40

J

E

50 

F3

50 45 N

Given F Rx = 6F x;

0

F1  F2 cos  D 1 sin  D 2  F 3 cos  D

F Ry = 6F y;

0

F2 cos  D 1 cos  D 2  F3 cos  E

F Rz = 6F z;

0

F2 sin  D 1  F3 cos  J

cos  D  cos  E  cos  J 2

§ F3 · ¨ ¸ ¨D¸ ¨E ¸ ¨ ¸ ©J ¹

Find  F 3  D  E  J

2

F3

250 N

2

1

§¨ D ·¸ ¨E ¸ ¨J ¸ © ¹

§ 87.0 · ¨ 142.9 ¸ deg ¨ ¸ © 53.1 ¹

Ans.

Ans: F3 = 250 N a = 87.0 b = 142.9 g = 53.1 90

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2–70. The screw eye is subjected to the two forces shown. Express each force in Cartesian vector form and then determine the resultant force. Find the magnitude and coordinate direction angles of the resultant force.

z

F1 = 300 N

60° 120°

45° y

45°

SOLUTION x

F1 = 300(- cos 60° sin 45° i + cos 60° cos 45° j + sin 60°k) = {- 106.07 i + 106.07 j + 259.81 k} N

60°

F2 = 500 N

= {-106 i + 106 j + 260 k} N

Ans.

F2 = 500(cos 60° i + cos 45° j + cos 120° k) = {250.0 i + 353.55 j - 250.0k} N = {250 i + 354 j - 250 k} N

Ans.

FR = F1 + F2 = - 106.07 i + 106.07 j + 259.81 k + 250.0 i + 353.55 j - 250.0 k = 143.93 i + 459.62 j + 9.81k = {144 i + 460 j + 9.81 k} N

Ans.

FR = 2143.932 + 459.622 + 9.812 = 481.73 N = 482 N uFR =

Ans.

143.93i + 459.62j + 9.81k FR = 0.2988i + 0.9541j + 0.02036k = FR 481.73

cos a = 0.2988

a = 72.6°

Ans.

cos b = 0.9541

b = 17.4°

Ans.

cos g

g

88.8°

Ans.

0.02036

Ans: F1 = { - 106i + 106j + 260k} N F2 = {250i + 354j - 250k} N FR = {144i + 460j + 9.81k} N FR = 482 N a = 72.6° b = 17.4° g = 88.8° 91

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2–71. Determine the coordinate direction angles of F1.

z

F1 = 300 N

60° 120°

45° y

SOLUTION 45°

F1 = 300( - cos 60° sin 45° i + cos 60° cos 45° j + sin 60° k) = { - 106.07 i + 106.07 j + 259.81 k} N

x

60°

= { - 106 i + 106 j + 260 k} N F2 = 500 N

F1 = - 0.3536 i + 0.3536 j + 0.8660 k u1 = 300 a1 = cos-1 (-0.3536) = 111°

Ans.

b 1 = cos-1 (0.3536) = 69.3°

Ans.

g1 = cos-1 (0.8660) = 30.0°

Ans.

Ans: a1 = 111° b1 = 69.3° g1 = 30.0° 92

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*2–72. z

Determine the magnitude and coordinate direction angles of the resultant force, and sketch this vector on the coordinate system.

F1 = 400 N 60

135 20

60

60

y

x F2  500 N

Solution Cartesian Vector Notation. For F1 and F2, F1 = 400 (sin 60° cos 20°i - sin 60° sin 20°j + cos 60°k) = {325.52i - 118.48j + 200k} N F2 = 500 (cos 60°i + cos 60°j + cos 135°k) = {250i + 250j - 353.55k} N Resultant Force. FR = F1 + F2 = (325.52i - 118.48j + 200k) + (250i + 250j - 353.55k) = {575.52i + 131.52j - 153.55 k} N The magnitude of the resultant force is



FR = 2(FR)2x + (FR)2y + (FR)2z = 2575.522 + 131.522 + ( -153.55)2 = 610.00 N = 610 N

 Ans.

The coordinate direction angles are

cos a =



cos b =



cos g =

(FR)x FR (FR)y FR (FR)z FR

=

575.52 610.00

a = 19.36° = 19.4°

Ans.

=

131.52 610.00

b = 77.549° = 77.5°

Ans.

=

- 153.55 610.00

g = 104.58° = 105°

Ans.

Ans: FR = a = b = g = 93

610 N 19.4° 77.5° 105°

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2–73. z

Determine the coordinate direction angles of force F1. F1 ⫽ 600 N

F2 ⫽ 450 N

5 3

4

45⬚ 30⬚

SOLUTION

x

y

Rectangular Components: By referring to Figs. a, the x, y, and z components of F1 can be written as 4 (F1)x = 600 a b cos 30° N 5

4 (F1)y = 600 a b sin 30° N 5

3 (F1)z = 600a b N 5

Thus, F1 expressed in Cartesian vector form can be written as F1 = 600 e

4 4 3 cos 30°(+i) + sin 30°(-j) + (+k) f N 5 5 5

= 600[0.6928i - 0.4j + 0.6k] N Therefore, the unit vector for F1 is given by uF1 =

F1 600(0.6928i - 0.4j + 0.6k = = 0.6928i - 0.4j + 0.6k F1 600

The coordinate direction angles of F1 are a = cos-1(uF1)x = cos-1(0.6928) = 46.1°

Ans.

b = cos-1(uF1)y = cos-1(-0.4) = 114°

Ans.

g = cos-1(uF1)z = cos-1(0.6) = 53.1°

Ans.

Ans: a = 46.1 b = 114 g = 53.1 94

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2–74. z

Determine the magnitude and coordinate direction angles of the resultant force acting on the eyebolt. F1 ⫽ 600 N

F2 ⫽ 450 N

5 3

4

45⬚ 30⬚

x

y

SOLUTION Force Vectors: By resolving F1 and F2 into their x, y, and z components, as shown in Figs. a and b, respectively, they are expressed in Cartesian vector form as 4 4 3 F1 = 600 a bcos 30°(+ i) + 600a bsin 30°( -j) + 600 a b(+k) 5 5 5 = 5415.69i - 240j + 360k6 N F2 = 0i + 450 cos 45°(+j) + 450 sin 45°( +k) = 5318.20j + 318.20k6 N Resultant Force: The resultant force acting on the eyebolt can be obtained by vectorally adding F1 and F2. Thus, FR = F1 + F2 = (415.69i - 240j + 360k) + (318.20j + 318.20k) = 5415.69i + 78.20j + 678.20k6 N The magnitude of FR is given by FR = 3(FR)x 2 + (FR)y 2 + (FR)z 2 = 3(415.69)2 + (78.20)2 + (678.20)2 = 799.29 N = 799 N

Ans.

The coordinate direction angles of FR are a = cos-1 c

(FR)x 415.69 d = cos-1 a b = 58.7° FR 799.29

b = cos-1 c

(FR)y

g = cos-1 c

(FR)z

FR FR

Ans.

d = cos-1 a

78.20 b = 84.4° 799.29

Ans.

d = cos-1 a

678.20 b = 32.0° 799.29

Ans.

Ans: FR = 799 N a = 58.7 b = 84.4 g = 32.0 95

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2–75. Express each force in Cartesian vector form.

F3  200 N

5

F1  90 N 3

F2  150 N

60

y

4

45

Solution x

Cartesian Vector Notation. For F1, F2 and F3, 4 3 F1 = 90 a i + kb = {72.0i + 54.0k} N 5 5

Ans.

F2 = 150 (cos 60° sin 45°i + cos 60° cos 45°j + sin 60°k) = {53.03i + 53.03j + 129.90k} N Ans.

= {53.0i + 53.0j + 130k} N F3 = {200 k}

Ans.

Ans: F1 = {72.0i + 54.0k} N F2 = {53.0i + 53.0j + 130k} N F3 = {200 k} 96

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*2–76. Determine the magnitude and coordinate direction angles of the resultant force, and sketch this vector on the coordinate system.

F3  200 N

5

F1  90 N 3

F2  150 N

60

y

4

45

Solution x

Cartesian Vector Notation. For F1, F2 and F3, 4 3 F1 = 90 a i + kb = {72.0i + 54.0k} N 5 5

F2 = 150 (cos 60° sin 45°i + cos 60° cos 45°j + sin 60°k) = {53.03i + 53.03j + 129.90k} N F3 = {200 k} N Resultant Force. F = F1 + F2 + F3 = (72.0i + 54.0k) + (53.03i + 53.03j + 129.90k) + (200k) = {125.03i + 53.03j + 383.90} N The magnitude of the resultant force is



FR = 2(FR)2x + (FR)2y + (FR)2z = 2125.032 + 53.032 + 383.902 = 407.22 N = 407 N



Ans.

And the coordinate direction angles are

cos a =



cos b =



cos g =

(FR)x FR (FR)y FR (FR)z FR

=

125.03 ; 407.22

a = 72.12° = 72.1°

Ans.

=

53.03 ; 407.22

b = 82.52° = 82.5°

Ans.

=

383.90 ; 407.22

g = 19.48° = 19.5°

Ans.

Ans: FR = a = b = g = 97

407 N 72.1° 82.5° 19.5°

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2–77. z

The cables attached to the screw eye are subjected to the three forces shown. Express each force in Cartesian vector form and determine the magnitude and coordinate direction angles of the resultant force.

F3 = 250 N

40°

60°

120°

45°

y

60°

SOLUTION Cartesian Vector Notation: x

F1 = 3505sin 40°j + cos 40°k6 N

F1 = 350 N

60° 45° F2 = 100 N

= 5224.98j + 268.12k6 N = 5225j + 268k6 N

Ans.

F2 = 1005cos 45°i + cos 60°j + cos 120°k6 N = 570.71i + 50.0j - 50.0k6 N

= 570.7i + 50.0j - 50.0k6 N

Ans.

F3 = 2505cos 60°i + cos 135°j + cos 60°k6 N = 5125.0i - 176.78j + 125.0k6 N

= 5125i - 177j + 125k6 N

Ans.

Resultant Force: FR = F1 + F2 + F3 = 5170.71 + 125.02i + 1224.98 + 50.0 - 176.782j + 1268.12 - 50.0 + 125.02k6 N

= 5195.71i + 98.20j + 343.12k6 N

The magnitude of the resultant force is FR = 2F2Rx + F2Ry + F2Rz = 2195.712 + 98.202 + 343.122 = 407.03 N = 407 N

Ans.

The coordinate direction angles are cos a = cos b = cos g =

FRx FR FRy FR FRz FR

=

195.71 407.03

a = 61.3°

Ans.

=

98.20 407.03

b = 76.0°

Ans.

=

343.12 407.03

g = 32.5°

Ans.

98

Ans: F1 = {225j + 268k} N F2 = {70.7i + 50.0j - 50.0k} N F3 = {125i - 177j + 125k} N FR = 407 N a = 61.3 b = 76.0 g = 32.5

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2–78. The mast is subjected to the three forces shown. Determine the coordinate direction angles a1, b 1, g1 of F1 so that the resultant force acting on the mast is FR = 5350i6 N.

z F1 g1 a1

b1

F3 ⫽ 300 N y x

F2 ⫽ 200 N

Solution F1 = 500 cos a1i + 500 cos b1j + 500 cos g1k FR = F1 + ( -300j) + ( - 200k) 350i = 500 cosa1i + (500 cosb1 - 300)j + (500 cosg1 - 200)k 350 = 500 cos a1;

a1 = 45.6°

Ans.

0 = 500 cos b1 - 300;

b1 = 53.1°

Ans.

0 = 500 cos g1 - 200;

g1 = 66.4°

Ans.

Ans: a1 = 45.6 b1 = 53.1 g1 = 66.4 99

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2–79. The mast is subjected to the three forces shown. Determine the coordinate direction angles a1, b 1, g1 of F1 so that the resultant force acting on the mast is zero.

z F1 g1 a1

b1

F3 ⫽ 300 N y x

F2 ⫽ 200 N

Solution F1 = 5500 cos a1i + 500 cos b1j + 500 cos g1k6N F2 = 5 - 200k6 N F3 = 5 - 300j 6 N

FR = F1 + F2 + F3 = 0 500 cos a1 = 0;

a1 = 90°

Ans.

500 cos b1 = 300;

b1 = 53.1°

Ans.

500 cos g1 = 200;

g1 = 66.4°

Ans.

Ans: a1 = 90 b1 = 53.1 g1 = 66.4 100

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*2–80. z

Express each force as a Cartesian vector.

30⬚ 30⬚

x

SOLUTION Rectangular Components: By referring to Figs. a and b, the x, y, and z components of F1 and F2 can be written as (F1)x = 300 cos 30° = 259.8 N

(F2)x = 500 cos 45° sin 30° = 176.78 N

(F1)y = 0

(F2)y = 500 cos 45° cos 30° = 306.19 N

(F1)t = 300 sin 30° = 150 N

(F2)z = 500 sin 45° = 353.55 N

F1 ⫽ 300 N

y

45⬚

F2 ⫽ 500 N

Thus, F1 and F2 can be written in Cartesian vector form as

F1 = 259.81(+i) + 0j + 150(-k) = {260i - 150k} N

Ans.

F2 = 176.78(+i) + 306.19(+j) + 353.55(-k) = 2{177i + 306j - 354k} N

Ans.

Ans:

F1 = {260i - 150k} N F2 = 2{177i + 306j - 354k} N 101

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2–81. z

Determine the magnitude and coordinate direction angles of the resultant force acting on the hook.

30⬚ 30⬚

x

SOLUTION Force Vectors: By resolving F1 and F2 into their x, y, and z components, as shown in Figs. a and b, respectively, F1 and F2 can be expessed in Cartesian vector form as

F1 ⫽ 300 N

y

45⬚

F1 = 300 cos 30°(+i) + 0j + 300 sin 30°( -k) F2 ⫽ 500 N

= {259.81i - 150k} N F2 = 500 cos 45°sin 30°( +i) + 500 cos 45° cos 30°(+j) + 500 sin 45°( -k) = {176.78i - 306.19j - 353.55k} N Resultant Force: The resultant force acting on the hook can be obtained by vectorally adding F1 and F2. Thus,

FR = F1 + F2 = (259.81i - 150k) + (176.78i + 306.19j - 353.55k) = {436.58i) + 306.19j - 503.55k} N The magnitude of FR is FR = 2(FR)x2 + (FR)y 2(FR)z 2 = 2(436.58)2 + (306.19)2 + (-503.55)2 = 733.43 N = 733 N The coordinate direction angles of FR are (FR)x 436.58 d = cos-1 a b = 53.5° ux = cos-1 c FR 733.43 uy = cos-1 c

(FR)y

uz = cos-1 c

(FR)z

FR FR

Ans.

Ans.

d = cos-1 a

306.19 b = 65.3° 733.43

Ans.

d = cos-1 a

-503.55 b = 133° 733.43

Ans.

Ans: FR = 733 N ux = 53.5 uy = 65.3 uz = 133 102

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2–82. Three forces act on the ring. If the resultant force FR has a magnitude and direction as shown, determine the magnitude and the coordinate direction angles of force F3.

z F3 F2

F1

SOLUTION

80 N

5 3

Cartesian Vector Notation:

FR

110 N

4

45

120 N

y 30

FR = 120{cos 45°sin 30°i + cos 45°cos 30°j + sin 45°k} N = {42.43i + 73.48j + 84.85k} N x

3 4 F1 = 80 b i + k r N = {64.0i + 48.0k} N 5 5 F2 = { - 110k} N F3 = {F3x i + F3y j + F3z k} N Resultant Force: FR = F1 + F2 + F3 {42.43i + 73.48j + 84.85k} =

E A 64.0 + F3 x B i + F3 y j + A 48.0 - 110 + F3 z B k F

Equating i, j and k components, we have 64.0 + F3 x = 42.43

F3x = -21.57 N F3 y = 73.48 N

48.0 - 110 + F3 z = 84.85

F3 z = 146.85 N

The magnitude of force F3 is F3 = 2F 23 x + F 23 y + F 23 z = 2(- 21.57)2 + 73.482 + 146.852 = 165.62 N = 166 N

Ans.

The coordinate direction angles for F3 are cos a = cos b =

F3 x F3 F3 y F3

cos g = =

=

- 21.57 165.62

a = 97.5°

Ans.

=

73.48 165.62

b = 63.7°

Ans.

g = 27.5°

Ans.

F3 z F3

=

146.85 165.62

Ans: F3 = 166 N a = 97.5 b = 63.7 g = 27.5 103

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2–83. Determine the coordinate direction angles of F1 and FR.

z F3 F2

FR

110 N

120 N

SOLUTION Unit Vector of F1 and FR : u F1

F1

80 N

5 3

3 4 = i + k = 0.8i + 0.6k 5 5

4

45

y 30

uR = cos 45° sin 30°i + cos 45° cos 30°j + sin 45°k = 0.3536i + 0.6124j + 0.7071k

x

Thus, the coordinate direction angles F1 and FR are cos aF1 = 0.8

aF1 = 36.9°

Ans.

cos b F1 = 0

b F1 = 90.0°

Ans.

cos gF1 = 0.6

gF1 = 53.1°

Ans.

cos aR = 0.3536

aR = 69.3°

Ans.

cos b R = 0.6124

b R = 52.2°

Ans.

cos gR = 0.7071

gR = 45.0°

Ans.

Ans: aF1 = 36.9 bF1 = 90.0 gF1 = 53.1 aR = 69.3 bR = 52.2 gR = 45.0 104

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*2–84. The pole is subjected to the force F, which has components acting along the x, y, z axes as shown. If the magnitude of F is 3 kN, b = 30°, and g = 75°, determine the magnitudes of its three components.

z Fz

F

g b

Fy

y

a

SOLUTION

Fx

cos2 a + cos2 b + cos2 g = 1

x

cos a + cos 30° + cos 75° = 1 2

2

2

a = 64.67° Fx = 3 cos 64.67° = 1.28 kN

Ans.

Fy = 3 cos 30° = 2.60 kN

Ans.

Fz = 3 cos 75° = 0.776 kN

Ans.

Ans: Fx = 1.28 kN Fy = 2.60 kN Fz = 0.776 kN 105

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2–85. The pole is subjected to the force F which has components Fx = 1.5 kN and Fz = 1.25 kN. If b = 75°, determine the magnitudes of F and Fy.

z Fz

F

g b

Fy

y

a

SOLUTION Fx

cos2 a + cos2 b + cos2 g = 1 a

1.25 2 1.5 2 b + cos2 75° + a b = 1 F F

x

F = 2.02 kN

Ans.

Fy = 2.02 cos 75° = 0.523 kN

Ans.

Ans: F = 2.02 kN, Fy = 0.523 kN 106

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2–86. Determine the lengths of wires AD, BD, and CD. The ring at D is midway between A and B.

z C

D

A

SOLUTION

2m

0.5 m

2 + 0 0 + 2 1.5 + 0.5 , , b m = D(1, 1, 1) m Da 2 2 2

B 0.5 m y

2m 1.5 m

rAD = (1 - 2)i + (1 - 0)j + (1 - 1.5)k = -1i + 1j - 0.5k

x

rBD = (1 - 0)i + (1 - 2)j + (1 - 0.5)k = 1i - 1j + 0.5k rCD = (1 - 0)i + (1 - 0)j + (1 - 2)k = 1i + 1j - 1k rAD = 2(-1)2 + 12 + (- 0.5)2 = 1.50 m

Ans.

rBD = 212 + ( - 1)2 + 0.52 = 1.50 m

Ans.

rCD = 212 + 12 + (- 1)2 = 1.73 m

Ans.

Ans: rAD = 1.50 m rBD = 1.50 m rCD = 1.73 m 107

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2–87. y

Determine the length of the connecting rod AB by first formulating a Cartesian position vector from A to B and then determining its magnitude.

B

300 mm

Solution O

Position Vector. The coordinates of points A and B are A( -150 cos 30°, - 150 sin 30°) mm and B(0, 300) mm respectively. Then

30

rAB = [0 - ( -150 cos 30°)]i + [300 - ( - 150 sin 30°)]j

x

A 150 mm

= {129.90i + 375j} mm Thus, the magnitude of rAB is

rAB = 2129.902 + 3752 = 396.86 mm = 397 mm

Ans.

Ans: rAB = 397 mm 108

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*2–88. z

If F = 5350i - 250j - 450k6 N and cable AB is 9 m long, determine the x, y, z coordinates of point A.

A F

B

SOLUTION Position Vector: The position vector rAB, directed from point A to point B, is given by

x

y

z

x y

rAB = [0 - ( -x)]i + (0 - y)j + (0 - z)k = xi - yj - zk Unit Vector: Knowing the magnitude of rAB is 9 m, the unit vector for rAB is given by uAB =

xi - yj - zk rAB = rAB 9

The unit vector for force F is uF =

F = F

350i - 250j - 450k 3502 + ( -250)2 + ( -450)2

= 0.5623i - 0.4016j - 0.7229k

Since force F is also directed from point A to point B, then uAB = uF xi - yj - zk = 0.5623i - 0.4016j - 0.7229k 9 Equating the i, j, and k components, x = 0.5623 9

x = 5.06 m

Ans.

-y = -0.4016 9

y = 3.61 m

Ans.

-z = 0.7229 9

z = 6.51 m

Ans.

Ans: x = 5.06 m y = 3.61 m z = 6.51 m 109

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2–89. z

Determine the magnitude and coordinate direction angles of the resultant force acting at A.

B

C

2m FB ⫽ 600 N

SOLUTION

0.5 m

Force Vectors: The unit vectors uB and uC of FB and FC must be determined first. From Fig. a

uC =

FC ⫽ 450 N 1.5 m

1m 1.5 m

(-1.5 - 0.5)i + [-2.5 - (-1.5)]j + (2 - 0)k rB uB = = rB 3(-1.5 - 0.5)2 + [-2.5 - (-1.5)]2 + (2 - 0)2 = -

A

3.5 m

0.5 m y

x

2 1 2 i - j + k 3 3 3

(-1.5 - 0.5)i + [0.5 - ( -1.5)]j + (3.5 - 0)k rC = rC 3(-1.5 - 0.5)2 + [0.5 - ( -1.5)]2 + (3.5 - 0)2 = -

4 4 7 i + j + k 9 9 9

Thus, the force vectors FB and FC are given by 2 1 2 FB = FB uB = 600a - i - j + kb = 5-400i - 200j + 400k6 N 3 3 3 4 4 7 FC = FC uC = 450 a - i + j + kb = 5 -200i + 200j + 350k6 N 9 9 9 Resultant Force: FR = FB + FC = (-400i - 200j + 400k) + (-200i + 200j + 350k) = 5-600i + 750k6 N The magnitude of FR is FR = 3(FR)x 2 + (FR)y 2 + (FR)z 2 = 3(-600)2 + 02 + 7502 = 960.47 N = 960 N The coordinate direction angles of FR are a = cos-1 c

(FR)x -600 d = cos-1 a b = 129° FR 960.47

b = cos-1 c

(FR)y

g = cos-1 c

(FR)z

FR FR

d = cos-1 a d = cos-1 a

Ans.

0 b = 90° 960.47

Ans.

760 b = 38.7° 960.47

Ans.

Ans: a = 129 b = 90 g = 38.7 110

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z

2–90. The door is held opened by means of two chains. If the tension in AB and CD is FA = 300 N and FC = 250 N, respectively, express each of these forces in Cartesian vector form.

C 1.5 m

2.5 m FC = 250 N A FA = 300 N

D

30 1m

x

0.5 m B y

Solution Unit Vector: First determine the position vector rAB and rCD. The coordinates of points A and C are A30, -(1 + 1.5 cos 30°), 1.5 sin 30° 4 m = A(0, -2.299, 0.750) m

Then

C[ - 2.50, - (1 + 1.5 cos 30°), 1.5 sin 30° 4 m = C( - 2.50, -2.299, 0.750) m rAB = 5 (0 - 0)i + 30 - ( - 2.299) 4j + (0 - 0.750k6 m = 5 2.299j - 0.750k6 m

rAB = 22.2992 + ( -0.750)2 = 2.418 m uAB =

2.299j - 0.750k rAB = = 0.9507j - 0.3101k rAB 2.418

rCD = 5[ -0.5 - ( - 2.5) 4i + 30 - ( - 2.299) 4j + (0 - 0.750)k6 m = 5 2.00i + 2.299j - 0.750k6 m

rCD = 22.002 + 2.2992 + ( - 0.750)2 = 3.318 m uCD =

2.00i + 2.299j - 0.750k rCD = = 0.6373i + 0.7326j - 0.2390k rCD 3.318

Force Vector: FA = FAuAB = 30050.9507j - 0.3101k} N = 5285.21j - 93.04k6 N

Ans.

= 5285j - 93.0k6 N

FC = FCuCD = 2505 0.6373i + 0.7326j - 0.2390k6 N = 5159.33i + 183.15j - 59.75k6 N

Ans.

= 5159i + 183j - 59.7k6 N

Ans: FA = {285j - 93.0k} N FC = {159i + 183j - 59.7k} N 111

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2–91. 560 N and 700 N, determine the magnitude If and coordinate direction angles of the resultant force acting on the flag pole.

6m

A FB FC

SOLUTION Force Vectors: The unit vectors u and u of F and F must be determined first. From Fig. a u u

(2

r

0)i

( 3 2

(2 0) (3 0)i

r

(0

6)k

2

0)2

(2

2 i 7

2

( 3 0) (0 6) (2 0)j (0 6)k

0)2

(3

0)j

6)2

(0

3 j 7

3 i 7

2 j 7

2m B

6 k 7

3m

6 k 7

x

3m

2m C

Thus, the force vectors F and F are given by F

u

560

2 i 7

3 j 7

6 k 7

160i

240j

480k N

F

u

700

3 i 7

2 j 7

6 k 7

300i

200j

600k N

Resultant Force: F

F

F

460i

(160i 40j

240j

480k)

(300i

200j

600k)

1080k N

The magnitude of F is (

)

2

(460)2

(

)

2

( 40)2

(

)

2

( 1080)2

1174.56 N

Ans.

1.17 kN

The coordinate direction angles of F are cos

1

cos

1

cos

1

(

)

(

)

(

)

cos

1

460 1174.56

66.9°

Ans.

cos

1

40 1174.56

92.0°

Ans.

cos

1

1080 1174.56

157°

Ans.

Ans: FR = 1.17 kN a = 66.9° b = 92.0° g = 157° 112

y

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*2–92. If 700 N, and 560 N, determine the magnitude and coordinate direction angles of the resultant force acting on the flag pole.

6m

A FB FC

SOLUTION Force Vectors: The unit vectors u and u of F and F must be determined first. From Fig. a u u

(2

r

0)i 0)2

(2 (3

r

( 3

0)i

(2

0)

(0

0)2

( 3

2

(3

0)j 0)j 0)

6)2

(0

(0 2

(2

6)k 6)k 2

(0

6)

2 i 7

3 j 7

3 i 7

2 j 7

2m B 3m

6 k 7 x

6 k 7

3m

2m C

Thus, the force vectors F and F are given by F

u

700

2 i 7

3 j 7

6 k 7

200i

300j

600k N

F

u

560

3 i 7

2 j 7

6 k 7

240i

160j

480k N

Resultant Force: F

F

F

440i

(200i 140j

300j

600k)

(240i

160j

480k)

1080k N

The magnitude of F is (

)

2

(440)2

(

)

2

( 140)2

(

)

2

( 1080)2

1174.56 N

1.17 kN

Ans.

The coordinate direction angles of F are cos

1

cos

1

cos

1

(

)

(

)

(

)

cos

1

440 1174.56

68.0°

Ans.

cos

1

140 1174.56

96.8°

Ans.

cos

1

1080 1174.56

157°

Ans.

Ans: FR = 1.17 kN a = 68.0° b = 96.8° g = 157° 113

y

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2–93. z

Express each of the forces in Cartesian vector form and determine the magnitude and coordinate direction angles of the resultant force.

0.75 m A FAB  250 N FAC  400 N

3m

y 40

2m 1m

Unit Vectors. The coordinates for points A, B and C are (0, - 0.75, 3) m, B(2 cos 40°, 2 sin 40°, 0) m and C(2, - 1, 0) m respectively. uAB =

B

C

Solution

2m

x

(2 cos 40° - 0)i + [2 sin 40° - ( - 0.75)]j + (0 - 3)k rAB = rAB 2(2 cos 40° - 0)2 + [2 sin 40° - ( - 0.75)]2 + (0 - 3)2 = 0.3893i + 0.5172j - 0.7622k

uAC =

(2 - 0)i + [ - 1 - ( - 0.75)]j + (0 - 3)k rAC = rAC 2(2 - 0)2 + [ - 1 - ( - 0.75)]2 + (0 - 3)2 = 0.5534i - 0.0692j - 0.8301k

Force Vectors FAB = FAB uAB = 250 (0.3893i + 0.5172j - 0.7622k) = {97.32i + 129.30j - 190.56k} N = {97.3i + 129j - 191k} N

Ans.



FAC = FAC uAC = 400 (0.5534i - 0.06917j - 0.8301k) = {221.35i - 27.67j - 332.02k} N = {221i - 27.7j - 332k} N

Ans.



Resultant Force FR = FAB + FAC = {97.32i + 129.30j - 190.56k} + {221.35i - 27.67j - 332.02k} = {318.67i + 101.63j - 522.58 k} N The magnitude of FR is FR = 2(FR)2x + (FR)2y + (FR)2z = 2318.672 + 101.632 + ( -522.58)2 = 620.46 N = 620 N

And its coordinate direction angles are

cos a =



cos b =



cos g =

(FR)x FR (FR)y FR (FR)z FR

=

318.67 ; 620.46

101.63 ; = 620.46 =

-522.58 ; 620.46

Ans.

a = 59.10° = 59.1° b = 80.57° = 80.6°

Ans.

g = 147.38° = 147°

Ans.

114

Ans: FAB FAC  FR cos a cos b cos g

= = = = = =

{97.3i - 129j - 191k} N {221i - 27.7j - 332k} N 620 N 59.1° 80.6° 147°

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2–94. The 8-m-long cable is anchored to the ground at A. If x = 4 m and y = 2 m, determine the coordinate z to the highest point of attachment along the column.

z

B

Solution

z

r = {4i + 2j + zk} m r = 2(4)2 + (2)2 + (z)2 = 8

y

Ans.

z = 6.63 m

x

y

A

x

Ans: 6.63 m 115

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2–95. z

The 8-m-long cable is anchored to the ground at A. If z = 5 m, determine the location + x, + y of point A. Choose a value such that x = y.

B

Solution

z

r = {xi + yj + 5k} m r = 2(x)2 + (y)2 + (5)2 = 8

y

x = y, thus

2x2 = 82 - 52

x

y

A

x

Ans.

x = y = 4.42 m

Ans:  4.42 m 116

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*2–96. z

Determine the magnitude and coordinate direction angles of the resultant force acting at point A on the post. FAC 150 N

C

4 5

A

FAB  200 N

3m 3m

3

y O

2m 4m

Solution Unit Vector. The coordinates for points A, B and C are A(0, 0, 3) m, B(2, 4, 0) m and C( - 3, - 4, 0) m respectively

B

x

rAB = (2 - 0)i + (4 - 0)j + (0 - 3)k = {2i + 4j - 3k} m uAB =

2i + 4j - 3k rAB 2 4 3 = = i + j k rAB 222 + 42 + ( - 3)2 229 229 229

rAC = ( -3 - 0)i + ( -4 - 0)j + (0 - 3)k = { -3i - 4j - 3k} m uAC =

-3i - 4j - 3k rAC 3 4 3 = = i j k 2 2 2 rAC 2( - 3) + ( - 4) + ( - 3) 234 234 234

Force Vectors

FAB = FAB uAB = 200 a

2 229

i +

4 229

j -

= {74.28i + 148.56j - 111.42k} N

FAC = FAC uAC = 150 a -

3 234

i -

4 234

= { - 77.17i - 102.90j - 77.17k} N

3 229

j -

kb

3 234

kb

Resultant Force FR = FAB + FAC = {74.28i + 148.56j - 111.42k} + { -77.17i - 102.90j - 77.17k} = { -2.896i + 45.66j - 188.59 k} N The magnitude of the resultant force is FR = 2(FR)2x + (FR)2y + (FR)2z = 2( - 2.896)2 + 45.662 + ( - 188.59)2

 Ans.

= 194.06 N = 194 N

And its coordinate direction angles are

cos a =



cos b =



cos g =

(FR)x FR (FR)y FR (FR)z FR

=

-2.896 ; 194.06

a = 90.86° = 90.9°

Ans.

=

45.66 ; 194.06

b = 76.39° = 76.4°

Ans.

=

-188.59 ; 194.06

g = 166.36° = 166°

Ans.

117

Ans: FR = a = b = g =

194 N 90.9° 76.4° 166°

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2–97. The cylindrical plate is subjected to the three cable forces which are concurrent at point D. Express each force which the cables exert on the plate as a Cartesian vector, and determine the magnitude and coordinate direction angles of the resultant force.

z D

FC ⫽ 5 kN

3m FB ⫽ 8 kN

C

B

45⬚

30⬚ y

Solution

A

rA = (0 - 0.75)i + (0 - 0)j + (3 - 0)k = 5 - 0.75i + 0j + 3k6 m

x

rA = 2( -0.75)2 + 02 + 32 = 3.0923 m FA = FAa

rA - 0.75i + 3k b = 6a b rA 3.0923

= 5 -1.4552i + 5.8209k6 kN

Ans.

= 5 -1.46i + 5.82k6 kN

rC = 30 - (0.75 sin 45°) 4i + 30 - ( -0.75 cos 45°) 4j + (3 - 0)k = 5 0.5303i + 0.5303j + 3k6 m

rC = 2(0.5303)2 + (0.5303)2 + 32 = 3.0923 m FC = FCa

0.5303i + 0.5303j + 3k rC b = 5a b rC 3.0923

= 5 0.8575i + 0.8575j + 4.8507k6 kN

Ans.

= 5 0.857i + 0.857j + 4.85k6 kN

rB = 30 - (0.75 sin 30°) 4i + (0 - 0.75 cos 30°)j + (3 - 0)k = 5 0.375i - 0.6495j + 3k6 m

rB = 2(0.375)2 + ( -0.6495)2 + 32 = 3.0923 m FB = FBa

0.375i - 0.6495j + 3k rB b = 8a b rB 3.0923

= 5 0.9701i - 1.6803j + 7.7611k6 kN

Ans.

= 5 0.970i - 1.68j + 7.76k6 kN

118

0.75 m FA ⫽ 6 kN

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2–97 (Continued)

FR = FA + FB + FC = 5 - 1.4552i + 5.8209k6 + 50.9701i - 1.6803j + 7.7611k6 + 5 0.8575i + 0.8575j + 4.8507k6 = 5 0.3724i - 0.8228j + 18.4327k6 kN

FR = 2(0.3724)2 + ( -0.8228)2 + (18.4327)2

Ans.

= 18.4548 kN = 18.5 kN

uR =

0.3724i - 0.8228j + 18.4327k FR = FR 18.4548

= 0.02018i - 0.04459j + 0.9988k cos a = 0.02018

a = 88.8°

Ans.

cos b = - 0.04458

b = 92.6°

Ans.

cos g = 0.9988

g = 2.81°

Ans.

Ans: FA = { - 1.46i + 5.82k} kN FC = {0.857i + 0.857j + 4.85k} kN FB = {0.970i - 1.68j + 7.76k} kN FR = 18.5 kN a = 88.8 b = 92.6 g = 2.81 119

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2–98. The cord exerts a force F = {12i + 9j – 8k} kN on the hook. If the cord is 4 m long, determine the location x,y of the point of attachment B, and the height z of the hook.

z

1m A F y 4m

z

x y B

x

Solution Given:

F =

§ 12 · ¨ 9 ¸N ¨ ¸ © 8 ¹

L = 4m a = 1m Initial guesses

Given

§x  a· ¨ y ¸ ¨ ¸ © z ¹

x = 1m

F = L˜ F

y = 1m

§x· ¨y¸ ¨ ¸ ©z¹

z = 1m

= Find ( x  y  z)

§x· ¨y¸ ¨ ¸ ©z¹

=

§ 3.82 · ¨ 2.12 ¸ m ¨ ¸ © 1.88 ¹

Ans.

Ans: x = 3.82 m y = 2.12 m z = 1.88 m 120

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2–99. The three supporting cables exert the forces shown on the sign. Represent each force as a Cartesian vector.

z C

2m E

2m B

FE  350 N 3m

FC  400 N FB  400 N

Solution

D 2m

rC = (0 - 5)i + ( - 2 - 0)j + (3 - 0)k = { - 5i - 2j + 3k} m

A

rC = 2( - 5)2 + ( - 2)2 + 32 = 238 m

y

3m

x

rB = (0 - 5)i + (2 - 0)j + (3 - 0)k = { -5i + 2j + 3k} m rB = 2( - 5)2 + 22 + 32 = 238 m

rE = (0 - 2)i + (0 - 0)j + (3 - 0)k = { -2i + 0j + 3k} m rE = 2( - 2)2 + 02 + 32 = 213 m r   F = Fu = F a b r FC = 400 a FB = 400 a FE = 350 a

- 5i - 2j + 3k 138

b = { - 324i - 130j + 195k} N

Ans.

138

b = { - 324i + 130j + 195k} N

Ans.

113

b = { - 194i + 291k} N

Ans.

- 5i + 2j + 3k - 2i + 0j + 3k

Ans: FC = { - 324i - 130j + 195k} N FB = { - 324i + 130j + 195k} N FE = { - 194i + 291k} N 121

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*2–100. z

Determine the magnitude and coordinate direction angles of the resultant force of the two forces acting on the sign at point A.

C

2m E

2m B

FE  350 N 3m

FC  400 N FB  400 N

Solution

2m

 rC = (0 - 5)i + ( - 2 - 0)j + (3 - 0)k = { -5i - 2j + 3k}  rC = 2( - 5)2 + ( - 2)2 + (3)2 = 238 m

D

A

y

3m

x

( -5i - 2j + 3k) rC FC = 400 a b = 400 a b rC 138

FC = ( -324.4428i - 129.777j + 194.666k) rB = (0 - 5)i + (2 - 0)j + (3 - 0)k = { - 5i + 2j + 3k}  rB = 2( - 5)2 + 22 + 32 = 238 m FB = 400 a

( -5i + 2j + 3k) rB b = 400 a b rB 138

FB = ( -324.443i + 129.777j + 194.666k) FR = FC + FB = ( -648.89i + 389.33k)

FR = 2( - 648.89)2 + (389.33)2 + 02 = 756.7242

Ans.

FR = 757 N a = cos-1 a b = cos-1 a

g = cos-1 a

- 648.89 b = 149.03 = 149° 756.7242

Ans.

0 b = 90.0° 756.7242

Ans.

389.33 b = 59.036 = 59.0° 756.7242

Ans.

Ans: FR = 757 N a = 149° b = 90.0° g = 59.0° 122

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2–101. The tower is held in place by three cables. If the force of each cable acting on the tower is shown, determine the magnitude and coordinate direction angles a, b, g, of the resultant force. Take x = 20 m, y =15 m.

z D 600 N 400 N

800 N

24 m

16 m

Solution FDA

18 m

2 15 24 = 400a i + j kb N 34.66 34.66 34.66

FDB = 800a

- 6 4 24 i + j kb N 25.06 25.06 25.06

FDC = 600a

16 18 24 i + j kb N 34 34 34

O

C

4m

B

6m x

y

y

A x

FR = FDA + FDB + FDC = 5 321.66i - 16.82j - 1466.71k6 N

FR = 2(321.66)2 + ( -16.82)2 + ( - 1466.71)2

Ans.

= 1501.66 N = 1.50 kN

a = cos - 1a

b = cos - 1a g = cos - 1a

321.66 b = 77.6° 1501.66

Ans.

-16.82 b = 90.6° 1501.66

Ans.

-1466.71 b = 168° 1501.66

Ans.

Ans: FR = 1.50 kN a = 77.6 b = 90.6 g = 168 123

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2–102. The guy wires are used to support the telephone pole. Represent the force in each wire in Cartesian vector form. Neglect the diameter of the pole.

z

B FB

175 N

SOLUTION 2m

Unit Vector: rAC = {( -1 - 0)i + (4 - 0)j + (0 - 4)k} m = { -1i + 4j - 4k} m

uAC =

2

2

FA

250 N

4m D 3m

rAC = 2( -1) + 4 + (-4) = 5.745 m 2

1.5 m A

4m C 1m

y

x

- 1i + 4j - 4k rAC = -0.1741i + 0.6963j - 0.6963k = rAC 5.745

rBD = {(2 - 0) i + (- 3 - 0)j + (0 - 5.5)k} m = {2i - 3j - 5.5k} m rBD = 22 2 + (-3)2 + (-5.5)2 = 6.576 m uBD =

2i - 3j - 5.5k rBD = 0.3041i - 0.4562j - 0.8363k = rBD 6.576

Force Vector: FA = FA uAC = 250{ -0.1741i + 0.6963j - 0.6963k} N = { -43.52i + 174.08j - 174.08k} N = { -43.5i + 174j - 174k} N

Ans.

FB = FBuBD = 175{0.3041i + 0.4562j - 0.8363k} N = {53.22i - 79.83j - 146.36k} N = {53.2i - 79.8j - 146k} N

Ans.

Ans: FA = { - 43.5i + 174j - 174k} N FB = {53.2i - 79.8j - 146k} N 124

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2–103. Determine the magnitude and coordinate direction angles of the resultant force acting at point A.

z A

1.5 m F1

F2

200 N

4m

SOLUTION rAC = {3i - 0.5j - 4k} m

F2 = 200a

60

3m

y

3m

B

C

|rAC| = 232 + (- 0.5)2 + ( -4)2 = 225.25 = 5.02494

150 N

2m

3i - 0.5j - 4k b = (119.4044i - 19.9007j - 159.2059k) 5.02494

x

rAB = (3 cos 60°i + (1.5 + 3 sin 60°) j - 4k) rAB = (1.5i + 4.0981j + 4k) |rAB| = 2(1.5)2 + (4.0981)2 + (-4)2 = 5.9198 F1 = 150a

1.5i + 4.0981j - 4k b = (38.0079i + 103.8396j - 101.3545k) 5.9198

FR = F1 + F2 = (157.4124i + 83.9389j - 260.5607k) FR = 2(157.4124)2 + (83.9389)2 + (- 260.5604)2 = 315.7786 = 316 N

Ans.

a = cos-1 a

157.4124 b = 60.100° = 60.1° 315.7786

Ans.

b = cos-1 a

83.9389 b = 74.585° = 74.6° 315.7786

Ans.

g = cos-1 a

-260.5607 b = 145.60° = 146° 315.7786

Ans.

Ans: FR = 316 N a = 60.1 b = 74.6 g = 146 125

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*2–104. Two cables are used to secure the overhang boom in position and support the 1500-N load. If the resultant force is directed along the boom from point A towards O, determine the magnitudes of the resultant force and forces FB and FC. Set x = 3 m and z = 2 m.

z x

2m B 3m

C

z

x

FB 6 m FC

A y

Solution Force Vectors: The unit vectors uB and uC must be determined first. From Fig. a, uB = uC =

1500 N

( -2 - 0)i + (0 - 6)j + (3 - 0)k rB 2 6 3 = = - i - j + k rB 7 7 7 2( -2 - 0)2 + (0 - 6)2 + (3 - 0)2 (3 - 0)i + (0 - 6)j + (2 - 0)k rC 3 6 2 = = i - j + k 2 2 2 rC 7 7 7 2(3 - 0) + (0 - 6) + (2 - 0)

Thus, the force vectors FB and FC are given by 2 6 3 FBi - FBj + FBk 7 7 7

FB = FBuB = FC = FCuC =

3 6 2 F i - FCj + FCk 7 C 7 7

Since the resultant force FR is directed along the negative y axis, and the load W is directed along the z axis, these two forces can be written as FR = - FRj

and

W = [ - 1500k] N

Resultant Force: The vector addition of FB, FC, and W is equal to FR. Thus, FR = FB + FC + W 2 6 3 3 6 2 - FRj = a- FBi - FBj + FBkb + a FCi - FCj + FCkb + ( - 1500k) 7 7 7 7 7 7 2 3 6 6 3 2 - FRj = a- FB + FC bi + a- FB - FC bj + a FB + FC - 1500bk 7 7 7 7 7 7

Equating the i, j, and k components, 2 3 0 = - FB + FC 7 7

(1)

6 6 - FR = - FB - FC 7 7

(2)

0 =

3 2 F + FC - 1500 7 B 7

(3)

Solving Eqs. (1), (2), and (3) yields FC = 1615.38 N = 1.62 kN

Ans.

FB = 2423.08 N = 2.42 kN

Ans.

FR = 3461.53 N = 3.46 kN

Ans. Ans: FC = 1.62 kN FB = 2.42 kN FR = 3.46 kN 126

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2–105. Two cables are used to secure the overhang boom in position and support the 1500-N load. If the resultant force is directed along the boom from point A towards O, determine the values of x and z for the coordinates of point C and the magnitude of the resultant force. Set FB = 1610 N and FC = 2400 N.

z x

2m B 3m

C

z

FB 6 m FC

x

A y

Solution 1500 N

Force Vectors: From Fig. a, ( - 2 - 0)i + (0 - 6)j + (3 - 0)k rB 2 6 3 = = - i - j + k rB 7 7 7 2( - 2 - 0)2 + (0 - 6)2 + (3 - 0)2

uB = uC = Thus,

(x - 0)i + (0 - 6)j + (z - 0)k rC x 6 z = = i j + k 2 2 2 2 2 2 2 2 rC 2(x - 0) + (0 - 6) + (z - 0) 2x + z + 36 2x + z + 36 2x + z2 + 36

2 6 3 FB = FBuB = 1610 a- i - j + kb = 3 -460i - 1380j + 690k4 N 7 7 7 x

FC = FCuC = 2400 a=

2

2

2x + z + 36

2400x

2x2 + z2 + 36

i -

6

i -

2

2

2x + z + 36

14400

2x2 + z2 + 36

j +

j +

z

2

2x + z2 + 36

2400z

b

2x2 + z2 + 36

Since the resultant force FR is directed along the negative y axis, and the load is directed along the z axis, these two forces can be written as FR = - FRj

W = [ - 1500k] N

and

Resultant Force: FR = FB + FC + W - FRj = ( - 460i - 1380j + 690k) + a - FRj = a

2400x

2x2 + z2 + 36

2400x

2400x 2

- 460bi - a

- 460

2

2x + z + 36

- FR = - a 0 = 690 +

14400

2x2 + z2 + 36 2400z

2

2

2x + z + 36

2

2x + z + 36

Equating the i, j, and k components, 0 =

2

14400

2x2 + z2 + 36

2400x 2

2x + z2 + 36

+ 1380b FR = a

- 1500

i -

2

2x + z + 36

2400z 2

2x + z2 + 36 2400z

2x2 + z2 + 36 (1)

14400 2x2 + z2 + 36

2x + z2 + 36

j +

+ 1380bj + a690 +

= 460

2400z

2

14400 2

+ 1380b

(2) (3)

= 810

127

kb + ( -1500k) - 1500bk

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2–105 (Continued)

Dividing Eq. (1) by Eq. (3), yields (4)

x = 0.5679 Substituting Eq. (4) into Eq. (1), and solving

Ans.

z = 2.197 m = 2.20 m Substituting z = 2.197 m into Eq. (4), yields,

Ans.

x = 1.248 m = 1.25 m Substituting x = 1.248 m and z = 2.197 m into Eq. (2), yields

Ans.

FR = 3591.85 N = 3.59 kN

Ans: Z = 2.20 m X = 1.25 m FR = 3.59 kN 128

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2–106. Given the three vectors A, B, and D, show that A # (B + D) = (A # B) + (A # D).

SOLUTION Since the component of (B + D) is equal to the sum of the components of B and D, then A # (B + D) = A # B + A # D

(QED)

Also, A # (B + D) = (A x i + A y j + A zk) # [(Bx + Dx)i + (By + Dy)j + (Bz + Dz)k] = A x (Bx + Dx) + A y (By + Dy) + A z (Bz + Dz) = (A xBx + A yBy + A zBz) + (A xDx + A yDy + A zDz) = (A # B) + (A # D)

(QED)

129

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2–107. Determine the design angle u (0° … u … 90°) for strut AB so that the 400-N horizontal force has a component of 500 N directed from A towards C. What is the component of force acting along member AB? Take f = 40°.

400 N A u f

B

C

Solution Parallelogram Law: The parallelogram law of addition is shown in Fig. (a). Trigonometry: Using law of sines [Fig. (b)], we have sin u sin 40° = 500 400 sin u = 0.8035 u = 53.46° = 53.5°

Ans.

Thus, f = 180° - 40° - 53.46° = 86.54°

N

Using law of sines [Fig. (b)] FAB 400 = sin 86.54° sin 40° Ans.

FAB = 621 N

N

N

N

Ans: u = 53.5 FAB = 621 N 130

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*2–108. z

Determine the projection of the force F along the pole.

F = {2i + 4j + 10k} kN

O

y

SOLUTION

2m

2 1 2 Proj F = F # ua = 12 i + 4 j + 10 k2 # a i + j - kb 3 3 3 Proj F = 0.667 kN

Ans.

2m x

1m

Ans: Proj F = 0.667 kN 131

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2–109. Determine the angle u between the sides of the triangular plate.

z

3m

B

4m

SOLUTION

A 1m

rAC = 53 i + 4 j - 1 k6 m

θ y

1m

rAC = 21322 + 1422 + 1-122 = 5.0990 m

rAB = 52 j + 3 k6 m

5m

rAB = 21222 + 1322 = 3.6056 m

3m C

x

rAC # rAB = 0 + 4122 + 1-12132 = 5 u = cos-1

rAC # rAB rACrAB

= cos-1

5 15.0990213.60562

u = 74.219° = 74.2°

Ans.

Ans: u = 74.2 132

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2–110. Determine the length of side BC of the triangular plate. Solve the problem by finding the magnitude of rBC; then check the result by first finding , rAB, and rAC and then using the cosine law.

z

3m

B

4m A 1m

SOLUTION

u y

1m

3m

rBC = {3 i + 2 j - 4 k} m rBC = 2(3)2 + (2)2 + ( - 4)2 = 5.39 m

5m

Ans.

C

x

Also, rAC = {3 i + 4 j - 1 k} m rAC = 2(3)2 + (4)2 + (- 1)2 = 5.0990 m rAB = {2 j + 3 k} m rAB = 2(2)2 + (3)2 = 3.6056 m rAC # rAB = 0 + 4(2) + (- 1)(3) = 5 u = cos-1 a

rAC # rAB 5 b = cos-1 rAC rAB (5.0990)(3.6056)

u = 74.219° rBC = 2(5.0990)2 + (3.6056)2 - 2(5.0990)(3.6056) cos 74.219° rBC = 5.39 m

Ans.

Ans: rBC = 5.39 m 133

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2–111. Determine the magnitude of the projected component of r1 along r2, and the projection of r2 along r1.

z r1 = 9 m 40

SOLUTION

120

Given: r1

9m

r2

6m

30

u 60

r2

x

D

60°

E

45°

J

120°

I

30°

H

40°

45

y 6m

Write the vectors and unit vectors

r1v

§¨ sin  H cos  I ¸· r1 ¨ sin  H sin  I ¸ ¨ cos  H ¸ © ¹

r2v

§¨ cos  D ¸· r2 ¨ cos  E ¸ ¨ cos  J ¸ © ¹

u1

r1v r1v

u2

r2v r2v

r1v

§ 5.01 · ¨ 2.89 ¸ m ¨ ¸ © 6.89 ¹

r2v

§ 3 · ¨ 4.24 ¸ m ¨ ¸ © 3 ¹

u1

§ 0.557 · ¨ 0.321 ¸ ¨ ¸ © 0.766 ¹

u2

§ 0.5 · ¨ 0.707 ¸ ¨ ¸ © 0.5 ¹

The magnitude of the projection of r1 along r2.

r1v˜ u2

2.99 m

Ans.

The magnitude of the projection of r2 along r1.

r2v˜ u1

1.99 m

Ans.

Ans: ƒr1 # u2 ƒ = 2.99 m, ƒr2 # u1 ƒ = 1.99 m 134

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*2–112.

Determine the angle T between the two cords. z

3m B

C

4m u

6m y

A

SOLUTION

3m

Given:

x

a

3m

b

2m

c

6m

d

3m

e

4m

rAC

2m

§b· ¨a¸ m ¨ ¸ ©c¹

rAB

§0· ¨ d ¸ m ¨ ¸ ©e ¹

T

§ rAC˜ rAB · ¸ © rAC rAB ¹

acos ¨

T

64.6°

Ans.

Ans:

T 135

64.6°

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2–113. Determine the magnitudes of the components of F = 600 N acting along and perpendicular to segment DE of the pipe assembly.

z A

2m

B 2m x

SOLUTION Unit Vectors: The unit vectors uEB and uED must be determined first. From Fig. a, uEB

(0 - 4)i + (2 - 5)j + [0 - (- 2)]k rEB = = = -0.7428i - 0.5571j + 0.3714k rEB 2(0 - 4)2 + (2 - 5)2 + [0 - (- 2)]2

y

2m 2m

C

D

F

600 N

3m E

uED = -j Thus, the force vector F is given by F = FuEB = 600 A -0.7428i - 0.5571j + 0.3714k) = [-445.66i - 334.25j + 222.83k] N Vector Dot Product: The magnitude of the component of F parallel to segment DE of the pipe assembly is (FED)paral = F # uED =

A -445.66i - 334.25j + 222.83k B # A -j B

= ( -445.66)(0) + ( -334.25)( - 1) + (222.83)(0) Ans.

= 334.25 = 334 N The component of F perpendicular to segment DE of the pipe assembly is (FED)per = 2F 2 - (FED)paral2 = 26002 - 334.252 = 498 N

Ans.

Ans: (FED) = 334 N (FED) # = 498 N 136

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2–114. z

Determine the angle u between the two cables.

C

F2  40 N

A

4m

u F1  70 N 2m

3m 2m

Solution Unit Vectors. Here, the coordinates of points A, B and C are A(2, -3, 3) m, B(0, 3, 0) and C( - 2, 3, 4) m respectively. Thus, the unit vectors along AB and AC are uAB =

uAC =

3m

B

y

3m x

(0 - 2)i + [3 - ( - 3)]j + (0 - 3)k

6 3 2 = - i + j - k 7 7 7 2(0 - 2) + [3 - ( - 3)] + (0 - 3) 2

2

2

( -2 - 2)i + [3 - ( - 3)]j + (4 - 3)k 2

2

2( - 2 - 2) + [3 - ( - 3)] + (4 - 3)

The Angle U Between AB and AC.

2

= -

4

253

i +

6

253

j +

1 253

k

2 6 3 4 6 1 i + j + kb uAB # uAC = a - i + j - kb # a 7 7 7 253 253 253 = a=

2 4 6 6 3 1 ba b + a b + a - ba b 7 7 253 7 253 253

41

7253

Then

u = cos - 1 ( uAB # uAC ) = cos-1a

41 7253

b = 36.43° = 36.4°

Ans.

Ans: u = 36.4° 137

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2–115. Determine the magnitude of the projection of the force F1 along cable AC.

z C

F2  40 N

A

4m

u F1  70 N 2m

3m 2m

Solution Unit Vectors. Here, the coordinates of points A, B and C are A(2, - 3, 3)m, B(0, 3, 0) and C( -2, 3, 4) m respectively. Thus, the unit vectors along AB and AC are uAB =

uAC =

3m

B

y

3m x

(0 - 2)i + [3 - ( - 3)]j + (0 - 3)k

6 3 2 = - i + j - k 7 7 7 2(0 - 2) + [3 - ( - 3)] + (0- 3) 2

2

( -2 - 2)i + [3 - ( - 3)]j + (4 - 3)k

2( - 2 - 2)2 + [3 - ( - 3)]2 + (4 - 3)2

2

4

= -

253

Force Vector, For F1,

i +

6

253

j +

1 253

k

2 6 3 F1 = F1 uAB = 70 a - i + j - kb = { -20i + 60j - 30k} N 7 7 7

Projected Component of F1. Along AC, it is

(F1)AC = F1 # uAC = ( - 20i + 60j - 30k) # a 4

b + 60a



= ( - 20)a -



= 56.32 N = 56.3 N

253

6

253

4 253

b + ( -30)a

i + 1

253

6 253

j +

1 253

b

kb

 Ans.

The positive sign indicates that this component points in the same direction as uAC.

Ans: (F1)AC = 56.3 N 138

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*2–116. A force of F = 80 N is applied to the handle of the wrench. Determine the angle u between the tail of the force and the handle AB.

z F

30

u

B

80 N

45 A 300 mm

x

Solution

y

500 mm

Given: a

300 mm

b

500 mm

F

80 N

T1

30°

T2

45°

Fv

§ cos  T 1 sin  T 2 · ¨ ¸ F ¨ cos  T 1 cos  T 2 ¸ ¨ ¸ sin  T 1 © ¹

T

§ Fv˜ uab · ¸ © F ¹

acos ¨

uab

T

§0· ¨ 1 ¸ ¨ ¸ ©0¹ 127.8°

Ans.

Ans:

T 139

127.8°

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2–117. Determine the angle u between the cables AB and AC.

z 1m B

1m

2m C

D F

u

3m

Solution Unit Vectors. Here, the coordinates of points A, B and C are A(6, 0, 0) m, x B(0, - 1, 2) m and C(0, 1, 3) respectively. Thus, the unit vectors along AB and AC are uAB =

uAC =

(0 - 6)i + ( - 1 - 0)j + (2 - 0)k 2

2

2(0 - 6) + ( - 1 - 0) + (2 - 0) (0 - 6)i + (1 - 0)j + (3 - 0)k 2

2

2(0 - 6) + (1 - 0) + (3 - 0)

2

= -

2

= -

The Angle u Between AB and AC. uAB # uAC = a -

6 241

= a= Then

6

241

41

i -

1 241

ba -

j +

6

246

2 241

b + a-

kb # a 1

241

ba

6 241 6

246 6 246 1

246

i -

i +

i +

b +

1 241 1

246 1 246 2

j +

j +

j + a

3

A

2 241 3

246 3 246

241 246

6m

y

k

k

kb

b

21886

u = cos - 1 ( UAB # UAC ) = cos-1a

41 21886

b = 19.24998° = 19.2°

Ans.

Ans: u = 19.2° 140

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2–118. z

Determine the magnitude of the projected component of the force F = {400i - 200j + 500k} N acting along the cable BA.

1m B

1m

2m C

D F

u

Solution Unit Vector. Here, the coordinates of points A and B are A(6, 0, 0) m and x B(0, -1, 2) m respectively. Thus the unit vector along BA is uBA =

A

6m

3m

y

(6 - 0)i + [0 - ( - 1)]j + (0 - 2)k rBA 6 1 2 = = i + j k rBA 2(6 - 0)2 + [0 - ( - 1)]2 + (0- 2)2 241 241 241

Projected component of F. Along BA, it is

FBA = F # uBA = (400i - 200j + 500k) # a



= 400 a

6

241

b + ( - 200)a

= 187.41 N = 187 N

6 241 1

241

i +

1 241

j -

b + 500a -

2

2 241

241

b

kb

 Ans.

The positive sign indicates that this component points in the same direction as uBA.

Ans: FBA = 187 N 141

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2–119. Determine the magnitude of the projected component of the force F = {400i - 200j + 500k} N acting along the cable CA.

z 1m B

1m

2m C

D F

u

3m

Solution Unit Vector. Here, the coordinates of points A and C are A(6, 0, 0) m and C(0, 1, 3) m respectively. Thus, the unit vector along CA is uCA =

x

A

6m

y

(6 - 0)i + (0 - 1)j + (0 - 3)k rCA 6 1 3 = = i j k 2 2 2 rCA 2(6 - 0) + (0 - 1) + (0 - 3) 246 246 246

Projected component of F. Along CA, it is

FCA = F # uCA = (400i - 200j + 500k) # a = 400a

6

246

6 246

b + ( - 200)a -

= 162.19 N = 162 N

i -

1

246

1 246

j -

b + 500 a -

3 246 3

246

b

kb

 Ans.

The positive sign indicates that this component points in the same direction as uCA.

Ans: FCA = 162 N 142

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z

*2–120. Determine the magnitude of the projected component of force FAB acting along the z axis. FAC FAB

3 kN

A 9m

3.5 kN

D 4.5 m 3m

O B

Solution

3m

Unit Vector: The unit vector uAB must be determined first. From Fig. a, uAB =

rAB rAB

=

(4.5 – 0)i + (–3 – 0)j + (0 – 9)k (4.5 – 0) 2 + (–3 – 0) 2 + (0 – 9) 2

3m C

30

y

x

3 2 6 = i– j– k 7 7 7

Thus, the force vector FAB is given by ⎛3 2 6 ⎞ FAB = FABuAB = 3.5 ⎜ i – j – k⎟ = {1.5i – 1j – 3k} kN 7 7 ⎠ ⎝7 Vector Dot Product: The projected component of FAB along the z axis is (FAB)z = FAB · k = (1.5i – 1j – 3k) · k = –3 kN The negative sign indicates that (FAB)z is directed towards the negative z axis. Thus (FAB)z = 3 kN

Ans.

A(0, 0, 9) m

B(4.5, –3, 0) m

Ans: (FAB)z = 3 kN

143

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z

2–121. Determine the magnitude of the projected component of force FAC acting along the z axis. FAC FAB

3 kN

A 9m

3.5 kN

D 4.5 m 3m

O B 3m

3m C

30

y

x

Solution Unit Vector: The unit vector uAC must be determined first. From Fig. a, uAC =

rAC rAC

=

(3 sin 30° – 0)i + (3 cos 30° – 0)j + (0 – 9)k

= 0.1581i + 0.2739j – 0.9487k

(3 sin 30° – 0) 2 + (3 cos 30° – 0) 2 + (0 – 9) 2

Thus, the force vector FAC is given by FAC = FACuAC = 3(0.1581i + 0.2739j – 0.9487k) = {0.4743i + 0.8217j – 2.8461k} kN Vector Dot Product: The projected component of FAC along the z axis is (FAC)z = FAC · k = (0.4743i + 0.8217j – 2.8461k) · k = –2.8461 kN The negative sign indicates that (FAC)z is directed towards the negative z axis. Thus (FAC)z = 2.846 kN

Ans.

A(0, 0, 9) m

C(3 sin 30°, 3 cos 30°, 0) m

Ans:

144

1 FAC 2 z

= 2.846 kN

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2–122. Determine the components of F that act along rod AC and perpendicular to it. Point B is located at the midpoint of the rod.

z

A

B

4m

O

SOLUTION rAC = (- 3i + 4j - 4k), rAB =

F

4m C

600 N

x

rAC = 2(- 3)2 + 42 + ( - 4)2 = 241 m

3m

6m

- 3i + 4j + 4k rAC = = -1.5i + 2j - 2k 2 2

D

4m

y

rAD = rAB + rBD rBD = rAD - rAB = (4i + 6j - 4k) - ( - 1.5i + 2j - 2k) = {5.5i + 4j - 2k} m rBD = 2(5.5)2 + (4)2 + (- 2)2 = 7.0887 m F = 600a

rBD b = 465.528i + 338.5659j - 169.2829k rBD

Component of F along rAC is F| | F| | =

(465.528i + 338.5659j - 169.2829k) # ( -3i + 4j - 4k) F # rAC = rAC 241

F| | = 99.1408 = 99.1 N

Ans.

Component of F perpendicular to rAC is F F2 + F2|| = F2 = 6002 F2 = 6002 - 99.14082 F

= 591.75 = 592 N

Ans.

Ans: F| | = 99.1 N F # = 592 N 145

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2–123. Determine the components of F that act along rod AC and perpendicular to it. Point B is located 3 m along the rod from end C.

z

A

B

4m

O

SOLUTION

4m C

600 N

x

rCA = 3i - 4j + 4k

3m

6m

rCA = 6.403124 rCB =

F

D

4m

y

3 (r ) = 1.40556i - 1.874085j + 1.874085k 6.403124 CA

rOB = rOC + rCB = -3i + 4j + r CB = -1.59444i + 2.1259j + 1.874085k rOD = rOB + rBD rBD = rOD - rOB = (4i + 6j) - rOB = 5.5944i + 3.8741j - 1.874085k rBD = 2(5.5944)2 + (3.8741)2 + ( -1.874085)2 = 7.0582 F = 600(

rBD ) = 475.568i + 329.326j - 159.311k rBD

rAC = (- 3i + 4j - 4k),

rAC = 241

Component of F along rAC is F| F| | =

|

(475.568i + 329.326j - 159.311k) # (- 3i + 4j - 4k) F # rAC = rAC 241

F| | = 82.4351 = 82.4 N

Ans.

Component of F perpendicular to rAC is F F2 + F2|| = F2 = 6002 F2 = 6002 - 82.43512 F

= 594 N

Ans.

Ans: F} = 82.4 N F # = 592 N 146

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*2–124. Determine the magnitudes of the projected components of the force F = [60i + 12j - 40k] N along the cables AB and AC.

0.75 m B 1m

z

1m C

1.5 m

u A

Solution

y

F

F = {60 i + 12 j - 40 k} N uAB =

3m

x

-3 i - 0.75 j + 1 k 1( - 3)2 + ( - 0.75)2 + (1)2

= -0.9231 i - 0.2308 j + 0.3077 k uAC =

-3 i + 1 j + 1.5 k 1( - 3)2 + (1)2 + (1.5)2

= -0.8571 i + 0.2857 j + 0.4286 k

Proj FAB = F # uAB = (60)( - 0.9231) + (12)( - 0.2308) + ( - 40)(0.3077)

= -70.46 N Ans.

|Proj FAB| = 70.5 N

Proj FAC = F # uAC = (60)( - 0.8571) + (12)(0.2857) + ( -40)(0.4286)

= -65.14 N Ans.

|Proj FAC| = 65.1 N

Ans: |Proj FAB| = 70.5 N |Proj FAC| = 65.1 N 147

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2–125. Determine the angle u between cables AB and AC.

0.75 m B 1m

z

1m C

1.5 m

u A 3m

x

Solution

y

F

rAB = { - 3 i - 0.75 j + 1 k} m rAB = 2( - 3)2 + ( - 0.75)2 + (1)2 = 3.25 m rAC = { - 3 i + 1 j + 1.5 k} m

rAC = 2( - 3)2 + (1)2 + (1.5)2 = 3.50 m

rAB # rAC = ( - 3)( -3) + ( -0.75)(1) + (1)(1.5) = 9.75 u = cos-1 a u = 31.0°

rAB # rAC 9.75 b = cos-1 a b rAB rAC (3.25)(3.50)

Ans.

Ans: u = 31.0° 148

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2–126. Determine the angles u and f made between the axes OA of the flag pole and AB and AC, respectively, of each cable.

z 1.5 m 2m

B

4m

SOLUTION rA C = {- 2i - 4j + 1k} m ;

C

FB

rA C = 4.58 m

rAB = {1.5i - 4j + 3k} m;

rAB = 5.22 m

rA O = { -4j - 3k} m;

rA O = 5.00 m

6m

u

O

= cos - 1 ¢

rAB # rAO ≤ rAB rAO

f

A

3m

rA B # rA O = (1.5)(0) + ( -4)(-4) + (3)(- 3) = 7 u = cos - 1 ¢

40 N

FC

55 N

4m x y

7 ≤ = 74.4° 5.22(5.00)

Ans.

rAC # rAO = (-2)(0) + ( - 4)(-4) + (1)(-3) = 13 f = cos - 1 a = cos - 1 a

rAC # rAO b rAC rAO 13 b = 55.4° 4.58(5.00)

Ans.

Ans: u = 74.4° f = 55.4° 149

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2–127. z

Determine the angle u between BA and BC. A

B

2m

u

5m 4m x

D

F  3 kN

3m C

y

1m

Solution Unit Vectors. Here, the coordinates of points A, B and C are A(0, -2, 0) m, B(0, 0, 0) m and C(3, 4, - 1) m respectively. Thus, the unit vectors along BA and BC are uBA = - j

uBE =

(3 - 0) i + (4 - 0) j + ( - 1 - 0) k 2

2

2(3 - 0) + (4 - 0) + ( - 1 - 0)

2

The Angle U Between BA and BC. uBA uBC = ( - j)

Then

= ( -1) a

#

a

4

3 226

226

u = cos - 1 (uBA # uBC) = cos - 1 a -

i +

b = 4 226

4 226

4

j -

=

3 226 1

226

i +

4 226

j-

1 226

k

kb

226

b = 141.67° = 142°

Ans.

Ans: u = 142° 150

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*2–128. z

Determine the magnitude of the projected component of the 3 kN force acting along the axis BC of the pipe. A

B

2m

u

5m 4m x

D

F  3 kN

3m C

y

1m

Solution Unit Vectors. Here, the coordinates of points B, C and D are B (0, 0, 0) m, C(3, 4, - 1) m and D(8, 0, 0). Thus the unit vectors along BC and CD are uBC =

uCD =

(3 - 0) i + (4 - 0) j + ( - 1- 0) k 2

2(3 - 0) + (4 - 0)

2

2

2

+ ( - 1 - 0)

2

=

(8 - 3) i + (0 - 4) j + [0 - ( - 1)] k

2(8 - 3) + (0 - 4)

+ [0 - ( - 1)]

2

3 226

5

=

F = FuCD = 3 a = a

5 242

15

242

i -

i -

Projected Component of F. Along BC, it is ` (FBC) ` = ` F # uBC ` = ` a



15 242 15

= `a

242

= `-

21092

6

12

i-

242

ba

3

226

j+

4 242

12

242 3 242

b + a-

1

j +

j +

3

kb # a

12

242

242

226

j +

k

1

242

k

kb kN

3 226

ba

4

1

kb

242

242

j -

226

i -

242

Force Vector. For F,

4

i +

i+

4

226

4 226

b +

j-

3

242

1 226

a-

1

kb `

226

b`

Ans.

` = ` - 0.1816 kN ` = 0.182 kN

The negative signs indicate that this component points in the direction opposite to that of uBC.

Ans: 0.182 kN 151

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2-129. Determine the angles T and I between the axis OA of the pole and each cable, AB and AC.

z 4m 6m

A F1

f 4m

Solution

x

Given:

u

F2

C

1m 5m

50 N

F2

35 N

a

1m

b

3m

c

2m

d

5m

e

4m

f

6m

g

4m

§0· ¨ g ¸ ¨ ¸ ©f ¹

rAB

§e· ¨a¸ ¨ ¸ ©f¹

rAC

3m

§ c · ¨a  b¸ ¨ ¸ © f ¹

T

52.4°

Ans.

§ rAO˜ rAC · ¸ © rAO rAC ¹

I

68.2°

Ans.

acos ¨

I

acos ¨

2m y

§ rAO˜ rAB · ¸ © rAO rAB ¹

T

50 N

35 N

B

F1

rAO

O

Ans: u = 52.4 f = 68.2 152

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2-130. The two cables exert the forces shown on the pole. Determine the magnitude of the projected component of each force acting along the axis OA of the pole.

z 4m 6m

A F1

f 4m

Solution

O

u

F2

35 N

C

1m

x

5m

3m B

Given: F1

50 N

F2

35 N

a

1m

b

3m

c

2m

d

5m

e

4m

f

6m

g

4m

rAB

§e· ¨a¸ ¨ ¸ ©f¹

F 1v

F1

F 2v

F2

rAC

rAC rAC rAB rAB

§ c · ¨a  b¸ ¨ ¸ © f ¹

rAO

§0· ¨ g ¸ ¨ ¸ ©f ¹

50 N

2m y

uAO

rAO rAO

F 1AO

F 1v˜ uAO

F 1AO

18.5 N

Ans.

F 2AO

F 2v˜ uAO

F 2AO

21.3 N

Ans.

Ans: F1AO = 18.5 N F2AO = 21.3 N 153

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2–131. z

Determine the magnitude of the projection of force F = 600 N along the u axis.

F

600 N

A 4m

SOLUTION

O

Unit Vectors: The unit vectors uOA and uu must be determined first. From Fig. a, ( -2 - 0)i + (4 - 0)j + (4 - 0)k rOA 2 2 1 = = - i + j + k uOA = rOA 3 3 3 2 2 2 3( -2 - 0) + (4 - 0) + (4 - 0)

4m 2m 30

x

u

uu = sin30°i + cos30°j Thus, the force vectors F is given by F = F uOA = 600 a -

2 2 1 i - j + kb = 5 -200i + 400j + 400k6 N 3 3 3

Vector Dot Product: The magnitude of the projected component of F along the u axis is Fu = F # uu = (-200i + 400j + 400k) # (sin30°i + cos 30°j) = ( -200)(sin30°) + 400(cos 30°) + 400(0) Ans.

= 246 N

Ans: Fu = 246 N 154

y

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*2–132.

Determine the projected component of the force F acting along the axis AB of the pipe. z

C B u x

y

4m

Solution

6m 3m

Given:

12 m

F

80 N

a

4m

b

3m

c

12 m

d

2m

e

6m

2m F

A 80 N

Find the force and the unit vector

rA

rAB

§ e · ¨ a  b ¸ ¨ ¸ © dc ¹ § e · ¨ b ¸ ¨ ¸ ©d¹

rA

§ 6 · ¨ 7 ¸ m ¨ ¸ © 10 ¹

rAB

§ 6 · ¨ 3 ¸ m ¨ ¸ ©2¹

Fv

uAB

F

rA

Fv

rA

rAB

uAB

rAB

§ 35.3 · ¨ 41.2 ¸ N ¨ ¸ © 58.8 ¹ § 0.9 · ¨ 0.4 ¸ ¨ ¸ © 0.3 ¹

Now find the projection using the Dot product. F AB

F v ˜ uAB

F AB

31.1 N

Ans.

Ans:

F AB 155

31.1 N

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2–133. The cables each exert a force of 400 N on the post. Determine the magnitude of the projected component of F1 along the line of action of F2.

z F1

400 N

35

120

20

SOLUTION Force Vector:

u 45

y

60

x

F2

uF1 = sin 35° cos 20°i - sin 35° sin 20°j + cos 35°k

400 N

= 0.5390i - 0.1962j + 0.8192k F1 = F1uF1 = 400(0.5390i - 0.1962j + 0.8192k) N = {215.59i - 78.47j + 327.66k} N Unit Vector: The unit vector along the line of action of F2 is uF2 = cos 45°i + cos 60°j + cos 120°k = 0.7071i + 0.5j - 0.5k Projected Component of F1 Along Line of Action of F2: (F1)F2 = F1 # uF2 = (215.59i - 78.47j + 327.66k) # (0.7071i + 0.5j - 0.5k) = (215.59)(0.7071) + (-78.47)(0.5) + (327.66)(- 0.5) = -50.6 N Negative sign indicates that the force component (F1)F2 acts in the opposite sense of direction to that of uF2. thus the magnitude is (F 1)F2 = 50.6 N

Ans.

Ans:

1 F1 2 F

2

156

= 50.6 N

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2–134. Determine the angle u between the two cables attached to the post.

z F1

400 N

35

120

SOLUTION

20

u 45

y

60

Unit Vector: x

uF1 = sin 35° cos 20°i - sin 35° sin 20°j + cos 35°k

F2

400 N

= 0.5390i - 0.1962j + 0.8192k uF2 = cos 45°i + cos 60°j + cos 120°k = 0.7071i + 0.5j - 0.5k The Angle Between Two Vectors u: The dot product of two unit vectors must be determined first. uF1 # uF2 = (0.5390i - 0.1962j + 0.8192k) # (0.7071i + 0.5j - 0.5k) = 0.5390(0.7071) + ( -0.1962)(0.5) + 0.8192(- 0.5) = -0.1265 Then, u = cos-1 A uF1 # uF2 B = cos-1(- 0.1265) = 97.3°

Ans.

Ans: u = 97.3 157

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2–135. If the force F = 100 N lies in the plane DBEC, which is parallel to the x–z plane, and makes an angle of 10° with the extended line DB as shown, determine the angle that F makes with the diagonal AB of the crate.

z¿ z 15 0.2 m

F

10 B 30

x A 15

F

2

E 0.2 m

D

6k

N

0.5 m

C y

Solution Use the x, y, z axes. uAB = a

- 0.5i + 0.2j + 0.2k 0.57446

b

= - 0.8704i + 0.3482j + 0.3482k

F = - 100 cos 10°i + 100 sin 10°k u = cos-1a

F # uAB b F uAB

= cos-1 a

- 100 (cos 10°)( -0.8704) + 0 + 100 sin 10° (0.3482) 100(1)

= cos-1 (0.9176) = 23.4°

b

Ans.

Ans: u = 23.4° 158

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*2–136. Determine the magnitudes of the projected components of the force F = 300 N acting along the x and y axes.

z

30

F

A

300 N

30 300 mm

O x

300 mm

300 mm y

SOLUTION Force Vector: The force vector F must be determined first. From Fig. a, F = -300 sin 30°sin 30°i + 300 cos 30°j + 300 sin 30°cos 30°k = [-75i + 259.81j + 129.90k] N Vector Dot Product: The magnitudes of the projected component of F along the x and y axes are Fx = F # i =

A -75i + 259.81j + 129.90k B # i

= - 75(1) + 259.81(0) + 129.90(0) = - 75 N Fy = F # j =

A - 75i + 259.81j + 129.90k B # j

= - 75(0) + 259.81(1) + 129.90(0) = 260 N The negative sign indicates that Fx is directed towards the negative x axis. Thus Fx = 75 N,

Ans.

Fy = 260 N

Ans: Fx = 75 N Fy = 260 N 159

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2–137. Determine the magnitude of the projected component of the force F = 300 N acting along line OA.

30

F

z A

300 N

30 300 mm

O

SOLUTION

x

300 mm

300 mm y

Force and Unit Vector: The force vector F and unit vector uOA must be determined first. From Fig. a F = (-300 sin 30° sin 30°i + 300 cos 30°j + 300 sin 30° cos 30°k) = { -75i + 259.81j + 129.90k} N uOA =

(- 0.45 - 0)i + (0.3 - 0)j + (0.2598 - 0)k rOA = = - 0.75i + 0.5j + 0.4330k rOA 2( -0.45 - 0)2 + (0.3 - 0)2 + (0.2598 - 0)2

Vector Dot Product: The magnitude of the projected component of F along line OA is FOA = F # uOA =

A - 75i + 259.81j + 129.90k B # A -0.75i + 0.5j + 0.4330k B

= ( - 75)(-0.75) + 259.81(0.5) + 129.90(0.4330) Ans.

= 242 N

Ans: FOA = 242 N 160

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z

2–138 . Cable OA is used to support column OB. Determine the angle u it makes with beam OC.

D 30

O f C

y 4m

u 8m

x 8m

B

A

Solution Unit Vector: uOC = 1i uOA = =

(4 - 0)i + (8 - 0)j + ( -8 - 0)k 2(4 - 0)2 + (8 - 0)2 + ( -8 - 0)2

1 2 2 i + j - k 3 3 3

The Angles Between Two Vectors U:

Then,

1 2 2 1 2 2 1 uOC # uOA = (1i) # a i + j - kb = 1a b + 0a b + (0)a- b = 3 3 3 3 3 3 3 U = cos - 1(uOC # uOA) = cos - 1

1 = 70.5° 3

Ans.

Ans: u = 70.5 161

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z

2–139 . Cable OA is used to support column OB. Determine the angle f it makes with beam OD.

D 30

O f C

y 4m

u 8m

x 8m

B

A

Solution Unit Vector: uOD = -sin 30°i + cos 30°j = -0.5i + 0.8660j uOA = =

(4 - 0)i + (8 - 0)j + ( - 8 - 0)k 2(4 - 0)2 + (8 - 0)2 + ( - 8 - 0)2

1 2 2 i + j - k 3 3 3

The Angles Between Two Vectors F: 1 2 2 uOD # uOA = ( - 0.5i + 0.8660j) # a i + j - kb 3 3 3

1 2 2 = ( -0.5)a b + (0.8660)a b + 0a - b 3 3 3 = 0.4107

Then, F = cos - 1(uOD # uOA) = cos - 10.4107 = 65.8°

Ans.

Ans: f = 65.8 162

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3–1. Determine the magnitude and direction u of F so that the particle is in equilibrium.

y 8 kN

30

x 5 kN

60

4 kN

Solution

u

Equations of Equilibrium. Referring to the FBD shown in Fig. a, + ΣFx = 0;   F sin u + 5 - 4 cos 60° - 8 cos 30° = 0    S

(1)

F sin u = 3.9282

F

   +cΣFy = 0;      8 sin 30° - 4 sin 60° - F cos u = 0

(2)

F cos u = 0.5359

Divide Eq (1) by (2), sin u = 7.3301 cos u sin u Realizing that tan u = , then cos u tan u = 7.3301

Ans.

u = 82.23° = 82.2°

Substitute this result into Eq. (1), F sin 82.23° = 3.9282

Ans.

F = 3.9646 kN = 3.96 kN

Ans:  u = 82.2° F = 3.96 kN 163

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3–2. The members of a truss are pin connected at joint O. Determine the magnitudes of F1 and F2 for equilibrium. Set u = 60°.

y

5 kN

F2

70 30

x O

SOLUTION + ©F = 0; : x

5 4

4 F2 sin 70° + F1 cos 60° - 5 cos 30° - (7) = 0 5

7 kN

u

3

F1

0.9397F2 + 0.5F1 = 9.930 + c ©Fy = 0;

F2 cos 70° + 5 sin 30° - F1 sin 60° -

3 (7) = 0 5

0.3420F2 - 0.8660F1 = 1.7 Solving: F2 = 9.60 kN

Ans.

F1 = 1.83 kN

Ans.

Ans:  F2 = 9.60 kN F1 = 1.83 kN 164

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3–3. The members of a truss are pin connected at joint O. Determine the magnitude of F1 and its angle u for equilibrium. Set F2 = 6 kN.

y

5 kN

F2

70 30

x O

SOLUTION + ©F = 0; : x

5 4

4 6 sin 70° + F1 cos u - 5 cos 30° - (7) = 0 5

7 kN

u

3

F1

F1 cos u = 4.2920 + c ©Fy = 0;

6 cos 70° + 5 sin 30° - F1 sin u -

3 (7) = 0 5

F1 sin u = 0.3521 Solving: u = 4.69°

Ans.

F1 = 4.31 kN

Ans.

Ans: u = 4.69° F1 = 4.31 kN 165

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*3–4. The members of a truss are connected to the gusset plate. If the forces are concurrent at point O, determine the magnitudes of F and T for equilibrium. Take u = 90°.

y

9 kN F A 5 3 B 4

SOLUTION 3 f = 90° - tan - 1 a b = 53.13° 4

O

+ ©F = 0; : x

4 T cos 53.13° - F a b = 0 5

+ c ©Fy = 0;

3 9 - T sin 53.13° - Fa b = 0 5

x

u

C T

Solving, T = 7.20 kN

Ans.

F = 5.40 kN

Ans.

Ans: T = 7.20 kN F = 5.40 kN 166

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3–5. The gusset plate is subjected to the forces of three members. Determine the tension force in member C and its angle u for equilibrium. The forces are concurrent at point O. Take F = 8 kN.

y

9 kN F A 5 3 B 4

SOLUTION + ©F = 0; : x

4 T cos f - 8a b = 0 5

(1)

+ c ©Fy = 0;

3 9 - 8a b - T sin f = 0 5

(2)

Rearrange then divide Eq. (1) into Eq. (2):

O

x

u

C T

tan f = 0.656, f = 33.27° T = 7.66 kN

Ans.

3 u = f + tan - 1 a b = 70.1° 4

Ans.

Ans: T = 7.66 kN u = 70.1° 167

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3–6.

The bearing consists of rollers, symmetrically confined within the housing. The bottom one is subjected to a 125-N force at its contact A due to the load on the shaft. Determine the normal reactions NB and NC on the bearing at its contact points B and C for equilibrium.

40°

SOLUTION + c ©Fy = 0;

NB

125 - NC cos 40° = 0 Ans.

NC = 163.176 = 163 N + ©F = 0; : x

NC

C

B A 125 N

NB - 163.176 sin 40° = 0 Ans.

NB = 105 N

Ans: NC = 163 N NB = 105 N 168

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3–7. Determine the tension developed in wires CA and CB required for equilibrium of the 10-kg cylinder. Take u = 40°.

B

A u

30° C

SOLUTION Equations of Equilibrium: Applying the equations of equilibrium along the x and y axes to the free-body diagram shown in Fig. a, + ©F = 0; : x

FCB cos 40° - FCA cos 30° = 0

(1)

+ c ©Fy = 0;

FCB sin 40° + FCA sin 30° - 10(9.81) = 0

(2)

Solving Eqs. (1) and (2), yields FCA = 80.0 N

FCB = 90.4 N

Ans.

Ans: FCA = 80.0 N FCB = 90.4 N 169

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*3–8. If cable CB is subjected to a tension that is twice that of cable CA, determine the angle u for equilibrium of the 10-kg cylinder. Also, what are the tensions in wires CA and CB?

B

A u

30° C

SOLUTION Equations of Equilibrium: Applying the equations of equilibrium along the x and y axes, + ©F = 0; : x

FCB cos u - FCA cos 30° = 0

(1)

+ c ©Fy = 0;

FCB sin u + FCA sin 30° - 10(9.81) = 0

(2)

However, it is required that (3)

FCB = 2FCA Solving Eqs. (1), (2), and (3), yields u = 64.3°

FCB = 85.2 N

FCA = 42.6 N

Ans.

Ans: u = 64.3° FCB = 85.2 N FCA = 42.6 N 170

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3–9. Determine the force in each cable and the force F needed to hold the 4-kg lamp in the position shown. Hint: First analyze the equilibrium at B; then, using the result for the force in BC, analyze the equilibrium at C.

A

60

D B 30 30

SOLUTION Initial guesses: TBC

30

F

TBA

1N

C

2N

Given At B:  6F x = 0; o

TBC cos  T 1  TBA cos  T 2



TBA sin  T 2  TBC sin  T 1  M g

n 6Fy = 0;

M

4 kg

T1

30°

T2

60°

T3

30°

§ TBC · ¨ ¸ © TBA ¹ At C:

§ TBC · ¨ ¸ © TBA ¹

Find  TBC  TBA

TCD

1N

F

0 0

§ 39.24 · ¨ ¸N © 67.97 ¹

Ans.

2N

Given  6F x = 0; o

TBC cos  T 1  TCD cos  T 3



TBC sin  T 1  TCD sin  T 3  F

n 6Fy = 0;

§ TCD · ¨ ¸ © F ¹

Find  TCD  F

§ TCD · ¨ ¸ © F ¹

0 0

§ 39.24 · ¨ ¸N © 39.24 ¹

Ans. Ans. T BC = 39.24 N T BA = 67.97 N TCD = 39.24 N F = 39.24 N 171

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3–10. The towing pendant AB is subjected to the force of 50 kN exerted by a tugboat. Determine the force in each of the bridles, BC and BD, if the ship is moving forward with constant velocity.

D

C 30⬚

20⬚

B

A 50 kN

Solution +

S ΣFx = 0; TBC sin 30° - TBD sin 20° = 0 + c ΣFy = 0; TBC cos 30° + TBD cos 20° - 50 = 0 Solving, TBC = 22.3 kN

Ans.

TBD = 32.6 kN

Ans.

Ans. TBC = 22.3 kN T BD = 32.6 kN 172

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3–11 . Two spheres A and B have an equal mass and are electrostatically charged such that the repulsive force acting between them has magnitude of 20 mN and is directed along line AB. Determine the angle T, the tension in cords AC and BC, and the mass m of each sphere.

C

Solution Guesses

TB

1 mN

m

1g

TA

1 mN

T

30°

u

30 B

Given: 20 mN

F

20 mN 20 mN

g

2

9.81 m > s

T1

30°

T2

30°

30

A

Given For B:  6F x = 0; o

F cos  T 2  TB sin  T 1



F sin  T 2  TB cos  T 1  m g

n 6Fy = 0;

0 0

For A:  6F x = 0; o

TA sin  T  F cos  T 2



TA cos  T  F sin  T 2  m g

n 6Fy = 0;

§ TA · ¨ ¸ ¨ TB ¸ ¨T ¸ ¨ ¸ ©m¹

Find  TA  TB  T  M

0

§ TA · ¨ ¸ © TB ¹

0

§ 52.92 · ¨ ¸ mN © 34.64 ¹

T

19.11

m

4.08 g

Ans.

Ans. T A = 52.92 mN, T B = 34.64 mN, u = 19.11, m = 4.08 g 173

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*3–12. Determine the stretch in each spring for equlibrium of the 2-kg block. The springs are shown in the equilibrium position.

3m

4m

C

3m

B kAC  20 N/m kAB  30 N/m

SOLUTION FAD = 2(9.81) = xAD(40) + ©F = 0; : x + c ©Fy = 0;

xAD = 0.4905 m

A

Ans.

4 1 FAB a b - FAC a b = 0 5 22 FAC a

kAD  40 N/m

1

3 b + FAB a b - 2(9.81) = 0 5 22

D

FAC = 15.86 N xAC =

15.86 = 0.793 m 20

Ans.

FAB = 14.01 N xAB =

14.01 = 0.467 m 30

Ans.

Ans: xAD = 0.4905 m xAC = 0.793 m xAB = 0.467 m 174

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3–13. The unstretched length of spring AB is 3 m. If the block is held in the equilibrium position shown, determine the mass of the block at D.

3m

4m

C

3m

B 20 N/m

kAC

kAB

SOLUTION

A

F = kx = 30(5 - 3) = 60 N + ©F = 0; : x

4 Tcos 45° - 60 a b = 0 5 T = 67.88 N

+ c ©Fy = 0;

30 N/m

D

3 -W + 67.88 sin 45° + 60 a b = 0 5 W = 84 N m =

84 = 8.56 kg 9.81

Ans.

Ans: m = 8.56 kg 175

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3–14. Determine the mass of each of the two cylinders if they cause a sag of s = 0.5 m when suspended from the rings at A and B. Note that s = 0 when the cylinders are removed.

2m

1.5 m

s

1m

2m D

C

k

SOLUTION

100 N/m

k A

100 N/m

B

TAC = 100 N>m (2.828 - 2.5) = 32.84 N + c ©Fy = 0;

32.84 sin 45° - m(9.81) = 0 Ans.

m = 2.37 kg

Ans: m = 2.37 kg 176

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3–15. Determine the magnitude and direction u of the equlibrium force FAB exerted along link AB by the tractive apparatus shown. The suspended mass is 10 kg. Neglect the size of the pulley at A.

75° A 45° B

θ

FAB

Solution Free Body Diagram: The tension in the cord is the same throughout the cord, that is 10(9.81) = 9.81 N. Equations of Equilibrium: +

S ΣFx = 0; FAB cos u - 98.1 cos 75° - 98.1 cos 45° = 0 FAB cos u = 94.757

[1]

+ c ΣFy = 0; 98.1 sin 75° - 98.1 sin 45° - FAB sin u = 0 FAB sin u = 25.390

[2]

Solving Eqs. [1]and [2] yields u = 15.0°

Ans.

FAB = 98.1 N

Ans: u = 15.0 FAB = 98.1 N 177

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*3–16. The lift sling is used to hoist a container having a mass of 500 kg. Determine the force in each of the cables AB and AC as a function of u. If the maximum tension allowed in each cable is 5 kN, determine the shortest length of cables AB and AC that can be used for the lift. The center of gravity of the container is located at G.

F A

B

SOLUTION

θ

θ

1.5 m

C

1.5 m

Free-Body Diagram: By observation, the force F1 has to support the entire weight of the container. Thus, F1 = 50019.812 = 4905 N. Equations of Equilibrium:

G

+ ©F = 0; : x

FAC cos u - FAB cos u = 0

+ c ©Fy = 0;

4905 - 2F sin u = 0

FAC = FAB = F

F = 52452.5 cos u6 N

Thus, Ans.

FAC = FAB = F = 52.45 cos u6 kN If the maximum allowable tension in the cable is 5 kN, then 2452.5 cos u = 5000 u = 29.37° From the geometry, l =

1.5 and u = 29.37°. Therefore cos u l =

1.5 = 1.72 m cos 29.37°

Ans.

Ans: FAC = {2.45 cos u} kN l = 1.72 m 178

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3–17. The springs BA and BC each have a stiffness of 500 N>m and an unstretched length of 3 m. Determine the horizontal force F applied to the cord which is attached to the small ring B so that the displacement of AB from the wall is d = 1.5 m.

A k

500 N/m B

6m

F

SOLUTION + ©F = 0; : x

k

1.5 211.25

500 N/m

(T)(2) - F = 0

C d

T = ks = 500(232 + (1.5)2 - 3) = 177.05 N Ans.

F = 158 N

Ans: F = 158 N 179

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3–18. The springs BA and BC each have a stiffness of 500 N>m and an unstretched length of 3 m. Determine the displacement d of the cord from the wall when a force F = 175 N is applied to the cord.

A k

500 N/m B

6m

F

SOLUTION + ©F = 0; : x

k

500 N/m

175 = 2T sin u

C

T sin u = 87.5 TC

d 2

23 + d 2

d

S = 87.5

T = ks = 500( 232 + d 2 - 3) d a1 -

3 29 + d2

b = 0.175

By trial and error:

175 N

Ans.

d = 1.56 m

Ans: d = 1.56 m 180

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3–19. Determine the tension developed in each wire used to support the 50-kg chandelier.

A 30⬚ B

C 30⬚

45⬚ D

Solution Equations of Equilibrium: First, we will apply the equations of equilibrium along the x and y axes to the free-body diagram of joint D shown in Fig. (a). +

S ΣFx = 0; FCD cos 30° - FBD cos 45° = 0

(1)

+ c ΣFy = 0; FCD sin 30° + FBD sin 45° - 50(9.81) = 0

(2)

Solving Eqs. (1) and (2), yields Ans.

FCD = 359 N FBD = 439.77 N = 440 N

Using the result FBD = 439.77 N and applying the equations of equilibrium along the x and y axes to the free-body diagram of joint B shown in Fig. (b), + c ΣFy = 0; FAB sin 30° - 439.77 sin 45° = 0 Ans.

FAB = 621.93N = 622 N +

S ΣFx = 0; FBC + 439.77 cos 45° - 621.93 cos 30° = 0 Ans.

FBC = 228 N

Ans: FBD = 440 N FAB = 622 N FBC = 228 N 181

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*3–20. If the tension developed in each of the four wires is not allowed to exceed 600 N, determine the maximum mass of the chandelier that can be supported.

A 30⬚ B

C 30⬚

45⬚ D

Solution Equations of Equilibrium: First, we will apply the equations of equilibrium along the x and y axes to the free-body diagram of joint D shown in Fig. (a). +

S ΣFx = 0; FCD cos 30° - FBD cos 45° = 0

(1)

+ c ΣFy = 0; FCD sin 30° + FBD sin 45° - m(9.81) = 0

(2)

Solving Eqs. (1) and (2), yields FCD = 7.1841m

FBD = 8.7954m

Using the result FBD = 8.7954 m and applying the equations of equilibrium along the x and y axes to the free-body diagram of joint B shown in Fig. (b), + c ΣFy = 0; FAB sin 30° - 8.7954m sin 45° = 0 FAB = 12.4386m +

S ΣFx = 0; FBC + 8.7954m cos 45° - 12.4386m cos 30° = 0 FBC = 4.5528m From this result, notice that cable AB is subjected to the greatest tensile force. Thus, it will achieve the maximum allowable tensile force first. FAB = 600 = 12.4386m Ans.

m = 48.2 kg

Ans: m = 48.2 kg 182

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3–21. If the spring DB has an unstretched length of 2 m, determine the stiffness of the spring to hold the 40-kg crate in the position shown.

2m

3m

C

B

2m

k

D

Solution

A

Equations of Equilibrium. Referring to the FBD shown in Fig. a,



+ ΣFx = 0;   TBD a 3 b - TCDa 1 b = 0 S 113 12 +cΣFy = 0;      TBD a

Solving Eqs (1) and (2)

(1)

2 1 b + TCDa b - 40(9.81) = 0 113 12

(2)

TBD = 282.96 N   TCD = 332.96 N The stretched length of the spring is

Then, x = l - l0 =

l = 232 + 22 = 213 m

( 113 - 2 ) m. Thus, Fsp = kx;



282.96 = k ( 113 - 2 )

Ans.

k = 176.24 N>m = 176 N>m

Ans: k = 176 N>m 183

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3–22. Determine the unstretched length of DB to hold the 40-kg crate in the position shown. Take k = 180 N>m.

2m

3m

C

B

2m

k

D

Solution

A

Equations of Equilibrium. Referring to the FBD shown in Fig. a,



+ ΣFx = 0;   TBD a 3 b - TCDa 1 b = 0 S 113 12 +cΣFy = 0;      TBD a

Solving Eqs (1) and (2)

(1)

2 1 b + TCDa b - 40(9.81) = 0 113 12

(2)

TBD = 282.96 N   TCD = 332.96 N The stretched length of the spring is l = 232 + 22 = 213 m

Then, x = l - l0 = 113 - l0. Thus Fsp = kx;



282.96 = 180 ( 113 - l0 )

Ans.

l0 = 2.034 m = 2.03 m

Ans: l0 = 2.03 m 184

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3–23.

Unstretched position

Determine the stiffness kT of the single spring such that the force F will stretch it by the same amount s as the force F stretches the two springs. Express kT in terms of stiffness k1 and k2 of the two springs.

kT F

s k1

k2 F

s

Solution F = ks s = s1 + s2 s =

F F F = + kT k1 k2

1 1 1 = + kT k1 k2

Ans.

Ans: 1 1 1 = + kT k1 k2 185

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*3–24. A 4-kg sphere rests on the smooth parabolic surface. Determine the normal force it exerts on the surface and the mass mB of block B needed to hold it in the equilibrium position shown.

y

B

60

SOLUTION Geometry: The angle u which the surface make with the horizontal is to be determined first. tan u `

dy = = 5.0x ` = 2.00 ` dx x = 0.4m x = 0.4 m

x = 0.4 m

A y  2.5x2 0.4 m x 0.4 m

u = 63.43° Free Body Diagram: The tension in the cord is the same throughout the cord and is equal to the weight of block B, WB = mB (9.81). Equations of Equilibrium: + ©F = 0; : x

mB (9.81) cos 60° - Nsin 63.43° = 0 [1]

N = 5.4840mB + c ©Fy = 0;

mB (9.81) sin 60° + Ncos 63.43° - 39.24 = 0 [2]

8.4957mB + 0.4472N = 39.24 Solving Eqs. [1] and [2] yields mB = 3.58 kg

Ans.

N = 19.7 N

Ans: mB = 3.58 kg N = 19.7 N 186

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3–25. 3.5 m x

Cable ABC has a length of 5 m. Determine the position x and the tension developed in ABC required for equilibrium of the 100-kg sack. Neglect the size of the pulley at B.

C

0.75 m A B

SOLUTION Equations of Equilibrium: Since cable ABC passes over the smooth pulley at B, the tension in the cable is constant throughout its entire length. Applying the equation of equilibrium along the y axis to the free-body diagram in Fig. a, we have + c ©Fy = 0;

2T sin f - 100(9.81) = 0

(1)

Geometry: Referring to Fig. b, we can write x 3.5 - x + = 5 cos f cos f f = cos - 1 a

3.5 b = 45.57° 5

Also, x tan 45.57° + 0.75 = (3.5 - x) tan 45.57° Ans.

x = 1.38 m Substituting f = 45.57° into Eq. (1), yields

Ans.

T = 687 N

Ans: x = 1.38 m T = 687 N 187

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3–26. The device shown is used to straighten the frames of wrecked autos. Determine the tension of each segment of the chain, i.e., AB and BC, if the force which the hydraulic cylinder DB exerts on point B is 3.50 kN, as shown.

B

A 3.50 kN D

C

450 mm

400 mm 250 mm

SOLUTION Equations of Equilibrium: A direct solution for FBC can be obtained by summing forces along the y axis. + c ©Fy = 0;

3.5 sin 48.37° - FBC sin 60.95° = 0 Ans.

FBC = 2.993 kN = 2.99 kN Using the result FBC = 2.993 kN and summing forces along x axis, we have + ©F = 0; : x

3.5 cos 48.37° + 2.993 cos 60.95° - FAB = 0 Ans.

FAB = 3.78 kN

Ans: FBC = 2.99 kN FAB = 3.78 kN 188

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3–27. Determine the force in each cord for equilibrium of the 200-kg crate. Cord BC remains horizontal due to the roller at C, and AB has a length of 1.5 m. Set y = 0.75 m.

2m A y

C

B

Solution Geometry: From the geometry of the figure, u = sin

-1

a

0.75 b = 30° 1.5

Equations of Equilibrium: Applying the equations of equilibrium to the free-body diagram in Fig. (a), + c ΣFy = 0; FBA sin 30° - 200(9.81) = 0 FBA = 3924 N = 3.92 kN +

S ΣFx = 0; 3924 cos 30° - FBC = 0 FBC = 3398.28 N = 3.40 kN

Ans. Ans.

Ans: FBA = 3.92 kN FBC = 3.40 kN 189

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*3–28. If the 1.5-m-long cord AB can withstand a maximum force of 3500 N, determine the force in cord BC and the distance y so that the 200-kg crate can be supported.

2m A y

C

B

Solution Equations of Equilibrium: Applying the equations of equilibrium along the x and y axes to the free-body diagram in Fig. (a), + c ΣFy = 0; 3500 sin u - 200(9.81) = 0 u = 34.10° +

S ΣFx = 0; 3500 cos 34.10° - FBC = 0 FBC = 2898.37 N = 2.90 kN

Ans. Ans.

y = 1.5 sin 34.10° = 0.841 m = 841 mm

Ans: FBC = 2.90 kN y = 841 mm 190

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3–29. Blocks D and E have a mass of 4 kg and 6 kg, respectively. If x = 2 m determine the force F and the sag s for equilibrium.

6m x

C

B

s A

D

Solution

F

E

Equations of Equilibrium. Referring to the geometry shown in Fig. a, cos f = cos u =

s 2

2

2s + 2 s

2

2

2s + 4

   sin u =

Referring to the FBD shown in Fig. b,

+ ΣFx = 0;     6(9.81)a S

2 2

2

2s + 2 3

2



2

2s + 2

2

2s + 22 4

2s + 42

b - 4(9.81)a =

4

2

4 2

2s + 42

b = 0

2

2s + 42

Ans.

s = 3.381 m = 3.38 m

  +cΣFy = 0;    6(9.81)a

2

   sin f =

3.381 2

2

23.381 + 2

b + 4(9.81)a

F = 75.99 N = 76.0 N

3.381 23.3812 + 42

b - F = 0

Ans.

Ans: s = 3.38 m F = 76.0 N 191

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3–30. Blocks D and E have a mass of 4 kg and 6 kg, respectively. If F = 80 N, determine the sag s and distance x for equilibrium.

6m x

C

B

s A

D

Solution

F

E

Equations of Equilibrium. Referring to the FBD shown in Fig. a,

+ ΣFx = 0;   6(9.81) sin f - 4(9.81) sin u = 0 S sin f =



2 sin u 3

(1)

+cΣFy = 0;      6(9.81) cos f + 4(9.81) cos u - 80 = 0 (2)

3 cos f + 2 cos u = 4.0775 Using Eq (1), the geometry shown in Fig. b can be constructed. Thus cos f =

29 - 4 sin2 u 3

Substitute this result into Eq. (2), 3a

29 - 4 sin2u b + 2 cos u = 4.0775 3

29 - 4 sin2 u = 4.0775 - 2 cos u

9 - 4 sin2 u = 4 cos2 u - 16.310 cos u + 16.6258 16.310 cos u = 4 ( cos2 u + sin2 u ) + 7.6258 Here, cos2 u + sin2 u = 1. Then cos u = 0.7128   u = 44.54° Substitute this result into Eq (1) 2 sin f = sin 44.54°   f = 27.88° 3 From Fig. c,

6 - x x = tan 44.54° and = tan 27.88°. s s

So then, 6 - x x + = tan 44.54° + tan 27.88° s s 6 = 1.5129 s Ans.

s = 3.9659 m = 3.97 m x = 3.9659 tan 27.88°

Ans.

= 2.0978 m = 2.10 m

Ans: s = 3.97 m x = 2.10 m 192

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3–31. A

Determine the tension developed in each cord required for equilibrium of the 20-kg lamp.

E 5

4 3

C

B

30°

D

45° F

SOLUTION Equations of Equilibrium: Applying the equations of equilibrium along the x and y axes to the free-body diagram of joint D shown in Fig. a, we have + ©F = 0; : x

FDE sin 30° - 20(9.81) = 0

FDE = 392.4 N = 392 N

+ c ©Fy = 0;

392.4 cos 30° - FCD = 0

FCD = 339.83 N = 340 N Ans.

Ans.

Using the result FCD = 339.83 N and applying the equations of equilibrium along the x and y axes to the free-body diagram of joint D shown in Fig. b, we have + ©F = 0; : x

3 339.83 - FCA a b - FCD cos 45° = 0 5

(1)

+ c ©Fy = 0;

4 FCA a b - FCB sin 45° = 0 5

(2)

Solving Eqs. (1) and (2), yields FCB = 275 N

Ans.

FCA = 243 N

Ans: FDE = FCD = FCB = FCA = 193

392 N 340 N 275 N 243 N

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*3–32. A

Determine the maximum mass of the lamp that the cord system can support so that no single cord develops a tension exceeding 400 N.

4

B

3

E

5

C

30°

D

45° F

SOLUTION Equations of Equilibrium: Applying the equations of equilibrium along the x and y axes to the free-body diagram of joint D shown in Fig. a, we have + c ©Fy = 0;

FDE sin 30° - m(9.81) = 0

FDE = 19.62m

+ ©F = 0; : x

19.62m cos 30° - FCD = 0

FCD = 16.99m

Using the result FCD = 16.99m and applying the equations of equilibrium along the x and y axes to the free-body diagram of joint D shown in Fig. b, we have + ©F = 0; : x + c ©Fy = 0;

3 16.99m - FCA a b - FCD cos 45° = 0 5 4 FCA a b - FCB sin 45° = 0 5

(1) (2)

Solving Eqs. (1) and (2), yields FCB = 13.73m

FCA = 12.14m

Notice that cord DE is subjected to the greatest tensile force, and so it will achieve the maximum allowable tensile force first. Thus FDE = 400 = 19.62m

m = 20.4 kg

Ans.

Ans: m = 20.4 kg 194

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3–33. A scale is constructed using the 10-kg mass, the 2-kg pan P, and the pulley and cord arrangement. Cord BCA is 2  m long. If s = 0.75 m, determine the mass D in the pan. Neglect the size of the pulley.

1.5 m A

C 1.5

0 s

m

B D

Solution

P +

S ΣFx = 0; 98.1 cos u - TAB cos f = 0

(1)

+ c ΣFy = 0; TAB sin f + 98.1 sin u - m(9.81) = 0

(2)

2

2

2

(1.5) = x + y

(1.25)2 = (1.5 - x)2 + y2 (1.25)2 = (1.5 - x)2 + (1.5)2 - x2 -3x + 2.9375 = 0 x = 0.9792 m y = 1.1363 m Thus, f = sin - 1 a

Solving Eq. (1) and (2),

u = sin - 1 a

1.1363 b = 49.25° 1.5

1.1363 b = 65.38° 1.25

TAB = 62.62 N m = 13.9 kg Therefore, mD = 13.9 kg - 2 kg = 11.9 kg

Ans.

Ans: m D = 11.9 kg 195

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3–34. The 30-kg pipe is supported at A by a system of five cords. Determine the force in each cord for equilibrium.

5

3

D

4

C B 60°

A

E

H

SOLUTION At H: + c ΣFy = 0;

THA - 30(9.81) = 0 Ans.

THA = 294 N At A: + c ΣFy = 0;

TAB sin 60° - 30(9.81) = 0 Ans.

TAB = 339.83 = 340 N + ΣFx = 0; S

TAE - 339.83 cos 60° = 0 Ans.

TAE = 170 N At B: + c ΣFy = 0;

3 TBD a b - 339.83 sin 60° = 0 5

Ans.

TBD = 490.50 = 490 N + ΣFx = 0; S

4 490.50 a b + 339.83 cos 60° - TBC = 0 5

Ans.

TBC = 562 N

Ans: THA = TAB = TAE = TBD = TBC = 196

294 N 340 N 170 N 490 N 562 N

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3–35. Each cord can sustain a maximum tension of 500 N. Determine the largest mass of pipe that can be supported.

5

3

D

4

C B 60°

SOLUTION

E

A

H

At H: + c ©Fy = 0;

FHA = W

At A: + c ©Fy = 0;

FAB sin 60° - W = 0 FAB = 1.1547 W

+ ©F = 0; : x

FAE - (1.1547 W) cos 60° = 0 FAE = 0.5774 W

At B: + c ©Fy = 0;

3 FBD a b - (1.1547 cos 30°)W = 0 5 FBD = 1.667 W

+ ©F = 0; : x

4 -FBC + 1.667 W a b + 1.1547 sin 30° = 0 5 FBC = 1.9107 W

By comparison, cord BC carries the largest load. Thus 500 = 1.9107 W W = 261.69 N m =

261.69 = 26.7 kg 9.81

Ans.

Ans: m = 26.7 kg 197

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*3–36. Determine the distances x and y for equilibrium if F1 = 800 N and F2 = 1000 N.

F1

D

C y

B

F2

2m

Solution Equations of Equilibrium. The tension throughout rope ABCD is constant, that is F1 = 800 N. Referring to the FBD shown in Fig. a,

A x

+cΣFy = 0;   800 sin f - 800 sin u = 0   f = 0

+ ΣFx = 0;        1000 - 2[800 cos u] = 0      u = 51.32° S Referring to the geometry shown in Fig. b,

y = 2 m

Ans.

2 = tan 51.32°;  x = 1.601 m = 1.60 m x

Ans.

and

Ans: y = 2m x = 1.60 m 198

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3–37. Determine the magnitude of F1 and the distance y if x = 1.5 m and F2 = 1000 N.

F1

D

C y

B

F2

2m

Solution Equations of Equilibrium. The tension throughout rope ABCD is constant, that is F1. Referring to the FBD shown in Fig. a,

+cΣFy = 0;      F1a

y 2

2

2y + 1.5 y

2



b - F1a 2

2y + 1.5

=

A x

2 b = 0 2.5

2 2.5

Ans.

y = 2 m

+ ΣFx = 0;         1000 - 2c F1a 1.5 b d = 0 S 2.5

Ans.

  F1 = 833.33 N = 833 N

Ans: y = 2m F1 = 833 N 199

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3–38. Determine the force in cables AB and AC necessary to support the 12-kg traffic light.

B

C 7

25 24

A

12°

Solution Equations of Equilibrium: +

S ΣFx = 0; FAB cos 12° - FAC a FAB = 0.9814FAC

+ c ΣFy = 0; FAB sin 12° + FAC a

24 b = 0 25

7 b - 117.72 = 0 25

0.2079FAB + 0.28FAC = 117.72

[1]

[2]

Solving Eqs. [1] and [2] yields Ans.

FAB = 239 N FAC = 243 N

Ans: FAB = 239 N FAC = 243 N 200

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C 2m

3–39. The pail and its contents have a mass of 60 kg. If the cable is 15 m long, determine the distance y of the pulley for equilibrium. Neglect the size of the pulley at A.

B y A

10 m

Solution Free Body Diagram: Since the pulley is smooth, the tension in the cable is the same throughout the cable. Equations of Equilibrium: +

S ΣFx = 0; T sin u - T sin f = 0 u = f Geometry: l1 = 2(10 - x)2 + (y - 2)2 l2 = 2x2 + y2

Since u = f, two triangles are similar.

2(10 - x)2 + (y - 2)2 y - 2 10 - x = = x y 2x2 + y2

Also,

[1]

l1 + l2 = 15 2

2(10 - x) + (y - 2)2 + 2x2 + y2 = 15

£

2x2 + y2

2x2 + y2

However, from Eq.[1]

≥ 2(10 - x)2 + (y - 2)2 + 2x2 + y2 = 15

2(10 - x)2 + (y - 2)2 2x2 + y2

=

10 - x , Eq.[2] becomes x

10 - x b + 2x2 + y2 = 15 x Dividing both sides of Eq.[3] by 2x2 + y2 yields 2x2 + y2 a

From Eq.[1]

10 15 = 2 x 2x + y2

[2]

[3]

x = 20.8y

[4]

y - 2 5y 10 - x = x = x y y - 1

[5]

Equating Eq.[1] and [5] yields 5y y - 1 y = 6.59m

20.8y =

Ans.

Ans: y = 6.59 m 201

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*3–40. Determine the forces in cables AC and AB needed to hold the 20-kg ball D in equilibrium. Take F = 300 N and d = 1 m.

B 1.5 m C

SOLUTION

d A

Equations of Equilibrium: : ©Fx = 0; +

+ c ©Fy = 0;

F

2m

4 2 b - FAC a b = 0 241 25 06247FAB + 0.8944FAC = 300

(1)

5 1 b + FAC a b - 196.2 = 0 241 25 0.7809FAB + 0.4472FAC = 196.2

(2)

300 - FAB a

D

FAB a

Solving Eqs. (1) and (2) yields FAB = 98.6 N

Ans.

FAC = 267 N

Ans: F = {73.6 sec u} N 202

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3–41. The ball D has a mass of 20 kg. If a force of F = 100 N is applied horizontally to the ring at A, determine the largest dimension d so that the force in cable AC is zero.

B 1.5 m C d

SOLUTION

A

Equations of Equilibrium: : ©Fx = 0;

100 - FAB cos u = 0

FAB cos u = 100

(1)

+ c ©Fy = 0;

FAB sin u - 196.2 = 0

FAB sin u = 196.2

(2)

+

F

2m D

Solving Eqs. (1) and (2) yields u = 62.99°

FAB = 220.21 N

From the geometry, d + 1.5 = 2 tan 62.99° Ans.

d = 2.42 m

Ans: d = 2.42 m 203

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3–42. The load has a mass of 15 kg and is lifted by the pulley system shown. Determine the force F in the cord as a function of the angle u. Plot the function of force F versus the angle u for 0 … u … 90°.

45°

θ

θ F

SOLUTION Free-Body Diagram: The tension force is the same throughout the cord. Equations of Equilibrium: + ©F = 0; : x

F sin u - F sin u = 0

+ c ©Fy = 0;

2F cos u - 147.15 = 0

(Satisfied!)

F = 573.6 sec u6 N

Ans.

Ans:

F = 573.6 sec u6 N

204

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3–43. The three cables are used to support the 40-kg flowerpot. Determine the force developed in each cable for equilibrium.

z

D

1.5 m y A

Solution

ΣFz = 0;   FAD a

B

2m

Equations of Equilibrium. Referring to the FBD shown in Fig. a, 1.5 2

21.5 + 22 + 1.52

x

1.5 m

C

b - 40(9.81) = 0 

FAD = 762.69 N = 763 N  Ans.

Using this result, ΣFx = 0;   FAC - 762.69 a

ΣFy = 0;   FAB - 762.69 a

1.5 2

21.5 + 22 + 1.52 2 2

21.5 + 22 + 1.52

b = 0  

  FAC = 392.4 N = 392 N  Ans.

b = 0

  FAB = 523.2 N = 523 N  Ans.

Ans: FAD = 763 N FAC = 392 N FAB = 523 N 205

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*3–44. Determine the magnitudes of F1, F2, and F3 for equilibrium of the particle.

z F2 4 kN

10 kN 25

30

24

7

y

30 F3

Solution

x

Equations of Equilibrium. Referring to the FBD shown, ΣFy = 0;  10 a

F1

24 b - 4 cos 30° - F2 cos 30° = 0  F2 = 7.085 kN = 7.09 kN  Ans. 25

ΣFx = 0;  F1 - 4 sin 30° - 10 a

7 b = 0 25

Ans.

F1 = 4.80 kN

Using the result of F2 = 7.085 kN, ΣFz = 0;  7.085 sin 30° - F3 = 0

F3 = 3.543 kN = 3.54 kN Ans.

Ans: F2 = 7.09 kN F1 = 4.80 kN F3 = 3.54 kN 206

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3–45. Determine the tension in the cables in order to support the 100-kg crate in the equilibrium position shown.

z

C

2m

D 1m

A 2.5 m

SOLUTION

B

Force Vectors: We can express each of the forces on the free-body diagram shown in Fig. (a) in Cartesian vector form as

x

2m

2m y

FAB = FAB i FAC = - FAC j FAD = FAD B

(- 2 - 0)i+ (2 - 0)j+ (1 - 0)k 2

2

2

2(- 2 - 0) + (2 - 0) + (1 - 0)

R = - FAD i + 2 3

2 1 F j + FAD k 3 AD 3

W = [ -100(9.81)k]N = [ - 981 k]N Equations of Equilibrium: Equilibrium requires ©F = 0;

FAB + FAC + FAD + W = 0

2 2 1 FAB i + ( -FAC j) + a - FAD i + FAD j + FAD kb + ( - 981k) = 0 3 3 3 a FAB -

2 2 1 F b i + a - FAC + FAD b j + a FAD - 981b k = 0 3 AD 3 3

Equating the i, j, and k components yields FAB -FAC +

2 F = 0 3 AD

(1)

2 F = 0 3 AD

(2)

1 F - 981 = 0 3 AD

(3)

Solving Eqs. (1) through (3) yields FAD = 2943 N = 2.94 kN

Ans.

FAB = FAC = 1962 N = 1.96 kN

Ans.

Ans: FAD = 2.94 kN FAB = 1.96 kN 207

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3–46. Determine the maximum mass of the crate so that the tension developed in any cable does not exceeded 3 kN.

z

C

2m

D 1m

A 2.5 m

SOLUTION

B

Force Vectors: We can express each of the forces on the free-body diagram shown in Fig. (a) in Cartesian vector form as

x

2m

2m y

FAB = FAB i FAC = - FAC j FAD = FAD B

(- 2 - 0)i + (2 - 0)j +(1 - 0)k 2(- 2 - 0)2 + (2 - 0)2 + (1 - 0)2

R = - FAD i + 2 3

2 1 F j + FAD 3 AD 3

W = [ -m(9.81)k] Equations of Equilibrium: Equilibrium requires ©F = 0;

FAB + FAC + FAD + W = 0

2 2 1 FAB i + ( -FAC j) + a - FAD i + FAD j + FAD kb + [- m(9.81)k] = 0 3 3 3 aFAB -

2 2 1 F b i + a - FAC + FAD b j + a FAD - 9.81m bk = 0 3 AD 3 3

Equating the i, j, and k components yields FAB -

2 F = 0 3 AD

(1)

2 F = 0 3 AD

(2)

1 - 9.81m = 0 F 3 AD

(3)

-FAC +

When cable AD is subjected to maximum tension, FAD = 3000 N. Thus, by substituting this value into Eqs. (1) through (3), we have FAB = FAC = 2000 N Ans.

m = 102 kg

Ans: m = 102 kg 208

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3–47. Determine the force in each cable needed to support the 20-kg flowerpot.

z B

6m

4m D

A 3m

Solution Equations of Equilibrium. ΣFz = 0;     FAB a ΣFx = 0;     FAC a

2m C x

2m

y

6 b - 20(9.81) = 0        FAB = 219.36 N = 219 N Ans. 145 2 2 b - FADa b = 0      FAC = FAD = F 120 120

Using the results of FAB = 219.36 N and FAC = FAD = F, ΣFy = 0;     2c F a



4 3 b d - 219.36 a b = 0 120 145

Ans.

FAC = FAD = F = 54.84 N = 54.8 N

Ans: FAB = 219 N FAC = FAD = 54.8 N 209

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*3–48. The lamp has a mass of 15 kg and is supported by a pole AO and cables AB and AC. If the force in the pole acts along its axis, determine the forces in AO, AB, and AC for equilibrium.

z 1.5 m

A

B

4m O

SOLUTION FAO = FAO {

C

y

2m 6m

2 6 1.5 j + i k} N 6.5 6.5 6.5

1.5 m 1.5 m

6 3 6 FAB = FAB {- i + j - k} N 9 9 9 x

2 3 6 FAC = FAC {- i + j - k} N 7 7 7 W = 15(9.81)k = {-147.15k} N ©Fx = 0;

0.3077FAO - 0.6667FAB - 0.2857FAC = 0

©Fy = 0;

-0.2308FAO + 0.3333FAB + 0.4286FAC = 0

©Fz = 0;

0.9231FAO - 0.667FAB - 0.8571FAC - 147.15 = 0 FAO = 319 N

Ans.

FAB = 110 N

Ans.

FAC = 85.8 N

Ans.

Ans: FAO = 319 N FAB = 110 N FAC = 85.8 N 210

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3–49. Cables AB and AC can sustain a maximum tension of 500 N, and the pole can support a maximum compression of 300 N. Determine the maximum weight of the lamp that can be supported in the position shown. The force in the pole acts along the axis of the pole.

z 1.5 m

A

B

4m O

SOLUTION

C

y

2m 6m

2 1.5 6 FAO = FAO { i j + k} N 6.5 6.5 6.5 3 6 6 FAB = FAB {- i + j - k } N 9 9 9 2 3 6 FAC = FAC {- i + j - k } N 7 7 7 W = {Wk} N 2 6 2 ©Fx = 0; F - FAB - FAC = 0 6.5 AO 9 7 1.5 3 3 F + FAB + FAC = 0 ©Fy = 0; 6.5 AO 9 7 6 6 6 F - FAB - FAC - W = 0 ©Fz = 0; 6.5 AO 9 7

1.5 m 1.5 m

x

1) Assume FAB = 500 N 2 FAO 6.5 1.5 F 6.5 AO 6 FAO 6.5

6 (500) 9 3 + (500) 9 6 (500) 9

2 FAC = 0 7 3 + FAC = 0 7 6 FAC - W = 0 7

Solving, FAO = 1444.444 N 7 300 N (N . G!) FAC = 388.889 N W = 666.667 N 2) Assume FAC = 500 N 2 F 6.5 AO 1.5 F 6.5 AO 6 F 6.5 AO

6 F 9 AB 3 + FAB 9 6 F 9 AB

2 (500) = 0 7 3 + (500) = 0 7 6 (500) - W = 0 7

Solving, FAO = 1857.143 N 7 300 N (N . G!) FAB = 642.857 N 7 500 N (N . G!) 3) Assume FAO = 300 N 2 (300) 6.5 1.5 (300) 6.5 6 (300) 6.5

6 F 9 AB 3 + FAB 9 6 F 9 AB

2 F = 0 7 AC 3 + FAC = 0 7 6 F - W = 0 7 AC

Solving, FAC = 80.8 N FAB = 104 N W = 138 N

Ans. 211

Ans: W = 138 N

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3–50.

z

If the balloon is subjected to a net uplift force of F = 800 N, determine the tension developed in ropes AB, AC, AD.

F

A

6m 1.5 m 2m

SOLUTION Force Vectors: We can express each of the forces on the free-body diagram shown in Fig. (a) in Cartesian vector form as

3m x

FAB

2m

C

B

2.5 m

( -1.5 - 0)i + (-2 - 0)j + (-6 - 0)k

3 4 12 = FAB C S = FAB i FAB j FAB k 2 2 2 13 13 13 2(-1.5 - 0) + (-2 - 0) + (-6 - 0)

FAC = FAC C

(2 - 0)i + (-3 - 0)j + (-6 - 0)k 2(2 - 0) + (-3 - 0) + (-6 - 0)

FAD = FAD C

2

2

2

(0 - 0)i + (2.5 - 0)j + ( -6 - 0)k 2(0 - 0) + (2.5 - 0) + (-6 - 0) 2

2

2

S =

2 3 6 F i - FAC j - FAC k 7 AC 7 7

S =

5 12 F j F k 13 AD 13 AD

W = {800k}N Equations of Equilibrium: Equilibrium requires gF = 0;

FAB + FAC + FAD + W = 0

¢-

3 4 12 2 3 6 5 12 F i F j F k ≤ + ¢ FAC i - FAC j - FAC k ≤ + ¢ FAD j F k ≤ + 800 k = 0 13 AB 13 AB 13 AB 7 7 7 13 13 AD

¢-

3 2 4 3 5 12 6 12 F + FAC ≤ i + ¢ F - FAC F ≤j + ¢ F - FAC F + 800 ≤ k = 0 13 AB 7 13 AB 7 13 AD 13 AB 7 13 AD

Equating the i, j, and k components yields

-

3 2 FAB + FAC = 0 13 7

(1)

-

4 3 5 FAB - FAC + F = 0 13 7 13 AD

(2)

-

12 6 12 FAB - FAC FAD + 800 = 0 13 7 13

(3)

Solving Eqs. (1) through (3) yields FAC = 203 N

Ans.

FAB = 251 N

Ans.

FAD = 427 N

Ans.

212

D

y

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3- 51. If each one of the ropes will break when it is subjected to a tensile force of 450 N, determine the maximum uplift force F the balloon can have before one of the ropes breaks.

F

A

6m 1.5 m 2m

SOLUTION

C

Force Vectors: We can express each of the forces on the free-body diagram shown in Fig. a in Cartesian vector form as

FAB = FAB C

FAC = FAC C

FAD = FAD C

( -1.5 - 0)i + (-2 - 0)j + (-6 - 0)k 2(-1.5 - 0) + (-2 - 0) + (-6 - 0) 2

2

2

(2 - 0)i + (-3 - 0)j + (-6 - 0)k 2(2 - 0) + (-3 - 0) + (-6 - 0) 2

2

2

(0 - 0)i + (2.5 - 0)j + ( -6 - 0)k 2(0 - 0)2 + (2.5 - 0)2 + (-6 - 0)2

S = -

3m x

2m

B

2.5 m

3 4 12 F iF j F k 13 AB 13 AB 13 AB

S =

2 3 6 FAC i - FAC j - FAC k 7 7 7

S =

5 12 F j F k 13 AD 13 AD

F=Fk Equations of Equilibrium: Equilibrium requires gF = 0;

FAB + FAC + FAD + F = 0

¢-

3 4 12 2 3 6 5 12 F i F j F k ≤ + ¢ FAC i - FAC j - FAC k ≤ + ¢ FAD j F k ≤ + Fk = 0 13 AB 13 AB 13 AB 7 7 7 13 13 AD

¢-

3 2 4 3 5 12 6 12 FAB + FAC ≤ i + ¢ FAB - FAC + FAD ≤ j + ¢ FAB - FAC FAD + F ≤ k = 0 13 7 13 7 13 13 7 13

Equating the i, j, and k components yields

-

3 2 FAB + FAC = 0 13 7

(1)

-

4 3 5 F - FAC + F = 0 13 AB 7 13 AD

(2)

-

12 6 12 F - FAC F + F = 0 13 AB 7 13 AD

(3)

Assume that cord AD will break first. Substituting FAD = 450 N into Eqs. (2) and (3) and solving Eqs. (1) through (3), yields FAB = 264.71 N FAC = 213.8 N F = 842.99 N = 843 N

Ans.

Since FAC = 213.8 N 6 450 N and FAB = 264.71 N 6 450 N, our assumption is correct.

213

D

y

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*3–52. The shear leg derrick is used to haul the 200-kg net of fish onto the dock. Determine the compressive force along each of the legs AB and CB and the tension in the winch cable DB. Assume the force in each leg acts along its axis.

z 5.6 m 4m B

D

4m

C

Solution

A

2 4 4 FAB = FABa- i + j + kb 6 6 6

x

2m

2m y

= -0.3333FABi + 0.6667FABj + 0.6667FABk

2 4 4 FCB = FCBa i + j + kb 6 6 6

= 0.3333FCBi + 0.6667FCBj + 0.6667FCBk

FBD = FBDa -

9.6 4 j kb 10.4 10.4

= -0.9231FBDj - 0.3846FBBk

W = - 1962 k ΣFx = 0;

- 0.3333FAB + 0.3333FCB = 0

ΣFy = 0; 0.6667FAB + 0.6667FCB - 0.9231FBD = 0 ΣFz = 0; 0.6667FAB + 0.6667FCB - 0.3846FBD - 1962 = 0 FAB = 2.52 kN

Ans.

FCB = 2.52 kN

Ans.

FBD = 3.64 kN

Ans.

Ans: FAB = 2.52 kN FCB = 2.52 kN FBD = 3.64 kN 214

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3–53. Determine the stretch in each of the two springs required to hold the 20-kg crate in the equilibrium position shown. Each spring has an unstretched length of 2 m and a stiffness of k = 360 N>m.

z C

B A O

12 m

SOLUTION x

Cartesian Vector Notation: FOC = FOC ¢

6i + 4j + 12k 2

2

26 + 4 + 12

FOA = -FOA j

2

4m

6m

≤ = FOCi + FOCj + FOCk 3 7

2 7

6 7

FOB = -FOB i

F = {-196.2k} N Equations of Equilibrium: ©F = 0;

FOC + FOA + FOB + F = 0

3 2 6 a FOC - FOB b i + a FOC - FOA bj + a FOC - 196.2b k = 0 7 7 7 Equating i, j, and k components, we have 3 F - FOB = 0 7 OC

(1)

2 F - FOA = 0 7 OC

(2)

6 FOC - 196.2 = 0 7

(3)

Solving Eqs. (1),(2) and (3) yields FOC = 228.9 N

FOB = 98.1 N

FOA = 65.4 N

Spring Elongation: Using spring formula, Eq. 3–2, the spring elongation is s =

F . k

sOB =

98.1 = 0.327 m = 327 mm 300

Ans.

sOA =

65.4 = 0.218 m = 218 mm 300

Ans.

Ans: sOB = 327 mm sOA = 218 mm 215

y

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3–54. z

The thin ring can be adjusted vertically between three equally long cables from which the 100-kg chandelier is suspended. If the ring remains in the horizontal plane and z = 600 mm, determine the tension in each cable.

0.5 m

C 120 D

120 120

B

y

x

SOLUTION

z

A

Geometry: Referring to the geometry of the free-body diagram shown in Fig. a, the lengths of cables AB, AC, and AD are all l = 20.52 + 0.62 = 20.61 m Equations of Equilibrium: Equilibrium requires ©Fx = 0; FAD ¢ ©Fy = 0; FAB ¢

0.5 cos 30° 20.61 0.5 20.61

≤ - FAC ¢

≤ - 2BF¢

0.5 cos 30° 20.61

0.5 sin 30° 20.61

≤ = 0 FAD = FAC = F

≤R = 0

FAB = F

Thus, cables AB, AC, and AD all develop the same tension. ©Fz = 0; 3F ¢

0.6 20.61

≤ - 100(9.81) = 0

FAB = FAC = FAD = 426 N

Ans.

Ans: FAB = FAC = FAD = 426 N 216

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3–55. z

The thin ring can be adjusted vertically between three equally long cables from which the 100-kg chandelier is suspended. If the ring remains in the horizontal plane and the tension in each cable is not allowed to exceed 1 kN, determine the smallest allowable distance z required for equilibrium.

0.5 m

C 120 D

120 120

B

y

x

SOLUTION

z

A

Geometry: Referring to the geometry of the free-body diagram shown in Fig. a, the lengths of cables AB, AC, and AD are all l = 20.52 + z2. Equations of Equilibrium: Equilibrium requires ©Fx = 0; FAD £

©Fy = 0; FAB £

0.5 cos 30° 20.5 + z 2

2

0.5 20.5 + z 2

2

≥ - FAC £

≥ - 2C F£

0.5 cos 30° 20.52 + z2 0.5 sin 30° 20.52 + z2

≥ = 0

FAD = FAC = F

≥S = 0

FAB = F

Thus, cables AB, AC, and AD all develop the same tension. ©Fz = 0; 3F £

z 20.52 + z2

≥ - 100(9.81) = 0

Cables AB, AC, and AD will also achieve maximum tension simultaneously. Substituting F = 1000 N, we obtain 3(1000) £

z 20.52 + z2

≥ - 100(9.81) = 0

z = 0.1730 m = 173 mm

Ans.

Ans: z = 173 mm 217

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*3–56. Determine the tension in each cable for equilibrium.

z 800 N A

D

3m

5m C

4m

2m O

5m

Solution

4m

  ΣFy = 0;  FABa

4 2 4 b - FAC a b - FADa b = 0 157 138 166

B

(1)

4 3 5 b + FAC a b - FADa b = 0 157 138 166

  ΣFz = 0;  - FABa

y

x

Equations of Equilibrium. Referring to the FBD shown in Fig. a,   ΣFx = 0;  FABa

4m

(2)

5 5 5 b - FAC a b - FADa b + 800 = 0 157 138 166

(3)

Solving Eqs (1), (2) and (3)

FAC = 85.77 N = 85.8 N

Ans.



FAB = 577.73 N = 578 N

Ans.



FAD = 565.15 N = 565 N

Ans.

Ans: FAC = 85.8 N FAB = 578 N FAD = 565 N 218

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3–57. The 25-kg flowerpot is supported at A by the three cords. Determine the force acting in each cord for equilibrium.

z C

D

B 60

30 45

SOLUTION

30

FAD = FAD (sin 30°i - cos 30° sin 60°j + cos 30° cos 60°k)

A

= 0.5FADi - 0.75FADj + 0.4330FAD k FAC = FAC ( - sin 30°i - cos 30° sin 60°j + cos 30° cos 60°k)

x

= -0.5FAC i - 0.75FACj + 0.4330FACk FAB = FAB(sin 45°j + cos 45°k) = 0.7071FAB j + 0.7071FAB k F = -25(9.81)k = {- 245.25k} N ©F = 0 ;

FAD + FAB + FAC + F = 0

(0.5FAD i - 0.75FAD j) + 0.4330FAD k + (0.7071FAB j + 0.7071FAB k) + ( -0.5FACi - 0.75FACj + 0.4330FACk) + ( -245.25k) = 0 (0.5FAD - 0.5FAC)i + ( - 0.75FAD + 0.7071FAB - 0.75FAC) j + (0.4330FAD + 0.7071FAB + 0.4330FAC - 245.25) k = 0 Thus, ©Fx = 0;

0.5FAD - 0.5FAC = 0

[1]

©Fy = 0;

-0.75FAD + 0.7071FAB - 0.75FAC = 0

[2]

©Fz = 0;

0.4330FAD + 0.7071FAB + 0.4330FAC - 245.25 = 0

[3]

Solving Eqs. [1], [2], and [3] yields: FAD = FAC = 104 N

Ans.

FAB = 220 N

Ans: FAD = FAC = 104 N FAB = 220 N 219

y

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3–58. If each cord can sustain a maximum tension of 50 N before it fails, determine the greatest weight of the flowerpot the cords can support.

z C

D

B 60

30 45

SOLUTION

30

FAD = FAD (sin 30°i - cos 30° sin 60°j + cos 30° cos 60°k)

A

= 0.5FAD i - 0.75FAD j + 0.4330FAD k

y

x

FAC = FAC (- sin 30°i - cos 30° sin 60° j + cos 30° cos 60° k) = -0.5FAC i - 0.75FAC j + 0.4330FAC k FAB = FAB (sin 45° j + cos 45° k) = 0.7071FAB j + 0.7071FAB k W = -Wk ©Fx = 0;

0.5FAD - 0.5FAC = 0 (1)

FAD = FAC ©Fy = 0;

-0.75FAD + 0.7071FAB - 0.75FAC = 0 (2)

0.7071FAB = 1.5FAC ©Fz = 0;

0.4330FAD + 0.7071FAB + 0.4330FAC - W = 0 0.8660FAC + 1.5FAC - W = 0 2.366FAC = W

Assume FAC = 50 N then FAB =

1.5(50) = 106.07 N 7 50 N (N . G!) 0.7071

Assume FAB = 50 N. Then FAC =

0.7071(50) = 23.57 N 6 50 N (O. K!) 1.5

Thus, Ans.

W = 2.366(23.57) = 55.767 = 55.8 N

Ans: W = 55.8 N 220

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z

3–59. If the mass of the flowerpot is 50 kg, determine the tension developed in each wire for equilibrium. Set x = 1.5 m and z = 2 m.

C

2m x

D z

3m

x

6m

Solution

A

B y

Force Vectors: We can express each of the forces on the free - body diagram shown in Fig. (a) in Cartesian vector from as FAB = FABj FAC = FAC c FAD = FAD c

(2 - 0)i + ( - 6 - 0)j + (3 - 0)k 2(2 - 0)2 + ( - 6 - 0)2 + (3 - 0)2

d =

( -1.5 - 0)i + ( -6 - 0)j + (2 - 0)k 2

2

2( - 1.5 - 0) + ( -6 - 0) + (2 - 0)

W = 3 -50(9.81)k4N = [ -490.5k4N

2

2 6 3 F i - FACj + FACk 7 AC 7 7 d =

3 12 4 F i F j + F k 13 AD 13 AD 13 AD

Equations of Equilibrium: Equilibrium requires a F = 0; FAB + FAC + FAD + W = 0

2 6 3 3 12 4 FABj + a FACi - FACj + FACkb + a - FADi F j + F kb + ( -490.5k) = 0 7 7 7 13 13 AD 13 AD 2 3 6 12 3 4 a FAC F bi + aFAB - FAC F bj + a FAC + F - 490.5bk = 0 7 13 AD 7 13 AD 7 13 AD

Equating i, j, and k components yields 2 3 F F = 0 7 AC 13 AD FAB -

(1)

6 12 F F = 0 7 AC 13 AD

(2)

3 4 F + F - 490.5 = 0 7 AC 13 AD

(3)

Solving Eqs. (1) through (3) yields FAB = 1211.82 N = 1.21 kN

Ans.

FAC = 606 N

Ans.

FAD = 750 N

Ans.

Ans: FAB = 1.21 kN FAC = 606 N FAD = 750 N 221

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z

*3–60. If the mass of the flowerpot is 50 kg, determine the tension developed in each wire for equilibrium. Set x = 2 m and z = 1.5 m.

C

2m x

D z

3m

x

Solution

6m

A

B y

Force Vectors: We can express each of the forces on the free - body diagram shown in Fig. (a) in Cartesian vector from as FAB = FABj FAC = FAC c FAD = FAD c

(2 - 0)i + ( -6 - 0)j + (3 - 0)k 2

2

2(2 - 0) + ( - 6 - 0) + (3 - 0)

2

d =

( - 2 - 0)i + ( -6 - 0)j + (1.5 - 0)k 2

2

2( -2 - 0) + ( -6 - 0) + (1.5 - 0)

W = 3 -50(9.81)k4N = 3 - 490.5k4N

2

2 6 3 FACi - FACj + FACk 7 7 7 d = -

4 12 3 F i F j + F k 13 AD 13 AD 13 AD

Equations of Equilibrium: Equilibrium requires ΣF = 0; FAB + FAC + FAD + W = 0 2 6 3 4 12 3 FABj + a FACi - FACj + FACkb + a - FADi F j + F kb + ( -490.5k) = 0 7 7 7 13 13 AD 13 AD 2 3 4 6 12 3 a FAC F bi + aFAB - FAC F bj + a FAC + F - 490.5bk = 0 7 13 AD 7 13 AD 7 13 AD

Equating i, j, and k components yields 2 4 F F = 0 7 AC 13 AD FAB -

(1)

6 12 F F = 0 7 AC 13 AD

(2)

3 4 F + F - 490.5 = 0 7 AC 13 AD

(3)

Solving Eqs. (1) through (3) yields FAB = 1308 N = 1.31 kN

Ans.

FAC = 763 N

Ans.

FAD = 708.5 N

Ans.

Ans: FAB = 1.31 kN FAC = 763 N FAD = 708.5 N 222

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3–61. Determine the tension developed in the three cables required to support the traffic light, which has a mass of 15 kg. Take h = 4 m.

z C 6m D A h

Solution

4m

ΣFx = 0;

ΣFy = 0; ΣFz = 0;

3 6 F - FAC + 5 AB 7 4 3 F - FAC 5 AB 7 2 F - 15(9.81) 7 AC

4m

3m

6 3 2 = e - i - j + kf 7 7 7

4 3 uAD = e i - j f 5 5

B

4m

3 4 uAB = e i + j f 5 5 uAC

3m

x

4m

6m 3m

y

4 F = 0 5 AD 3 F = 0 5 AD = 0



FAB = 441 N

Ans.



FAC = 515 N

Ans.



FAD = 221 N

Ans.

Ans: FAB = 441 N FAC = 515 N FAD = 221 N 223

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3–62.

z

Determine the tension developed in the three cables required to support the traffic light, which has a mass of 20 kg. Take h = 3.5 m.

C 6m D A h

3m B

4m

Solution uAB = uAC = uAD =

4m

3i + 4 j + 0.5k 2

2

23 + 4 + (0.5)

2

3i + 4 j + 0.5k

=

225.25

- 6i - 3 j + 2.5k

2( - 6)2 + ( - 3)2 + 2.52 4i - 3 j + 0.5k

=

242 + ( - 3)2 + 0.52

ΣFx = 0; ΣFy = 0; ΣFz = 0; Solving,

3

225.25 4

225.25 0.5

225.25

FAB -

FAB FAB +

=

4m

3m x

-6i - 3 j + 2.5k

4m

6m 3m

y

251.25

4i - 3 j + 0.5k 6

225.25

251.25 3

251.25 2.5

251.25

FAC +

FAC FAC +

4 225.25 3

225.25 0.5

225.25

FAD = 0 FAD = 0 FAD - 20(9.81) = 0



FAB = 348 N

Ans.



FAC = 413 N

Ans.



FAD = 174 N

Ans.

Ans: FAB = 348 N FAC = 413 N FAD = 174 N 224

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3–63. The crate has a mass of 130 kg. Determine the tension developed in each cable for equilibrium.

z

A x

1m 1m

4m

B

3m C

D 2m

1m

y

Solution Equations of Equilibrium. Referring to the FBD shown in Fig. a,   ΣFx = 0;  FADa

2 2 2 b - FBD a b - FCD a b = 0 3 16 16

  ΣFy = 0;  - FADa   ΣFz = 0;  FADa

(1)

1 1 2 b - FBD a b + FCD a b = 0 3 16 16

(2)

1 1 1 b + FBD a b + FCD a b - 130(9.81) = 0 3 16 16

(3)

Solving Eqs (1), (2) and (3)

FAD = 1561.92 N = 1.56 kN

Ans.



FBD = 520.64 N = 521 N

Ans.



FCD = 1275.3 N = 1.28 kN

Ans.

Ans: FAD = 1.56 kN FBD = 521 N FCD = 1.28 kN 225

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*3–64. If the maximum force in each rod can not exceed 1500 N, determine the greatest mass of the crate that can be supported.

z

C

B

3m

2m

2m 1m

A

2m

2m O

3m y

1m 3m

Solution Equations of Equilibrium. Referring to the FBD shown in Fig. a,

x

2 3 1 b - FOC a b + FOB a b = 0 (1)   ΣFx = 0;  FOAa 3 114 122   ΣFy = 0;  -FOAa   ΣFz = 0;  FOAa

3 2 2 b + FOC a b + FOB a b = 0 3 114 122

(2)

1 3 2 b + FOC a b - FOB a b - m(9.81) = 0 3 114 122

(3)

Solving Eqs (1), (2) and (3),

FOC = 16.95m  FOA = 15.46m  FOB = 7.745m Since link OC subjected to the greatest force, it will reach the limiting force first, that is FOC = 1500 N. Then

1500 = 16.95 m



m = 88.48 kg = 88.5 kg

Ans.

Ans: m = 88.5 kg 226

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3–65. Determine the force in each cable needed to support the 17.5-kN (≈ 1750-kg) platform. Set d = 1.2 m.

Solution Cartesian Vector Notation : ⎛ ⎞ 1.2i – 0.9j – 3k FAB = FAB ⎜ ⎟ = 0.3578FABi – 0.2683FAB j – 0.8944FABk ⎜⎝ 1.2 2 + (–0.9) 2 + (–3) 2 ⎟⎠ ⎛ FAC = FAC ⎜ ⎜⎝

⎞ ⎟ = 0.2873FAC j – 0.9578FACk 0.9 2 + (–3) 2 ⎟⎠ 0.9j – 3k

⎛ ⎞ –1.2i + 0.3j – 3k FAD = FAD ⎜ ⎟ = –0.3698FADi + 0.09245FAD j – 0.9245FADk ⎜⎝ (–1.2) 2 + 0.3 2 + (–3) 2 ⎟⎠ F = {17.5k} kN Equations of Equilibrium :

F = 17.5 kN

ΣF = 0; FAB + FAC + FAD + F = 0

0.3 m

(0.3578FAB – 0.3698FAD)i + (–0.2683FAB + 0.2873FAC + 0.09245FAD)j 1.2 m

+ (–0.8944FAB – 0.9578FAC – 0.9245FAD + 17.5)k = 0 1.2 m

Equating the i, j and k components, we have

0.9 m

0.3578FAB – 0.3698FAD = 0

[1]

–0.2683FAB + 0.2873FAC + 0.09245FAD = 0

[2]

–0.8944FAB – 0.9578FAC – 0.9245FAD + 17.5 = 0

[3]

3m

3m 3m

Solving Eqs. [1], [2] and [3] yields

0.9 m

FAB = 7.337 kN

Ans.

FAC = 4.568 kN

Ans.

FAD = 7.098 kN

Ans.

Ans: FAB = 7.337 kN FAC = 4.568 kN FAD = 7.098 kN 227

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z

3–66. The ends of the three cables are attached to a ring at A and to the edge of the uniform plate. Determine the largest mass the plate can have if each cable can support a maximum tension of 15 kN.

A

2m C

Solution B

2m 4m

6m

6m

x

6 4 12 i j kb 14 14 14

FD = FDa -

6m

2m

4 6 12 i j kb 14 14 14

FC = FCa -

4 6 12 i + j kb 14 14 14

ΣFx = 0;

4 6 4 FB FC F = 0 14 14 14 D

ΣFy = 0;

-

ΣFz = 0;

D

12 m

y

W = Wk FB = FBa

10 m

6 4 6 F F + F = 0 14 B 14 C 14 D

-

12 12 12 F F F + W = 0 14 B 14 C 14 D

Assume FB = 15 kN. Solving, FC = 0 6 15 kN (OK) FD = 15 kN (OK) Thus, -

12 12 (15) - 0 (15) + W = 0 14 14

W = 25.714 kN m =

W 25.714 = = 2.62 Mg g 9.81

Ans.

Ans: m = 2.62 Mg 228

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3–67. A small peg P rests on a spring that is contained inside  the smooth pipe. When the spring is compressed so that s = 0.15 m, the spring exerts an upward force of 60 N on the peg. Determine the point of attachment A(x, y, 0) of cord PA so that the tension in cords PB and PC equals 30 N and 50 N, respectively.

z

s

P

0.2 m C 0.3 m

Solution

B

Cartesian Vector Notation:

FPC = 50£

x

0.4 m y

FPA = (FPA)xi + (FPA)yj + (FPA)zk FPB = 30£

y

- 0.4j - 0.15k 2( - 0.4)2 + ( -0.15)2

x

A

≥ = 5 - 28.09j - 10.53k6 N

-0.3i + 0.2j - 0.15k

2( - 0.3)2 + 0.22 + ( - 0.15)2

≥ = 5 - 38.41i + 25.61j - 19.21k6 N

F = 560k6 N

Equations of Equilibrium: ΣF = 0; FPA + FPB + FPC + F = 0 3(FPA)x - 38.414i + 3(FPA)y - 28.09 + 25.614j + 3(FPA)z - 10.53 - 19.21 + 604k = 0

Equating i, j, and k components, we have (FPA)x - 38.41 = 0

(FPA)x = 38.41 N

(FPA)y - 28.09 + 25.61 = 0

(FPA)y = 2.48 N

(FPA)z - 10.53 - 19.21 + 60 = 0

(FPA)z = - 30.26 N

The magnitude of FPA is FPA = 2(FPA)2x + (FPA)2y + (FPA)2z = 238.412 + 2.482 + ( - 30.26)2 = 48.96 N

The coordinate direction angles are a = cos - 1 c b = cos - 1 c g = cos - 1 c

(FPA)x FPA (FPA)y FPA (FPA)z FPA

The wire PA has a length of PA =

(PA)z cos g

d = cos - 1a d = cos - 1a d = cos - 1a

=

38.41 b = 38.32° 48.96 2.48 b = 87.09° 48.96

-30.26 b = 128.17° 48.96

- 0.15 = 0.2427 m cos 128.17°

Thus, x = PA cos a = 0.2427 cos 38.32° = 0.190 m

Ans.

y = PA cos b = 0.2427 cos 87.09° = 0.0123 m

Ans. 229

Ans: x = 0.190 m y = 0.0123 m

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*3–68. Determine the height d of cable AB so that the force in cables AD and AC is one-half as great as the force in cable AB. What is the force in each cable for this case? The flower pot has a mass of 50 kg.

z 2m C

2m D 3m

SOLUTION

6m A

Cartesian Vector Notation:

y

B

FAB = (FAB)x i + (FAB)y k

6m

FAC =

- 6i - 2j + 3k FAB 3 1 3 FAB k ¢ ≤ = - FABi - FAB j + 2 2 2 2 7 7 14 2(- 6) + ( -2) + 3

FAD =

-6i + 2j + 3k FAB 3 1 3 FAB k ¢ ≤ = - FABi + FAB j + 2 2 2 2 7 7 14 2(- 6) + 2 + 3

d x

F = { - 490.5k} N Equations of Equilibrium: FAB + FAC + FAD + F = 0

©F = 0;

3 3 1 1 a (FAB)x - FAB - FAB b i + a - FAB + FAB bj 7 7 7 7 3 3 F + F + a (FAB)z + - 490.5bk = 0 14 AB 14 AB Equating i, j, and k components, we have (FAB)x -

3 3 FAB - FAB = 0 7 7

1 1 FAB + FAB = 0 7 7

(FAB)z +

(FAB)x =

6 F 7 AB

(1)

(Satisfied!)

3 3 F F + - 490.5 = 0 14 AB 14 AB

(FAB)z = 490.5 -

3 F 7 AB

(2)

However, F2AB = (FAB)2x + (FAB)2z, then substitute Eqs. (1) and (2) into this expression yields 2 2 6 3 F2AB = a FAB b + a490.5 - FAB b 7 7

Solving for positive root, we have FAB = 519.79 N = 520 N FAC = FAD =

Thus,

Ans.

1 (519.79) = 260 N 2

Ans.

Also, (FAB)x =

6 (519.79) = 445.53 N 7

(FAB)z = 490.5 then,

u = tan - 1 c

(FAB)z (FAB)x

3 (519.79) = 267.73 N 7

d = tan - 1 a

Ans:

267.73 b = 31.00° 445.53

d = 6 tan u = 6 tan 31.00° = 3.61 m

Ans.

230

FAB = 520 N FAC = 260 N d = 3.61 m

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4–1. If A, B, and D are given vectors, prove the distributive law for the vector cross product, i.e., A : (B + D) = (A : B) + (A : D).

SOLUTION Consider the three vectors; with A vertical. Note obd is perpendicular to A. od = ƒ A * (B + D) ƒ = ƒ A ƒ ƒ B + D ƒ sin u3 ob = ƒ A * B ƒ = ƒ A ƒ ƒ B ƒ sin u1 bd = ƒ A * D ƒ = ƒ A ƒ ƒ D ƒ sin u2 Also, these three cross products all lie in the plane obd since they are all perpendicular to A. As noted the magnitude of each cross product is proportional to the length of each side of the triangle. The three vector cross products also form a closed triangle o¿b¿d¿ which is similar to triangle obd. Thus from the figure, A * (B + D) = (A * B) + (A * D)

(QED)

Note also, A = Ax i + Ay j + Az k B = Bx i + By j + Bz k D = Dx i + Dy j + Dz k A * (B + D) = 3

i Ax Bx + Dx

j Ay By + Dy

k Az 3 Bz + Dz

= [A y (Bz + Dz) - A z(By + Dy)]i - [A x(Bz + Dz) - A z(Bx + Dx)]j + [A x(By + Dy) - A y(Bx + Dx)]k = [(A y Bz - A zBy)i - (A x Bz - A z Bx)]j + (A x By - A y Bx)k + [(A y Dz - A z Dy)i - (A x Dz - A z Dx)j + (A x Dy - A y Dx)k i = 3 Ax Bx

j Ay By

k i Az 3 + 3 Ax Bz Dx

j Ay Dy

k Az 3 Dz

= (A * B) + (A * D)

(QED)

231

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4–2. Prove the triple scalar A # (B : C) = (A : B) # C.

product

identity

SOLUTION As shown in the figure Area = B(C sin u) = |B * C| Thus, Volume of parallelepiped is |B * C||h| But, |h| = |A # u(B * C)| = ` A # a

B * C b` |B * C|

Thus, Volume = |A # (B * C)| Since |(A * B) # C| represents this same volume then A # (B : C) = (A : B) # C

(QED)

Also, LHS = A # (B : C) i

= (A x i + A y j +

A z k) # 3 Bx Cx

j By Cy

k Bz 3 Cz

= A x (ByCz - BzCy) - A y (BxCz - BzCx) + A z (BxCy - ByCx) = A xByCz - A xBzCy - A yBxCz + A yBzCx + A zBxCy - A zByCx RHS = (A : B) # C i = 3 Ax Bx

j Ay By

k A z 3 # (Cx i + Cy j + Cz k) Bz

= Cx(A y Bz - A zBy) - Cy(A xBz - A zBx) + Cz(A xBy - A yBx) = A xByCz - A xBzCy - A yBxCz + A yBzCx + A zBxCy - A zByCx Thus, LHS = RHS A # (B : C) = (A : B) # C

(QED)

232

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4–3. Given the three nonzero vectors A, B, and C, show that if A # (B : C) = 0, the three vectors must lie in the same plane.

SOLUTION Consider, |A # (B * C)| = |A| |B * C | cos u = (|A| cos u)|B * C| = |h| |B * C| = BC |h| sin f = volume of parallelepiped. If A # (B * C) = 0, then the volume equals zero, so that A, B, and C are coplanar.

233

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*4–4. Determine the magnitude and directional sense of the moment of the force at A about point O.

y P

520 N 30˚ 4m

13

12 5

O

A

6m

x

Solution a+ M0 = 520a

12 b(6) = 2880 N # m d 13

= 2.88 kN # m d

Ans.

Ans:

a+ M0 = 2.88 kN # m d 234

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y

4–5. Determine the magnitude and directional sense of the moment of the force at A about point P. P

520 N 30˚ 4m

13

12 5

O

A

6m

x

Solution a+ MP = 520a

12 5 b 16 + 4 sin 30° 2 - 520a b 14 cos 30° 2 13 13

= 3147 N # m

= 3.15 kN # m

Ans.

1Counterclockwise2

Ans:

#

a + M P = 3.15 kN m (Counterclockwise)

235

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4–6. The crane can be adjusted for any angle 0° … u … 90° and any extension 0 … x … 5 m. For a suspended mass of 120 kg, determine the moment developed at A as a function of x and u. What values of both x and u develop the maximum possible moment at A? Compute this moment. Neglect the size of the pulley at B.

x 9m

B

1.5 m

θ A

SOLUTION a + MA = - 12019.81217.5 + x2 cos u = 5-1177.2 cos u17.5 + x26 N # m = 51.18 cos u17.5 + x26 kN # m (Clockwise)

Ans.

The maximum moment at A occurs when u = 0° and x = 5 m.

Ans.

a + 1MA2max = 5-1177.2 cos 0°17.5 + 526 N # m = - 14 715 N # m = 14.7 kN # m (Clockwise)

Ans.

Ans:

a + M A = {1.18 cos u(7.5 + x)} kN # m (Clockwise)

The maximum moment at A occurs when u = 0 and x = 5 m. a +(M A )max = 14.7 kN # m (Clockwise) 236

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4–7. The towline exerts a force of P = 6 kN at the end of the 8-m-long crane boom. If u = 30°, determine the placement x of the hook at B so that this force creates a maximum moment about point O. What is this moment?

A P  6 kN 8m u

O

1m

B x

Solution In order to produce the maximum moment about point O, P must act perpendicular to the boom’s axis OA as shown in Fig. a. Thus

a+ (MO)max = 6 (8) = 48.0 kN # m (counterclockwise)

Ans.

Referring to the geometry of Fig. a,

x = x' + x" =

8 + tan 30° = 9.814 m = 9.81 m cos 30°

Ans.

Ans: (MO)max = 48.0 kN # m d x = 9.81 m 237

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*4–8. A

The towline exerts a force of P = 6 kN at the end of the 8-m-long crane boom. If x = 10 m, determine the position u of the boom so that this force creates a maximum moment about point O. What is this moment?

P  6 kN 8m u

O

1m

B x

Solution In order to produce the maximum moment about point O, P must act perpendicular to the boom’s axis OA as shown in Fig. a. Thus,

a+ (MO)max = 6 (8) = 48.0 kN # m (counterclockwise)

Ans.

Referring to the geometry of Fig. a, 10 =

8 + tan u cos u



10 =

8 sin u + cos u cos u



10 cos u - sin u = 8



x = x' + x";

10 1 8 cos u sin u =  1101 1101 1101



(1)

From the geometry shown in Fig. b,

a = tan-1 a sin a =

Then Eq (1) becomes

1 b = 5.711° 10

1 1101

cos a =

10 1101

cos u cos 5.711° - sin u sin 5.711° =

8 1101

Referring that cos (u + 5.711°) = cos u cos 5.711° - sin u sin 5.711°

cos (u + 5.711°) =

8 1101

u + 5.711° = 37.247° Ans.

u = 31.54° = 31.5°

Ans: (MO)max = 48.0 kN # m (counterclockwise) u = 31.5° 238

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4–9. z

The 20-N horizontal force acts on the handle of the socket wrench. What is the moment of this force about point B. Specify the coordinate direction angles a, b, g of the moment axis.

20 N B

200 mm A

60 10 mm

50 mm O

y

x

Solution Force Vector And Position Vector. Referring to Fig. a, F = 20 (sin 60°i - cos 60°j) = {17.32i - 10j} N

rBA = { - 0.01i + 0.2j} m

Moment of Force F about point B. MB = rBA * F  

i = † - 0.01 17.32

j 0.2 - 10

k 0† 0

= { -3.3641 k} N # m = { -3.36 k} N # m

Ans.

Here the unit vector for MB is u = - k. Thus, the coordinate direction angles of MB are

a = cos-1 0 = 90°

Ans.



b = cos-1 0 = 90°

Ans.



-1

Ans.

g = cos

( -1) = 180°

Ans: MB = { -3.36 k} N # m a = 90° b = 90° g = 180° 239

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4–10. z

The 20-N horizontal force acts on the handle of the socket wrench. Determine the moment of this force about point O. Specify the coordinate direction angles a, b, g of the moment axis.

20 N B

200 mm A

60 10 mm

50 mm O

y

x

Solution Force Vector And Position Vector. Referring to Fig. a, F = 20 (sin 60°i - cos 60°j) = {17.32i - 10j} N

rOA = { -0.01i + 0.2j + 0.05k} m

Moment of F About point O.

MO = rOA * F i = † -0.01 17.32

j 0.2 - 10

k 0.05 † 0

= {0.5i + 0.8660j - 3.3641k} N # m = {0.5i + 0.866j - 3.36k} N # m

Ans.

The magnitude of MO is MO = 2(MO)2x + (MO)2y + (MO)2z = 20.52 + 0.86602 + ( - 3.3641)2 = 3.5096 N # m

Thus, the coordinate direction angles of MO are

a = cos-1 c



b = cos-1 c



g = cos-1 c

(MO)x MO (MO)y MO (MO)z MO

d = cos-1 a d = cos-1 a d = cos-1 a

0.5 b = 81.81° = 81.8° 3.5096 0.8660 b = 75.71° = 75.7° 3.5096

-3.3641 b = 163.45° = 163° 3.5096

Ans. Ans. Ans.

Ans: MO = {0.5i + 0.866j - 3.36k} N # m a = 81.8° b = 75.7° g = 163° 240

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4–11. Determine the moment of each of the three forces about point A.

F1

F2

250 N 30

300 N

60

A 2m

3m

4m

SOLUTION The moment arm measured perpendicular to each force from point A is d1 = 2 sin 60° = 1.732 m

B

4

5 3

d2 = 5 sin 60° = 4.330 m

F3

500 N

d3 = 2 sin 53.13° = 1.60 m Using each force where MA = Fd, we have a + 1MF12A = - 25011.7322 = - 433 N # m = 433 N # m (Clockwise)

Ans.

a + 1MF22A = - 30014.3302 = - 1299 N # m = 1.30 kN # m (Clockwise)

Ans.

a + 1MF32A = - 50011.602 = - 800 N # m = 800 N # m (Clockwise)

Ans.

Ans:

241

a + 1 M F1 2 A = -433 N # m = 433 N # m (Clockwise) a + 1 M F2 2 A = - 1299 N # m = 1.30 kN # m (Clockwise) a + 1 M F3 2 A = - 800 N # m = 800 kN # m (Clockwise)

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*4–12. Determine the moment of each of the three forces about point B.

F2 ⫽ 300 N

F1 ⫽ 250 N 30⬚

60⬚

A 2m

3m

SOLUTION

4m

The forces are resolved into horizontal and vertical component as shown in Fig. a. For F1, a + MB = 250 cos 30°(3) - 250 sin 30°(4) = 149.51 N # m = 150 N # m d

B

Ans.

4

5 3

For F2,

F3 ⫽ 500 N

a + MB = 300 sin 60°(0) + 300 cos 60°(4) = 600 N # m d

Ans.

Since the line of action of F3 passes through B, its moment arm about point B is zero. Thus MB = 0

Ans.

Ans:

a + MB = 150 N # m d a + MB = 600 N # m d MB = 0 242

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4–13. The connected bar BC is used to increase the lever arm of the crescent wrench as shown. If a clockwise moment of MA = 120 N # m is needed to tighten the bolt at A and the force F = 200 N, determine the required extension d in order to develop this moment.

C

d

15⬚

F

30⬚ 300 mm B

A

SOLUTION By resolving the 200-N force into components parallel and perpendicular to the box wrench BC, Fig. a, the moment can be obtained by adding algebraically the moments of these two components about point A in accordance with the principle of moments.

a +(MR)A = ©Fd; -120 = 200 sin 15°(0.3 sin 30°) - 200 cos 15°(0.3 cos 30° + d) d = 0.4016 m = 402 mm

Ans.

Ans: d = 402 mm 243

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4–14. The connected bar BC is used to increase the lever arm of the crescent wrench as shown. If a clockwise moment of MA = 120 N # m is needed to tighten the nut at A and the extension d = 300 mm, determine the required force F in order to develop this moment.

C

d

15⬚

F

30⬚ 300 mm B

SOLUTION A

By resolving force F into components parallel and perpendicular to the box wrench BC, Fig. a, the moment of F can be obtained by adding algebraically the moments of these two components about point A in accordance with the principle of moments.

a +(MR)A = ©Fd;

-120 = F sin 15°(0.3 sin 30°) - F cos 15°(0.3 cos 30° + 0.3) F = 239 N

Ans.

Ans: F = 239 N 244

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4–15. The railway crossing gate consists of the 100-kg gate arm having a center of mass at Ga and the 250-kg counterweight having a center of mass at GW. Determine the magnitude and directional sense of the resultant moment produced by the weights about point A.

A

2.5 m Ga

SOLUTION

0.75 m

GW 1m 0.5 m B

0.25 m

a +(MR)A = gFd; (MR)A = 100(9.81)(2.5 + 0.25) - 250(9.81)(0.5 - 0.25) = 2084.625 N # m = 2.08 kN # m (Counterclockwise)

Ans.

Ans:

245

1 MR 2 A

= 2.08 kN # m (Counterclockwise)

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*4–16. The railway crossing gate consists of the 100-kg gate arm having a center of mass at Ga and the 250-kg counterweight having a center of mass at GW. Determine the magnitude and directional sense of the resultant moment produced by the weights about point B.

A

2.5 m

SOLUTION

Ga

0.75 m

a +(MR)B = gFd; (MR)B = 100(9.81)(2.5) - 250(9.81)(0.5) = 1226.25 N # m = 1.23 kN # m (Counterclockwise)

GW 1m 0.5 m B

Ans. 0.25 m

Ans: a +(MR)B = 1.23 kN # m (Counterclockwise) 246

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4–17. The torque wrench ABC is used to measure the moment or torque applied to a bolt when the bolt is located at A and a force is applied to the handle at C. The mechanic reads the torque on the scale at B. If an extension AO of length d is used on the wrench, determine the required scale reading if the desired torque on the bolt at O is to be M.

F M

A

O d

B

l

C

Solution Moment at A = m = Fl Moment at O = M = (d + l)F

M = (d + l)



m = a

m l

l bM d + l

Ans.

Ans: m = a 247

l bM d + l

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A

4–18. The 70-N force acts on the end of the pipe at B. Determine (a) the moment of this force about point A, and (b) the magnitude and direction of a horizontal force, applied at C, which produces the same moment. Take u = 60. 0.9 m

70 N

u C

0.3 m

0.7 m

B

SOLUTION Given:

(a) (b)

F

70 N

a

0.9 m

b

0.3 m

c

0.7 m

T

60 ° c +

MA

F sin  T c  F cos  T a

F C ( a)

MA

FC

MA a

MA

73.9 N˜ m

Ans.

FC

82.2 N

Ans.

Ans: a. c + M A = 73.9 N # m b. FC = 82.2 N 248

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4–19. The 70-N force acts on the end of the pipe at B. Determine the angles u (0 … u … 180) of the force that will produce maximum and minimum moments about point A. What are the magnitudes of these moments? 0.9 m

70 N

u

SOLUTION

C

0.3 m

B

0.7 m

Given: F

70 N

a

0.9 m

b

0.3 m

c

0.7 m

c +

MA

F sin  T c  F cos  T a

For maximum moment

T max

MAmax

MAmin

dT

MA

c F cos  T  a F sin  T

0

§c· ¸ © a¹

atan ¨

T max

F sin  T max c  F cos  T max a

For minimum moment

T min

d

180 

F sin  T c  F cos  T a

MA

§ a · ¸ ©c¹

 atan ¨

F c sin  T min  F ( a) cos  T min

MAmax

37.9 

79.812 N˜ m

Ans.

Ans.

0

T min

MAmin

128 

0 N˜ m

Ans.

Ans.

Ans: umax = 37.9 MAmax = 79.812 N # m umin = 128 MAmin = 0 N # m 249

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*4–20. z

Determine the moment of the force F about point O. Express the result as a Cartesian vector.

F  {–6i + 4 j  8k} kN

A

4m P 3m

6m O 1m

y

2m

Solution x

Position Vector. The coordinates of point A are (1, -2, 6) m. Thus,

rOA = {i - 2j + 6k} m

The moment of F about Point O.

MO = rOA * F



i = † 1 -6



j -2 4

k 6† 8

= { - 40i - 44j - 8k} kN # m

Ans.

Ans: MO = { - 40i - 44j - 8k} kN # m 250

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4–21. z

Determine the moment of the force F about point P. Express the result as a Cartesian vector.

F  {–6i + 4 j  8k} kN

A

4m P 3m

6m O 1m 2m

Solution Position Vector. The coordinates of points A and P are A (1, -2, 6) m and P (0, 4, 3) m, respectively. Thus

y

x

rPA = (1 - 0)i + ( -2 - 4)j + (6 - 3)k = {i - 6j + 3k} m

The moment of F about Point P.

MP = rPA * F



i = † 1 -6



j -6 4

k 3† 8

= { -60i - 26j - 32k} kN # m

Ans.

Ans: MP = { - 60i - 26j - 32k} kN # m 251

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4–22. Determine the magnitude of the force F that should be applied at the end of the lever such that this force creates a clockwise moment of 15 N # m about point O when u = 30.

y F u 50 mm

SOLUTION

300 mm 60

Given:

O

M

15 N # m

I

60 °

T

30°

a

50 mm

b

300 mm

c +

M

F

x

F cos  T  a  b sin  I  F sin  T  b cos  I M

cos  T  a  b sin  I  sin  T  b cos  I

F

77.6 N

Ans.

Ans: F = 77.6 N 252

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y

4–23. If the force F = 100 N, determine the angle u (0 … u … 90) so that the force develops a clockwise moment about point O of 20 N # m.

F u 50 mm

300 mm 60

SOLUTION Initial Guess

O

T

x

30°

Given: F M

100 N 20 N˜ m

I a

60° 50 mm

b

300 mm

Given M

F cos  T  a  b sin  I  F sin  T  b cos  I

T

Find  T

T

28.6°

Ans.

Ans: u = 28.6 253

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*4–24. The Achilles tendon force of Ft = 650 N is mobilized when the man tries to stand on his toes. As this is done, each of his feet is subjected to a reactive force of Nf = 400 N. Determine the resultant moment of Ft and Nf about the ankle joint A.

Ft

5

A

200 mm

SOLUTION Referring to Fig. a, a +(MR)A = ©Fd;

65 mm

(MR)A = 400(0.1) - 650(0.065) cos 5°

100 mm

Nf

400 N

= - 2.09 N # m = 2.09 N # m (Clockwise) Ans.

Ans:

a +(MR)A = 2.09 N # m (Clockwise) 254

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4–25. The Achilles tendon force Ft is mobilized when the man tries to stand on his toes. As this is done, each of his feet is subjected to a reactive force of Nt = 400 N. If the resultant moment produced by forces Ft and Nf about the ankle joint A is required to be zero, determine the magnitude of Ff.

Ft

5

A

200 mm

SOLUTION Referring to Fig. a, a +(MR)A = ©Fd;

65 mm

100 mm

Nf

400 N

0 = 400(0.1) - F cos 5°(0.065) F = 618 N

Ans.

Ans: F = 618 N 255

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4–26. y

Old clocks were constructed using a fusee B to drive the gears and watch hands. The purpose of the fusee is to increase the leverage developed by the mainspring A as it uncoils and thereby loses some of its tension. The mainspring can develop a torque (moment) Ts = ku, where k = 0.015 N # m>rad is the torsional stiffness and u is the angle of twist of the spring in radians. If the torque Tf developed by the fusee is to remain constant as the mainspring winds down, and x = 10 mm when u = 4 rad, determine the required radius of the fusee when u = 3 rad.

x

A

B y

t

x 12 mm Ts

Tf

Solution When u = 4 rad, r = 10 mm Ts = 0.015(4) = 0.06 N # m F =

0.06 = 5N 0.012

Tf = 5(0.010) = 0.05 N # m (constant) When u = 3 rad,

Ts = 0.015(3) = 0.045 N # m F =

0.045 = 3.75 N 0.012

For the fusee require 0.05 = 3.75 r Ans.

r = 0.0133 m = 13.3 mm

Ans: r = 13.3 mm 256

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4–27. 4m

The tower crane is used to hoist the 2-Mg load upward at constant velocity. The 1.5-Mg jib BD, 0.5-Mg jib BC, and 6-Mg counterweight C have centers of mass at G1 , G2 , and G3 , respectively. Determine the resultant moment produced by the load and the weights of the tower crane jibs about point A and about point B.

G2

9.5m B

D

C G3

7.5 m

12.5 m

G1

23 m

SOLUTION Since the moment arms of the weights and the load measured to points A and B are the same, the resultant moments produced by the load and the weight about points A and B are the same. a + (MR)A = (MR)B = ©Fd;

A

(MR)A = (MR)B = 6000(9.81)(7.5) + 500(9.81)(4) - 1500(9.81)(9.5)

- 2000(9.81)(12.5) = 76 027.5 N # m = 76.0 kN # m (Counterclockwise)

Ans.

Ans: (MR)A = (MR)B = 76.0 kN # md 257

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*4–28. 4m

The tower crane is used to hoist a 2-Mg load upward at constant velocity. The 1.5-Mg jib BD and 0.5-Mg jib BC have centers of mass at G1 and G2, respectively. Determine the required mass of the counterweight C so that the resultant moment produced by the load and the weight of the tower crane jibs about point A is zero. The center of mass for the counterweight is located at G3.

G2

9.5m B

D

C G3

7.5 m

12.5 m

G1

23 m

SOLUTION a + (MR)A = ©Fd;

A

0 = MC(9.81)(7.5) + 500(9.81)(4) - 1500(9.81)(9.5) - 2000(9.81)(12.5) MC = 4966.67 kg = 4.97 Mg

Ans.

Ans: MC = 4.97 Mg 258

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4–29. The force F = 5600i + 300j - 600k6 N acts at the end of the beam. Determine the moment of the force about point A.

z

A

x

SOLUTION

1.2 m

r = {0.2i + 1.2j} m i M O = r * F = 3 0.2 600

O

F

B 0.4 m

j 1.2 300

k 0 3 - 600

0.2 m

M O = { - 720i + 120j - 660k} N # m

y

Ans.

Ans: Mo = { -720i + 120j - 660k} N # m 259

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4–30. z

Determine the moment of the force F about point P. Express the result as a Cartesian vector.

P 2m

2m 3m

y

O

3m

3m

Solution

x

Position Vector. The coordinates of points A and P are A (3, 3, -1) m and P ( -2, -3, 2) m respectively. Thus,

1m A

F  {2i  4j  6k} kN

rPA = [3 - ( - 2)]i + [3 - ( - 3)] j + ( - 1 - 2)k = {5i + 6j - 3k} m

Moment of F About Point P.

MP = rAP * F



i = †5 2



j 6 4

k -3 † -6

= { -24i + 24j + 8k} kN # m

Ans.

Ans: MP = { -24i + 24j + 8k} kN # m 260

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4–31. The curved rod lies in the x–y plane and has a radius of 3 m. If a force of F = 80 N acts at its end as shown, determine the moment of this force about point O.

z

O y B 3m 45

3m

SOLUTION A

1m

rAC = {1i - 3j - 2k} m

F

rAC = 2(1)2 + (-3)2 + (- 2)2 = 3.742 m M O = rOC * F = 3

i 4 1 3.742 (80)

j 0 3 - 3.742 (80)

k 3 -2 2 - 3.742 (80)

80 N

2m x

M O = {- 128i + 128j - 257k} N # m

C

Ans.

Ans: Mo = { -128i + 128j - 257k} N # m 261

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*4–32. The curved rod lies in the x–y plane and has a radius of 3 m. If a force of F = 80 N acts at its end as shown, determine the moment of this force about point B.

z

O y B 3m 45

SOLUTION rAC = {1i - 3j - 2k} m

A

1m

rAC = 2(1) + (- 3) + (- 2) = 3.742 m 2

M B = rBA

2

i * F = 3 3 cos 45° 1 3.742 (80)

3m

2

F

j k 3 (3 - 3 sin 45°) 0 3 2 - 3.742(80) - 3.742(80)

80 N

2m x

M B = { -37.6i + 90.7j - 155k} N # m

C

Ans.

Ans:

M B = { -37.6i + 90.7j - 155k} N # m

262

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4–33. A 20-N horizontal force is applied perpendicular to the handle of the socket wrench. Determine the magnitude and the coordinate direction angles of the moment created by this force about point O.

z

200 mm

75 mm

A 20 N

SOLUTION

O

rA = 0.2 sin 15°i + 0.2 cos 15°j + 0.075k = 0.05176 i + 0.1932 j + 0.075 k

15

x

F = - 20 cos 15°i + 20 sin 15°j = - 19.32 i + 5.176 j i MO = rA * F = 3 0.05176 - 19.32

j 0.1932 5.176

k 0.075 3 0

= { -0.3882 i - 1.449 j + 4.00 k} N # m MO = 4.272 = 4.27 N # m

Ans.

a = cos -1 a

-0.3882 b = 95.2° 4.272

Ans.

b = cos -1 a

-1.449 b = 110° 4.272

Ans.

g = cos -1 a

4 b = 20.6° 4.272

Ans.

Ans: MO = 4.27 N # m a = 95.2° b = 110° g = 20.6° 263

y

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4–34. Determine the coordinate direction angles a, b, g of force F, so that the moment of F about O is zero.

z

O

y 0.4 m

A 0.5 m

0.3 m

x

Solution

F

Position And Force Vectors. The coordinates of point A are A (0.4, 0.5, - 0.3) m. Thus,

rOA = {0.4i + 0.5j - 0.3k} m



u OA =



rAO = { -0.4i - 0.5j + 0.3k} m



u AO =

0.4i + 0.5j - 0.3k rOA 4 5 3 = = i + j k 2 2 2 rOA 150 150 150 20.4 + 0.5 + ( - 0.3)

- 0.4i - 0.5j + 0.3k rAO 4 5 3 = = i j + k 2 2 2 rAO 150 150 150 2( - 0.4) + ( - 0.5) + 0.3

Moment of F About Point O. To produce zero moment about point O, the line of action of F must pass through point O. Thus, F must directed from O to A (direction defined by uOA). Thus,

cos a = -



cos b = -



cos g =

4 ;  a = 55.56° = 55.6° 150

Ans.

5 ;  b = 45° 150

Ans.

-3 ;  g = 115.10° = 115° 150

Ans.

OR F must directed from A to O (direction defined by uAO). Thus

cos a = -

4 ;  a = 124.44° = 124° 150

Ans.

5 ;  b = 135° 150 3 ;  g = 64.90° = 64.9° cos g = 150 cos b = -

Ans. Ans.

Ans: a = 55.6° b = 45° g = 115° OR a = 124° b = 135° g = 64.9° 264

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4–35. z

Determine the moment of force F about point O. The force has a magnitude of 800 N and coordinate direction angles of a = 60°, b = 120°, g = 45°. Express the result as a Cartesian vector.

O

y 0.4 m

A 0.5 m

0.3 m

x

Solution

F

Position And Force Vectors. The coordinates of point A are A (0.4, 0.5, - 0.3) m. Thus

rOA = {0.4i + 0.5j - 0.3k} m



F = FuF = 800 (cos 60°i + cos 120°j + cos 45°k)



= {400i - 400j + 565.69k} N

Moment of F About Point O.

MO = rOA * F



i = † 0.4 400

j 0.5 - 400

k -0.3 † 565.69



= {162.84 i - 346.27j - 360 k} N # m



= {163i - 346j - 360k} N # m

Ans.

Ans: MO = {163i - 346j - 360k} N # m 265

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*4–36. Determine the moment produced by force FB about point O. Express the result as a Cartesian vector.

z

A

6m FC

420 N FB

780 N

SOLUTION Position Vector and Force Vectors: Either position vector rOA or rOB can be used to determine the moment of FB about point O. rOA = [6k] m

rOB = [2.5j] m

2.5 m

C 3m

The force vector FB is given by FB = FB u FB = 780 B

2m

O B y

x

(0 - 0)i + (2.5 - 0)j + (0 - 6)k 2(0 - 0)2 + (2.5 - 0)2 + (0 - 6)2

R = [300j - 720k] N

Vector Cross Product: The moment of FB about point O is given by i M O = rOA * FB = 3 0 0

j 0 300

k 6 3 = [ -1800i] N # m = [- 1.80i] kN # m - 720

Ans.

j 2.5 300

k 0 3 = [ -1800i] N # m = [ -1.80i] kN # m - 720

Ans.

or i M O = rOB * FB = 3 0 0

Ans:

M O = [ -1.80i] kN # m

266

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4–37.

Determine the moment produced by force FC about point O. .Express the result as a Cartesian vector

z

A

6m FC

420 N FB

780 N

SOLUTION Position Vector and Force Vectors: Either position vector rOA or rOC can be used to determine the moment of FC about point O. rOA = {6k} m

2m 2.5 m

C 3m

rOC = (2 - 0)i + (- 3 - 0)j + (0 - 0)k = [2i - 3j] m

x

O B y

The force vector FC is given by FC = FCuFC = 420 B

(2 - 0)i + (- 3 - 0)j + (0 - 6)k 2(2 - 0)2 + (- 3 - 0)2 + (0 - 6)2

R = [120i - 180j - 360k] N

Vector Cross Product: The moment of FC about point O is given by i M O = rOA * FC = 3 0 120

j 0 - 180

k 6 3 = [1080i + 720j] N # m -360

Ans.

j -3 -180

k 0 3 = [1080i + 720j] N # m -360

Ans.

or i M O = rOC * FC = 3 2 120

Ans: MO = rOA * FC = {1080i + 720j} N # m Or MO = rOC * FC = {1080i + 720j} N # m 267

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4–38. Determine the resultant moment produced by forces FB and FC about point O. Express the result as a Cartesian vector.

z

A

6m FC

420 N FB

780 N

SOLUTION Position Vector and Force Vectors: The position vector rOA and force vectors FB and FC, Fig. a, must be determined first. rOA = {6k} m

2m 2.5 m

C 3m

FB = FB uFB = 780 B FC = FCuFC = 420 B

(0 - 0)i + (2.5 - 0)j + (0 - 6)k 2(0 - 0)2 + (2.5 - 0)2 + (0 - 6)2 (2 - 0)i + (-3 - 0)j + (0 - 6)k

2(2 - 0)2 + (- 3 - 0)2 + (0 - 6)2

R = [300j - 720k] N

x

O B y

R = [120i - 180j - 360k] N

Resultant Moment: The resultant moment of FB and FC about point O is given by MO = rOA * FB + rOA * FC i = 30 0

j 0 300

i k 6 3 + 3 0 120 - 720

j 0 - 180

k 6 3 - 360

= [- 720i + 720j] N # m

Ans.

Ans: MO = { - 720i + 720j} N # m 268

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4–39. z

The pipe assembly is subjected to the force of F = {600i + 800j - 500k} N. Determine the moment of this force about point A. A 0.5 m B x

0.3 m

Solution

0.3 m

Position Vector. The coordinates of point C are C (0.5, 0.7, -0.3) m. Thus

C

rAC = {0.5i + 0.7 j - 0.3k} m

Moment of Force F About Point A.

MA = rAC * F



i = † 0.5 600



y

0.4 m

j 0.7 800

F

k - 0.3 † -500

= { -110i + 70j - 20k} N # m

Ans.

Ans: MA = { -110i + 70j - 20k} N # m 269

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*4–40. z

The pipe assembly is subjected to the force of F = {600i + 800j - 500k} N. Determine the moment of this force about point B. A 0.5 m B x

0.3 m

Solution Position Vector. The coordinates of points B and C are B (0.5, 0, 0) m and C (0.5, 0.7, - 0.3) m, respectively. Thus,

y

0.4 m

0.3 m C

rBC = (0.5 - 0.5)i + (0.7 - 0) j + ( - 0.3 - 0)k = {0.7j - 0.3k} m

F

Moment of Force F About Point B. Applying Eq. 4

MB = rBC * F



i = † 0 600



j 0.7 800

k -0.3 † - 500

= { - 110i - 180j - 420k} N # m

Ans.

Ans: MB = { -110i - 180j - 420k} N # m 270

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4–41. z

Determine the moment of the force of F = 600 N about point A.

A

45

B

4m

4m

x 6m

F C

Solution

6m

Position Vectors And Force Vector. The coordinates of points A, B and C are A (0, 0, 4) m, B (4 sin 45°, 0, 4 cos 45°) m and C (6, 6, 0) m, respectively. Thus

y

rAB = (4 sin 45° - 0)i + (0 - 0)j + (4 cos 45° - 4)k

= {2.8284i - 1.1716k} m rAC = (6 - 0)i + (6 - 0)j + (0 - 4)k = {6i + 6j - 4k} m rBC = (6 - 4 sin 45°)i + (6 - 0)j + (0 - 4 cos 45°)k = {3.1716i + 6j - 2.8284k} m F = Fa

3.1716i + 6j - 2.8284k rBC b = 600£ ≥ rBC 23.17162 + 62 + ( -2.8284)2

= {258.82i + 489.63j - 230.81k} N

The Moment of Force F About Point A.

MA = rAB * F i = † 2.8284 258.82

j 0 489.63

k -1.1716 † - 230.81

= {573.64i + 349.62j + 1384.89k} N # m = {574i + 350j + 1385k} N # m

Ans.

OR

MA = rAC * F = †

i 6 258.82

j 6 489.63

k -4 † - 230.81

= {573.64i + 349.62j + 1384.89k} N # m = {574i + 350j + 1385k}N # m

Ans.

Ans: MA = {574i + 350j + 1385k} N # m 271

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4–42. z

Determine the smallest force F that must be applied along the rope in order to cause the curved rod, which has a radius of 4 m, to fail at the support A. This requires a moment of M = 1500 N # m to be developed at A.

A

45

B

4m

4m

x 6m

F C

Solution

6m

Position Vectors And Force Vector. The coordinates of points A, B and C are A (0, 0, 4) m, B (4 sin 45°, 0, 4 cos 45°) m and C (6, 6, 0) m, respectively.

y

Thus, rAB = (4 sin 45° - 0)i + (0 - 0)j + (4 cos 45° - 4)k

= {2.8284i - 1.1716k} m rAC = (6 - 0)i + (6 - 0)j + (0 - 4)k = {6i + 6j - 4k} m rBC = (6 - 4 sin 45°)i + (6 - 0)j + (0 - 4 cos 45°)k = {3.1716i + 6j - 2.8284k} m F = Fa

3.1716i + 6j - 2.8284k rBC b = F£ ≥ rBC 23.17162 + 62 + ( -2.8284)2

= 0.4314F i + 0.8161Fj - 0.3847F k

The Moment of Force F About Point A.

MA = rAB * F



i = † 2.8284 0.4314F



j 0 0.8161F

k -1.1716 † - 0.3847F

= 0.9561F i + 0.5827Fj + 2.3081F k

OR

MA = rAC * F = †

i 6 0.4314F

j 6 0.8161F

k -4 † - 0.3847F

= 0.9561F i + 0.5827F j + 2.3081F k

The magnitude of MA is MA = 2(MA)2x + (MA)2y + (MA)2z = 2(0.9561F)2 + (0.5827F)2 + (2.3081F)2

= 2.5654F It is required that MA = 1500 N # m, then

1500 = 2.5654F



F = 584.71 N = 585 N

Ans. Ans: F = 585 N 272

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4–43. The pipe assembly is subjected to the 80-N force. Determine the moment of this force about point A.

z

A 400 mm B x

SOLUTION

300 mm

Position Vector And Force Vector:

200 mm

rAC = {(0.55 - 0)i + (0.4 - 0)j + ( -0.2 - 0)k} m

200 mm

C

250 mm

= {0.55i + 0.4j - 0.2k} m 40

F = 80(cos 30° sin 40°i + cos 30° cos 40°j - sin 30°k) N

30

= (44.53i + 53.07j - 40.0k} N

F

80 N

Moment of Force F About Point A: Applying Eq. 4–7, we have MA = rAC * F i = 3 0.55 44.53

j 0.4 53.07

k -0.2 3 - 40.0

= {-5.39i + 13.1j + 11.4k} N # m

Ans.

Ans: MA = {- 5.39i + 13.1j + 11.4k} N # m 273

y

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*4–44. The pipe assembly is subjected to the 80-N force. Determine the moment of this force about point B.

z

A 400 mm B x

SOLUTION

300 mm

y

Position Vector And Force Vector: 200 mm

rBC = {(0.55 - 0) i + (0.4 - 0.4)j + (- 0.2 - 0)k} m

200 mm C

250 mm

= {0.55i - 0.2k} m 40

F = 80 (cos 30° sin 40°i + cos 30° cos 40°j - sin 30°k) N

30

= (44.53i + 53.07j - 40.0k} N

F

80 N

Moment of Force F About Point B: Applying Eq. 4–7, we have MB = rBC * F i = 3 0.55 44.53

j 0 53.07

k - 0.2 3 - 40.0

= {10.6i + 13.1j + 29.2k} N # m

Ans.

Ans: MB = {10.6i + 13.1j + 29.2k} N # m 274

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4–45. A force of F = {6i - 2j + 1k} kN produces a moment of MO = {4i + 5j - 14k} kN # m about the origin, point O. If the force acts at a point having an x coordinate of x = 1 m, determine the y and z coordinates. Note: The figure shows F and MO in an arbitrary position.

z F

P MO z

d

y

O 1m

SOLUTION y

MO = r * F i 4i + 5j - 14k = 3 1 6

j y -2

x

k z3 1

4 = y + 2z 5 = - 1 + 6z -14 = - 2 - 6y y = 2m

Ans.

z = 1m

Ans.

Ans: y = 2m z = 1m 275

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4–46. The force F = 56i + 8j + 10k6 N creates a moment about point O of M O = 5 -14i + 8j + 2k6 N # m. If the force passes through a point having an x coordinate of 1 m, determine the y and z coordinates of the point.Also, realizing that MO = Fd, determine the perpendicular distance d from point O to the line of action of F. Note: The figure shows F and MO in an arbitrary position.

z F

P MO z

d

y

O 1m y x

SOLUTION i - 14i + 8j + 2k = 3 1 6

j y 8

k z 3 10

- 14 = 10y - 8z 8 = -10 + 6z 2 = 8 - 6y y = 1m

Ans.

z = 3m

Ans.

MO = 2( -14)2 + (8)2 + (2)2 = 16.25 N # m F = 2(6)2 + (8)2 + (10)2 = 14.14 N d =

16.25 = 1.15 m 14.14

Ans.

Ans: y = 1m z = 3m d = 1.15 m 276

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4–47. A force F having a magnitude of F = 100 N acts along the diagonal of the parallelepiped. Determine the moment of F about point A, using M A = rB : F and M A = rC : F.

z

F

C

200 mm rC

SOLUTION F = 100 a

400 mm B

-0.4 i + 0.6 j + 0.2 k b 0.7483

F

rB 600 mm

A

x

F = 5 -53.5 i + 80.2 j + 26.7 k6 N M A = rB * F = 3

i 0 -53.5

j -0.6 80.2

i - 0.4 -53.5

j 0 80.2

k 0 3 = 5- 16.0 i - 32.1 k6 N # m 26.7

Ans.

Also, M A = rC * F =

k 0.2 = 26.7

-16.0 i - 32.1 k N # m

Ans.

Ans: MA = { - 16.0i - 32.1k} N # m 277

y

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*4–48. Force F acts perpendicular to the inclined plane. Determine the moment produced by F about point A. Express the result as a Cartesian vector.

z 3m

A F

400 N

3m B

SOLUTION

x

C

4m

Force Vector: Since force F is perpendicular to the inclined plane, its unit vector uF is equal to the unit vector of the cross product, b = rAC * rBC, Fig. a. Here rAC = (0 - 0)i + (4 - 0)j + (0 - 3)k = [4j - 3k] m rBC = (0 - 3)i + (4 - 0)j + (0 - 0)k = [-3i + 4j] m Thus, i b = rCA * rCB = 3 0 -3

j 4 4

k -3 3 0

= [12i + 9j + 12k] m2 Then, uF =

12i + 9j + 12k b = = 0.6247i + 0.4685j + 0.6247k b 212 2 + 92 + 12 2

And finally F = FuF = 400(0.6247i + 0.4685j + 0.6247k) = [249.88i + 187.41j + 249.88k] N Vector Cross Product: The moment of F about point A is M A = rAC

* F = 3

i 0 249.88

j 4 187.41

k -3 3 249.88

= [1.56i - 0.750j - 1.00k] kN # m

Ans.

Ans: MA = [1.56i - 0.750j - 1.00k] kN # m 278

y

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4–49. Force F acts perpendicular to the inclined plane. Determine the moment produced by F about point B. Express the result as a Cartesian vector.

z 3m

A F

400 N

3m B

SOLUTION

x

C

4m

Force Vector: Since force F is perpendicular to the inclined plane, its unit vector uF is equal to the unit vector of the cross product, b = rAC * rBC, Fig. a. Here rAC = (0 - 0)i + (4 - 0)j + (0 - 3)k = [4j - 3k] m rBC = (0 - 3)i + (4 - 0)j + (0 - 0)k = [ -3k + 4j] m Thus, i b = rCA * rCB = 3 0 -3

j 4 4

k -3 3 = [12i + 9j + 12k] m2 0

Then, uF =

12i + 9j + 12k b = = 0.6247i + 0.4685j + 0.6247k b 2122 + 92 + 122

And finally F = FuF = 400(0.6247i + 0.4685j + 0.6247k) = [249.88i + 187.41j + 249.88k] N Vector Cross Product: The moment of F about point B is MB = rBC

* F = 3

i -3 249.88

j 4 187.41

k 0 3 249.88

= [1.00i + 0.750j - 1.56k] kN # m

Ans.

Ans: MB = {1.00i + 0.750j - 1.56k} kN # m 279

y

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4–50. Strut AB of the 1-m-diameter hatch door exerts a force of 450 N on point B. Determine the moment of this force about point O.

z

B 30°

0.5 m

F = 450 N

O

SOLUTION

y 0.5 m

Position Vector And Force Vector:

30°

rOB = 510 - 02i + 11 cos 30° - 02j + 11 sin 30° - 02k6 m

A

x

= 50.8660j + 0.5k6 m rOA = 510.5 sin 30° - 02i + 10.5 + 0.5 cos 30° - 02j + 10 - 02k6 m = 50.250i + 0.9330j6 m F = 450 ¢

10 - 0.5 sin 30°2i + 31 cos 30° - 10.5 + 0.5 cos 30°24j + 11 sin 30° - 02k 210 - 0.5 sin 30°22 + 31 cos 30° - 10.5 + 0.5 cos 30°242 + 11 sin 30° - 022

≤N

= 5-199.82i - 53.54j + 399.63k6 N Moment of Force F About Point O: Applying Eq. 4–7, we have M O = rOB * F = 3

i 0 - 199.82

j 0.8660 - 53.54

k 0.5 3 399.63

= 5373i - 99.9j + 173k6 N # m

Ans.

Or M O = rOA * F =

i 0.250 - 199.82

j 0.9330 - 53.54

k 0 399.63

=

373i - 99.9j + 173k N # m

Ans.

Ans: MO = {373i - 99.9j + 173k} N # m 280

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4–51. z

Using a ring collar the 75-N force can act in the vertical plane at various angles u. Determine the magnitude of the moment it produces about point A, plot the result of M (ordinate) versus u (abscissa) for 0° … u … 180°, and specify the angles that give the maximum and minimum moment.

A 2m

y

x

SOLUTION MA

1.5 m

i = 32 0

j 1.5 75 cos u

75 N

θ

k 3 0 75 sin u

= 112.5 sin u i - 150 sin u j + 150 cos u k MA = 21112.5 sin u22 + 1-150 sin u22 + 1150 cos u22 = 212 656.25 sin2 u + 22 500 dMA 1 1 = 112 656.25 sin2 u + 22 5002- 2 112 656.25212 sin u cos u2 = 0 du 2 sin u cos u = 0;

u = 0°, 90°, 180°

Ans.

umax = 187.5 N # m at u = 90° umin = 150 N # m at u = 0°, 180°

Ans: umax = 90° umin = 0°, 180° 281

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*4–52. Determine the magnitude of the moment of the force F = {50i - 20j - 80k} N about the base line AB of the tripod.

z F D 4m

2m 1.5 m

Solution uAB =

C

A

{3.5i + 0.5j}

y

2(3.5)2 + (0.5)2

x

uAB = {0.9899i + 0.1414j} MAB = uAB # ( rAD MAB = 136 N # m

0.9899 * F ) = † 2.5 50

0.1414 0 - 20

2.5 m

2m 0.5 m

0 4 † - 80

1m B

Ans.

Ans: MAB = 136 N # m 282

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4–53. Determine the magnitude of the moment of the force F = {50i - 20j - 80k} N about the base line BC of the tripod.

z F D 4m

2m 1.5 m

Solution uBC =

C

A

{ -1.5i - 2.5j}

y

2( - 1.5)2 + ( - 2.5)2

x

2m

uBC = { - 0.5145i - 0.8575j} MBC = uBC # ( rCD MBC = 165 N # m

- 0.5145 * F ) = † 0.5 50

- 0.8575 2 - 20

2.5 m

0.5 m

0 4 † - 80

1m B

Ans.

Ans: MBC = 165 N # m 283

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4–54. Determine the magnitude of the moment of the force F = {50i - 20j - 80k} N about the base line CA of the tripod.

z F D 4m

2m 1.5 m

Solution uCA =

C

A

{ -2i + 2j}

y

2( - 2)2 + (2)2

x

uCA = { - 0.707i + 0.707j} MCA = uCA # ( rAD MCA = 226 N # m

- 0.707 * F ) = † 2.5 50

0.707 0 - 20

2.5 m

2m 0.5 m

0 4 † - 80

1m B

Ans.

Ans: MCA = 226 N # m 284

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4–55. Determine the moment of this force F about an axis extending between A and C. Express the result as a Cartesian vector.

z A 4m y

Solution

x

3m

Position Vector: rCB = 5 - 2k6 m

C 2m

rAB = 5 14 - 02i + 13 - 02j + 1 - 2 -02k6m = 5 4i + 3j - 2k6 m

B

Unit Vector Along AC Axis: uAC =

14 - 02i + 13 - 02j

214 - 02 2 + 13 - 02 2

F  {4i  12j  3k} kN

= 0.8i + 0.6j

Moment of Force F about AC Axis: With F = {4i + 12j – 3k} kN. applying Eq. 4–7, we have MAC = uAC # 1rCB * F 2 0.8 = † 0 4

0.6 0 12

0 -2 † -3

= 0.83 102 1 -32 - 121 -22 4 -0.6301 -32 -41 - 22 4 + 0 = 14.4 kN # m

MAC = uAC # 1rAB * F 2 0.8 = † 4 4

0.6 3 12

0 -2 † -3

= 0.83 132 1 - 32 - 121 -22 4 -0.6341 -32 - 41 - 22 4 + 0

= 14.4 kN # m

Expressing MAC as a Cartesian vector yields MAC = MACuAC = 14.410.8i + 0.6j2

= 511.5i + 8.64j6kN # m

Ans.

Ans: MAC = {11.5i + 8.64j} kN # m 285

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*4–56. A vertical force of F = 60 N is applied to the handle of the pipe wrench. Determine the moment that this force exerts along the axis AB (x axis) of the pipe assembly. Both the wrench and pipe assembly ABC lie in the x-y plane. Suggestion: Use a scalar analysis.

z

A y 500 mm

F 150 mm

B x

45⬚ 200 mm C

Solution Scalar Analysis: From the geometry, the perpendicular distance from x axis to force F is d = 0.15sin 45° + 0.2sin 45° = 0.2475 m. MX = ΣMX; MX = -Fd = - 6010.2752 = -14.8 N # m Negative sign indicates that Mx is directed toward negative x axis. Mx = 14.8 N # m

Ans.

.

Ans: Mx = 14.8 N # m 286

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4–57. Determine the magnitude of the vertical force F acting on the handle of the wrench so that this force produces a component of moment along the AB axis (x axis) of the pipe assembly of (MA)x = 5 -5i6 N # m. Both the pipe assembly ABC and the wrench lie in the x-y plane. Suggestion: Use a scalar analysis.

z

A y 500 mm

F 150 mm

B x

45⬚ 200 mm C

Solution Scalar Analysis: From the geometry, the perpendicular distance from x axis to force F is d = 0.15 sin 45° + 0.2 sin 45° = 0.2475 m. Mx = ΣMx;

-5 = -F(0.2475) Ans.

F = 20.2 N

Ans: F = 20.2 N 287

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4–58. z

The board is used to hold the end of a four-way lug wrench in the position shown when the man applies a force of F = 100 N. Determine the magnitude of the moment produced by this force about the x axis. Force F lies in a vertical plane.

F 60

SOLUTION

x

250 mm y

250 mm

Vector Analysis Moment About the x Axis: The position vector rAB, Fig. a, will be used to determine the moment of F about the x axis. rAB = (0.25 - 0.25)i + (0.25 - 0)j + (0 - 0)k = {0.25j} m The force vector F, Fig. a, can be written as F = 100(cos 60°j - sin 60°k) = {50j - 86.60k} N Knowing that the unit vector of the x axis is i, the magnitude of the moment of F about the x axis is given by 1 Mx = i # rAB * F = 3 0 0

0 0.25 50

0 0 3 -86.60

= 1[0.25( -86.60) - 50(0)] + 0 + 0 = - 21.7 N # m

Ans.

The negative sign indicates that Mx is directed towards the negative x axis. Scalar Analysis This problem can be solved by summing the moment about the x axis Mx = ©Mx;

Mx = -100 sin 60°(0.25) + 100 cos 60°(0) = - 21.7 N # m Ans.

Ans: Mx = 21.7 Ν # m 288

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4–59. The board is used to hold the end of a four-way lug wrench in position. If a torque of 30 N # m about the x axis is required to tighten the nut, determine the required magnitude of the force F that the man’s foot must apply on the end of the wrench in order to turn it. Force F lies in a vertical plane.

z

F 60

250 mm y

x 250 mm

SOLUTION Vector Analysis Moment About the x Axis: The position vector rAB, Fig. a, will be used to determine the moment of F about the x axis. rAB = (0.25 - 0.25)i + (0.25 - 0)j + (0 - 0)k = {0.25j} m The force vector F, Fig. a, can be written as F = F(cos 60°j - sin 60°k) = 0.5Fj - 0.8660Fk Knowing that the unit vector of the x axis is i, the magnitude of the moment of F about the x axis is given by 1 Mx = i # rAB * F = 3 0 0

0 0.25 0.5F

0 3 0 - 0.8660F

= 1[0.25(- 0.8660F) - 0.5F(0)] + 0 + 0 = -0.2165F

Ans.

The negative sign indicates that Mx is directed towards the negative x axis. The magnitude of F required to produce Mx = 30 N # m can be determined from 30 = 0.2165F F = 139 N

Ans.

Scalar Analysis This problem can be solved by summing the moment about the x axis Mx = ©Mx;

- 30 = - F sin 60°(0.25) + F cos 60°(0) F = 139 N

Ans.

Ans: F = 139 Ν 289

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*4–60. z

The lug nut on the wheel of the automobile is to be removed using the wrench and applying the vertical force of F = 30 N at A. Determine if this force is adequate, provided 14 N # m of torque about the x axis is initially required to turn the nut. If the 30-N force can be applied at A in any other direction, will it be possible to turn the nut?

B

0.25 m

A

0.3 m

SOLUTION 2

2

Mx = 30 A 2(0.5) - (0.3)

B =

12 N #

m 6

14 N #

0.5 m y

m,

No

0.1 m

Ans. x

For (Mx)max , apply force perpendicular to the handle and the x - axis. (Mx)max = 30 (0.5) = 15 N # m 7 14 N # m,

30 N

F

Yes

Ans.

Ans: No Yes 290

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4–61. Solve Prob. 4–60 if the cheater pipe AB is slipped over the handle of the wrench and the 30-N force can be applied at any point and in any direction on the assembly.

z 30 N

F B

0.25 m

A

0.3 m

SOLUTION 4 Mx = 30 (0.75)a b = 18 N # m 7 14 N # m, 5

0.5 m y 0.1 m

Yes

Ans. x

(Mx)max occurs when force is applied perpendicular to both the handle and the x - axis. (Mx)max = 30(0.75) = 22.5 N # m 7 14N # m,

Yes

Ans.

Ans: Yes Yes 291

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4–62. If F = 450 N, determine the magnitude of the moment produced by this force about the x axis.

z F 45⬚ A

B

60⬚ 60⬚ 100 mm

Solution

x

Moment About the x axis: Either position vector rAB or rCB can be used to determine the moment of F about the x axis

300 mm

150 mm y

rAB = ( - 0.15- 0)i + (0.3 - 0)j + (0.1 - 0)k = [ -0.15i + 0.3j + 0.1k] m rCB = 3( -1.5 - ( -0.15) 4i + (0.3 - 0)j + (0.1 - 0)k = 30.3j + 0.1k4m

The force vector F is given by

F = 450( - cos 60°i + cos 60°j + cos 45°k) = 3 -225i + 225j + 318.20k]N

Knowing that the unit vector of the x axis is i, the magnitude of the moment of F about the x axis is given by 1 Mx = i # rAB * F = † -0.15 - 225

or

0 0.3 225

0 0.1 † 318.20

= 130.3(318.20) - (225)(0.1) 4 + 0 + 0 = 73.0 N # m

Mx = i # rCB * F = †

1 0 -225

0 0.3 225

Ans.

0 0.1 † 318.20

= 130.3(318.20) - (225)(0.1) 4 + 0 + 0 = 73.0 N # m

Ans.

Ans: M x = 73.0 N # m 292

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4–63. The friction at sleeve A can provide a maximum resisting moment of 125 N # m about the x axis. Determine the largest magnitude of force F that can be applied to the bracket so that the bracket will not turn.

z F 45⬚ A

B

60⬚ 60⬚ 100 mm

x

Solution

300 mm

150 mm y

Moment About the x axis: The position vector rAB, Fig. a, will be used to determine the moment of F about the x axis rAB = ( -0.15 - 0)i + (0.3 - 0)j + (0.1 - 0)k = [ -0.15i + 0.3j + 0.1k] m The force vector F is given by F = F( - cos 60°i + cos 60°j + cos 45°k) = -0.5Fi + 0.5Fj + 0.7071Fk Knowing that the unit vector of the x axis is i, the magnitude of the moment of F about the x axis is given by Mx = i # rAB *

1 F = † -0.15 - 0.5F

0 0.3 0.5F

0 0.1 † 0.7071F

= 130.3(0.7071F) - 0.5F(0.1) 4 + 0 + 0 = 0.1621F

Since the friction at sleeve A can resist a moment of Mx = 125N · m, the maximum allowable magnitude of F is given by 125 = 0.1612F Ans.

F = 771N

Ans: F = 771 N 293

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*4–64 .

z

The wrench A is used to hold the pipe in a stationary position while wrench B is used to tighten the elbow fitting. If FB = 150 N, determine the magnitude of the moment produced by this force about the y axis. Also, what is the magnitude of force FA in order to counteract this moment?

50 mm 50 mm

y

SOLUTION Vector Analysis Moment of FB About the y Axis: The position vector rCB, Fig. a, will be used to determine the moment of FB about the y axis. rCB = (-0.15 - 0)j + (0.05 - 0.05)j + (-0.2598 - 0)k = {-0.15i - 0.2598k} m

x

300 mm 300 mm 30⬚ FA 135⬚ 120⬚ A

Referring to Fig. a, the force vector FB can be written as

30⬚

B

FB

FB = 150(cos 60°i - sin 60°k) = {75i - 129.90k} N Knowing that the unit vector of the y axis is j, the magnitude of the moment of FB about the y axis is given by My = j # rCB * FB =

0 3 - 0.15

1 0 0

75

0 -0.2598 3 -129.90

= 0 - 1[-0.15(-129.90) - 75( -0.2598)] + 0

= - 38.97 N # m = 39.0 N # m

Ans.

The negative sign indicates that My is directed towards the negative y axis. Moment of FA About the y Axis: The position vector rDA, Fig. a, will be used to determine the moment of FA about the y axis. rDA = (0.15 - 0)i + [-0.05 - (- 0.05)]j + (-0.2598 - 0)k = {0.15i - 0.2598k} m Referring to Fig. a, the force vector FA can be written as FA = FA(-cos 15°i + sin 15°k) = - 0.9659FAi + 0.2588FAk Since the moment of FA about the y axis is required to counter that of FB about the same axis, FA must produce a moment of equal magnitude but in the opposite sense to that of FA. Mx = j # rDA * FB

+ 0.38.97 = 3

0 0.15 - 0.9659 FA

1 0 0

0 - 0.2598 3 0.2588FA

+ 0.38.97 = 0 - 1[0.15(0.2588FA) - (-0.9659FA)(-0.2598)] + 0 FA = 184 N

Ans.

Scalar Analysis This problem can be solved by first taking the moments of FB and then FA about the y axis. For FB we can write My = ©My;

My = -150 cos 60°(0.3 cos 30°) - 150 sin 60°(0.3 sin 30°) = -38.97 N # m

Ans.

The moment of FA, about the y axis also must be equal in magnitude but opposite in sense to that of FB about the same axis My = ©My;

38.97 = FA cos 15°(0.3 cos 30°) - FA sin 15°(0.3 sin 30°) FA = 184 N

Ans. 294

Ans: FA = 184 N My = -38.97 N # m FA = 184 N

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4–65.

z

The wrench A is used to hold the pipe in a stationary position while wrench B is used to tighten the elbow fitting. Determine the magnitude of force FB in order to develop a torque of 50 N # m about the y axis. Also, what is the required magnitude of force FA in order to counteract this moment?

50 mm 50 mm

y

x

SOLUTION

300 mm 300 mm

Vector Analysis Moment of FB About the y Axis: The position vector rCB, Fig. a, will be used to determine the moment of FB about the y axis. rCB = (-0.15 - 0)i + (0.05 - 0.05)j + (-0.2598 - 0)k = {-0.15i - 0.2598k} m

30⬚

30⬚

FA 135⬚ 120⬚ A

B

FB

Referring to Fig. a, the force vector FB can be written as FB = FB(cos 60°i - sin 60°k) = 0.5FBi - 0.8660FBk Knowing that the unit vector of the y axis is j, the moment of FB about the y axis is required to be equal to -50 N # m, which is given by My = j # rCB * FB 0 -50i = † -0.15 0.5FB

1 0 0

0 -0.2598 † -0.8660FB

-50 = 0 - 1[-0.15(-0.8660FB) - 0.5FB( -0.2598)] + 0 FB = 192 N

Ans.

Moment of F A About the y Axis: The position vector rDA, Fig. a, will be used to determine the moment of FA about the y axis. rDA = (0.15 - 0)i + [-0.05 - (- 0.05)]j + (-0.2598 - 0)k = {-0.15i - 0.2598k} m Referring to Fig. a, the force vector FA can be written as FA = FA(-cos 15°i + sin 15°k) = -0.9659FAi + 0.2588FAk Since the moment of FA about the y axis is required to produce a countermoment of 50 N # m about the y axis, we can write My = j # rDA * FA 50 = †

0 0.15 - 0.9659FA

1 0 0

0 -0.2598 † 0.2588FA

50 = 0 - 1[0.15(0.2588FA) - ( -0.9659FA)(-0.2598)] + 0 FA = 236 N

Ans.

Scalar Analysis This problem can be solved by first taking the moments of FB and then FA about the y axis. For FB we can write My = ©My;

-50 = -FB cos 60°(0.3 cos 30°) - FB sin 60°(0.3 sin 30°) FB = 192 N

Ans.

For FA, we can write My = ©My;

50 = FA cos 15°(0.3 cos 30°) - FA sin 15°(0.3 sin 30°) FA = 236 N

Ans. 295

Ans: FB = 192 N FA = 236 N

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4–66. z

The force of F = 30 N acts on the bracket as shown. Determine the moment of the force about the a-a axis of the pipe if a = 60°, b = 60°, and g = 45°. Also, determine the coordinate direction angles of F in order to produce the maximum moment about the a-a axis.What is this moment?

F  30 N y

g b a

50 mm x

100 mm

100 mm

SOLUTION

a

F = 30 1cos 60° i + cos 60° j + cos 45° k2 = 515 i + 15 j + 21.21 k6 N

a

r = 5 -0.1 i + 0.15 k6 m

u = j

0 Ma = 3 -0.1 15

1 0 15

0 0.15 3 = 4.37 N # m 21.21

Ans.

F must be perpendicular to u and r. uF =

0.1 0.15 i + k 0.1803 0.1803

= 0.8321i + 0.5547k a = cos-1 0.8321 = 33.7°

Ans.

b = cos-1 0 = 90°

Ans.

g = cos-1 0.5547 = 56.3°

Ans.

M = 30 0.1803 = 5.41 N # m

Ans.

Ans: Ma = 4.37 N # m a = 33.7° b = 90° g = 56.3° M = 5.41 N # m 296

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4–67. A clockwise couple M = 5 N # m is resisted by the shaft of the electric motor. Determine the magnitude of the reactive forces -R and R which act at supports A and B so that the resultant of the two couples is zero.

M

150 mm 60

A

Solution

R

60

B

R

a+MC = - 5 + R (2(0.15)>tan 60°) = 0 Ans.

R = 28.9 N

Ans: R = 28.9 N 297

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*4–68. A twist of 4 N # m is applied to the handle of the screwdriver. Resolve this couple moment into a pair of couple forces F exerted on the handle and P exerted on the blade.

–F –P P 5 mm

4 N·m

30 mm

F

SOLUTION For the handle MC = ©Mx ;

F10.032 = 4 F = 133 N

Ans.

For the blade, MC = ©Mx ;

P10.0052 = 4 P = 800 N

Ans.

Ans: F = 133 N P = 800 N 298

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4–69. If the resultant couple of the three couples acting on the triangular block is to be zero, determine the magnitude of forces F and P.

z F C

F 150 N B

SOLUTION

300 mm

BA = 0.5 m The couple created by the 150 - N forces is

x

400 mm

D 500 mm

150 N A

P

600 mm

P

MC1 = 150 (0.5) = 75 N # m Then 3 4 MC1 = 75 a b j + 75a b k 5 5 = 45 j + 60 k MC2 = - P (0.6) k MC3 = - F (0.6) j Require MC1 + MC2 + MC3 = 0 45 j + 60 k - P (0.6) k - F (0.6) j = 0 Equate the j and k components 45 - F (0.6) = 0 Ans.

F = 75 N 60 - P (0.6) = 0

Ans.

P = 100 N

Ans: F = 75 N P = 100 N 299

y

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4–70. Two couples act on the cantilever beam. If F = 6 kN, determine the resultant couple moment.

3m

3m

5 kN

F 5

4 3

A

SOLUTION

30

B

30

0.5 m

0.5 m 5

4

F

3

5 kN

a) By resolving the 6-kN and 5-kN couples into their x and y components, Fig. a, the couple moments (Mc)1 and (Mc)2 produced by the 6-kN and 5-kN couples, respectively, are given by a + (MC)1 = 6 sin 30°(3) - 6 cos 30°(0.5 + 0.5) = 3.804 kN # m 4 3 a + (MC)2 = 5a b(0.5 + 0.5) - 5a b (3) = -9 kN # m 5 5 Thus, the resultant couple moment can be determined from (MC)R = (MC)1 + (MC)2 = 3.804 - 9 = -5.196 kN # m = 5.20 kN # m (Clockwise)

Ans.

b) By resolving the 6-kN and 5-kN couples into their x and y components, Fig. a, and summing the moments of these force components about point A, we can write a + (MC)R = ©MA ;

4 3 (MC)R = 5a b (0.5) + 5a b (3) - 6 cos 30°(0.5) - 6 sin 30°(3) 5 5

4 3 + 6 sin 30°(6) - 6 cos 30°(0.5) + 5a b(0.5) - 5a b (6) 5 5 = -5.196 kN # m = 5.20 kN # m (Clockwise)

Ans.

Ans:

300

a. a + 1 M C 2 R = 5.20 kN # m (Clockwise) b. a + 1 M C2 R = 5.20 kN # m (Clockwise)

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4–71. Determine the required magnitude of force F, if the resultant couple moment on the beam is to be zero.

3m

3m

5 kN

F 5

4 3

A

SOLUTION

30

B

30

0.5 m

0.5 m 5

4

F

3

5 kN

By resolving F and the 5-kN couple into their x and y components, Fig. a, the couple moments (Mc)1 and (Mc)2 produced by F and the 5-kN couple, respectively, are given by a + (MC)1 = F sin 30°(3) - F cos 30°(1) = 0.6340F 4 3 a + (MC)2 = 5a b (1) - 5a b(3) = -9 kN 5 5 The resultant couple moment acting on the beam is required to be zero. Thus, (MC)R = (MC)1 + (MC)2 0 = 0.6340F - 9 F = 14.2 kN

Ans.

Ans: F = 14.2 kN 301

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*4–72. If u = 30, determine the magnitude of force F so that the resultant couple moment is 100 N # m, clockwise. The disk has a radius of 300 mm.

300 N 300 mm

15⬚

u

30⬚

⫺F F 30⬚

15⬚

u 300 N

Solution By resolving F and the 300-N couple into their radial and tangential components, Fig. a, and summing the moment of these two force components about point O, a+ (Mc)R = ΣMO;

-100 = F sin 45°(0.3) + F cos 15°(0.3) - 2(300 cos 30°)(0.3) Ans.

F = 111 N

Note: Since the line of action of the radial component of the forces pass through point O, no moment is produced about this point.

Ans: F = 111 N 302

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4–73. If F = 200 N, determine the required angle u so that the resultant couple moment in zero. The disk has a radius of 300 mm.

300 N 300 mm

15⬚

u

30⬚

⫺F F 30⬚

15⬚

u 300 N

Solution By resolving the 300-N and 200-N couples into their radial and tangential components, Fig. a, and summing the moment of these two force components about point O, a+ (Mc)R = ΣMO; 0 = 200 sin 45°(0.3) + 200 cos 15°(0.3) - 300 cos u(0.3) - 300 cos u(0.3) Ans.

u = 56.1°

Note: Since the line of action of the radial component of the forces pass through point O, no moment is produced about this point.

Ans: u = 56.1 303

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4–74. The cord passing over the two small pegs A and B of the square board is subjected to a tension of 100 N. Determine the required tension P acting on the cord that passes over pegs C and D so that the resultant couple produced by the two couples is 15 N # m acting clockwise. Take u = 15°.

300 mm

C

B

u

30⬚

⫺P 100 N

300 mm

100 N P

45⬚

30⬚ A

u

D

Solution c+ MR = 100 cos 30°(0.3) + 100 sin 30°(0.3) - P sin 15°(0.3) - P cos 15°(0.3) = 15 Ans.

P = 70.7 N

Ans: P = 70.7 N 304

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4–75. The cord passing over the two small pegs A and B of the board is subjected to a tension of 100 N. Determine the minimum tension P and the orientation u of the cord passing over pegs C and D, so that the resultant couple moment produced by the two cords is 20 N # m, clockwise.

300 mm

C

B

u

30⬚

⫺P 100 N

300 mm

100 N P

45⬚

30⬚ A

u

D

Solution Ans.

For minimum P require u = 45° c+ MR = 100 cos 30°(0.3) + 100 sin 30° - P a P = 49.5 N

0.3 b = 20 cos 45°

Ans.

Ans: P = 49.5 N 305

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*4–76. A twist of 4 N # m is applied to the handle of the screwdriver. Resolve this couple moment into a pair of couple forces F exerted on the handle and P exerted on the blade.

–F –P P 5 mm

4 N·m

30 mm

F

SOLUTION For the handle MC = ©Mx ;

F10.032 = 4 F = 133 N

Ans.

For the blade, MC = ©Mx ;

P10.0052 = 4 P = 800 N

Ans.

Ans: F = 133 N P = 800 N 306

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4–77. The ends of the triangular plate are subjected to three couples. Determine the magnitude of the force F so that the resultant couple moment is 400 N # m clockwise.

F

600 N

F

600 N 40

40

SOLUTION a + MR = ©M;

-400 = 600 a

1m

0.5 0.5 b -Fa b - 250(1) cos 40° cos 40°

250 N

F = 830 N

250 N

Ans.

Ans: F = 830 N 307

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4–78. The man tries to open the valve by applying the couple forces of F = 75 N to the wheel. Determine the couple moment produced.

150 mm

150 mm

F

SOLUTION a + Mc = ©M;

Mc = - 75(0.15 + 0.15) = - 22.5 N # m = 22.5 N # m b

Ans. F

Ans: MC = 22.5 N # mb 308

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4–79. If the valve can be opened with a couple moment of 25 N # m, determine the required magnitude of each couple force which must be applied to the wheel.

150 mm

150 mm

F

SOLUTION a +Mc = ©M;

-25 = -F(0.15 + 0.15) F = 83.3 N

Ans. F

Ans: F = 83.3 N 309

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*4–80. Determine the magnitude of F so that the resultant couple moment is 12 kN # m, counterclockwise. Where on the beam does the resultant couple moment act?

F

F 30 30

8 kN

0.3 m

1.2 m 0.4 m

Solution

a + MR = ΣMC;  12 = (F cos 30°)(0.3) + 8(1.2)

8 kN

Ans.

F = 9.238 kN = 9.24 kN

Since the couple moment is a free vector, the resultant couple moment can act at any point on or off the beam.

Ans: F = 9.24 kN 310

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4–81.

z

If F = 80 N, determine the magnitude and coordinate direction angles of the couple moment. The pipe assembly lies in the x–y plane.

F

300 mm 300 mm F

x 200 mm

SOLUTION It is easiest to find the couple moment of F by taking the moment of F or –F about point A or B, respectively, Fig. a. Here the position vectors rAB and rBA must be determined first.

200 mm

300 mm

y

rAB = (0.3 - 0.2)i + (0.8 - 0.3)j + (0 - 0)k = [0.1i + 0.5j] m rBA = (0.2 - 0.3)i + (0.3 - 0.8)j + (0 - 0)k = [ -0.1i - 0.5j] m The force vectors F and –F can be written as F = {80 k} N and - F = [ -80 k] N Thus, the couple moment of F can be determined from i M c = rAB * F = 3 0.1 0

j 0.5 0

k 0 3 = [40i - 8j] N # m 80

or i Mc = rBA * -F = 3 - 0.1 0

j - 0.5 0

k 0 3 = [40i - 8j] N # m - 80

The magnitude of Mc is given by Mc = 2Mx 2 + My 2 + Mz 2 = 2402 + (- 8)2 + 02 = 40.79 N # m = 40.8 N # m

Ans.

The coordinate angles of Mc are a = cos

-1

b = cos

-1

g = cos

-1

¢ ¢

¢

Mx 40 ≤ = cos ¢ ≤ = 11.3° M 40.79 My M Mz M

-8 ≤ = 101° 40.79

Ans.

0 ≤ = 90° 40.79

Ans.

≤ = cos ¢

≤ = cos ¢

Ans.

Ans: M C = 40.8 N # m 311

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4–82.

z

If the magnitude of the couple moment acting on the pipe assembly is 50 N # m, determine the magnitude of the couple forces applied to each wrench. The pipe assembly lies in the x–y plane.

F

300 mm 300 mm F

x 200 mm

SOLUTION It is easiest to find the couple moment of F by taking the moment of either F or –F about point A or B, respectively, Fig. a. Here the position vectors rAB and rBA must be determined first.

200 mm

300 mm

y

rAB = (0.3 - 0.2)i + (0.8 - 0.3)j + (0 - 0)k = [0.1i + 0.5j] m rBA = (0.2 - 0.3)i + (0.3 - 0.8)j + (0 - 0)k = [ -0.1i - 0.5j] m The force vectors F and –F can be written as F = {Fk} N and - F = [ -Fk]N Thus, the couple moment of F can be determined from i M c = rAB * F = 3 0.1 0

j 0.5 0

k 0 3 = 0.5Fi - 0.1Fj F

The magnitude of Mc is given by Mc = 2Mx 2 + My 2 + Mz 2 = 2(0.5F)2 + (0.1F)2 + 02 = 0.5099F Since Mc is required to equal 50 N # m, 50 = 0.5099F F = 98.1 N

Ans.

Ans: F = 98.1 N 312

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4–83.

z

{35k} N

Determine the resultant couple moment of the two couples that act on the pipe assembly. The distance from A to B is d = 400 mm. Express the result as a Cartesian vector.

B 250 mm d

{ 50i} N C

30

{ 35k} N

SOLUTION

350 mm

Vector Analysis

x

y

A {50i} N

Position Vector: rAB = {(0.35 - 0.35)i + (-0.4 cos 30° - 0)j + (0.4 sin 30° - 0)k} m = { -0.3464j + 0.20k} m Couple Moments: With F1 = {35k} N and F2 = { - 50i} N, applying Eq. 4–15, we have (M C)1 = rAB * F1 i = 30 0

j -0.3464 0

k 0.20 3 = {- 12.12i} N # m 35

(M C)2 = rAB * F2 i = 3 0 - 50

j - 0.3464 0

k 0.20 3 = { - 10.0j - 17.32k} N # m 0

Resultant Couple Moment: M R = ©M;

M R = (M C)1 + (M C)2 = {- 12.1i - 10.0j - 17.3k}N # m

Ans.

Scalar Analysis: Summing moments about x, y, and z axes, we have (MR)x = ©Mx ;

(MR)x = -35(0.4 cos 30°) = - 12.12 N # m

(MR)y = ©My ;

(MR)y = -50(0.4 sin 30°) = -10.0 N # m

(MR)z = ©Mz ;

(MR)z = -50(0.4 cos 30°) = -17.32 N # m

Express M R as a Cartesian vector, we have M R = { - 12.1i - 10.0j - 17.3k} N # m

Ans: MR = { -12.1i - 10.0j - 17.3k} N # m 313

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* *4–84.

z

{35k} N

Determine the distance d between A and B so that the resultant couple moment has a magnitude of MR = 20 N # m.

B 250 mm d

{ 50i} N C

30

{ 35k} N

SOLUTION

350 mm

Position Vector:

x

y

A {50i} N

rAB = {(0.35 - 0.35)i + (-d cos 30° - 0)j + (d sin 30° - 0)k} m = {- 0.8660d j + 0.50d k} m Couple Moments: With F1 = {35k} N and F2 = {- 50i} N, applying Eq. 4–15, we have (M C)1 = rAB * F1 i 3 = 0 0

j - 0.8660d 0

k 0.50d 3 = { - 30.31d i} N # m 35

(M C)2 = rAB * F2 i = 3 0 -50

j -0.8660d 0

k 0.50d 3 = {- 25.0d j - 43.30d k} N # m 0

Resultant Couple Moment: M R = ©M;

M R = (M C)1 + (M C)2 = {- 30.31d i - 25.0d j - 43.30d k} N # m

The magnitude of M R is 20 N # m, thus 20 = 2(-30.31d)2 + (- 25.0d)2 + (43.30d)2 d = 0.3421 m = 342 mm

Ans.

Ans: d = 342 mm 314

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4–85. Express the moment of the couple acting on the pipe in Cartesian vector form. What is the magnitude of the couple moment? Take F = 125 N.

z O –F

y

150 mm 600 mm A

SOLUTION

B

MC = rAB * (125 k)

200 mm

x 150 mm

MC = (0.2i + 0.3j) * (125 k)

F

MC = {37.5i - 25j} N # m MC = 2(37.5)2 + ( -25)2 = 45.1 N # m

Ans.

Ans: MC = 45.1 N # m 315

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4–86. If the couple moment acting on the pipe has a magnitude of 300 N # m, determine the magnitude F of the forces applied to the wrenches.

z O –F

y

150 mm 600 mm A

SOLUTION

B

MC = rAB * (F k)

200 mm

x 150 mm

= (0.2i + 0.3j) * (F k)

F

= {0.2Fi - 0.3Fj} N # m MC = F 2(0.2F)2 + (- 0.3F) = 0.3606 F 300 = 0.3606 F Ans.

F = 832 N

Ans: F = 832 N 316

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z

4–87. If F = 80 N, determine the magnitude and coordinate direction angles of the couple moment. The pipe assembly lies in the x–y plane.

F

300 mm 300 mm F

x 200 mm

SOLUTION It is easiest to find the couple moment of F by taking the moment of F or –F about point A or B, respectively, Fig. a. Here the position vectors rAB and rBA must be determined first.

200 mm

rAB = (0.3 - 0.2)i + (0.8 - 0.3)j + (0 - 0)k = [0.1i + 0.5j] m rBA = (0.2 - 0.3)i + (0.3 - 0.8)j + (0 - 0)k = [ -0.1i - 0.5j] m The force vectors F and –F can be written as F = {80 k} N and - F = [ -80 k] N Thus, the couple moment of F can be determined from i M c = rAB * F = 3 0.1 0

j 0.5 0

k 0 3 = [40i - 8j] N # m 80

or i Mc = rBA * -F = 3 - 0.1 0

j - 0.5 0

k 0 3 = [40i - 8j] N # m - 80

The magnitude of Mc is given by Mc = 2Mx 2 + My 2 + Mz 2 = 2402 + (- 8)2 + 02 = 40.79 N # m = 40.8 N # m The coordinate angles of Mc are a = cos

-1

b = cos

-1

g = cos

-1

¢ ¢

¢

Mx 40 ≤ = cos ¢ ≤ = 11.3° M 40.79 My M Mz M

-8 ≤ = 101° 40.79

Ans.

0 ≤ = 90° 40.79

Ans.

≤ = cos ¢

≤ = cos ¢

Ans.

317

Ans.

300 mm

y

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*4–88.

z

If the magnitude of the couple moment acting on the pipe assembly is 50 N # m, determine the magnitude of the couple forces applied to each wrench. The pipe assembly lies in the x–y plane.

F

300 mm 300 mm F

x 200 mm

SOLUTION It is easiest to find the couple moment of F by taking the moment of either F or –F about point A or B, respectively, Fig. a. Here the position vectors rAB and rBA must be determined first.

200 mm

300 mm

y

rAB = (0.3 - 0.2)i + (0.8 - 0.3)j + (0 - 0)k = [0.1i + 0.5j] m rBA = (0.2 - 0.3)i + (0.3 - 0.8)j + (0 - 0)k = [ -0.1i - 0.5j] m The force vectors F and –F can be written as F = {Fk} N and - F = [ -Fk]N Thus, the couple moment of F can be determined from i M c = rAB * F = 3 0.1 0

j 0.5 0

k 0 3 = 0.5Fi - 0.1Fj F

The magnitude of Mc is given by Mc = 2Mx 2 + My 2 + Mz 2 = 2(0.5F)2 + (0.1F)2 + 02 = 0.5099F Since Mc is required to equal 50 N # m, 50 = 0.5099F F = 98.1 N

Ans.

Ans: F = 98.1 N 318

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4–89.

z M2 = 20 N · m

The meshed gears are subjected to the couple moments shown. Determine the magnitude of the resultant couple moment and specify its coordinate direction angles.

20° 30°

SOLUTION

y

M 1 = 550k6 N # m M 2 = 201-cos 20° sin 30°i - cos 20° cos 30°j + sin 20°k2 N # m = 5-9.397i - 16.276j + 6.840k6 N # m

x M1 = 50 N · m

Resultant Couple Moment: M R = ©M;

MR = M1 + M2 = 5-9.397i - 16.276j + 150 + 6.8402k6 N # m = 5-9.397i - 16.276j + 56.840k6 N # m

The magnitude of the resultant couple moment is MR = 21-9.39722 + 1-16.27622 + 156.84022 = 59.867 N # m = 59.9 N # m

Ans.

The coordinate direction angles are a = cos-1 a

-9.397 b = 99.0° 59.867

Ans.

b = cos-1 a

-16.276 b = 106° 59.867

Ans.

g = cos-1

56.840 59.867

Ans.

= 18.3°

Ans: M R = 59.9 N # m a = 99.0 b = 106 g = 18.3 319

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4–90. Determine the required magnitude of the couple moments M2 and M3 so that the resultant couple moment is zero.

M2 45

SOLUTION

M3

Since the couple moment is the free vector, it can act at any point without altering its effect. Thus, the couple moments M1, M2, and M3 can be simplified as shown in Fig. a. Since the resultant of M1, M2, and M3 is required to be zero, (MR)y = ©My ;

M1  300 Nm

0 = M2 sin 45° - 300 M2 = 424.26 N # m = 424 N # m

(MR)x = ©Mx ;

Ans.

0 = 424.26 cos 45° - M3 M3 = 300 N # m

Ans.

Ans: M2 = 424 N # m M3 = 300 N # m 320

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4–91.

z

A couple acts on each of the handles of the minidual valve. Determine the magnitude and coordinate direction angles of the resultant couple moment. 35 N 25 N 60

SOLUTION

y

Mx = - 35(0.35) - 25(0.35) cos 60° = - 16.625

175 mm x

My = - 25(0.35) sin 60° = - 7.5777 N # m |M| = 2(- 16.625) + (-7.5777) = 2

2

175 mm 25 N

18.2705 = 18.3 N # m

Ans.

a = cos-1 a

- 16.625 b = 155° 18.2705

Ans.

b = cos-1 a

- 7.5777 b = 115° 18.2705

Ans.

g = cos-1 a

0 b = 90° 18.2705

Ans.

35 N

Ans: M  a = b = g = 321

= 18.3 N # m 155 115 90

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*4–92. Express the moment of the couple acting on the frame in Cartesian vector form. The forces are applied perpendicular to the frame. What is the magnitude of the couple moment? Take F = 50 N.

z

O F

y 3m

30

1.5 m

SOLUTION

x

MC = 80(1.5) = 75 N # m

F

Ans.

MC = - 75(cos 30° i + cos 60° k) = {- 65.0i - 37.5k} N # m

Ans.

Ans: MC = { -65.0i - 37.5k} N # m 322

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4–93. In order to turn over the frame, a couple moment is applied as shown. If the component of this couple moment along the x axis is Mx = 5 - 20i6 N # m, determine the magnitude F of the couple forces.

z

O F

y 3m

SOLUTION MC = F (1.5)

30

1.5 m

Thus

x

F

20 = F (1.5) cos 30° Ans.

F = 15.4 N

Ans: F = 15.4 N 323

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4–94. z

Express the moment of the couple acting on the rod in Cartesian vector form. What is the magnitude of the couple moment?

F  { 4i  3j  4k} kN

A 1m x 2m

Position Vector. The coordinates of points A and B are A (0, 0, 1) m and B (3, 2, -1) m, respectively. Thus,

3m

1m

Solution

y

B F  {– 4i + 3j  4k} kN

rAB = (3 - 0)i + (2 - 0)j + ( -1 - 1)k = {3i + 2j - 2k} m

Couple Moment.

MC = rAB * F



i = † 3 -4



j 2 3

k -2 † -4

= { -2i + 20j + 17k} kN # m

Ans.

The magnitude of MC is

MC = 2(MC)2x + (MC)2y + (MC)2z = 2( - 2)2 + 202 + 172

= 26.32 kN # m = 26.3 kN # m

Ans.

Ans: MC = { - 2i + 20j + 17k} kN # m MC = 26.3 kN # m

324

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4–95. If F1 = 100 N, F2 = 120 N and F3 = 80 N, determine the magnitude and coordinate direction angles of the resultant couple moment.

z

–F4

[ 150 k] N

0.3 m

0.2 m

0.2 m

SOLUTION Couple Moment: The position vectors r1, r2 , r3 , and r4 , Fig. a, must be determined first. r1 = {0.2i} m

r2 = {0.2j} m

0.3 m

0.2 m F1 0.2 m – F1

x – F2 0.2 m

30

F4

[150 k] N

y

F2

– F3

r3 = {0.2j} m

0.2 m

From the geometry of Figs. b and c, we obtain

F3

r4 = 0.3 cos 30° cos 45°i + 0.3 cos 30° sin 45°j - 0.3 sin 30°k = {0.1837i + 0.1837j - 0.15k} m The force vectors F1 , F2 , and F3 are given by F1 = {100k} N

F2 = {120k} N

F3 = {80i} N

Thus, M 1 = r1 * F1 = (0.2i) * (100k) = {- 20j} N # m M 2 = r2 * F2 = (0.2j) * (120k) = {24i} N # m

M 3 = r3 * F3 = (0.2j) * (80i) = { -16k} N # m

M 4 = r4 * F4 = (0.1837i + 0.1837j - 0.15k) * (150k) = {27.56i - 27.56j} N # m

Resultant Moment: The resultant couple moment is given by (M c)R =

M c;

(M c)R = M 1 + M 2 + M 3 + M 4 = ( -20j) + (24i) + ( -16k) + (27.56i -27.56j)

= {51.56i - 47.56j - 16k} N # m The magnitude of the couple moment is (M c)R = 2[(M c)R]x2 + [(M c)R]y 2 + [(M c)R]z 2 = 2(51.56)2 + ( -47.56)2 + ( -16)2 = 71.94 N # m = 71.9 N # m

Ans.

The coordinate angles of (Mc)R are a = cos -1 a

b = cos -1 a

g = cos -1 a

[(Mc)R]x 51.56 b = 44.2° b = cos a (Mc)R 71.94 [(Mc)R]y (Mc)R [(Mc)R]z (Mc)R

b = cos a

b = cos a

Ans.

- 47.56 b = 131° 71.94

Ans.

- 16 b = 103° 71.94

Ans.

325

Ans: (MC)R = 71.9 Ν # m a = 44.2° b = 131° g = 103°

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*4–96.

z

–F4 ⫽ [⫺150 k] N

Determine the required magnitude of F1, F2, and F3 so that the resultant couple moment is (Mc)R = [50 i - 45 j - 20 k] N # m.

0.3 m

0.2 m

0.2 m

Couple Moment: The position vectors r1, r2, r3, and r4, Fig. a, must be determined first. r1 = {0.2i} m

0.3 m

0.2 m

SOLUTION

r2 = {0.2j} m

F1 0.2 m – F1

x

30⬚

F4 ⫽ [150 k] N y

– F2 0.2 m

r3 = {0.2j} m

F2 – F3

From the geometry of Figs. b and c, we obtain

0.2 m

r4 = 0.3 cos 30° cos 45°i + 0.3 cos 30° sin 45°j - 0.3 sin 30°k

F3

= {0.1837i + 0.1837j - 0.15k} m The force vectors F1, F2, and F3 are given by F1 = F1k

F2 = F2k

F3 = F3i

Thus, M 1 = r1 * F1 = (0.2i) * (F1k) = -0.2 F1j M 2 = r2 * F2 = (0.2j) * (F2k) = 0.2 F2i M 3 = r3 * F3 = (0.2j) * (F3i) = -0.2 F3k

M 4 = r4 * F4 = (0.1837i + 0.1837j - 0.15k) * (150k) = {27.56i - 27.56j} N # m Resultant Moment: The resultant couple moment required to equal (M c)R = {50i - 45j - 20k} N # m. Thus, (M c)R = ©M c;

(M c)R = M 1 + M 2 + M 3 + M 4 50i - 45j - 20k = (-0.2F1j) + (0.2F2i) + ( -0.2F3k) + (27.56i - 27.56j) 50i - 45j - 20k = (0.2F2 + 27.56)i + (-0.2F1 - 27.56)j - 0.2F3k

Equating the i, j, and k components yields 50 = 0.2F2 + 27.56

F2 = 112 N

Ans.

-45 = -0.2F1 - 27.56

F1 = 87.2 N

Ans.

-20 = -0.2F3

F3 = 100 N

Ans.

Ans: F2 = 112 N F1 = 87.2 N F3 = 100 N 326

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4–97. Replace the force system by an equivalent resultant force and couple moment at point O.

y 455 N

12 13

5

2m

2.5 m O

x

0.75 m 60

Solution Equivalent Resultant Force And Couple Moment At O. 12 + (FR)x = ΣFx;  (FR)x = 600 cos 60° - 455 a b = - 120 N = 120 N d S 13 + c (FR)y = ΣFy;  (FR)y = 455 a

As indicated in Fig. a And Also,

(FR)y (FR)x

d = tan-1 a

a+(MR)O = ΣMO;  (MR)O = 455 a



P 1m

600 N

5 b - 600 sin 60° = -344.62 N = 344.62 NT 13

FR = 2 (FR)2x + (FR)2y = 21202 + 344.622 = 364.91 N = 365 N u = tan-1 c

0.75 m

344.62 b = 70.80° = 70.8° d 120

Ans.

Ans.

12 b(2) + 600 cos 60° (0.75) + 600 sin 60° (2.5) 13

= 2364.04 N # m

= 2364 N # m (counterclockwise)

Ans.

Ans: FR = 365 N u = 70.8° d

(MR)O = 2364 N # m (counterclockwise)

327

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4–98. Replace the force system by an equivalent resultant force and couple moment at point P.

y 455 N

12 13

5

2m

2.5 m O

x

0.75 m

0.75 m 60

Solution Equivalent Resultant Force And Couple Moment At P. 12 + (FR)x = ΣFx;  (FR)x = 600 cos 60° - 455 a b = -120 N = 120 N d S 13 5 + c (FR)y = ΣFy;  (FR)y = 455 a b - 600 sin 60° = -344.62 N = 344.62 N T 13

P 1m

600 N

As indicated in Fig. a,

FR = 2 (FR)2x + (FR)2y = 21202 + 344.622 = 364.91 N = 365 N

Ans.

And

u = tan-1 c

Also,



(FR)y (FR)x

d = tan-1 a

344.62 b = 70.80° = 70.8° d 120

a+ (MR)P = ΣMP;  (MR)P = 455 a

Ans.

12 5 b(2.75) - 455 a b(1) + 600 sin 60° (3.5) 13 13

= 2798.65 N # m

= 2799 N # m (counterclockwise)

Ans.

Ans: FR = 365 N u = 70.8° d

(MR)P = 2799 N # m (counterclockwise)

328

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4–99. Replace the loading system acting on the post by an equivalent resultant force and couple moment at point A.

650 N 30

500 N

300 N 1500 N  m

60 B

A

3m

5m

2m

Solution Equivalent Resultant Force And Couple Moment at Point A. + (FR)x = ΣFx ;  (FR)x = 650 sin 30° - 500 cos 60° = 75 N S S + c (FR)y = ΣFy ;  (FR)y = -650 cos 30° - 300 - 500 sin 60° = -1295.93 N = 1295.93 NT

As indicated in Fig. a, And Also,

FR = 2(FR)2x + (FR)2y = 2752 + 1295.932 = 1298.10 N = 1.30 kN Ans. u = tan-1 c

(FR)y (FR)x

d = tan-1 a

1295.93 b = 86.69° = 86.7° c 75

Ans.

a+ (MR)A = ΣMA;  (MR)A = 650 cos 30° (3) + 1500 - 500 sin 60° (5)

= 1023.69 N # m

= 1.02 kN # m (counter clockwise)

Ans.

Ans: FR = 1.30 kN u = 86.7° c

(MR)A = 1.02 kN # m (counterclockwise)

329

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*4–100. Replace the loading system acting on the post by an equivalent resultant force and couple moment at point B.

650 N 30

500 N

300 N 1500 N  m

60 B

A

3m

5m

2m

Solution Equivalent Resultant Force And Couple Moment At Point B. + (FR)x = ΣFx;  (FR)x = 650 sin 30° - 500 cos 60° = 75 N S S + c (FR)y = ΣFy;  (FR)y = - 650 cos 30° - 300 - 500 sin 60° = - 1295.93 N = 1295.93 NT As indicated in Fig. a, And Also,

FR = 2 (FR)2x + (FR)2y = 2752 + 1295.932 = 1298.10 N = 1.30 kNAns. u = tan-1 c

(FR)y (FR)x

d = tan-1 a

1295.93 b = 86.69° = 86.7° c 75

Ans.

a+ (MR)B = ΣMB;  (MR)B = 650 cos 30° (10) + 300(7) + 500 sin 60°(2) + 1500

= 10,095.19 N # m

= 10.1 kN # m (counterclockwise)

Ans.

Ans: FR = 1.30 kN u = 86.7° c

(MR)B = 1.01 kN # m (counterclockwise)

330

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4–101. 5 kN

Replace the force system acting on the frame by a resultant force and couple moment at point A.

3

3 kN

2 kN 13

5

1m

4m

4

B

C

1m

12

5

D

5m

SOLUTION Equivalent Resultant Force: Resolving F1, F2, and F3 into their x and y components, Fig. a, and summing these force components algebraically along the x and y axes, we have + ©(F ) = ©F ; : R x x + c (FR)y = ©Fy;

A

4 5 (FR)x = 5 a b - 3 a b = 2.846 kN : 5 13 3 12 (FR)y = - 5a b - 2 - 3 a b = -7.769 kN = 7.769 kN T 5 13

The magnitude of the resultant force FR is given by FR = 2(FR)x2 + (FR)y2 = 22.8462 + 7.7692 = 8.274 kN = 8.27 kN

Ans.

The angle u of FR is u = tan-1 c

(FR)y (FR)x

d = tan-1 c

7.769 d = 69.88° = 69.9° c 2.846

Ans.

Equivalent Couple Moment: Applying the principle of moments and summing the moments of the force components algebraically about point A, we can write a + (MR)A = ©MA;

3 4 12 5 (MR)A = 5 a b(4) - 5 a b(5) - 2(1) - 3a b (2) + 3 a b (5) 5 5 13 13 = -9.768 kN # m = 9.77 kN # m (Clockwise)

331

Ans.

Ans: FR = 8.27 kN u = 69.9 d 1 M R 2 A = 9.77 kN # m (Clockwise)

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4–102. Replace the force system acting on the frame by an equivalent resultant force and couple moment acting at point A.

A 300 N

0.5 m

30

1m

500 N

Solution Equivalent Resultant Force And Couple Moment At A. + (FR)x = ΣFx;    (FR)x = 300 cos 30° + 500 = 759.81 N   S

0.5 m

S

  +c (FR)y = ΣFy;    (FR)y = - 300 sin 30° - 400 = -550 N = 550 N

0.3 m 400 N

T

As indicated in Fig. a,   FR = 2(FR)2x + (FR)2y = 2759.812 + 5502 = 937.98 N = 938 N

Ans.

And

    u = tan-1 c

(FR)y

550 b = 35.90° = 35.9°  c 759.81

Also;

d = tan-1a

a+(MR)A = ΣMA;  

(MR)A = 300 cos 30°(0.5) + 500(1.5) - 400(0.5)

(FR)x

Ans.

= 679.90 N # m

= 680 N # m (counterclockwise)

Ans.

Ans: FR = 938 N u = 35.9° c

(MR)A = 680 N # m (counterclockwise)

332

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4–103.

3 kN 2.5 kN 1.5 kN 30

Replace the force system acting on the beam by an equivalent force and couple moment at point A.

5

3

4

B

A 2m

4m

2m

SOLUTION 4 FRx = 1.5 sin 30° - 2.5a b 5

+ F = ©F ; : Rx x

= - 1.25 kN = 1.25 kN ; 3 FRy = - 1.5 cos 30° - 2.5 a b - 3 5

+ c FRy = ©Fy ;

= - 5.799 kN = 5.799 kN T Thus, FR = 2F 2Rx + F 2Ry = 21.252 + 5.7992 = 5.93 kN

Ans.

and u = tan

a + MRA = ©MA ;

-1

¢

FRy FRx

≤ = tan

-1

a

5.799 b = 77.8° d 1.25

Ans.

3 MRA = -2.5a b (2) - 1.5 cos 30°(6) - 3(8) 5

= -34.8 kN # m = 34.8 kN # m (Clockwise)

Ans.

Ans: FR = 5.93 kN u = 77.8 d a + M R A = 34.8 kN # m (Clockwise) 333

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*4–104.

3 kN 2.5 kN 1.5 kN 30

Replace the force system acting on the beam by an equivalent force and couple moment at point B.

5

3

4

B

A 2m

4m

2m

SOLUTION 4 FRx = 1.5 sin 30° - 2.5a b 5

+ F = ©F ; : Rx x

= -1.25 kN = 1.25 kN ; 3 FRy = -1.5 cos 30° - 2.5a b - 3 5

+ c FRy = ©Fy ;

= -5.799 kN = 5.799 kN T Thus, FR = 2F 2Rx + F 2Ry = 21.252 + 5.7992 = 5.93 kN

Ans.

and u = tan

-1

¢

FRy FRx

a + MRB = ©MRB ;

≤ = tan

-1

a

5.799 b = 77.8° d 1.25

Ans.

3 MB = 1.5cos 30°(2) + 2.5a b (6) 5 = 11.6 kN # m (Counterclockwise)

Ans.

Ans: FR = 5.93 kN u = 77.8° d

a + MRB = 11.6 kN # m (Counterclockwise) 334

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4–105.

y 450 N

Replace the loading system acting on the beam by an equivalent resultant force and couple moment at point O.

30 200 N  m

0.2 m

x

O 1.5 m

2m

1.5 m 200 N

Solution + d FRx = ΣFx ;

FRx = 450 sin 30° = 225.0

+ T FRy = ΣFy ;

FRy = 450 cos 30° - 200 = 189.7

FR = 2(225)2 + (189.7)2 = 294 N

Ans.

u = tan-1 a

Ans.

189.7 b = 40.1° d 225

c+MRO = ΣMO ;  MRO = 450 cos 30° (1.5) - 450 (sin 30°)(0.2) - 200 (3.5) + 200 MRO = 39.6 N # m b

Ans.

Ans: FR = 294 N u = 40.1° d MRO = 39.6 N # m b 335

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4–106. The forces F1 = { - 4i + 2j - 3k} kN and F2 = {3i - 4j 2k} kN act on the end of the beam. Replace these forces by an equivalent force and couple moment acting at point O.

z

F1

150 mm F 150 mm 2

O

250 mm y 4m

Solution

x

FR = F1 + F2 = { -1i - 2j - 5k} kN

Ans.

MRO = r1 * F1 + r2 * F2 i = † 4 -4

j - 0.15 2

k i 0.25 † + † 4 -3 3

j 0.15 -4

k 0.25 † -2

= ( - 0.05i + 11j + 7.4k) + (0.7i + 8.75j - 16.45k) = (0.65i + 19.75j - 9.05k)

MRO = {0.650i + 19.75j - 9.05k} kN # m

Ans.



Ans: MRO = {0.650i + 19.75j - 9.05k} kN # m 336

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4–107. A biomechanical model of the lumbar region of the human trunk is shown. The forces acting in the four muscle groups consist of FR = 35 N for the rectus, FO = 45 N for the oblique, FL = 23 N for the lumbar latissimus dorsi, and FE = 32 N for the erector spinae. These loadings are symmetric with respect to the y–z plane. Replace this system of parallel forces by an equivalent force and couple moment acting at the spine, point O. Express the results in Cartesian vector form.

z

FR

FR

FO FE

FL

FE FO

FL O

75 mm

15 mm 45 mm

50 mm

30 mm 40 mm

x

SOLUTION FR = ©Fz ;

FR = {2(35 + 45 + 23 + 32)k } = {270k} N

Ans.

MROx = ©MOx ;

MR O = [ - 2(35)(0.075) + 2(32)(0.015) + 2(23)(0.045)]i MR O = { - 2.22i} N # m

Ans.

337

Ans: FR = 5270k6 N MRO = 5 -2.22i6 N # m

y

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*4–108. z

Replace the force system by an equivalent resultant force  and couple moment at point O. Take F3 = { - 200i + 500j - 300k} N.

F1 = 300 N

O 2m

x F3

1.5 m

Solution

y

1.5 m

Position And Force Vectors.

F2 = 200 N

  r1 = {2j} m   r2 = {1.5i + 3.5j}   r3 = {1.5i + 2j} m   F1 = { - 300k} N  F2 = {200j} N       F3 = { -200i + 500j - 300k} N Equivalent Resultant Force And Couple Moment At Point O.   FR = ΣF;       FR = F1 + F2 + F3

= ( -300k) + 200j + ( - 200i + 500j - 300k)



= { -200i + 700j - 600k} N

Ans.

  (MR)O = ΣMO;  (MR)O = r1 * F1 + r2 * F2 + r3 * F3

i = 30 0



= ( -600i) + (300k) + ( - 600i + 450j + 1150k)



j 2 0

k i 0 3 + 3 1.5 - 300 0

j 3.5 200

k i 0 3 + 3 1.5 - 200 0

= { - 1200i + 450j + 1450k} N # m

j 2 500

k 0 3 - 300

Ans.

Ans: FR = { - 200i + 700j - 600k} N

(MR)O = { -1200i + 450j + 1450k} N # m

338

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4–109.

z

Replace the loading by an equivalent resultant force and couple moment at point O.

O

x 0.5 m

y

0.7 m F2 = {–2 i + 5 j – 3 k} kN

0.8 m F1  {8 i – 2 k} kN

Solution Position Vectors. The required position vectors are r1 = {0.8i - 1.2k} m  r2 = { -0.5k} m Equivalent Resultant Force And Couple Moment At Point O.   FR = ΣF;       FR = F1 + F2

= (8i - 2k) + ( - 2i + 5j - 3k)



= {6i + 5j - 5k} kN

Ans.

  (MR)O = ΣMO;  (MR)O = r1 * F1 + r2 * F2 i = 3 0.8 8

j 0 0

k i -1.2 3 + 3 0 -2 -2

j 0 5

k -0.5 3 -3

= ( -8j) + (2.5i + j) = {2.5i - 7j} kN # m

Ans.

Ans: FR = {6i + 5j - 5k} kN

(MR)O = {2.5i - 7j} kN # m

339

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4–110.

z

Replace the force of F = 80 N acting on the pipe assembly by an equivalent resultant force and couple moment at point A.

A 400 mm B [

300 mm

y

200 mm

Solution

200 mm

250 mm

FR = ΣF ;

40

FR = 80 cos 30° sin 40° i + 80 cos 30° cos 40° j - 80 sin 30° k

30

= 44.53 i + 53.07 j - 40 k

F

= {44.5 i + 53.1 j - 40 k} N i MRA = ΣMA ;  MRA = † 0.55 44.53

j 0.4 53.07

80 N

Ans.

 k -0.2 † -40

= { - 5.39 i + 13.1 j + 11.4 k} N # m

Ans.

Ans: FR = {44.5 i + 53.1 j + 40 k} N MRA = { - 5.39 i + 13.1 j + 11.4 k} N # m 340

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4–111. The belt passing over the pulley is subjected to forces F1 and F2, each having a magnitude of 40 N. F1 acts in the - k direction. Replace these forces by an equivalent force and couple moment at point A. Express the result in Cartesian vector form. Set u = 0° so that F2 acts in the - j direction.

z 

r  80 mm

y

300 mm A x

F2

SOLUTION

F1

FR = F1 + F2 FR = { -40j - 40 k} N

Ans.

M RA = ©(r * F) i = 3 -0.3 0

j 0 - 40

k i 0.08 3 + 3 - 0.3 0 0

j 0.08 0

k 0 3 -40

MRA = { -12j + 12k} N # m

Ans.

Ans:

FR = 5 -40j - 40k 6 N 341

MRA = 5 -12j + 12k6 N # m

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*4–112. The belt passing over the pulley is subjected to two forces F1 and F2, each having a magnitude of 40 N. F1 acts in the - k direction. Replace these forces by an equivalent force and couple moment at point A. Express the result in Cartesian vector form. Take u = 45°.

z

r

80 mm

y

300 mm A

SOLUTION

x

FR = F1 + F2 = -40 cos 45°j + ( -40 - 40 sin 45°)k

F2

FR = {- 28.3j - 68.3k} N

Ans.

F1

rAF1 = {- 0.3i + 0.08j} m rAF2 = -0.3i - 0.08 sin 45°j + 0.08 cos 45°k = {- 0.3i - 0.0566j + 0.0566k} m MRA = (rAF1 * F1) + (rAF2 * F2) i = 3 - 0.3 0

j 0.08 0

i k 0 3 + 3 -0.3 0 - 40

j - 0.0566 -40 cos 45°

k 0.0566 3 -40 sin 45°

MRA = {- 20.5j + 8.49k} N # m

Ans.

Also, MRAx = ©MAx MRAx = 28.28(0.0566) + 28.28(0.0566) - 40(0.08) MRAx = 0 MRAy = ©MAy MRAy = -28.28(0.3) - 40(0.3) MRAy = -20.5 N # m MRAz = ©MAz MRAz = 28.28(0.3) MRAz = 8.49 N # m MRA = {- 20.5j + 8.49k} N # m

Ans.

Ans:

FR = 5-28.3j - 68.3k6 N 342

MRA = 5-20.5j + 8.49k6 N # m

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4–113.

700 N 450 N

Replace the loading acting on the beam by a single resultant force. Specify where the force acts, measured from end A.

30

300 N

60 B A 2m

4m

3m

3000 N m

SOLUTION + F = ©F ; : Rx x

FRx = 450 cos 60° - 700 sin 30° = - 125 N = 125 N

+ c FRy = ©Fy ;

FRy = - 450 sin 60° - 700 cos 30° - 300 = - 1296 N = 1296 N

; T

F = 2(-125)2 + ( -1296)2 = 1302 N

Ans.

1296 b = 84.5° 125

Ans.

u = tan-1 a

c + MRA = ©MA ;

d

1296(x) = 450 sin 60°(2) + 300(6) + 700 cos 30°(9) + 3000 x = 8.51 m

Ans.

Ans: F = 1302 N u = 84.5 d x = 8.51 m 343

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4–114. Replace the loading acting on the beam by a single resultant force. Specify where the force acts, measured from B.

700 N 450 N

30

300 N

60 B A 2m

4m

3m

3000 N m

SOLUTION + F = ©F ; : Rx x

FRx = 450 cos 60° - 700 sin 30° = - 125 N = 125 N

+ c FRy = ©Fy ;

FRy = - 450 sin 60° - 700 cos 30° - 300 = - 1296 N = 1296 N

F = 2(- 125)2 + ( - 1296)2 = 1302 N u = tan-1 a

T Ans.

1296 b = 84.5° d 125

c + MRB = ©MB ;

;

Ans.

1296(x) = - 450 sin 60°(4) + 700 cos 30°(3) + 3000 x = 2.52 m (to the right)

Ans.

Ans: F = 1302 N u = 84.5 d x = 2.52 m (to the right) 344

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5 kN

4–115. Determine the magnitude and direction u of force F and its placement d on the beam so that the loading system is equivalent to a resultant force of 12 kN acting vertically downward at point A and a clockwise couple moment of 50 kN # m.

3 kN 24

25

F

7

A

d

3m

4m

6m

SOLUTION Initial guesses:

F

T

1 kN

30 °

d

2m

Given: F1

5 kN

a

3m

F2

3 kN

b

4m

FR

12 kN

c

6m

M

50 kN˜ m

e

7

Given

f

§ e · F  F cos  T ¨ 2 2¸ 1 © e f ¹

24

0

§  f · F  F sin  T  F 2 ¨ 2 2¸ 1  f e © ¹

F R

f § · ¨ 2 2 ¸ F1 a  F sin  T ( a  b  d)  F2 ( a  b) © e f ¹

§F· ¨T ¸ ¨ ¸ ©d¹

Find  F  T  d

F

4.427 kN

T

71.565 °

M

d

3.524 m

Ans.

,

Ans: F = 4.427 kN u = 71.565 d = 3.524 m 345

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5 kN

*4–116. Determine the magnitude and direction u of force F and its placement d on the beam so that the loading system is equivalent to a resultant force of 10 kN acting vertically downward at point A and a clockwise couple moment of 45 kN # m.

3 kN 24

25

F

7

A

d

3m

4m

6m

SOLUTION Initial guesses:

F

1 kN

T

30 °

d

1m

Given: F1

5 kN

a

3m

F2

3 kN

b

4m

FR

10 kN

c

6m

M

45 kN˜ m e f

Given

7 24

§ e · F  F cos  T ¨ 2 2¸ 1 © e f ¹

0

§  f · F  F sin  T  F 2 ¨ 2 2¸ 1 © e f ¹

F R

f § · ¨ 2 2 ¸ F1 a  F sin  T ( a  b  d)  F2 ( a  b) © e f ¹

§F· ¨T ¸ ¨ ¸ ©d¹

Find  F  T  d

F

2.608 kN

T

57.529°

M

d

2.636 m

Ans.

Ans:

F

T d 346

2.608 kN 57.529  2.636 m

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4–117. Replace the force system acting on the post by a resultant force, and specify where its line of action intersects the post AB measured from point A.

0.5 m B

1m

500 N 0.2 m

30

5

3

4

250 N

1m

SOLUTION

300 N

Equivalent Resultant Force: Forces F1 and F2 are resolved into their x and y components, Fig. a. Summing these force components algebraically along the x and y axes, + (F ) = ©F ; : R x x

4 (FR)x = 250 a b - 500 cos 30° - 300 = - 533.01 N = 533.01 N ; 5

+ c (FR)y = ©Fy;

3 (FR)y = 500 sin 30° - 250 a b = 100 N c 5

1m A

The magnitude of the resultant force FR is given by FR = 2(FR)x 2 + (FR)y 2 = 2533.012 + 1002 = 542.31 N = 542 N

Ans.

The angle u of FR is u = tan

-1

B

(FR)y (FR)x

R = tan

-1

c

100 d = 10.63° = 10.6° b 533.01

Ans.

Location of the Resultant Force: Applying the principle of moments, Figs. a and b, and summing the moments of the force components algebraically about point A, a +(MR)A = ©MA;

4 3 533.01(d) = 500 cos 30°(2) - 500 sin 30°(0.2) - 250 a b(0.5) - 250 a b(3) + 300(1) 5 5 d = 0.8274 mm = 827 mm

Ans.

Ans: FR = 542 N u = 10.6° b d = 0.827 m 347

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4–118. Replace the force system acting on the post by a resultant force, and specify where its line of action intersects the post AB measured from point B.

0.5 m B 1m

500 N 0.2 m

30

5

3

4

250 N

1m

SOLUTION

300 N

Equivalent Resultant Force: Forces F1 and F2 are resolved into their x and y components, Fig. a. Summing these force components algebraically along the x and y axes,

1m A

4 (FR)x = 250 a b - 500 cos 30° - 300 = -533.01N = 533.01 N ; 5

+ ©(F ) = ©F ; : R x x

3 (FR)y = 500 sin 30° - 250 a b = 100 N c 5

+ c (FR)y = ©Fy;

The magnitude of the resultant force FR is given by FR = 2(FR)x 2 + (FR)y 2 = 2533.012 + 1002 = 542.31 N = 542 N

Ans.

The angle u of FR is u = tan

-1

B

(FR)y (FR)x

R = tan

-1

c

100 d = 10.63° = 10.6° b 533.01

Ans.

Location of the Resultant Force: Applying the principle of moments, Figs. a and b, and summing the moments of the force components algebraically about point B, a +(MR)B = ©Mb;

3 -533.01(d) = -500 cos 30°(1) - 500 sin 30°(0.2) - 250 a b (0.5) - 300(2) 5 d = 2.17 m

Ans.

Ans: FR = 542 N u = 10.6° b d = 2.17 m 348

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4–119. y

Replace the loading on the frame by a single resultant force. Specify where its line of action intersects a vertical line along member AB, measured from A.

1m 600 N

0.5 m B 0.5 m

400 N

3

1.5 m

4

5

900 N

Solution

1m

S

3 4    + c (FR)y = ΣFy;   (FR)y = 600 + 400 a b - 400 - 900 a b 5 5 = - 280 N = 280 N T

As indicated in Fig. a,   FR = 2(FR)2x + (FR)2y = 22202 + 2802 = 356.09 N = 356 N

Ans.

And

u = tan-1 c

(FR)y (FR)x

d = tan-1a

280 b = 51.84° = 51.8° 220

Ans.

Location of Resultant Force. Referring to Fig. a

3   a+ (MR)A = ΣMA;   280 a - 220 b = 400(1.5) - 600(0.5) - 900 a b(2.5) 5 4 + 400 a b(1) 5



400 N x

+ (FR)x = ΣFx;     (FR)x = 900 a 3 b - 400 a 4 b = 220 N    S 5 5



5 4

A

Equivalent Resultant Force. Referring to Fig. a



3

(1)

220 b - 280 a = 730

349

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4–119. Continued

Along AB, a = 0. Then Eq (1) becomes 220 b - 280(0) = 730 b = 3.318 m Thus, the intersection point of line of action of FR on AB measured upward from point A is

Ans.

d = b = 3.32 m

Ans: FR = 356 N u = 51.8° d = b = 3.32 m 350

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*4–120. y

Replace the loading on the frame by a single resultant force. Specify where its line of action intersects a horizontal line along member CB, measured from end C.

1m 600 N

0.5 m B 0.5 m

400 N

3

1.5 m

4

5

900 N

Solution

1m

S

3 4    + c (FR)y = ΣFy;         (FR)y = 600 + 400 a b - 400 - 900 a b 5 5 = -280 N = 280 N T

As indicated in Fig. a,   FR = 2(FR)2x + (FR)2y = 22202 + 2802 = 356.09 N = 356 N

Ans.

And

u = tan-1 c

(FR)y (FR)x

d = tan-1a

280 b = 51.84° = 51.8° 220

Ans.

Location of Resultant Force. Referring to Fig. a

3    a+ (MR)A = ΣMA;   280 a - 220 b = 400(1.5) - 600(0.5) - 900 a b(2.5) 5 4 + 400 a b(1) 5



400 N x

+ (FR)x = ΣFx;   (FR)x = 900 a 3 b - 400 a 4 b = 220 N    S 5 5



5 4

A

Equivalent Resultant Force. Referring to Fig. a



3

(1)

220 b - 280 a = 730

351

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*4–120. Continued

Along BC, b = 3 m. Then Eq (1) becomes 220(3) - 280 a = 730 a = -0.25 m Thus, the intersection point of line of action of FR on CB measured to the right of point C is

Ans.

d = 1.5 - ( - 0.25) = 1.75 m

Ans: FR = 356 N u = 51.8° d = 1.75 m 352

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4–121. Replace the loading acting on the beam by a single resultant force. Specify where the force acts, measured from end A.

700 N 450 N

30

300 N

60 B A 2m

4m

3m

1500 N m

SOLUTION + F = ©F ; : Rx x

FRx = 450 cos 60° - 700 sin 30° = - 125 N = 125 N

+ c FRy = ©Fy ;

FRy = - 450 sin 60° - 700 cos 30° - 300 = - 1296 N = 1296 N

; T

F = 2(-125)2 + ( -1296)2 = 1302 N

Ans.

1296 b = 84.5° 125

Ans.

u = tan-1 a

c + MRA = ©MA ;

d

1296(x) = 450 sin 60°(2) + 300(6) + 700 cos 30°(9) + 1500 x = 7.36 m

Ans.

Ans: F = 1302 N u = 84.5° d x = 7.36 m 353

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4–122. Replace the loading acting on the beam by a single resultant force. Specify where the force acts, measured from B.

700 N 450 N

30

300 N

60 B A 2m

4m

3m

1500 N m

SOLUTION + F = ©F ; : Rx x

FRx = 450 cos 60° - 700 sin 30° = - 125 N = 125 N

+ c FRy = ©Fy ;

FRy = - 450 sin 60° - 700 cos 30° - 300 = - 1296 N = 1296 N

F = 2(- 125)2 + ( -1296)2 = 1302 N u = tan-1 a

T Ans.

1296 b = 84.5° d 125

c + MRB = ©MB ;

;

Ans.

1296(x) = - 450 sin 60°(4) + 700 cos 30°(3) + 1500 x = 1.36 m (to the right)

Ans.

Ans: F = 1302 N u = 84.5° d x = 1.36 m (to the right) 354

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4–123. Replace the loading on the frame by a single resultant force. Specify where its line of action intersects a vertical line along member AB, measured from A.

400 N

200 N 0.5 m

200 N 0.5 m 600 N

B C

1.5 m

Solution A

Equivalent Resultant Force. Referring to Fig. a, + (FR)x = ΣFx;          S

(FR)x = 600 N S

   + c (FR)y = ΣFy;        (FR)y = -200 - 400 - 200 = -800 N = 800 NT As indicated in Fig. a,   FR = 2(FR)2x + (FR)2y = 26002 + 8002 = 1000 N

Ans.

And

u = tan-1 c

(FR)y (FR)x

d = tan-1a

800 b = 53.13° = 53.1°  c 600

Ans.

Location of Resultant Force. Along AB,

  a+ (MR)B = ΣMB;   600(1.5 - d) = -400(0.5) - 200(1)

Ans.

d = 2.1667 m = 2.17 m

Ans: FR = 1000 N u = 53.1° c d = 2.17 m 355

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*4–124. Replace the parallel force system acting on the plate by a resultant force and specify its location on the x–z plane.

z 0.5 m 1m

SOLUTION

2 kN 5 kN

Resultant Force: Summing the forces acting on the plate, (FR)y = ©Fy;

1m

FR = - 5 kN - 2 kN - 3 kN Ans.

= - 10 kN The negative sign indicates that FR acts along the negative y axis.

0.5 m

Resultant Moment: Using the right-hand rule, and equating the moment of FR to the sum of the moments of the force system about the x and z axes, (MR)x = ©Mx;

y 3 kN

x

(10 kN)(z) = (3 kN)(0.5 m) + (5 kN)(1.5 m) + 2 kN(2.5 m) Ans.

z = 1.40 m (MR)z = ©Mz;

1m

-(10 kN)(x) = - (5 kN)(0.5 m) - (2 kN)(1.5 m) - (3 kN)(1.5 m) Ans.

x = 1.00 m

Ans: FR = -10 kN 356

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300 N

4–125.

250 N

Replace the loading on the frame by a single resultant force. Specify where its line of action intersects member AB, measured from A.

1m C

2m

3m

5

B

3

4

D 2m

400 N m 60

SOLUTION + ©F = F ; : x Rx

3m

500 N

4 FRx = -250a b - 500(cos 60°) = -450 N = 450 N ; 5 A

+ c ©Fy = ©Fy ;

FRy

3 = -300 - 250a b - 500 sin 60° = -883.0127 N = 883.0127 N T 5

FR = 2(- 450)2 + (- 883.0127)2 = 991 N u = tan-1 a

Ans.

883.0127 b = 63.0° d 450

a + MRA = ©MA ;

3 4 450y = 400 + (500 cos 60°)(3) + 250a b(5) - 300(2) - 250a b(5) 5 5 y =

800 = 1.78 m 450

Ans.

Ans: FR = 991 N y = 1.78 m 357

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4–126.

300 N 250 N

Replace the loading on the frame by a single resultant force. Specify where its line of action intersects member CD, measured from end C.

1m C

2m

2m

60 3m

500 N

FRx

4 = -250a b - 500(cos 60°) = -450 N = 450 N ; 5

FRy

3 = -300 - 250a b - 500 sin 60° = -883.0127 N = 883.0127 N T 5

FR = 2(- 450)2 + (- 883.0127)2 = 991 N u = tan-1 a

3

4

D

SOLUTION

+ c ©Fy = ©Fy ;

5

B

400 N m

+ ©F = F ; : x Rx

3m

A

Ans.

883.0127 b = 63.0° d 450

c + MRA = ©MC ;

3 883.0127x = - 400 + 300(3) + 250 a b(6) + 500 cos 60°(2) + (500 sin 60°)(1) 5 x =

2333 = 2.64 m 883.0127

Ans.

Ans: FR = 991 N u = 63.0 d x = 2.64 m 358

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100 N · m

4–127. Replace the two wrenches and the force, acting on the pipe assembly, by an equivalent resultant force and couple moment at point O.

300 N z

C

SOLUTION

O 0.5 m

Force And Moment Vectors:

A 0.6 m

B 0.8 m

100 N

x

F1 = 5300k6 N

45°

F3 = 5100j6 N

F2 = 2005cos 45°i - sin 45°k6 N

200 N 180 N · m

= 5141.42i - 141.42k6 N M 1 = 5100k6 N # m M 2 = 1805cos 45°i - sin 45°k6 N # m = 5127.28i - 127.28k6 N # m Equivalent Force and Couple Moment At Point O: FR = ©F;

FR = F1 + F2 + F3 = 141.42i + 100.0j + 1300 - 141.422k = 5141i + 100j + 159k6 N

Ans.

The position vectors are r1 = 50.5j6 m and r2 = 51.1j6 m. M RO = ©M O ;

M RO = r1 * F1 + r2 * F2 + M 1 + M 2 i = 30 0

j 0.5 0

i 0 + 141.42

k 0 3 300 j 1.1 0

k 0 -141.42

+ 100k + 127.28i - 127.28k =

122i - 183k N # m

Ans.

359

Ans: FR = 5 141i + 100j + 159k 6 N MR O = 5 122i - 183k 6 N # m

y

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*4–128. z

Replace the force system by a wrench and specify the magnitude of the force and couple moment of the wrench and the point where the wrench intersects the x–z plane.

200 N 400 N 200 N 4

3m

2m x

Resultant Force. Referring to Fig. a 3 4 FR = e c 200 a b - 400 d i - 200j + 200 a bk f 5 5 = { - 280i - 200j + 160k} N

The magnitude of FR is   FR = 2 ( - 280 ) 2 +

( - 200 ) 2 + 1602 = 379.47 N = 379 N

Ans.

The direction of FR is defined by uFR =

- 280i - 200j + 160k FR = -0.7379i - 0.5270j + 0.4216k = FR 379.47

Resultant Moment. The line of action of MR of the wrench is parallel to that of FR. Also, assume that MR and FR have the same sense. Then uMR = -0.7379i - 0.5270j + 0.4216k

360

y

5 3

Solution

0.5 m

O

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*4–128. Continued

Referring to Fig. a, where the origin of the x′, y′, z′ axes is the point where the wrench intersects the xz plane, (MR)x' = ΣMx'; - 0.7379 MR = - 200(z - 0.5)

 (1)

2 4 (MR)y' = ΣMy'; -0.5270 MR = - 200 a b(z - 0.5) - 200 a b(3 - x) + 400(z - 0.5)(2) 5 5

(MR)z' = ΣMz'; 0.4216 MR = 200 x + 400(2)

 (3)

Solving Eqs (1), (2) and (3) MR = 590.29 N # m = 590 N # m

Ans.

z = 2.6778 m = 2.68 m



Ans.

x = - 2.7556 m = -2.76 m



Ans.

Ans: FR = 379 N

MR = 590 N # m z = 2.68 m x = - 2.76 m

361

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4–129. The tube supports the four parallel forces. Determine the magnitudes of forces FC and FD acting at C and D so that the equivalent resultant force of the force system acts through the midpoint O of the tube.

FD

z

600 N D FC

A 400 mm

SOLUTION Since the resultant force passes through point O, the resultant moment components about x and y axes are both zero. ©Mx = 0;

500 N C

400 mm x

z B

200 mm 200 mm y

FD(0.4) + 600(0.4) - FC(0.4) - 500(0.4) = 0 FC - FD = 100

©My = 0;

O

(1)

500(0.2) + 600(0.2) - FC(0.2) - FD(0.2) = 0 FC + FD = 1100

(2)

Solving Eqs. (1) and (2) yields: FC = 600 N

FD = 500 N

Ans.

Ans: FC = 600 N FD = 500 N 362

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4–130. If FA = 7 kN and FB = 5 kN, represent the force system acting on the corbels by a resultant force, and specify its location on the x–y plane.

z 150 mm

6 kN

100 mm

FB 750 mm FA

650 mm

SOLUTION

x

8kN 700 mm

O 100 mm 600 mm

150 mm

y

Equivalent Resultant Force: By equating the sum of the forces in Fig. a along the z axis to the resultant force FR, Fig. b, + c FR = ©Fz;

- FR = -6 - 5 - 7- 8 FR = 26 kN

Ans.

Point of Application: By equating the moment of the forces shown in Fig. a and FR, Fig. b, about the x and y axes, (MR)x = ©Mx;

-26(y) = 6(650) + 5(750) - 7(600) - 8(700) y = 82.7 mm

(MR)y = ©My;

Ans.

26(x) = 6(100) + 7(150) - 5(150) - 8(100) x = 3.85 mm

Ans.

Ans: FR = 26 kN y = 82.7 mm x = 3.85 mm 363

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4–131. Determine the magnitudes of FA and FB so that the resultant force passes through point O of the column.

z 150 mm

6 kN

100 mm 650 mm

SOLUTION

x

FB 750 mm O

FA

8kN 700 mm 100 mm

600 mm

150 mm

y

Equivalent Resultant Force: By equating the sum of the forces in Fig. a along the z axis to the resultant force FR, Fig. b, + c FR = ©Fz;

- FR = - FA - FB - 8 - 6 (1)

FR = FA + FB + 14

Point of Application: Since FR is required to pass through point O, the moment of FR about the x and y axes are equal to zero. Thus, (MR)x = ©Mx;

0 = FB (750) + 6(650) - FA (600) - 8(700) (2)

750FB - 600FA - 1700 = 0 (MR)y = ©My;

0 = FA (150) + 6(100) - FB (150) - 8(100) (3)

159FA - 150FB + 200 = 0 Solving Eqs. (1) through (3) yields FA = 18.0 kN

FB = 16.7 kN

FR = 48.7 kN

Ans.

Ans: FA = 18.0 kN FB = 16.7 kN FR = 48.7 kN 364

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*4–132. If FA = 40 kN and FB = 35 kN, determine the magnitude of the resultant force and specify the location of its point of application (x, y) on the slab.

z 30 kN FB

0.75 m 2.5 m

90 kN 20 kN

2.5 m 0.75 m FA

0.75 m

SOLUTION

x

Equivalent Resultant Force: By equating the sum of the forces along the z axis to the resultant force FR, Fig. b, + c FR = ©Fz;

y

3m 3m 0.75 m

- FR = - 30 - 20 - 90 - 35 - 40 FR = 215 kN

Ans.

Point of Application: By equating the moment of the forces and FR, about the x and y axes, (MR)x = ©Mx;

- 215(y) = - 35(0.75) - 30(0.75) - 90(3.75) - 20(6.75) - 40(6.75) y = 3.68 m

(MR)y = ©My;

Ans.

215(x) = 30(0.75) + 20(0.75) + 90(3.25) + 35(5.75) + 40(5.75) x = 3.54 m

Ans.

Ans: FR = 215 kN y = 3.68 m x = 3.54 m 365

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4–133. If the resultant force is required to act at the center of the slab, determine the magnitude of the column loadings FA and FB and the magnitude of the resultant force.

z 30 kN FB

0.75 m 2.5 m

90 kN 20 kN

2.5 m 0.75 m FA

0.75 m x

SOLUTION

3m

Equivalent Resultant Force: By equating the sum of the forces along the z axis to the resultant force FR, + c FR = ©Fz;

y

3m 0.75 m

- FR = - 30 - 20 - 90 - FA - FB FR = 140 + FA + FB

(1)

Point of Application: By equating the moment of the forces and FR, about the x and y axes, (MR)x = ©Mx;

- FR(3.75) = - FB(0.75) - 30(0.75) - 90(3.75) - 20(6.75) - FA(6.75) FR = 0.2FB + 1.8FA + 132

(MR)y = ©My;

(2)

FR(3.25) = 30(0.75) + 20(0.75) + 90(3.25) + FA(5.75) + FB(5.75) FR = 1.769FA + 1.769FB + 101.54

(3)

Solving Eqs.(1) through (3) yields FA = 30 kN

FB = 20 kN

FR = 190 kN

Ans.

Ans: FA = 30 kN FB = 20 kN FR = 190 kN 366

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4–134. z

The building slab is subjected to four parallel column loadings. Determine the equivalent resultant force and specify its location (x, y) on the slab. Take F1 = 8 kN and F2 = 9 kN.

12 kN F1

F2

6 kN

x

8m 16 m

12 m

Solution

6m

y

4m

Equivalent Resultant Force. Sum the forces along z axis by referring to Fig. a      + c (FR)z = ΣFz;    -FR = - 8 - 6 - 12 - 9  FR = 35 kN Ans. Location of the Resultant Force. Sum the moments about the x and y axes by referring to Fig. a, (MR)x = ΣMx;     -35 y = - 12(8) - 6(20) - 9(20) Ans.

y = 11.31 m = 11.3 m (MR)y = ΣMy;   

  35 x = 12(6) + 8(22) + 6(26) Ans.

       x = 11.54 m = 11.5 m

Ans: FR = 35 kN y = 11.3 m x = 11.5 m 367

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4–135. z

The building slab is subjected to four parallel column loadings. Determine F1 and F2 if the resultant force acts through point (12 m, 10 m).

12 kN F1

F2

6 kN

x

8m 16 m

12 m

Solution

6m

y

4m

Equivalent Resultant Force. Sum the forces along z axis by referring to Fig. a, + c (FR)z = ΣFz;    - FR = -F1 - F2 - 12 - 6  FR = F1 + F2 + 18 Location of the Resultant Force. Sum the moments about the x and y axes by referring to Fig. a,   (MR)x = ΣMx;    - (F1 + F2 + 18)(10) = -12(8) - 6(20) - F2(20)

(1)

10F1 - 10F2 = 36

  (MR)y = ΣMy;   (F1 + F2 + 18)(12) = 12(6) + 6(26) + F1(22)

(2)

12F2 - 10F1 = 12

Solving Eqs (1) and (2),

Ans.

F1 = 27.6 kN  F2 = 24.0 kN

Ans: F1 = 27.6 kN F2 = 24.0 kN 368

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*4–136. Replace the five forces acting on the plate by a wrench. Specify the magnitude of the force and couple moment for the wrench and the point P(x, z) where the wrench intersects the x–z plane.

z

800 N 4m

2m 4m

2m

200 N 400 N

Solution x

Resultant Force. Referring to Fig. a FR = { - 600i - (300 + 200 + 400)j - 800k} N = 5-600i - 900j - 800k6 N

Then the magnitude of FR is

FR = 2( - 600)2 + ( - 900)2 + ( - 800)2 = 1345.36 N = 1.35 kN

Ans.

The direction of FR is defined by uFR =

- 600i - 900j - 800k FR = = -0.4460i - 0.6690j - 0.5946k FR 1345.36

Resultant Moment. The line of action of MR of the wrench is parallel to that of FR. Also, assume that both MR and FR have the same sense. Then uMR = -0.4460i - 0.6690j - 0.5946k

369

600 N

300 N

y

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*4–136. Continued

Referring to Fig. a, (MR)x′ = ΣMx′;

-0.4460 MR = - 300z - 200(z - 2) - 400z(1)

(MR)y′ = ΣMy′;

-0.6690 MR = 800(4 - x) + 600z(2)

(MR)z′ = ΣMz′;

-0.5946 MR = 200(x - 2) + 400x - 300(4 - x)(3)

Solving Eqs (1), (2) and (3) MR = - 1367.66 N # m = - 1.37 kN # m

Ans.

x = 2.681 m = 2.68 m



Ans.

z = - 0.2333 m = -0.233 m



Ans.

The negative sign indicates that the line of action of MR is directed in the opposite sense to that of FR.

Ans: MR = -1.37 kN # m x = 2.68 m z = - 0.233 m 370

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4–137. z

Replace the three forces acting on the plate by a wrench. Specify the magnitude of the force and couple moment for the wrench and the point P(x, y) where the wrench intersects the plate.

FB  { 300k} N FC  {200j} N C

y

B x

Solution

5m

Resultant Force. Referring to Fig. a, FA  {400i} N

Then, the magnitude of FR is FR = 24002 + 2002 + ( - 300)2 = 538.52 N = 539 N

Ans.

The direction of FR is defined by uFR =

3m A

FR = {400i + 200j - 300k} N



x P

400i + 200j - 300k FR = = 0.7428i + 0.3714j - 0.5571k FR 538.52

Resultant Moment. The line of action of MR of the wrench is parallel to that of FR. Also, assume that both MR and FR have the same sense. Then uMR = 0.7428i + 0.3714j - 0.5571k

371

y

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4–137. Continued

Referring to Fig. a, (MR)x′ = ΣMx′; 0.7428 MR = 300y(1) (MR)y′ = ΣMy′; 0.3714 MR = 300 (3 - x)(2) (MR)z′ = ΣMz; -0.5571 MR = - 200x - 400 (5 - y)(3) Solving Eqs (1), (2) and (3) MR = 1448.42 N # m = 1.45 kN # m

Ans.

x = 1.2069 m = 1.21 m



Ans.

y = 3.5862 m = 3.59 m



Ans.

Ans: FR = 539 N

MR = 1.45 kN # m x = 1.21 m y = 3.59 m

372

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4–138. Replace the loading by an equivalent resultant force and couple moment acting at point O.

8 kN/m 5 kN/m O

1.5 m

0.75 m

0.75 m

Solution Equivalent Resultant Force And Couple Moment About Point O. Summing the forces along the y axis by referring to Fig. a, 1 1 +c (FR)y = ΣFy;  FR = - (3)(1.5) - 5(2.25) - (5)(0.75) 2 2 = -15.375 kN = 15.4 kN T

Ans.

Summing the Moment about point O, 1 a + (MR)O = ΣMO;  (MR)O = - (3)(1.5)(0.5) - 5(2.25)(1.125) 2 1 - (5)(0.75)(2.5) 2

= -18.46875 kN # m = 18.5 kN # m (clockwise) Ans.

Ans: FR = 15.4 kN

(MR)O = 18.5 kN # m (clockwise)

373

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4–139. Replace this loading by an equivalent resultant force and specify its location, measured from point O.

6 kN/m 4 kN/m

O 2m

1.5 m

Solution Equivalent Resultant Force. Summing the forces along the y axis by referring to Fig. a, +c (FR)y = ΣFy;

-FR = - 4(2) -

1 (6)(1.5) 2 Ans.

FR = 12.5 kN Location of the Resultant Force. Summing the Moment about point O, a + (MR)O = ΣMO;

- 12.5(d) = -4(2)(1) -

1 (6)(1.5)(2.5) 2 Ans.

d = 1.54 m

Ans: FR = 12.5 kN d = 1.54 m 374

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*4–140. The loading on the bookshelf is distributed as shown. Determine the magnitude of the equivalent resultant location, measured from point O. 30 N/ m

50 N/ m O

A 1m 1.5 m

0.5 m

Solution Given: w1 = 30 N>m w2 = 50 N>m a = 1m b = 1.5 m c = 0.5 m Guesses R = 1 N

d = 1m

Given w1 # b + w2 # c = R w1 # b # aa -

b c b - w2 # c # a + b - ab = - d # R 2 2

R a b: = Find1R, d2 d

R = 70 N

d = 0.107 m

Ans:

5 G 375

70 1 0.107 P

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4–141. w

Replace the loading by an equivalent resultant force and specify its location on the beam, measured from point A.

5 kN/m 2 kN/m A

x

B 4m

2m

Solution Equivalent Resultant Force. Summing the forces along the y axis by referring to Fig. a +c (FR)y = ΣFy;

- FR = - 2(6) -

1 (3)(6) 2

Ans. Ans.

FR = 21.0 kNT Location of the Resultant Force. Summing the moments about point A, 1 a + (MR)A = ΣMA; - 21.0(d) = -2(6)(3) - (3)(6)(4) 2 d = 3.429 m = 3.43 m

Ans.

Ans: FR = 21.0 kN d = 3.43 m 376

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4–142. Replace the distributed loading with an equivalent resultant force, and specify its location on the beam measured from point O.

3 kN/m

O

3m

SOLUTION

1.5 m

Loading: The distributed loading can be divided into two parts as shown in Fig. a. Equations of Equilibrium: Equating the forces along the y axis of Figs. a and b, we have + T FR = ©F;

FR =

1 1 (3)(3) + (3)(1.5) = 6.75 kN T 2 2

Ans.

If we equate the moment of FR, Fig. b, to the sum of the moment of the forces in Fig. a about point O, we have a + (MR)O = ©MO;

1 1 (3)(3)(2) - (3)(1.5)(3.5) 2 2 x = 2.5 m

- 6.75(x) = -

Ans.

Ans: FR = 6.75 kN x = 2.5 m 377

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4–143. O

The masonry support creates the loading distribution acting on the end of the beam. Simplify this load to a single resultant force and specify its location measured from point O.

0.3 m

1 kN/m 2.5 kN/m

SOLUTION Equivalent Resultant Force: + c FR = ©Fy ;

FR = 1(0.3) +

1 (2.5 - 1)(0.3) = 0.525 kN c 2

Ans.

Location of Equivalent Resultant Force: a + 1MR2O = ©MO ;

0.5251d2 = 0.30010.152 + 0.22510.22 d = 0.171 m

Ans.

Ans: FR = 0.525 kN c d = 0.171 m 378

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*4–144. Determine the length b of the triangular load and its position a on the beam such that the equivalent resultant force is zero and the resultant couple moment is 8 kN # m clockwise.

a

b

4 kN/m

A

2.5 kN/m 9m

Solution + c FR = 0 = ΣFy ;  0 = a+ MRA = ΣMA; 

1 1 (2.5)(9) - (4)(b)  b = 5.625 m 2 2

1 1 2   -8 = - (2.5)(9)(6) + (4)(5.625) aa + (5.625) b 2 2 3 a = 1.54 m

Ans.

Ans.

Ans: b = 5.625 m a = 1.54 m 379

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4–145. 6 kN/m

Replace the distributed loading by an equivalent resultant force and couple moment acting at point A.

6 kN/m 3 kN/m

A

B 3m

3m

Solution Equivalent Resultant Force And Couple Moment About Point A. Summing the forces along the y axis by referring to Fig. a, 1 1 + c (FR)y = ΣFy;  FR = - (3)(3) - 3(6) - (3)(3) 2 2

Ans.

= - 27.0 kN = 27.0 kN T 

Summing the moments about point A, 1 1 a+ (MR)A = ΣMA;  (MR)A = - (3)(3)(1) - 3(6)(3) - (3)(3)(5) 2 2

= - 81.0 kN # m = 81.0 kN # m (clockwise)

Ans.

Ans: FR = 27.0 kN

(MR)A = 81.0 kN # m (clockwise)

380

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1 kN/m

4–146. Replace the loading by an equivalent resultant force and couple moment at point A.

1 kN/m B 1.2 m

1.8 m

2 kN/m 60 A

F3 = 1.2 kN

F1 = 0.9 kN F2 = 1.8 kN 0.9 m

Solution F1 =

0.6 m

FRx = 2.338 kN

1 (1.8) (1) = 0.9 kN 2

F2 = (1.8) (1) = 1.8 kN F3 = (1.2) (1) = 1.2 kN = ΣFx;

+↓FRy = ΣFy; FR =

FRy = 2.55 kN

FRy = 0.9 cos 60° + 1.8 cos 60° + 1.2 = 2.55 kN

(2.338) 2 + (2.55) 2 = 3.460 kN

⎛ 2.55 ⎞ ␪ = tan–1 ⎜ = 47.5° ⎝ 2.338 ⎟⎠

哭

+ MRA = ΣMA;

(1.8 cos 60° + 0.6) m

FRx = 0.9 sin 60° + 1.8 sin 60° = 2.338 kN

Ans.

Ans.

MRA = 0.9 (0.6) + 1.8 (0.9) + 1.2 (1.8 cos 60° + 0.6) = 3.96 kN · m

哭

+ → FRx

Ans.

Ans: FR = 3.460 kN ␪ = 47.5° M RA = 3.96 kN # m b 381

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4–147. Replace the distributed loading by an equivalent resultant force, and specify its location on the beam, measured from the pin at A.

4 kN/m

2 kN/m

A

Solution

B

3m

3m

Equivalent Resultant Force. Summing the forces along the y axis by referring to Fig. a, +c (FR)y = ΣFy;

-FR = -2(6) -

1 (2)(3) 2 Ans.

FR = 15.0 kN T Location of the Resultant Force. Summing the Moments about point A, a + (MR)A = ΣMA;

- 15.0(d) = -2(6)(3) -

1 (2)(3)(5) 2 Ans.

d = 3.40 m

Ans: FR = 15.0 kN d = 3.40 m 382

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*4–148. Determine the length b of the triangular load and its position a on the beam such that the equivalent resultant force is zero and the resultant couple moment is 8 kN # m clockwise.

a

b 6 kN/m

A

2 kN/m

4m

Solution Equivalent Resultant Force And Couple Moment At Point A. Summing the forces along the y axis by referring to Fig. a, with the requirement that FR = 0, + c (FR)y = ΣFy;  0 = 2(a + b) -

1 (6)(b) 2

2a - b = 0(1)



Summing the moments about point A, with the requirement that (MR)A = 8 kN # m,



a+ (MR)A = ΣMA;  -8 = 2(a + b) c4 -

1 1 1 (a + b)d - (6)(b) a4 - bb 2 2 3

- 8 = 8a - 4b - 2ab -a2(2)

Solving Eqs (1) and (2),

a = 1.264 m = 1.26 m

Ans.



b = 2.530 m = 2.53 m 

Ans.

Ans: a = 1.26 m b = 2.53 m 383

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4–149. Replace the distributed loading by an equivalent resultant force and specify where its line of action intersects a horizontal line along member AB, measured from A.

3 kN/m B

A 3m

2 kN/m

4m

Solution Equivalent Resultant Force. Summing the forces along the x and y axes by referring to Fig. a, + (FR)x = ΣFx;  (FR)x = -2(4) = -8 kN = 8 kN d S + c (FR)y = ΣFy;  (FR)y = - 3(3) = - 9 kN = 9 kN T Then And

FR = 2(FR)2x + (FR)2y = 282 + 92 = 12.04 kN = 12.0 kN u = tan-1 c

(FR)y (FR)x

Ans.

9 d = tan-1 a b = 48.37° = 48.4° dAns. 8

Location of the Resultant Force. Summing the moments about point A, by referring to Fig. a, a+ (MR)A = ΣMA ;   - 8x - 9y = - 3(3)(1.5) - 2(4)(2)

8x + 9y = 29.5

(1)

384

C

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4–149. Continued

Along AB, x = 0.  Then Eq (1) becomes 8(0) + 9y = 29.5 y = 3.278 m Thus, the inter section point of line of action of FR on AB measured to the right from point A is

Ans.

d = y = 3.28 m

Ans: FR = 12.0 kN u = 48.4° d d = 3.28 m 385

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4–150. Replace the distributed loading by an equivalent resultant force and specify where its line of action intersects a vertical line along member BC, measured from C.

3 kN/m B

A 3m

2 kN/m

4m

Solution Equivalent Resultant Force. Summing the forces along the x and y axes by referring to Fig. a, + (FR)x = ΣFx;  (FR)x = -2(4) = -8 kN = 8 kN d S + c (FR)y = ΣFy;  (FR)y = - 3(3) = - 9 kN = 9 kN T Then And

FR = 2(FR)2x + (FR)2y = 282 + 92 = 12.04 kN = 12.0 kN u = tan-1 c

(FR)y (FR)x

Ans.

9 d = tan-1 a b = 48.37° = 48.4° dAns. 8

Location of the Resultant Force. Summing the moments about point A, by referring to Fig. a, a+ (MR)A = ΣMA;   -8x - 9y = -3(3)(1.5) - 2(4)(2)

8x + 9y = 29.5

(1)

386

C

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4–150. Continued

Along BC, y = 3 m.  Then Eq (1) becomes 8x + 9(3) = 29.5 x = 0.3125 m Thus, the intersection point of line of action of FR on BC measured upward from point C is

d = 4 - x = 4 - 0.3125 = 3.6875 m = 3.69 m

Ans.

Ans: FR = 12.0 kN u = 48.4° d d = 3.69 m 387

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4–151. If the soil exerts a trapezoidal distribution of load on the bottom of the footing, determine the intensities w1 and w2 of this distribution needed to support the column loadings.

80 kN

60 kN

1m

50 kN 2.5 m

SOLUTION

3.5 m

1m

w2

Loading: The trapezoidal reactive distributed load can be divided into two parts as shown on the free-body diagram of the footing, Fig. a. The magnitude and location measured from point A of the resultant force of each part are also indicated in Fig. a.

w1

Equations of Equilibrium: Writing the moment equation of equilibrium about point B, we have

a + ©MB = 0; w2(8) ¢ 4 -

8 8 8 8 ≤ + 60 ¢ - 1 ≤ - 80 ¢ 3.5 - ≤ - 50 ¢ 7 - ≤ = 0 3 3 3 3

w2 = 17.1875 kN>m = 17.2 kN>m

Ans.

Using the result of w2 and writing the force equation of equilibrium along the y axis, we obtain + c ©Fy = 0;

1 (w - 17.1875)8 + 17.1875(8) - 60 - 80 - 50 = 0 2 1 w1 = 30.3125 kN>m = 30.3 kN>m

Ans.

Ans: w2 = 17.2 kN>m w1 = 30.3 kN>m 388

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*4–152. The form is used to cast a concrete wall having a width of 5 m. Determine the equivalent resultant force the wet concrete exerts on the form AB if the pressure distribution due to the concrete can be approximated as shown. Specify the location of the resultant force, measured from point B.

B

p

1

p  (4z 2 ) kPa 4m

Solution L

dA =

L0

4

A

1 2

8 kPa

4z dz

4 3 2 = c (4)z2 d 3 0

z

= 21.33 kN>m Ans.

FR = 21.33(5) = 107 kN L

zdA =

L0

4

3

4z2 dz

4 5 2 = c (4)z2 d 5 0

= 51.2 kN z =

51.2 = 2.40 m 21.33

Ans.

Also, from the back of the book, A =

2 2 ab = (8)(4) = 21.33 3 3

FR = 21.33 (5) = 107 kN



Ans.

z = 4 - 1.6 = 2.40 m



Ans.

Ans: FR = 107 kN z = 2.40 m 389

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4–153. Replace the loading by an equivalent resultant force and couple moment acting at point A.

400 N/m

B

A 3m

3m

Solution Equivalent Resultant Force And Couple Moment At Point A. Summing the forces along the y axis by referring to Fig. a, + c (FR)y = ΣFy;  FR = - 400(3)

1 (400)(3) 2 Ans.

= - 1800 N = 1.80 kN T 

Summing the moment about point A, a+ (MR)A = ΣMA;  (MR)A = - 400(3)(1.5) -

1 (400)(3)(4) 2

= - 4200 N # m = 4.20 kN # m (clockwise) Ans.

Ans: FR = 1.80 kN

(MR)A = 4.20 kN # m (clockwise)

390

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4–154. Replace the loading by a single resultant force, and specify its location on the beam measured from point A.

400 N/m

B

A 3m

3m

Solution Equivalent Resultant Force. Summing the forces along the y axis by referring to Fig. a, + c (FR)y = ΣFy;  -FR = - 400(3)

1 (400)(3) 2 Ans.

FR = 1800 N = 1.80 kN T 

Location of Resultant Force. Summing the moment about point A by referring to Fig. a, a+ (MR)A = ΣMA;  -1800 d = - 400(3)(1.5)

1 (400)(3)(4) 2 Ans.

d = 2.333 m = 2.33 m

Ans: FR = 1.80 kN d = 2.33 m 391

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4–155. Replace the loading by an equivalent force and couple moment acting at point O.

6 kN/m

15 kN

500 kN m O

7.5 m

4.5 m

SOLUTION + c FR = ©Fy ;

FR = -22.5 - 13.5 - 15.0 = - 51.0 kN = 51.0 kN T

a + MRo = ©Mo ;

Ans.

MRo = - 500 - 22.5(5) - 13.5(9) - 15(12) = - 914 kN # m = 914 kN # m (Clockwise)

Ans.

Ans: FR = 51.0 kN T MRO = 914 kN # m b 392

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*4–156. Replace the loading by a single resultant force, and specify the location of the force measured from point O.

6 kN/m

15 kN

500 kN m O

7.5 m

4.5 m

SOLUTION Equivalent Resultant Force: + c FR = ©Fy ;

- FR = - 22.5 - 13.5 - 15 FR = 51.0 kN T

Ans.

Location of Equivalent Resultant Force: a + (MR)O = ©MO ;

- 51.0(d) = -500 - 22.5(5) - 13.5(9) - 15(12) d = 17.9 m

Ans.

Ans: FR = 51.0 kN T d = 17.9 m 393

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4–157. w

Determine the equivalent resultant force and couple moment at point O.

9 kN/m w  ( 13 x3 ) kN/m

O

x

3m

Solution Equivalent Resultant Force And Couple Moment About Point O. The differential 1 force indicated in Fig. a is dFR = w dx = x3dx. Thus, summing the forces along the 3 y axis, 3m 1 3 x dx + c (FR)y = ΣFy;   FR = - dFR = L L0 3 = -



1 4 3m x 12 L 0 Ans.

= - 6.75 kN = 6.75 kNT 

Summing the moments about point O, a+ (MR)O = ΣMO;  (MR)O =

L L0

3m

L0

3m



=



=



= a



(3 - x)dFR 1 (3 - x) a x3dxb 3 ax3 -

1 4 x b dx 3

x4 1 5 3m x b` 4 15 0

= 4.05 kN # m (counterclockwise)

Ans.

Ans: FR = 6.75 kNT

(MR)O = 4.05 kN # m (counterclockwise)

394

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w

4–158. Determine the equivalent resultant force of the distributed loading and its location, measured from point A. Evaluate the integrals using Simpson’s rule.

w=

5x + (16 + x2)1/2 kN/m

5.07 kN/m

2 kN/m A

x

B 3m

1m

Solution FR =

L

L0

wdx =

4 1

25x + (16 + x2)2 dx Ans.

FR = 14.9 kN L0

4

x dF =

L0

4

1 2

(x) 25x + (16 + x2) dx

= 33.74 kN # m x =

33.74 = 2.27 m 14.9

Ans.

Ans: FR = 14.9 kN x = 2.27 m 395

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4–159. w

Replace the loading by an equivalent resultant force and couple moment acting at point O.

p x w  w0 cos ( 2L (

x

O L

Solution Equivalent Resultant Force And Couple Moment About Point O. The differential p xbdx. Thus, summing the force indicated in Fig. a is dFR = w dx = aw0 cos 2L forces along the y axis, + c (FR)y = ΣFy;  FR =

= -



= -

L

dFR = -

L

LO

¢w0 cos

p x≤dx 2L

L 2Lw0 p asin xb ` p 2L O

2Lw0 2Lw0 = T p p

Ans.

Summing the moments about point O, a+ (MR)O = ΣMO;  (MR)O = -

L

xdFR

L

LO

x aw0 cos



= -



= -w0 a



= -a



= a

p xb dx 2L

L 4L2 2L p p x + 2 cos xb ` x sin p 2L 2L p O

2p - 4 bw0L2 p2

2p - 4 bw0L2 (clockwise) p2

Ans.

Ans: FR =

2Lw0 p

(MR)O = a 396

2p - 4 bw0L2 (clockwise) p2

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*4–160. Replace the distributed loading with an equivalent resultant force, and specify its location on the beam measured from point A.

10 kN/m

1 (⫺x2 ⫺ 4x ⫹ 60) kN/m w ⫽ –– 6

A

B

x

6m

SOLUTION Resultant: The magnitude of the differential force dFR is equal to the area of the element shown shaded in Fig. a. Thus, dFR = w dx =

1 ( - x2 - 4x + 60)dx 6

Integrating dFR over the entire length of the beam gives the resultant force FR. 6m

FR =

+T

LL

dFR =

L0

6m 1 1 x3 (-x2 - 4x + 60)dx = B - 2x2 + 60x R ` 6 6 3 0

= 36 kN T

Ans.

Location: The location of dFR on the beam is xc = x, measured from point A. Thus the location x of FR measured from point A is 6m

x =

LL

xcdFR

LL

= dFR

L0

1 x B (- x2 - 4x + 60) R dx 6 36

6m

=

L0

6m x4 4x3 1 1 + 30x2 ≤ ` ¢- (-x3 - 4x2 + 60x)dx 6 4 3 0 6 = 36 36

= 2.17 m

Ans.

Ans: FR = 36 kN T x = 2.17 m 397

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4–161.

p x) –– w ⫽ w0 sin (2L

Replace the distributed loading with an equivalent resultant force, and specify its location on the beam measured from point A.

w0

x

A

L

SOLUTION Resultant: The magnitude of the differential force dFR is equal to the area of the element shown shaded in Fig. a. Thus, dFR = w dx = ¢ w0 sin

p x ≤ dx 2L

Integrating dFR over the entire length of the beam gives the resultant force FR. L

+T

FR =

LL

dFR =

L0

¢ w0 sin

L 2w0L 2w0L p p cos T x ≤ dx = ¢x≤ ` = p p 2L 2L 0

Ans.

Location: The location of dFR on the beam is xc = x measured from point A. Thus, the location x of FR measured from point A is given by L

x =

LL

xcdFR

LL

= dFR

L0

x ¢ w0 sin

p x ≤ dx 2L

2w0L p

4w0L2 =

2L p2 = 2w0L p p

Ans.

Ans: 2w 0L T p 2L x = p FR =

398

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4–162. Wet concrete exerts a pressure distribution along the wall of the form. Determine the resultant force of this distribution and specify the height h where the bracing strut should be placed so that it lies through the line of action of the resultant force. The wall has a width of 5 m.

p

4m

p

1

(4 z /2) kPa

SOLUTION Equivalent Resultant Force: h

z

+ F = ©F ; : R x

-FR = - LdA = 4m

FR =

L0

L0

8 kPa

wdz

1

a20z2 b A 103 B dz

z

= 106.67 A 103 B N = 107 kN ;

Ans.

Location of Equivalent Resultant Force: z

z =

LA

zdA = dA

LA

zwdz

L0

z

L0

wdz

4m

=

L0

1

zc A 20z2 B (103) ddz

4m

L0 4m

=

L0

1

A 20z2 B (103)dz 3

c A 20z2 B (10 3) ddz

4m

L0

1

A 20z2 B (103)dz

= 2.40 m Thus,

h = 4 - z = 4 - 2.40 = 1.60 m

Ans.

Ans: FR = 107 kN h = 1.60 m 399

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5–10. Determine the reactions at the supports.

5 kN

2m B

A 6 kN 2m

2m

8 kN 2m

Solution Equations of Equilibrium. Ay and NB can be determined by writing the moment equations of equilibrium about points B and A, respectively, by referring to the FBD of the truss shown in Fig. a. a+ ΣMB = 0;  8(2) + 6(4) - 5(2) - Ay(6) = 0 Ans.

Ay = 5.00 kN a+ ΣMA = 0;  NB(6) - 8(4) - 6(2) - 5(2) = 0

Ans.

NB = 9.00 kN

Also, Ax can be determined directly by writing the force equation of equilibrium along x axis. + ΣFx = 0;  5 - Ax = 0  Ax = 5.00 kN S

Ans.

Ans: Ay = 5.00 kN NB = 9.00 kN Ax = 5.00 kN 400

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5–11. 4 kN

Determine the horizontal and vertical components of reaction at the pin A and the reaction of the rocker B on the beam.

B

A

30 6m

2m

SOLUTION Equations of Equilibrium: From the free-body diagram of the beam, Fig. a, NB can be obtained by writing the moment equation of equilibrium about point A. a+ ©MA = 0;

NB cos 30°(8) - 4(6) = 0 Ans.

NB = 3.464 kN = 3.46 kN

Using this result and writing the force equations of equilibrium along the x and y axes, we have + ©F = 0; : x

A x - 3.464 sin 30° = 0 Ans.

A x = 1.73 kN + c ©Fy = 0;

A y + 3.464 cos 30° - 4 = 0 Ans.

A y = 1.00 kN

Ans: NB = 3.46 kN Ax = 1.73 kN Ay = 1.00 kN 401

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*5–12. Determine the reactions at the supports.

400 N/m 5

3 4

B A 3m

3m

Solution Equations of Equilibrium. NA and By can be determined directly by writing the moment equations of equilibrium about points B and A, respectively, by referring to the beam’s FBD shown in Fig. a. 4 1 a+ ΣMB = 0;   (400)(6)(3) - NA a b(6) = 0 2 5 NA = 750 N

a+ ΣMA = 0;  By(6) -

Ans.

1 (400)(6)(3) = 0 2 Ans.

By = 600 N

Using the result of NA to write the force equation of equilibrium along the x axis, + ΣFx = 0;  750 a 3 b - Bx = 0 S 5

Ans.

Bx = 450 N

Ans: NA = 750 N By = 600 N Bx = 450 N 402

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5–13. Determine the components of the support reactions at the fixed support A on the cantilevered beam.

6 kN

30 30 A

SOLUTION

1.5 m 1.5 m

1.5 m

4 kN

Equations of Equilibrium: From the free-body diagram of the cantilever beam, Fig. a, Ax, Ay, and MA can be obtained by writing the moment equation of equilibrium about point A. + ©F = 0; : x

4 cos 30° - A x = 0 Ans.

A x = 3.46 kN + c ©Fy = 0;

A y - 6 - 4 sin 30° = 0 Ans.

A y = 8 kN

a+ ©MA = 0;MA - 6(1.5) - 4 cos 30° (1.5 sin 30°) - 4 sin 30°(3 + 1.5 cos 30°) = 0 MA = 20.2 kN # m

Ans.

Ans: Ax = 3.46 kN Ay = 8 kN MA = 20.2 kN # m 403

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5–14. Determine the reactions at the supports.

900 N/m 600 N/m

B

A 3m

3m

Solution Equations of Equilibrium. NA and By can be determined directly by writing the moment equations of equilibrium about points B and A, respectively, by referring to the FBD of the beam shown in Fig. a. a+ ΣMB = 0;  600(6)(3) +

1 (300)(3)(5) - NA(6) = 0 2 Ans.

NA = 2175 N = 2.175 kN a+ ΣMA = 0;  By(6) -

1 (300)(3)(1) - 600(6)(3) = 0 2 Ans.

By = 1875 N = 1.875 kN

Also, Bx can be determined directly by writing the force equation of equilibrium along the x axis. + ΣFx = 0;      Bx = 0 S

Ans.

Ans: NA = 2.175 kN By = 1.875 kN Bx = 0 404

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5–15. Determine the reactions at the supports. 800 N/m A

3m

B 1m

3m

Solution Equations of Equilibrium. NA can be determined directly by writing the moment equation of equilibrium about point B by referring to the FBD of the beam shown in Fig. a. a+ ΣMB = 0;  800(5)(2.5) - NA(3) = 0 Ans.

NA = 3333.33 N = 3.33 kN

Using this result to write the force equations of equilibrium along the x and y axes, + ΣFx = 0;  Bx - 800(5) a 3 b = 0 S 5

Ans.

Bx = 2400 N = 2.40 kN

4 + c ΣFy = 0;  3333.33 - 800 (5)a b - By = 0 5

Ans.

By = 133.33 N = 133 N

Ans: NA = 3.33 kN Bx = 2.40 kN By = 133 N 405

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*5–16. The man has a weight W and stands at the center of the plank. If the planes at A and B are smooth, determine the tension in the cord in terms of W and u.

B

SOLUTION a + ©MB = 0;

L Wa cos f b -NA(L cos f ) = 0 2

+ ©F = 0; : x

T cos u - NB sin u = 0

+ c ©Fy = 0;

T sin u + NB cos u +

NA

f

W = 2

L

u

A

(1)

W - W= 0 2

(2)

Solving Eqs. (1) and (2) yields: T= NB =

W sin u 2

Ans.

W cos u 2

Ans: W T= sin u 2 406

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5–17. The uniform rod AB has a mass of 40 kg. Determine the force in the cable when the rod is in the position shown. There is a smooth collar at A.

A

3m

60 C

Solution

B

Equations of Equilibrium. TBC can be determined by writing the moment equation of equilibrium about point O by referring to the FBD of the rod shown in Fig. a. a+ ΣMO = 0;  40(9.81)(1.5 cos 600°) - TBC (3 sin 60°) = 0 Ans.

TBC = 113.28 N = 113 N

Ans: TBC = 113 N 407

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5–18. A uniform glass rod having a length L is placed in the smooth hemispherical bowl having a radius r. Determine the angle of inclination u for equilibrium.

u B r A

SOLUTION By observation f = u. Equilibrium: a + ©MA = 0;

NB (2r cos u) - Wa

L cos u b = 0 2

+Q ©Fx = 0;

NA cos u - W sin u = 0

+a©Fy = 0;

(W tan u) sin u +

WL 4r

NA = W tan u

WL - W cos u = 0 4r

sin2 u - cos2 u +

L cos u = 0 4r

(1 - cos2 u) - cos2 u + 2 cos2 u cos u =

NB =

L cos u = 0 4r

L cos u - 1 = 0 4r

L ; 2L2 + 128r2 16r

Take the positive root cos u =

L + 2L2 + 128r2 16r

u = cos - 1 ¢

L + 2L2 + 128r2 ≤ 16r

Ans.

Ans:

u = cos - 1 ¢

408

L + 2L2 + 128r2 ≤ 16r

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5–19. Determine the components of reaction at the supports A and B on the rod.

P L –– 2

A

L –– 2 B

SOLUTION Equations of Equilibrium: Since the roller at A offers no resistance to vertical movement, the vertical component of reaction at support A is equal to zero. From the free-body diagram, Ax, By, and MA can be obtained by writing the force equations of equilibrium along the x and y axes and the moment equation of equilibrium about point B, respectively. + ©F = 0; : x

Ax = 0

+ c ©Fy = 0;

By - P = 0

Ans.

Ans.

By = P a + ©MB = 0;

Pa

L b - MA = 0 2

MA =

PL 2

Ans.

Ans: Ax = 0 By = P PL MA = 2 409

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*5–20.

A

The 75-kg gate has a center of mass located at G. If A supports only a horizontal force and B can be assumed as a pin, determine the components of reaction at these supports.

1.25 m G

1m B

SOLUTION Equations of Equilibrium: From the free-body diagram of the gate, Fig. a, By and Ax can be obtained by writing the force equation of equilibrium along the y axis and the moment equation of equilibrium about point B. + c ©Fy = 0;

By - 75(9.81) = 0 Ans.

By = 735.75 N = 736 N a+©MB = 0;

A x(1) - 75(9.81)(1.25) = 0 Ans.

A x = 919.69 N = 920 N

Using the result Ax = 919.69 N and writing the force equation of equilibrium along the x axis, we have + ©F = 0; : x

Bx - 919.69 = 0 Ans.

Bx = 919.69 N = 920 N

Ans: By = 736 N A x = 920 N Bx = 920 N 410

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5–21. The woman exercises on the rowing machine. If she exerts a holding force of F = 200 N on handle ABC, determine the horizontal and vertical components of reaction at pin C and the force developed along the hydraulic cylinder BD on the handle.

F ⫽ 200 N 30⬚ 0.25 m

A B

0.25 m C

D 0.75 m

SOLUTION

0.15 m

0.15 m

Equations of Equilibrium: Since the hydraulic cylinder is pinned at both ends, it can be considered as a two-force member and therefore exerts a force FBD directed along its axis on the handle, as shown on the free-body diagram in Fig. a. From the free-body diagram, FBD can be obtained by writing the moment equation of equilibrium about point C. a+ ©MC = 0;

FBD cos 15.52°(250) + FBD sin 15.52°(150) - 200 cos 30°(250 + 250) -200 sin 30°(750 + 150) = 0 Ans.

FBD = 628.42 N = 628 N

Using the above result and writing the force equations of equilibrium along the x and y axes, we have + : ©Fx = 0;

Cx + 200 cos 30° - 628.42 cos 15.52° = 0 Ans.

Cx = 432.29 N = 432 N + c ©Fy = 0;

200 sin 30° - 628.42 sin 15.52° + Cy = 0 Ans.

Cy = 68.19 N = 68.2 N

Ans: FBD = 628 N Cx = 432 N Cy = 68.2 N 411

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5–22. If the intensity of the distributed load acting on the beam is w = 3 kN>m, determine the reactions at the roller A and pin B.

A w 30

B

3m 4m

Solution Equations of Equilibrium. NA can be determined directly by writing the moment equation of equilibrium about point B by referring to the FBD of the beam shown in Fig. a. a+ ΣMB = 0;  3(4)(2) - NA sin 30° (3 sin 30°) - NA cos 30° (3 cos 30° + 4) = 0 Ans.

NA = 3.713 kN = 3.71 kN

Using this result to write the force equation of equilibrium along the x and y axes, + ΣFx = 0;  3.713 sin 30° - Bx = 0 S Ans.

Bx = 1.856 kN = 1.86 kN + c ΣFy = 0;  By + 3.713 cos 30° - 3(4) = 0

Ans.

By = 8.7846 kN = 8.78 kN

Ans: NA = 3.71 kN Bx = 1.86 kN By = 8.78 kN 412

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5–23. If the roller at A and the pin at B can support a load up to 4 kN and 8 kN, respectively, determine the maximum intensity of the distributed load w, measured in kN>m, so that failure of the supports does not occur.

A w 30

B

3m 4m

Solution Equations of Equilibrium. NA can be determined directly by writing the moment equation of equilibrium about point B by referring to the FBD of the beam shown in Fig. a. a+ ΣMB = 0;  w(4)(2) - NA sin 30° (3 sin 30°) - NA cos 30° (3 cos 30° + 4) = 0 NA = 1.2376 w Using this result to write the force equation of equilibrium along x and y axes, + ΣFx = 0;  1.2376 w sin 30° - Bx = 0 S

Bx = 0.6188 w

+ c ΣFy = 0;  By + 1.2376 w cos 30° - w(4) = 0

By = 2.9282 w

Thus, FB = 2Bx2 + By2 = 2(0.6188 w)2 + (2.9282 w)2 = 2.9929 w

It is required that FB 6 8 kN;

2.9929 w 6 8

w 6 2.673 kN>m

1.2376 w 6 4

w 6 3.232 kN>m

And NA 6 4 kN;

Thus, the maximum intensity of the distributed load is Ans.

w = 2.673 kN>m = 2.67 kN>m

Ans: w = 2.67 kN>m 413

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*5–24. The relay regulates voltage and current. Determine the force in the spring CD, which has a stiffness of k 120 N m, so that it will allow the armature to make contact at A in figure (a) with a vertical force of 0.4 N. Also, determine the force in the spring when the coil is energized and attracts the armature to E, figure (b), thereby breaking contact at A.

50 mm 50 mm 30 mm 10°

A

B

A

C

E

B

C k

k

D

D

SOLUTION From Fig. (a): a + ©MB = 0;

0.4(100 cos 10°) - Fs (30 cos 10°) = 0

(a)

Ans.

Fs = 1.333 N = 1.33 N Fs = kx;

(b)

1.333 = 120 x x = 0.01111 m = 11.11 mm

From Fig (b), energizing the coil requires the spring to be stretched an additional amount ¢x = 30 sin 10° = 5.209 mm. Thus x¿ = 11.11 + 5.209 = 16.32 mm Ans.

Fs = 120 (0.01632) = 1.96 N

Ans: Fs = 1.33 N Fs = 1.96 N 414

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5–25. Determine the reactions on the bent rod which is supported by a smooth surface at B and by a collar at A, which is fixed to the rod and is free to slide over the fixed inclined rod.

100 N 0.3 m

0.3 m

20 Nm

A

0.2 m

3 5

4

B

13 12

5

Solution Initial Guesses: NA = 20 N

NB = 10 N

MA = 30 N # m

Given: F = 100 N

M = 20 N # m a = 0.3 m b = 0.3 m c = 0.2 m d = 3 e = 4 f = 12 g = 5 Given ΣMA = 0;

MA - F # a - M + NB #

ΣFx = 0;

NA #

ΣFy = 0;

NA #

e 2

2e + d 2 d 2e 2 + d 2

- NB # + NB #

NA ° NB ¢ = Find 1 NA, NB, MA 2 MA

a

f 2

2f + g g

2f 2 + g2 f 2f 2 + g2

2

# 1a +

b2 - NB #

= 0

g 2

2f + g2

#c = 0

- F = 0

NA 39.7 b = a b N MA = 10.6 N # m NB 82.5

Ans.

Ans: N A = 39.7 N N B = 82.5 N M A = 10.6 N # m 415

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5–26. The mobile crane is symmetrically supported by two outriggers at A and two at B in order to relieve the suspension of the truck upon which it rests and to provide greater stability. If the crane boom and truck have a mass of 18 Mg and center of mass at G1, and the boom has a mass of 1.8 Mg and a center of mass at G2, determine the vertical reactions at each of the four outriggers as a function of the boom angle u when the boom is supporting a load having a mass of 1.2 Mg. Plot the results measured from u = 0° to the critical angle where tipping starts to occur.

6.25 m

G2 6m

SOLUTION + ©MB = 0;

-NA (4) + 18 A 103 B (9.81)(1) + 1.8 A 103 B (9.81) (2 - 6 sin u)

θ

+ 1.2 A 103 B (9.81) (2 - 12.25 sin u) = 0 NA = 58 860 - 62 539 sin u

G1 A 2m

Tipping occurs when NA = 0, or

1m 1m

Ans.

u = 70.3° + c ©Fy = 0;

B

NB + 58 860 - 62 539 sin u - (18 + 1.8 + 1.2) A 103 B (9.81) = 0 NB = 147 150 + 62 539 sin u

Since there are two outriggers on each side of the crane, NA ¿ = (29.4 - 31.3 sin u) kN = NA 2 NB¿ =

Ans.

NB = (73.6 + 31.3 sin u) kN 2

Ans.

Ans: u = 70.3° = NA = (29.4 - 31.3 sin u) kN NB= = (73.6 + 31.3 sin u) kN 416

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5–27. Determine the reactions acting on the smooth uniform bar, which has a mass of 20 kg.

B

4m

A

30º

60º

Solution Equations of Equilibrium. NB can be determined directly by writing the moment equation of equilibrium about point A by referring to the FBD of the bar shown in Fig. a. a+ ΣMA = 0;  NB cos 30°(4) - 20(9.81) cos 30°(2) = 0

Ans.

NB = 98.1 N

Using this result to write the force equation of equilibrium along the x and y axes, + ΣFx = 0;  Ax - 98.1 sin 60° = 0 S

Ax = 84.96 N = 85.0 N

Ans.

+ c ΣFy = 0;  Ay + 98.1 cos 60° - 20(9.81) = 0 Ans.

Ay = 147.15 N = 147 N

Ans: NB = 98.1 N Ax = 85.0 N Ay = 147 N 417

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*5–28. A linear torsional spring deforms such that an applied couple moment M is related to the spring’s rotation u in radians by the equation M = (20 u) N # m. If such a spring is attached to the end of a pin-connected uniform 10-kg rod, determine the angle u for equilibrium. The spring is undeformed when u = 0°.

A

u

M  (20 u) N  m 0.5 m

Solution a+ ΣMA = 0;   -98.1 (0.25 cos u) + 20(u) = 0 Solving for u, Ans.

u = 47.5°

Ans: u = 47.5° 418

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5–29. Determine the force P needed to pull the 50-kg roller over the smooth step. Take u = 30°.

P

u A

50 mm

300 mm

B

Solution Equations of Equilibrium. P can be determined directly by writing the moment equation of Equilibrium about point B, by referring to the FBD of the roller shown in Fig. a. a+ ΣMB = 0;  P cos 30°(0.25) + P sin 30° ( 20.32 - 0.252 2 - 50(9.81) 20.32 - 0.252 = 0 Ans.

P = 271.66 N = 272 N

Ans: P = 272 N 419

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5–30. Determine the magnitude and direction u of the minimum force P needed to pull the 50-kg roller over the smooth step.

P

u A

50 mm

300 mm

B

Solution Equations of Equilibrium. P will be minimum if its orientation produces the greatest moment about point B. This happens when it acts perpendicular to AB as shown in Fig. a. Thus u = f = cos-1 a

0.25 b = 33.56° = 33.6° 0.3

Ans.

Pmin can be determined by writing the moment equation of equilibrium about point B by referring to the FBD of the roller shown in Fig. b. a+ ΣMB = 0;  Pmin (0.3) - 50(9.81)(0.3 sin 33.56°) = 0 Ans.

Pmin = 271.13 N = 271 N

Ans: Pmin = 271 N 420

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5–31. The operation of the fuel pump for an automobile depends on the reciprocating action of the rocker arm ABC, which is pinned at B and is spring loaded at A and D. When the smooth cam C is in the position shown, determine the horizontal and vertical components of force at the pin and the force along the spring DF for equilibrium. The vertical force acting on the rocker arm at A is FA = 60 N, and at C it is FC = 125 N.

E

30° F FC = 125 N

FA = 60 N

B A

C

D

SOLUTION a + ©MB = 0;

- 60(50) - FB cos 30°(10) + 125(30) = 0

50 mm

- Bx + 86.6025 sin 30° = 0 Ans.

Bx = 43.3 N + c ©Fy = 0;

20 mm

Ans.

FB = 86.6025 = 86.6 N + ©F = 0; : x

10 mm

60 - By - 86.6025 cos 30° + 125 = 0 Ans.

By = 110 N

Ans: FB = 86.6 N Bx = 43.3 N By = 110 N 421

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*5–32. The beam of negligible weight is supported horizontally by two springs. If the beam is horizontal and the springs are unstretched when the load is removed, determine the angle of tilt of the beam when the load is applied.

B

A 600 N/m

kA = 1 kN/m

kB = 1.5 kN/m D

C 3m

3m

Solution Equations of Equilibrium. FA and FB can be determined directly by writing the moment equations of equilibrium about points B and A, respectively, by referring to the FBD of the beam shown in Fig. a. Assuming that the angle of tilt is small, a+ ΣMA = 0;  FB(6) -

1 (600)(3)(2) = 0 FB = 300 N 2

1 a+ ΣMB = 0;   (600)(3)(4) - FA(6) = 0 FA = 600 N 2 Thus, the stretches of springs A and B can be determined from FA = kAxA;

600 = 1000 xA

xA = 0.6 m

FB = kB xB;

300 = 1500 xB

xB = 0.2 m

From the geometry shown in Fig. b u = sin-1 a

0.4 b = 3.82° 6

Ans.

The assumption of small u is confirmed.

Ans: u = 3.82° 422

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5–33. The dimensions of a jib crane, which is manufactured by the Basick Co., are given in the figure. If the crane has a mass of 800 kg and a center of mass at G, and the maximum rated force at its end is F 15 kN, determine the reactions at its bearings. The bearing at A is a journal bearing and supports only a horizontal force, whereas the bearing at B is a thrust bearing that supports both horizontal and vertical components.

3m A 0.75 m 2m

G F B

SOLUTION a + ©MB = 0;

Ax (2) - 800 (9.81) (0.75) - 15 000(3) = 0 Ans.

Ax = 25.4 kN + c ©Fy = 0;

By - 800 (9.81) - 15 000 = 0 Ans.

By = 22.8 kN + ©F = 0; : x

Bx - 25.4 = 0 Ans.

Bx = 25.4 kN

Ans: Ax = 25.4 kN By = 22.8 kN Bx = 25.4 kN 423

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5–34. The dimensions of a jib crane, which is manufactured by the Basick Co., are given in the figure. The crane has a mass of 800 kg and a center of mass at G. The bearing at A is a journal bearing and can support a horizontal force, whereas the bearing at B is a thrust bearing that supports both horizontal and vertical components. Determine the maximum load F that can be suspended from its end if the selected bearings at A and B can sustain a maximum resultant load of 24 kN and 34 kN, respectively.

3m A 0.75 m 2m

G F B

SOLUTION a + ©MB = 0;

Ax (2) - 800 (9.81) (0.75) - F (3) = 0

+ c ©Fy = 0;

By - 800 (9.81) - F = 0

+ ©F = 0; : x

Bx - Ax = 0

Assume Ax = 24 000 N. Solving, Bx = 24 kN By = 21.9 kN Ans.

F = 14.0 kN FB =

(24)2 + (21.9)2 = 32.5 kN 6 34 kN

OK

Ans: F = 14.0 kN 424

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5–35. The upper portion of the crane boom consists of the jib AB, which is supported by the pin at A, the guy line BC, and the backstay CD, each cable being separately attached to the mast at C. If the 5-kN load is supported by the hoist line, which passes over the pulley at B, determine the magnitude of the resultant force the pin exerts on the jib at A for equilibrium, the tension in the guy line BC, and the tension T in the hoist line. Neglect the weight of the jib. The pulley at B has a radius of 0.1 m.

C 1.5 m 0.1 m

B

A

r  0.1 m

5m T

D

5 kN

SOLUTION From pulley, tension in the hoist line is a + ©MB = 0;

T(0.1) - 5(0.1) = 0; T = 5 kN

Ans.

From the jib, a + ©MA = 0;

-5(5) + TBC a

1.6 227.56

b (5) = 0

TBC = 16.4055 = 16.4 kN + c ©Fy = 0;

-Ay + (16.4055)a

1.6 227.56

Ans.

b - 5 = 0

Ay = 0 + ©F = 0; : x

Ax - 16.4055a

5 227.56

b - 5 = 0

FA = Fx = 20.6 kN

Ans.

Ans: T = 5 kN T BC = 16.4 kN FA = Fx = 20.6 kN 425

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*5–36. The smooth pipe rests against the opening at the points of contact A, B, and C. Determine the reactions at these points needed to support the force of 300 N. Neglect the pipe’s thickness in the calculation.

A C

30 300 N

0.5 m

30 0.5 m

0.26 m

B

0.15 m

Solution Equations of Equilibrium. NA can be determined directly by writing the force equation of equilibrium along the x axis by referring to the FBD of the pipe shown in Fig. a. + ΣFx = 0;  NA cos 30° - 300 sin 30° = 0  NA = 173.21 N = 173 N S

Ans.

Using this result to write the moment equations of equilibrium about points B and C, a+ ΣMB = 0;  300 cos 30°(1) - 173.21 cos 30°(0.26) - 173.21 sin 30°(0.15) - NC (0.5) = 0 Ans.

NC = 415.63 N = 416 N

a+ ΣMC = 0;  300 cos 30°(0.5) - 173.21 cos 30°(0.26) - 173.21 sin 30°(0.65) - NB(0.5) = 0 Ans.

NB = 69.22 N = 69.2 N

Ans: NA = 173 N NC = 416 N NB = 69.2 N 426

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5–37. The boom supports the two vertical loads. Neglect the size of the collars at D and B and the thickness of the boom, and compute the horizontal and vertical components of force at the pin A and the force in cable CB. Set F1 = 800 N and F2 = 350 N.

C

3

5 4

SOLUTION a + ©MA = 0;

1m

- 800(1.5 cos 30°) - 350(2.5 cos 30°) +

3 4 F (2.5 sin 30°) + FCB(2.5 cos 30°) = 0 5 CB 5

4 Ax - (781.6) = 0 5

Ay - 800 - 350 +

30 A

F1

Ans.

Ax = 625 N + c ©Fy = 0;

D F2

Ans.

FCB = 781.6 = 782 N + ©F = 0; : x

1.5 m

B

3 (781.6) = 0 5 Ans.

Ay = 681 N

Ans: FCB = 782 N Ax = 625 N Ay = 681 N 427

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5–38. The boom is intended to support two vertical loads, F1 and F2. If the cable CB can sustain a maximum load of 1500 N before it fails, determine the critical loads if F1 = 2F2. Also, what is the magnitude of the maximum reaction at pin A?

C

3

5 4

SOLUTION a + ©MA = 0;

1m

- 2F2(1.5 cos 30°) - F2(2.5 cos 30°) +

4 3 (1500)(2.5 sin 30°) + (1500)(2.5 cos 30°) = 0 5 5

1.5 m

F1 = 2F2 = 1448 N

Ax -

30 A

F1

Ans.

F1 = 1.45 kN + ©F = 0; : x

D F2

Ans.

F2 = 724 N

B

4 (1500) = 0 5

Ax = 1200 N + c ©Fy = 0;

Ay - 724 - 1448 +

3 (1500) = 0 5

Ay = 1272 N FA = 2(1200)2 + (1272)2 = 1749 N = 1.75 kN

Ans.

Ans: F2 = 724 N F1 = 1.45 kN FA = 1.75 kN 428

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5–39. The bulk head AD is subjected to both water and soilbackfill pressures. Assuming AD is “pinned” to the ground at A, determine the horizontal and vertical reactions there and also the required tension in the ground anchor BC necessary for equilibrium. The bulk head has a mass of 800 kg.

D 0.5 m

B

C

F

6m 4m

118 kN/m A

SOLUTION

310 kN/m

Equations of Equilibrium: The force in ground anchor BC can be obtained directly by summing moments about point A. a + ©MA = 0;

1007.512.1672 - 23611.3332 - F162 = 0 Ans.

F = 311.375 kN = 311 kN + ©F = 0; : x

Ax + 311.375 + 236 - 1007.5 = 0 Ans.

Ax = 460 kN + c ©Fy = 0;

Ay - 7.848 = 0

Ans.

Ay = 7.85 kN

Ans: F = 311 kN Ax = 460 kN Ay = 7.85 kN 429

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*5–40. The bar of negligible weight is supported by two springs, each having a stiffness k = 100 N>m. If the springs are originally unstretched, and the force is vertical as shown, determine the angle u the bar makes with the horizontal, when the 30-N force is applied to the bar.

k 2m

1m C 30 N

B k

Solution Equations of Equilibrium. FA and FB can be determined directly by writing the moment equation of equilibrium about points B and A respectively by referring to the FBD of the bar shown in Fig. a. a+ ΣMB = 0;  30(1) - FA(2) = 0  FA = 15 N

A

a+ ΣMA = 0;  30(3) - FB(2) = 0  FB = 45 N Thus, the stretches of springs A and B can be determined from FA = kxA;

15 = 100 xA

xA = 0.15 m

FB = k xB;

45 = 100 xB

xB = 0.45 m

From the geometry shown in Fig. b, d 2 - d = ; 0.45 0.15

d = 1.5 m

Thus

u = sin-1 a

0.45 b = 17.46° = 17.5° 1.5

Ans.

Note: The moment equations are set up assuming small u, but even with non-small u the reactions come out with the same FA, FB, and then the rest of the solution goes through as before.

Ans: u = 17.5° 430

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5–41. Determine the stiffness k of each spring so that the  30-N force causes the bar to tip u = 15° when the force is applied. Originally the bar is horizontal and the springs are unstretched. Neglect the weight of the bar.

k 2m

1m C 30 N

B k

Solution Equations of Equilibrium. FA and FB can be determined directly by writing the moment equation of equilibrium about points B and A respectively by referring to the FBD of the bar shown in Fig. a. a+ ΣMB = 0;  30(1) - FA(2) = 0  FA = 15 N

A

a+ ΣMA = 0;  30(3) - FB(2) = 0  FB = 45 N Thus, the stretches of springs A and B can be determined from FA = kxA;

15 = kxA

xA =

15 k

FB = k xB;

45 = kxB

xB =

45 k

From the geometry shown in Fig. b d 2 - d = ; 45>k 15>k

d = 1.5 m

Thus, sin 15° =

45>k 1.5

Ans.

k = 115.91 N>m = 116 N>m

Note: The moment equations are set up assuming small u, but even with non-small u the reactions come out with the same FA, FB, and then the rest of the solution goes through as before.

Ans: k = 116 N>m 431

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5–42. The airstroke actuator at D is used to apply a force of F = 200 N on the member at B. Determine the horizontal and vertical components of reaction at the pin A and the force of the smooth shaft at C on the member.

C

15⬚ 600 mm

B A

60⬚

200 mm

Solution

600 mm

D

F

Equations of Equilibrium: From the free-body diagram of member ABC, Fig. a, NC can be obtained by writing the moment equation of equilibrium about point A. a+ ΣMA = 0;

200 sin 60° 18002 - NC1600 + 200 sin 15°) = 0 NC = 212.60 N = 213 N

Ans.

Using this result and writing the force equations of equilibrium along the x and y axes, +

S ΣFx = 0; + c ΣFy = 0;

- Ax + 212.60 sin 15° - 200 sin 60° = 0 Ax = 105 N

Ans.

-Ay - 212.60 cos 15° + 200 cos 60° = 0 Ay = 118 N

Ans.

Ans: N C = 213 N A x = 105 N A y = 118 N 432

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5–43 . The airstroke actuator at D is used to apply a force of F on the member at B. The normal reaction of the smooth shaft at C on the member is 300 N. Determine the magnitude of F and the horizontal and vertical components of reaction at pin A.

C

15⬚ 600 mm

B A

60⬚

200 mm

600 mm

D

F

Solution Equations of Equilibrium: From the free-body diagram of member ABC, Fig. a, force F can be obtained by writing the moment equation of equilibrium about point A. a+ ΣMA = 0;

F sin 60° 18002 - 3001600 + 200 sin 15° 2 = 0 F = 282.22 N = 282 N

Ans.

Using this result and writing the force equations of equilibrium along the x and y axes, +

S ΣFx = 0; + c ΣFy = 0;

- Ax + 300 cos 15° - 282.22 cos 60° = 0 Ax = 149 N

Ans.

- Ay - 300 sin 15° + 282.22 sin 60° = 0 Ay = 167 N

Ans.

Ans: F = 282 N A x = 149 N A y = 167 N 433

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*5–44. The 10-kg uniform rod is pinned at end A. If it is also subjected to a couple moment of 50 N # m, determine the smallest angle u for equilibrium. The spring is unstretched when u = 0, and has a stiffness of k = 60 N>m.

B k  60 N/m

u

2m

0.5 m

A

Solution

50 N  m

Equations of Equilibrium. Here the spring stretches x = 2 sin u. The force in the spring is Fsp = kx = 60 (2 sin u) = 120 sin u. Write the moment equation of equilibrium about point A by referring to the FBD of the rod shown in Fig. a, a+ ΣMA = 0;  120 sin u cos u (2) - 10(9.81) sin u (1) - 50 = 0

240 sin u cos u - 98.1 sin u - 50 = 0

Solve numerically

Ans.

u = 24.598° = 24.6°

Ans: u = 24.6° 434

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5–45. The man uses the hand truck to move material up the step. If the truck and its contents have a mass of 50 kg with center of gravity at G, determine the normal reaction on both wheels and the magnitude and direction of the minimum force required at the grip B needed to lift the load.

0.4 m B 0.5 m 0.2 m

G

0.4 m

60 0.4 m

Solution

A

0.1 m

Equations of Equilibriums. Py can be determined directly by writing the force equation of equilibrium along y axis by referring to the FBD of the hand truck shown in Fig. a. + c ΣFy = 0;  Py - 50(9.81) = 0

Py = 490.5 N

Using this result to write the moment equation of equilibrium about point A, a+ ΣMA = 0;  Px sin 60°(1.3) - Px cos 60°(0.1) - 490.5 cos 30°(0.1)

- 490.5 sin 30°(1.3) - 50(9.81) sin 60°(0.5)



+50(9.81) cos 60°(0.4) = 0



Px = 442.07 N

Thus, the magnitude of minimum force P, Fig. b, is P = 2Px2 + Py2 = 2442.072 + 490.52 = 660.32 N = 660 N

Ans.

and the angle is u = tan-1 a

490.5 b = 47.97° = 48.0° b   442.07

Ans.

Write the force equation of equilibrium along x axis, + ΣFx = 0;  NA - 442.07 = 0  NA = 442.07 N = 442 N S

Ans.

Ans:  P = 660 N NA = 442 N       u = 48.0° b 435

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5–46. Three uniform books, each having a weight W and length a, are stacked as shown. Determine the maximum distance d that the top book can extend out from the bottom one so the stack does not topple over.

a

d

SOLUTION Equilibrium: For top two books, the upper book will topple when the center of gravity of this book is to the right of point A. Therefore, the maximum distance from the right edge of this book to point A is a/2. Equation of Equilibrium: For the entire three books, the top two books will topple about point B. a + ©MB = 0;

a W(a-d) -W ad- b = 0 2 d =

3a 4

Ans.

Ans: d = 436

3a 4

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5–47. Determine the reactions at the pin A and the tension in cord BC. Set F = 40 kN. Neglect the thickness of the beam.

F

26 kN

C 13

12

5

5

3

4

A

B

2m

4m

Solution a+ ΣMA = 0;   - 26 a



12 3 b(2) - 40(6) + FBC(6) = 0 13 5

Ans.

FBC = 80 kN

+ ΣFx = 0;         80 a 4 b - Ax - 26 a 5 b = 0 S 5 13



+ c ΣFy = 0;            Ay - 26 a



Ans.

Ax = 54 kN

12 3 b - 40 + 80 a b = 0 13 5

Ans.

Ay = 16 kN

Ans: FBC = 80 kN Ax = 54 kN Ay = 16 kN 437

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*5–48. If rope BC will fail when the tension becomes  50  kN, determine the greatest vertical load F that can be applied to the beam at B. What is the magnitude of the reaction at A for this loading? Neglect the thickness of the beam.

F

26 kN

C 13

12

5

5

3

4

A

B

2m

4m

Solution a+ ΣMA = 0;   -26 a



12 3 b(2) - F(6) + (50)(6) = 0 13 5

Ans.

F = 22 kN

+ ΣFx = 0;         50 a 4 b - Ax - 26 a 5 b = 0 S 5 13



+ c ΣFy = 0;            Ay - 26 a



Ans.

Ax = 30 kN

12 3 b - 22 + 50 a b = 0 13 5

Ans.

Ay = 16 kN

Ans: F = 22 kN Ax = 30 kN Ay = 16 kN 438

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5–49. The rigid metal strip of negligible weight is used as part of an electromagnetic switch. If the stiffness of the springs at A and B is k = 5 N>m, and the strip is originally horizontal when the springs are unstretched, determine the smallest force needed to close the contact gap at C.

50 mm

50 mm

k

B C

A k

10 mm

SOLUTION ©MB = 0;

FA = FC = F

©Fy = 0;

FB = 2F

x 50 -x = yB yA

(1)

kyB 2F = F kyA (2)

2yA = yB Substituting into Eq. (1): 50 - x x = yA 2yA 2x = 50 - x x =

50 = 16.67 mm 3

x 100 - x = yA 10 Set x = 16.67, then yA = 2 mm From Eq. (2), yB = 4 mm Ans.

FC = FA = kyA = (5)(0.002) = 10 mN

Ans: FC = 10 mN 439

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5–50. The rigid metal strip of negligible weight is used as part of an electromagnetic switch. Determine the maximum stiffness k of the springs at A and B so that the contact at C closes when the vertical force developed there is 0.5 N. Originally the strip is horizontal as shown.

50 mm

50 mm

k

B C

A k

10 mm

SOLUTION ©MB = 0;

FA = FC = F

©Fy = 0;

FB = 2F

x 50 - x = yA yB

(1)

kyB 2F = F kyA (2)

2yA = yB Substituting into Eq. (1): 50 - x x = yA 2yA 2x = 50 - x x =

50 = 16.67 mm 3

x 100 - x = yA 10 Set x = 16.67, then yA = 2 mm From Eq. (2), yB = 4 mm FC = FA = kyA 0.5 = k(0.002) Ans.

k = 250 N/m

Ans: k = 250 N>m 440

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5–51. The device is used to hold an elevator door open. If the spring has a stiffness of k = 40 N>m and it is compressed 0.2 m, determine the horizontal and vertical components of reaction at the pin A and the resultant force at the wheel bearing B.

150 mm

125 mm

k A 100 mm

SOLUTION

B 30

Fs = ks = (40)(0.2) = 8 N a + ©MA = 0;

- (8)(150) + FB(cos 30°)(275) - FB(sin 30°)(100) = 0 FB = 6.37765 N = 6.38 N

+ ©F = 0; : x

Ans.

Ax - 6.37765 sin 30° = 0 Ax = 3.19 N

+ c ©Fy = 0;

Ans.

Ay - 8 + 6.37765 cos 30° = 0 Ay = 2.48 N

Ans.

Ans: FB = 6.38 N A x = 3.19 N A y = 2.48 N 441

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*5–52. The uniform beam has a weight W and length l and is supported by a pin at A and a cable BC. Determine the horizontal and vertical components of reaction at A and the tension in the cable necessary to hold the beam in the position shown.

C l B A

SOLUTION Equations of Equilibrium: The tension in the cable can be obtained directly by summing moments about point A. a + ©MA = 0;

l T sin 1f - u2l - W cos u a b = 0 2 T =

Using the result T = + ©F = 0; : x

a

Ans.

W cos u 2 sin 1f - u2 W cos u bcos f - Ax = 0 2 sin 1f - u2 Ax =

+ c ©Fy = 0;

W cos u 2 sin 1f - u2

Ay + a Ay =

W cos f cos u 2 sin 1f - u2

Ans.

W cos u bsin f - W = 0 2 sin 1f - u2

W1sin f cos u - 2 cos f sin u2

Ans.

2 sin f - u

Ans: W cos u 2 sin(f - u) Wcos f cos u Ax = 2 sin(f - u) W(sin f cos u - 2 cos f sin u) Ay = 2 sin (f - u) T =

442

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5–53. A boy stands out at the end of the diving board, which is supported by two springs A and B, each having a stiffness of k = 15kN>m. In the position shown the board is horizontal. If the boy has a mass of 40 kg, determine the angle of tilt which the board makes with the horizontal after he jumps off. Neglect the weight of the board and assume it is rigid.

1m A

3m B

SOLUTION Equations of Equilibrium: The spring force at A and B can be obtained directly by summing moments about points B and A, respectively. a + ©MB = 0;

FA (1) - 392.4(3) = 0

FA = 1177.2 N

a + ©MA = 0;

FB (1) - 392.4(4) = 0

FB = 1569.6 N

Spring Formula: Applying ¢ = ¢A =

F , we have k

1177.2 = 0.07848 m 15(103)

¢B =

1569.6 = 0.10464 m 15(103)

Geometry: The angle of tilt a is a = tan - 1 a

0.10464 + 0.07848 b = 10.4° 1

Ans.

Ans: a = 10.4° 443

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5–54. The platform assembly has a weight of 1000 N (≈ 100 kg) and center of gravity at G1. If it is intended to  support a maximum load of 1600 N placed, at point G2, determine the smallest counterweight W that should be placed at B in order to prevent the platform from tipping over.

G2 1m

3m

G1

Solution Given: W1 = 1000 N

4m

B

W2 = 1600 N a = 0.5 m

C

D

b = 3m 0.5 m

c = 0.5 m

3m

0.5 m

d = 4m e = 3m f = 1m When tipping occurs, Rc = 0 a+ ΣMD = 0;

- W2 # f + W1 # c + WB # 1b + c2 = 0 W2 # f - W1 # c WB = b + c WB = 314 N

Ans.

Ans: WB = 314 N 444

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5–55. The uniform rod of length L and weight W is supported on the smooth planes. Determine its position u for equilibrium. Neglect the thickness of the rod.

L u

c

f

SOLUTION a + ©MB = 0;

- Wa

L cos u b + NA cos f (L cos u) + NA sin f (L sin u) = 0 2 NA =

W cos u 2 cos (f - u)

+ ©F = 0; : x

NB sin c - NA sin f = 0

+ c ©Fy = 0;

NB cos c + NA cos f - W = 0 NB =

(1) (2)

W - NA cos f cos c

(3)

Substituting Eqs. (1) and (3) into Eq. (2): aW -

W cos u sin f W cos u cos f b tan c = 0 2 cos (f - u) 2 cos (f - u)

2 cos (f - u) tan c - cos u tan c cos f - cos u sin f = 0 sin u (2 sin f tan c) - cos u (sin f - cos f tan c) = 0 tan u =

sin f - cos f tan c 2 sin f tan c

u = tan - 1 a

1 1 cot c - cot f b 2 2

Ans.

Ans: 1 1 u = tan - 1 a cot c - cot f b 2 2 445

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*5–56. The uniform rod has a length l and weight W. It is supported at one end A by a smooth wall and the other end by a cord of length s which is attached to the wall as shown. Determine the placement h for equilibrium.

C

h s A

SOLUTION Equations of Equilibrium: The tension in the cable can be obtained directly by summing moments about point A.

l B

a + ©MA = 0;

l T sin f(l) - W sin u a b = 0 2 T =

Using the result T = + c ©Fy = 0;

W sin u 2 sin f

W sin u , 2 sin f

W sin u cos (u - f) - W = 0 2 sin f (1)

sin u cos (u - f) - 2 sin f = 0 Geometry: Applying the sine law with sin (180° - u) = sin u, we have sin f sin u = s h

sin f =

h sin u s

(2)

Substituting Eq. (2) into (1) yields cos (u - f) =

2h s

(3)

Using the cosine law, l2 = h2 + s2 - 2hs cos (u - f) cos (u - f) =

h2 + s2 - l2 2hs

(4)

Equating Eqs. (3) and (4) yields h2 + s 2 - l2 2h = s 2hs h =

A

s2 - l2 3

Ans.

Ans: h = 446

s2 - l 2 A 3

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5–57. The 30-N uniform rod has a length of l = 1 m. If s = 1.5 m, determine the distance h of placement at the end A along the smooth wall for equilibrium.

C h s

SOLUTION Equations of Equilibrium: Referring to the FBD of the rod shown in Fig. a, write the moment equation of equilibrium about point A. a + ©MA = 0;

A l

T sin f(1) - 3 sin u(0.5) = 0 T =

B

1.5 sin u sin f

Using this result to write the force equation of equilibrium along y axis, a

+ c ©Fy = 0;

15 sin u b cos (u - f) - 3 = 0 sin f (1)

sin u cos (u - f) - 2 sin f = 0

Geometry: Applying the sine law with sin (180° - u) = sin u by referring to Fig. b, sin f sin u = ; h 1.5

sin u = a

h b sin u 1.5

(2)

Substituting Eq. (2) into (1) yields sin u[cos (u - f) -

4 h] = 0 3

since sin u Z 0, then cos (u - f) - (4>3)h

(3)

cos (u - f) = (4>3)h

Again, applying law of cosine by referring to Fig. b, l2 = h2 + 1.52 - 2(h)(1.5) cos (u - f) cos (u - f) =

h2 + 1.25 3h

(4)

Equating Eqs. (3) and (4) yields h2 + 1.25 4 h = 3 3h 3h2 = 1.25 Ans.

h = 0.645 m

Ans: h = 0.645 m 447

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5–58. If d = 1 m, and u = 30°, determine the normal reaction at the smooth supports and the required distance a for the placement of the roller if P = 600 N. Neglect the weight of the bar.

d

u

P

a

SOLUTION Equations of Equilibrium: Referring to the FBD of the rod shown in Fig. a, a+ ©MA = 0;

a b - 600 cos 30°(1) = 0 cos 30° 450 NB = a NB = a

a+ ©Fy¿ = 0;

NB - NA sin 30° - 600 cos 30° = 0 NB - 0.5NA = 600 cos 30°

+Q ©F = 0; x¿

NA cos 30° - 600 sin 30° = 0 NA = 346.41 N = 346 N

(1) .

(2) Ans.

.

Substitute this result into Eq (2), NB - 0.5(346.41) = 600 cos 30° NB = 692.82 N = 693 N

Ans.

Substitute this result into Eq (1), 450 692.82 = a a = 0.6495 m

.

Ans.

a = 0.650 m

Ans: NA = 346 N NB = 693 N a = 0.650 m 448

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5–59. Determine the distance d for placement of the load P for equilibrium of the smooth bar in the position u as shown. Neglect the weight of the bar.

d P

u a

SOLUTION + c ©Fy = 0;

R cos u - P = 0

a + ©MA = 0;

- P(d cos u) + Ra Rd cos2 u = Ra d =

a b = 0 cos u

a b cos u

a cos3 u

Ans.

Also; Require forces to be concurrent at point O. AO = d cos u =

a>cos u cos u

Thus, d =

a cos3 u

Ans.

Ans: d = 449

a cos3 u

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*5–60. The rod supports a cylinder of mass 50 kg and is pinned at its end A. If it is also subjected to a couple moment of 600  N # m, determine the angle u for equilibrium. The spring has an unstretched length of 1 m and a stiffness of k = 600 N>m.

600 N  m A

1m k  600 N/m 1.5 m

u 1.5 m

B

Solution Equation of Equilibrium: At equilibrium position, the spring stretches x = 3 sin u m. Thus, the force in the spring is Fs = kx = 600 (3 sin u) = 1800 sin u N. Write the moment equation of equilibrium about point A by referring to the FBD of the rod, Fig. a, a + ΣMA = 0; 1800 sin u cos u(3) - [50(9.81) cosu](1.5) - 600 = 0 5400 sin u cos u - 735.75 cos u - 600 = 0 The numerical solution gives Ans.

u = 14.54° = 14.5° and u = 82.54 = 82.5°

Ans: u = 14.5° u = 85.2° 450

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5–61. The beam is subjected to the two concentrated loads. Assuming that the foundation exerts a linearly varying load distribution on its bottom, determine the load intensities w1 and w2 for equilibrium in terms of the parameters shown.

P L –– 3

2P L –– 3

L –– 3

w1 w2

SOLUTION Equations of Equilibrium: The load intensity w1 can be determined directly by summing moments about point A. a + ©MA = 0;

Pa

L L b - w1La b = 0 3 6 w1 =

+ c ©Fy = 0;

2P L

Ans.

2P 2P 1 aw2 bL + 1L2 - 3P = 0 2 L L w2 =

4P L

Ans.

Ans: w1 = 451

2P 4P ,w = L 2 L

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5–62. z

The uniform concrete slab has a mass of 2400 kg. Determine the tension in each of the three parallel supporting cables when the slab is held in the horizontal plane as shown.

TA

TB 15 kN

x

0.5 m

TC A

B

2m C

1m

1m

2m

y

Solution Equations of Equilibrium. Referring to the FBD of the slab shown in Fig. a, we notice that TC can be obtained directly by writing the moment equation of equilibrium about the x axis. ΣMx = 0; TC (2.5) - 2400(9.81)(1.25) - 15 ( 103 ) (0.5) = 0

Ans.

TC = 14,772 N = 14.8 kN

Using this result to write moment equation of equilibrium about y axis and force equation of equilibrium along z axis, ΣMy = 0; TB (2) + 14,772(4) - 2400(9.81)(2) - 15 ( 103 ) (3) = 0 Ans.

TB = 16,500 N = 16.5 kN ΣFz = 0; TA + 16,500 + 14,772 - 2400(9.81) - 15 ( 103 ) = 0

Ans.

TA = 7272 N = 7.27 kN

Ans: TC = 14.8 kN TB = 16.5 kN TA = 7.27 kN 452

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5–63. z

The smooth uniform rod AB is supported by a ball-and-socket joint at A, the wall at B, and cable BC. Determine the components of reaction at A, the tension in the cable, and the normal reaction at B if the rod has a mass of 20 kg.

C 0.5 m B

2m

Solution A

Force And Position Vectors. The coordinates of points A, B and G are A(1.5, 0, 0) m, B(0, 1, 2) m, C(0, 0, 2.5) m and G(0.75, 0.5, 1) m x

FA = - Axi + Ay j + Azk TBC = TBC a NB = NBi

1m

(0 - 1)j + (2.5 - 2)k rBC 1 0.5 b = TBC c = TBC j + TBC k 2 2 rBC 11.25 11.25 2(0 - 1) + (2.5 - 2)

W = { - 20(9.81)k} N rAG = (0.75 - 1.5)i + (0.5 - 0)j + (1 - 0)k = { -0.75i + 0.5j + k} m rAB = (0 - 1.5)i + (1 - 0)j + (2 - 0)k = { -1.5i + j + 2k} m Equations of Equilibrium. Referring to the FBD of the rod shown in Fig. a, the force equation of equilibrium gives ΣF = 0;  FA + TBC + NB + W = 0 ( -Ax + NB)i + aAy -

1 0.5 TBC bj + c Az + TBC - 20 (9.81)d k = 0 11.25 11.25

Equating i, j and k components,

(1)

   - Ax + NB = 0   Ay   Az +

1 TBC = 0 11.25

(2)

0.5 TBC - 20(9.81) = 0 11.25

(3)

The moment equation of equilibrium gives ΣMA = 0;  rAG * W + rAB * (TBC + NB) = 0 i † -0.75 0 a

j 0.5 0

k i 1 † + † - 1.5 - 20(9.81) NB

j 1 -

1 11.25

k 2 TBC

0.5 11.25

TBC

† =0

0.5 2 0.75 1.5 TBC + TBC - 98.1bi + a TBC + 2NB - 147.15bj + a TBC - NB bk = 0 11.25 11.25 11.25 11.25

453

1.5 m

y

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5–63. Continued

Equating i, j and k Components

0.5 2 TBC + TBC - 98.1 = 0 11.25 11.25

(4)

0.75 TBC + 2NB - 147.15 = 0 11.25

(5)

1.5 TBC - NB = 0 11.25

(6)

Solving Eqs. (1) to (6)

TBC = 43.87 N = 43.9 N

Ans.



NB = 58.86 N = 58.9 N

Ans.



Ax = 58.86 N = 58.9 N

Ans.



Ay = 39.24 N = 39.2 N

Ans.



Az = 176.58 N = 177 N

Ans.

Note: One of the equations (4), (5) and (6) is redundant that will be satisfied automatically.

Ans: TBC = NB =  Ax =  Ay =  Az = 454

43.9 N 58.9 N 58.9 N 39.2 N 177 N

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*5–64. Determine the tension in each cable and the components of reaction at D needed to support the load.

z B 3m

6m

C 2m D

x

A y

Solution

30

Force And Position Vectors. The coordinates of points A, B, and C are A(6, 0, 0) m, B(0, -3, 2) m and C(0, 0, 2) m respectively. FAB

(0 - 6)i + ( -3 - 0)j + (2 - 0)k rAB 6 3 2 = FAB a b = FAB c d = - FABi - FAB j + FABk rAB 7 7 7 1(0 - 6)2 + ( - 3 - 0)2 + (2 - 0)2

FAC = FAC a

(0 - 6)i + (2 - 0)k rAC 6 2 b = FAC c d = FAC i + FAC k 2 2 rAC 140 140 1(0 - 6) + (2 - 0)

F = 400 (sin 30°j - cos 30°k) = {200j - 346.41k}N FD = Dxi + Dy j + Dzk rDA = {6i} m

Referring to the FBD of the rod shown in Fig. a, the force equation of equilibrium gives ΣF = 0;  FAB + FAC + F + FD = 0 6 6 3 a- FAB FAC + Dx bi + a- FAB + Dy + 200bj 7 7 140

2 2 + a FAB + FAC + Dz - 346.41bk = 0 7 140

455

400 N

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5–64. Continued

Equating i, j and k components, 6 6 - FAB FAC + Dx = 0 7 140

(1)

3 - FAB + Dy + 200 = 0 7

(2)

2 2 F + FAC + Dz - 346.41 = 0 7 AB 140

(3)

Moment equation of equilibrium gives

ΣMD = 0;  rDA * (FAB + FAC + F) = 0

5

i 6

j 0

k 0

6 6 a- FAB F b 140 AC 7

3 a- FAB + 200b 7

2 2 FAC - 346.41b a FAB + 140 7

5 = 0

2 2 3 -6 a FAB + FAC - 346.41bj + 6 a- FAB + 200b k = 0 7 7 140

Equating j and k Components,

2 2 -6 a FAB + FAC - 346.41b = 0 7 140

(4)

3 6 a - FAB + 200b = 0 7

(5)

Solving Eqs. (1) to (5)

FAB = 466.67 N = 467 N

Ans.

FAC = 673.81 N = 674 N

Ans.

Dx = 1039.23 N = 1.04 kN

Ans.

Dy = 0



Ans.

Dz = 0



Ans.

Ans: FAB = FAC = Dx = Dy = Dz = 456

467 N 674 N 1.04 kN 0 0

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5–65. The cart supports the uniform crate having a mass of 85 kg. Determine the vertical reactions on the three casters at A, B, and C. The caster at B is not shown. Neglect the mass of the cart.

B

0.1 m

A 0.2 m

0.4 m 0.2 m

0.5 m 0.6 m

C

0.35 m 0.35 m

Solution Equations of Equilibrium: The normal reaction NC can be obtained directly by Sigming moments about x axis. ΣMx = 0; ΣMy = 0; ΣFz = 0;

NC 11.32 - 833.8510.452 = 0 NC = 288.64 N = 289 N

Ans.

833.8510.32 - 288.6410.352 - NA 10.72 = 0 NA = 213.04 N = 213 N

Ans.

NB + 288.64 + 213.04 - 833.85 = 0 NB = 332 N

Ans.

Ans: N C = 289 N N A = 213 N N B = 332 N 457

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5–66. z

The wing of the jet aircraft is subjected to a thrust of T = 8 kN from its engine and the resultant lift force L = 45 kN. If the mass of the wing is 2.1 Mg and the mass center is at G, determine the x, y, z components of reaction where the wing is fixed to the fuselage at A.

A

5m G

3m

7m

x

©Fx = 0;

T

- Ax + 8000 = 0 Ax = 8.00 kN

Ans.

©Fy = 0;

Ay = 0

Ans.

©Fz = 0;

- Az - 20 601 + 45 000 = 0

45 kN

Ans.

45 000(15) - 20 601(5) - Mx = 0 Mx = 572 kN # m

©Mz = 0;

L

My - 2.5(8000) = 0 My = 20.0 kN # m

©Mx = 0;

8 kN

Ans.

Az = 24.4 kN ©My = 0;

y

2.5 m

SOLUTION

Ans.

Mz - 8000(8) = 0 Mz = 64.0 kN # m

Ans.

Ans: A x = 8.00 kN Ay = 0 A z = 24.4 kN M y = 20.0 kN # m M x = 572 kN # m M z = 64.0 kN # m 458

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5–67. z

Determine the components of reaction at the fixed support A. The 400 N, 500 N, and 600 N forces are parallel to the x, y, and z axes, respectively.

600 N

1m 400 N

0.5 m

0.75 m

0.75 m

Solution

A

500 N

x

Equations of Equilibrium. Referring to the FBD of the rod shown in Fig. a ΣFx = 0;  Ax - 400 = 0  Ax = 400 N

Ans.

ΣFy = 0;  500 - Ay = 0  Ay = 500 N

Ans.

ΣFz = 0;  Az - 600 = 0  Az = 600 N

Ans.

y

ΣMx = 0;  (MA)x - 500(1.25) - 600(1) = 0 (MA)x = 1225 N # m = 1.225 kN # m

Ans.

ΣMy = 0;  (MA)y - 400(0.75) - 600(0.75) = 0 (MA)y = 750 N # m

Ans.

ΣMz = 0;  (MA)z = 0

Ans.

Ans: Ax = 400 N Ay = 500 N Az = 600 N (MA)x = 1.225 kN # m (MA)y = 750 N # m (MA)z = 0 459

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*5–68. Due to an unequal distribution of fuel in the wing tanks, the centers of gravity for the airplane fuselage A and wings B and C are located as shown. If these components have weights WA = 225 kN, and WB = 40 kN, WC = 30 kN, determine the normal reactions of the wheels D, E, and F on the ground.

z

D

B

A C E

2.4 m 1.8 m

Solution

x

Given:

F 2.4 m 1.8 m

6m

1.2 m 0.9 m

WA = 225 kN WB = 40 kN WC = 30 kN a b c d

= = = =

2.4 m 1.8 m 2.4 m 1.8 m

ΣMx = 0; ΣMy = 0; ΣFz = 0;

e = 6m f = 1.2 m g = 0.9 m WB # b - RD # 1a + b2 - WC # c + RE # 1c + d2 = 0 WB # f + WA # 1g + f 2 + WCf - RF # 1e + g + f 2 = 0 RD + RE + RF - WA - WB - WC = 0

RD £ RE ≥ = Find 1 RD, RE, RF 2 RF

RD 113.1 £ RE ≥ = £ 113.1 ≥ kN RF 68.7

Ans.

Ans: RD = 113.1 kN RE = 113.1 kN RF = 68.7 kN 460

y

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5–69. F

The uniform load has a mass of 600 kg and is lifted using a uniform 30-kg strongback beam and four wire ropes as shown. Determine the tension in each segment of rope and the force that must be applied to the sling at A.

1.25 m B

1.25 m A C

2m

SOLUTION

1.5 m

Equations of Equilibrium: Due to symmetry, all wires are subjected to the same tension. This condition statisfies moment equilibrium about the x and y axes and force equilibrium along y axis. ©Fz = 0;

1.5 m

4 4Ta b - 5886 = 0 5 Ans.

T = 1839.375 N = 1.84 kN

The force F applied to the sling A must support the weight of the load and strongback beam. Hence ©Fz = 0;

F - 60019.812 - 3019.812 = 0 Ans.

F = 6180.3 N = 6.18 kN

Ans: T = 1.84 kN F = 6.18 kN 461

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5–70. z

The member is supported by a square rod which fits loosely through the smooth square hole of the attached collar at A and by a roller at B. Determine the components of reaction at these supports when the member is subjected to the loading shown.

A

x

B

1m

2m y

2m

Solution

C

Force And Position Vectors. The coordinates of points B and C are B(2,0,0) m and C(3,0,- 2) m.

300 N

FA = -Ax i - Ay j F = {300i + 500j - 400k} N NB = NB k MA = - ( MA ) x i + ( MA ) y j - ( MA ) z k rAB = {2i} m  rAC = {3i - 2k} m Equations of Equilibrium. Referring to the FBD of the member shown in Fig. a, the force equation of equilibrium gives ΣF = 0;  FA + F + NB = 0

( 300 - Ax ) i + ( 500 - Ay ) j + ( NB - 400 ) k = 0

462

500 N 400 N

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5–70. Continued

Equating i, j and k components, 300 - Ax = 0  Ax = 300 N

Ans.

500 - Ay = 0  Ay = 500 N

Ans.

NB - 400 = 0       NB = 400 N

Ans.

The moment equation of equilibrium gives ΣMA = 0;  MA + rAB * NB + rAC * F = 0 i - ( MA ) x i + ( MA ) y j - ( MA ) z k + † 2 0

3 1000

- ( MA ) x 4 i +

3 ( MA ) y

Equating i, j and k components,

j 0 0

- 200 4 j +

k i 0 † + † 3 400 300

3 1500

j 0 500

- ( MA ) z 4 k = 0

1000 - ( MA ) x = 0   ( MA ) x = 1000 N # m = 1.00 kN # m

( MA ) y - 200 = 0   ( MA ) y = 200 N # m 1500 - ( MA ) z = 0   ( MA ) z = 1500 N # m =

k -2 † = 0 - 400

1.50 kN # m

Ans. Ans. Ans.

Ans: Ax = 300 N Ay = 500 N NB = 400 N ( MA ) x = 1.00 kN # m ( MA ) y = 200 N # m ( MA ) z = 1.50 kN # m 463

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5–71. z

Member AB is supported by a cable BC and at A by a square rod which fits loosely through the square hole in the collar fixed to the member as shown. Determine the components of reaction at A and the tension in the cable needed to hold the rod in equilibrium.

1.5 m 400 N

3m

A y C

200 N

Solution x

Force And Position Vectors. The coordinates of points B and C are B(3, 0, - 1) m C(0, 1.5, 0) m, respectively. TBC = TBC a

B

(0 - 3)i + (1.5 - 0)j + [0 - ( -1)]k rBC b = TBC • ¶ rBC 2(0 - 3)2 + (1.5 - 0)2 + [0 - ( -1)]2

6 3 2 = - TBC i + TBC j + T k 7 7 7 BC

F = {200j - 400k} N FA = Ax i + Ay j MA = (MA)x i + (MA)y j + (MA)z k r1{3 i} m r2 = {1.5 j} m Equations of Equilibrium. Referring to the FBD of member AB shown in Fig. a, the force equation of equilibrium gives ΣF = 0; TBC + F + FA = 0 6 3 2 a - TBC + Ax bi + a TBC + 200 + Ayb j + a TBC - 400bk = 0 7 7 7

Equating i, j and k components -

6 T + Ax = 0 7 BC

(1)

3 T + 200 + Ay = 0 7 BC

(2)

2 T - 400 = 0 7 BC

(3)

464

1m

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5–71. Continued

The moment equation of equilibrium gives ΣMA = O;  MA + r1 * F + r2 * TBC = 0 i

( MA ) x i + ( MA ) y j + ( MA ) zk + † 3 0

j 0 200

k 0 † + 5 - 400

i 0 6 - TBC 7

j 1.5 3 TBC 7

k 0 2 TBC 7

3 9 T d i + 3 ( MA ) y + 1200 4 j + c ( MA ) z + TBC + 600 d k = 0 7 BC 7 Equating i, j, and k components, 3 ( MA ) x + TBC = 0 7

5 = 0

c ( MA ) x +

( MA ) y + 1200 = 0

(4) (5)

9 TBC + 600 = 0 7 Solving Eqs. (1) to (6),

( MA ) z +

(6)

Ans.

TBC = 1400 N = 1.40 kN Ay = 800 N

Ans.

Ax = 1200 N = 1.20 kN

Ans.

( MA ) x = 600 N # m

Ans.

( MA ) y = - 1200 N # m = 1.20 kN # m

Ans.

( MA ) z = - 2400 N # m = 2.40 kN # m

Ans.

The negative signs indicate that Ay, ( MA ) x, ( MA ) y and ( MA ) z are directed in sense opposite to those shown in FBD.

Ans: TBC = 1.40 kN Ay = 800 N Ax = 1.20 kN ( MA ) x = 600 N # m ( MA ) y = 1.20 kN # m ( MA ) z = 2.40 kN # m 465

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*5–72. z

Determine the components of reaction at the ball-and-socket joint A and the tension in each cable necessary for equilibrium of the rod.

C

2m 2m D A

x

Solution

B

3m

Force And Position Vectors. The coordinates of points A, B, C, D and E are A(0, 0, 0), B(6, 0, 0), C(0, -2, 3) m, D(0, 2, 3) m and E(3, 0, 0) m respectively. FBC = FBC a

(0 - 6)i + ( -2 - 0)j + (3 - 0)k rBC 6 2 3 b = FBC £ § = - FBCi - FBC j + FBCk 2 2 2 rBC 7 7 7 2(0 - 6) + ( - 2 - 0) + (3 - 0)

FBD = FBD a

(0 - 6)i + (2 - 0)j + (3 - 0)k rBD 6 2 3 b = FBD £ § = - FBDi + FBD j + FBDk 2 2 2 rBD 7 7 7 2(0 - 6) + (2 - 0) + (3 - 0)

FA = Axi + Ay j + Azk F = { -600k} N rAB = {6i} m

3m E

rAE = {3i} m

Equations of Equilibrium. Referring to the FBD of the rod shown in Fig. a, the force equation of equilibrium gives ΣF = 0;  FBC + FBD + FA + F = 0 6 6 2 2 3 3 a - FBC - FBD + Ax bi + a FBD - FBC + Ay b j + a FBC + FBD + Az - 600bk = 0 7 7 7 7 7 7

466

600 N

3m y

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5–72. Continued

Equating i, j and k components, 6 6 - FBC - FBD + Ax = 0 7 7

(1)

2 2 F - FBC + Ay = 0 7 BD 7

(2)

3 3 F + FBD + Az - 600 = 0 7 BC 7

(3)

The moment equation of equilibrium gives ΣMA = 0;  rAE * F + rAB * (FBC + FBD ) = 0 i †3 0

j 0 0

k 0 † + 5 - 600

c 1800 -

i 6

j 0

k 0

6 - (FBC + FBD) 7

2 (F - FBC) 7 BD

3 (F + FBD) 7 BC

5 = 0

18 12 (FBC + FBD) d j + (F - FBC)k = 0 7 7 BD

Equating j and k components, 1800 -

18 (F + FBD) = 0 7 BC

(4)

12 (F - FBC) = 0 7 BD

(5)

Solving Eqs. (1) to (5), FBD = FBC = 350 N

Ans.

Ax = 600 N

Ans.

Ay = 0

Ans.

Az = 300 N

Ans.

Ans: FBD = Ax = Ay = Az = 467

FBC = 350 N 600 N 0 300 N

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5–73. z

The stiff-leg derrick used on ships is supported by a ball-andsocket joint at D and two cables BA and BC. The cables are attached to a smooth collar ring at B, which allows rotation of the derrick about z axis. If the derrick supports a crate having a mass of 200 kg, determine the tension in the cables and the x, y, z components of reaction at D.

B

6m

7.5 m

C 6m D

Solution 2 6 T - TBC 7 BA 9 3 3 ΣFy = 0;      Dy - TBA - TBC 7 9 6 6 ΣFz = 0;      Dz - TBA - TBC 7 9 3 3 ΣMx = 0;   TBA(6) + TBC(6) 7 9 2 6 ΣMy = 0;   TBA(6) - TBC(6) 7 9 ΣFx = 0;      Dx +

1m

2m

y

A

= 0

3m

4m

x

= 0 - 200(9.81) = 0 - 200(9.81)(4) = 0 + 200(9.81)(1) = 0



TBA = 2.00 kN

Ans.



TBC = 1.35 kN

Ans.



Dx = 0.327 kN

Ans.



Dy = 1.31 kN

Ans.



Dz = 4.58 kN

Ans.

Ans: TBA = 2.00 kN TBC = 1.35 kN Dx = 0.327 kN Dy = 1.31 kN Dz = 4.58 kN 468

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5–74. z

The bent rod is supported at A, B, and C by smooth journal bearings. Determine the components of reaction at the bearings if the rod is subjected to the force F = 800 N. The bearings are in proper alignment and exert only force reactions on the rod.

C A

2m

2m

x

Solution

B 0.75 m

1m

30 60

Equations of Equilibrium. The x, y and z components of force F are

F

Fx = 800 cos 60° cos 30° = 346.41 N

y

Fy = 800 cos 60° sin 30° = 200 N Fz = 800 sin 60° = 692.82 N Referring to the FBD of the bent rod shown in Fig. a, ΣMx = 0;

-Cy(2) + Bz(2) - 692.82 (2) = 0

(1)

ΣMy = 0;

Bz(1) + Cx(2) = 0

(2)

ΣMz = 0;

-Cy(1.75) - Cx(2) - By(1) - 346.41(2) = 0

(3)

ΣFx = 0;

Ax + Cx + 346.41 = 0

(4)

ΣFy = 0;

200 + By + Cy = 0

(5)

ΣFz = 0;

Az + Bz - 692.82 = 0

(6)

Solving Eqs. (1) to (6) Cy = 800 N Bz = - 107.18 N = 107 N By = 600 N

Ans.

Cx = 53.59 N = 53.6 N Ax = 400 N Az = 800 N

Ans.

The negative signs indicate that Cy, Bz and Az are directed in the senses opposite to those shown in FBD.

Ans: Cy = 800 N, B z = 107 N, B y = 600 N, Cx = 53.6 N, A x = 400 N, A z = 800 N 469

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5–75. z

Determine the components of reaction at the ball-andsocket joint A and the tension in the supporting cables DB and DC.

1.5 m B 1.5 m C D

Solution

1.5 m

Force And Position Vectors. The coordinates of points A, B, C, and D are A(0, 0, 0), B(0, - 1.5, 3) m, C(0, 1.5, 3) m and D(1, 0, 1) m, respectively. x FDC

(0 - 1)i + (1.5 - 0)j + (3 - 1)k rDC = FDC a b = FDC £ § rDC 2(0 - 1)2 + (1.5 - 0)2 + (3 - 1)2 = -

FDB = FDB a

1 1.5 2 FCDi + FDC j + FDC k 17.25 17.25 17.25

(0 - 1)i + ( -1.5 - 0)j + (3 - 1)k rDB b = FDB £ § rDB 2(0 - 1)2 + ( - 1.5 - 0)2 + (3 - 1)2 = -

1 1.5 2 FDBi + FDB j + FDBk 17.25 17.25 17.25

FA = Axi + Ay j + Azk F = { - 2400k} N

rAD = (1 - 0)i + (1 - 0)k = {i + k} m rF = {4i} m

470

800 N/m

3m

1m

1.5 m

A

3m

1m y

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5–75. Continued

Equations of Equilibrium. Referring to the FBD of the assembly shown in Fig. a. Force equation of equilibrium gives ΣF = 0;  FDC + FDB + FA + F = 0 a-

1.5 1 1 1.5 FDC FDB + Ax bi + a FDC FDB + Ay bj 17.25 17.25 17.25 17.25 + a

2 2 FDC + FDB + Az - 2400bk = 0 17.25 17.25

Equating i, j and k components, -

1 1 FDC FDB + Ax = 0 17.25 17.25

(1)

1.5 1.5 FDC F + Ay = 0 7.25 DB 17.25

(2)

2 2 FDC + FDB + Az - 2400 = 0 17.25 17.25 Moment equation of equilibrium gives

(3)

ΣMA = 0;  rF * F + rAD * (FDB + FDC) = 0 i †4 0

-

j 0 0

k 0 † + 5 - 2400

i 1 -

1 (FDB + FDC) 17.25

j 0

k 1

1.5 (FDC - FDB) 17.25

2 (FDC + FDB) 17.25

5 = 0

3 1.5 1.5 (FDC - FDB)i + c 9600 (FDC + FDB) d j + (FDC + FDB)k = 0 17.25 17.25 17.25

Equating i, j and k Components -

1.5 (FDC - FDB) = 0 17.25

9600 -

(4)

3 (FDC + FDB) = 0 17.25

(5)

1.5 (FDC - FDB) = 0 17.25

(6)

Solving Eqs. (1) to (6)

FDC = FDB = 4308.13 N = 4.31 kN

Ans.

Ax = 3200 N = 3.20 kN

Ans.

Ay = 0

Ans.

Az = -4000 N = -4 kN

Ans.

Negative sign indicates that Az directed in the sense opposite to that shown in FBD.

471

Ans: FDC = Ax = Ay = Az =

FDB = 4.31 kN 3.20 kN 0 -4 kN

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*5–76. The bent rod is supported at A, B, and C by smooth journal bearings. Determine the magnitude of F which will cause the positive x component of reaction at the bearing C to be Cx = 50 N. The bearings are in proper alignment and exert only force reactions on the rod.

z

C A

2m

2m

Solution

x

B 0.75 m

1m

30 60

Equations of Equilibrium. The x, y and z components of force F are

F

Fx = F cos 60° cos 30° = 0.4330 F

y

Fy = F cos 60° sin 30° = 0.25 F Fz = F sin 60° = 0.8660 F Here, it is required that Cx = 50. Thus, by referring to the FBD of the beat rod shown in Fig. a, ΣMx = 0;

(1)

-Cy(2) + Bz(2) - 0.8660 F(2) = 0

ΣMy = 0; ΣMz = 0;

(2)

Bz(1) + 50(2) = 0

(3)

- Cy(1.75) - 50(2) - By(1) - 0.4330 F(2) = 0

ΣFy = 0;

(4)

0.25 F + By + Cy = 0

Solving Eqs. (1) to (4) Ans.

F = 746.41 N = 746 N Cy = -746.41 N Bz = -100 N By = 559.81 N

Ans: F = 746 N 472

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5–77. z

The member is supported by a pin at A and cable  BC. Determine the components of reaction at these supports if the cylinder has a mass of 40 kg.

0.5 m B

1m A D

Solution

y 1m

Force And Position Vectors. The coordinates of points B, C and D are B(0, -0.5, 1) m, x C(3, 1, 0) m and D(3, - 1, 0) m, respectively. FCB = FCB a

(0 - 3)i + ( -0.5 - 1)j + (1 - 0)k rCB b = FCB c d rCB 2(0 - 3)2 + ( - 0.5 - 1)2 + (1 - 0)2 6 3 2 = - FCBi - FCBj + FCBk 7 7 7

W = { -40(9.81)k} N = { -392.4k} N. FA = Ax i + Ay j + Az k MA = ( MA ) x i + ( MA ) z k rAC = {3i + j} m  rAD = {3i - j} m Equations of Equilibrium. Referring to the FBD of the assembly shown in Fig. a. the force equation of equilibrium gives ΣF = 0;  FCB + W + FA = 0; 6 3 2 a - FCB + Ax bi + a - FCB + Ay bj + a FCB + Az - 392.4bk = 0 7 7 7

Equating i, j and k components

6 - FCB + Ax = 0 7 3 - FCB + Ay = 0 7 2 FCB + Az - 392.4 = 0 7 The moment equation of equilibrium gives

(1) (2) (3)

ΣMA = 0;  rAC * FCB + rAD * W + MA = 0 i 3 5 6 - FCB 7

1m

j 1 3 - FCB 7

k i 0 5 + †3 2 0 FCB 7

j -1 0

k 0 † + ( MA ) x i + ( MA ) Z k = 0 -392.4

2 6 9 6 c FCB + 392.4 + ( MA ) x d i + a - FCB + 1177.2bj + c - FCB + FCB + ( MA ) z d k = 0 7 7 7 7

473

C

3m

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5–77. Continued

Equating i, j and k components, 2 F + 392.4 + ( MA ) x = 0 7 CB 6 - FCB + 1177.2 = 0 7 9 6 - FCB + FCB + ( MA ) z = 0 7 7 Solving Eqs (1) to (6),

(4) (5) (6)

Ans.

FCB = 1373.4 N = 1.37 kN

( MA ) x = - 784.8 N # m = 785 N # m

Ans.

( MA ) z = 588.6 N # m = 589 N # m

Ans.

Ax = 1177.2 N = 1.18 kN

Ans.

Ay = 588.6 N = 589 N

Ans.

Az = 0

Ans.

Ans: FCB = 1.37 kN 1 M A 2 x = 785 N # m

1 MA 2z

= Ax = Ay = Az =

474

589 N # m 1.18 kN 589 N 0

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5–78. The platform has mass of 3 Mg and center of mass located at G. If it is lifted with constant velocity using the three cables, determine the force in each of these cables.

z

E

C

6m

B x

Given:

3m

2m

G

A

Solution

3m D

4m

y

4m

3m

M = 3 Mg a = 4m b = 3m c = 3m d = 4m e = 2m b # FAC 2

2

2a + b

-

c # FBC 2a2 + c 2

M # g # e - FAC # a # a 2

2a + c 2

= 0

d + e 2

2

2a + b

# FBC 1b + c2

- FBC #

a # 1d + r 2 2a2 + c 2

= 0

- M # g # b + FDE # b = 0

FAC £ FBC ≥ = Find 1 FAC, FBC, FDE 2 FDE

FAC 6.13 £ FBC ≥ = £ 6.13 ≥ kN FDE 19.62

Ans.

Ans: FAC = 6.13 kN FBC = 6.13 kN FDE = 19.62 kN 475

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5–79. The platform has a mass of 2 Mg and center of mass located at G. If it is lifted using the three cables, determine the force in each of the cables. Solve for each force by using a single moment equation of equilibrium.

z

E

C

6m

B x

3m

Given: M = 2 Mg a = 4m b = 3m

y

4m

3m

c = 3m d = 4m e = 2m

0 rBC = £ -c ≥ a

rAD = £

2m

G

A

Solution

3m D

4m

-e -d b ≥ 0

0 rAC = ° b ¢ a rBD = £

-d -e -c ≥ 0

First find FDE. ΣMy′ = 0;

FDE # 1d + e2 - M # g # d = 0

FDE =

M#g#d FDE = 13.1 kN d + e

Next find FBC. Guess FBC := 1kN e 0 e + d rBC b S # rAD = 0 FBC = Find1FBC) Given C £ 0 ≥ * £ 0 ≥ * ° c ¢ * aFBC #  r BC  # 0 -M g 0 FBC = 4.09 kN

Ans.

Ans: FBC = 4.09 kN 476

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*5–80. The boom is supported by a ball-and-socket joint at A and a guy wire at B. If the 5-kN loads lie in a plane which is parallel to the x–y plane, determine the x, y, z components of reaction at A and the tension in the cable at B.

z 5 kN 30° 5 kN

30° 3m

Solution Equations of Equilibrium: ΣMx = 0; ΣMy = 0; ΣFx = 0; ΣFy = 0; ΣFz = 0;

235 sin 30° 152 4 - TB 11.52 = 0 TB = 16.67 kN = 16.7 kN

Ans.

5 cos 30° 152 - 5 cos 30° 152 = 01Statisfied! 2

Ax + 5 cos 30° - 5 cos 30° = 0 Ax = 0

A

Ans.

Ay - 215 sin 30° 2 = 0 Ay = 5.00 kN

Ans.

Az - 16.67 = 0

Ans.

Az = 16.7 kN

2m B

x

y

1.5 m

Ans: TB = 16.7 kN Ax = 0 Az = 16.7 kN 477

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5–81. The shaft is supported by three smooth journal bearings at A, B, and C. Determine the components of reaction at these bearings.

900 N 600 N

450 N 0.9 m

0.6 m x

0.9 m 500 N

C

0.6 m

A

0.9 m

B

0.9 m

0.9 m y

Solution Equations of Equilibrium: From the free-body diagram, Fig. a, Cy and Cz, can be obtained by writing the force equation of equilibrium along the y axis and the moment equation of equilibrium about the y axis. ΣFy = 0; Cy - 450 = 0 Ans.

Cy = 450 N ΣMy = 0;

Cz 10.9 + 0.92 - 90010.92 + 60010.62 = 0 Cz = 250 N

Ans.

Using the above results ΣMx = 0; Bz(0.9 + 0.9) + 250(0.9 + 0.9 + 0.9) + 450(0.6) - 900(0.9 + 0.9 + 0.9) - 600(0.9) = 0 Ans.

Bz = 1125 N = 1.125 kN ΣMx′ = 0;

60010.92 + 45010.62 - 90010.92 + 250(0.9) - Az 10.9 + 0.92 = 0 Ans. Az = 125 N

ΣMz = 0;

-Bx 10.9 + 0.92 + 50010.92 + 45010.92 - 45010.9 + 0.92 = 0 Bx = 25 N

ΣFx = 0;

Ax + 25 - 500 = 0 Ax = 475 N

Ans. Ans.

Ans: Cy = 450 N, Cz = 250 N, B z = 1.125 kN, A z = 125 N, B x = 25 N, A x = 475 N 478

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5–82. z

Both pulleys are fixed to the shaft and as the shaft turns with constant angular velocity, the power of pulley A is transmitted to pulley B. Determine the horizontal tension T in the belt on pulley B and the x, y, z components of reaction at the journal bearing C and thrust bearing D if u = 45°. The bearings are in proper alignment and exert only force reactions on the shaft.

200 mm 250 mm

50 N D u

300 mm C 80 mm A

x

150 mm B

y T

65 N 80 N

SOLUTION Equations of Equilibrium: ©Mx = 0;

6510.082 - 8010.082 + T10.152 - 5010.152 = 0 Ans.

T = 58.0 N ©My = 0;

165 + 80210.452 - 50 sin 45°10.22 - Cz 10.752 = 0 Ans.

Cz = 77.57 N = 77.6 N ©Mz = 0;

©Fx = 0; ©Fy = 0;

58.010.22 + 50 cos 45°10.22 - Cy 10.752 = 0 Cy = 24.89 N = 24.9 N

Ans.

Dx = 0

Ans.

Dy + 24.89 - 50 cos 45° - 58.0 = 0 Ans.

Dy = 68.5 N ©Fz = 0;

Dz + 77.57 + 50 sin 45° - 80 - 65 = 0 Ans.

Dz = 32.1 N

Ans: T = Cz = Cy = Dy = Dz = 479

58.0 N 77.6 N 24.9 N 68.5 N 32.1 N

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5–83. Determine the tension in cables BD and CD and the x, y, z components of reaction at the ball-and-socket joint at A.

z

D

SOLUTION

3m

Given:

300 N

F

300 N

a

3m

b

1m

c

0.5 m

d

1.5 m

rBD

§ b · ¨d¸ ¨ ¸ ©a¹

rCD

§ b · ¨ d ¸ ¨ ¸ ©a¹

B

x

0.5 m C

§ Ax · § 0 · ¨ ¸ rBD rCD  TCD ¨ 0 ¸ ¨ Ay ¸  TBD ¨ ¸ rBD rCD ¨A ¸ © F ¹ © z¹

Find  TBD  TCD  A x  A y  A z

1m

y

0

§d· §d· §d  c· § 0 · rBD · ¨ ¸ § rCD · ¨ ¨ d ¸ u § T  d u T  0 ¸u¨ 0 ¸ ¨ ¸ ¨© BD rBD ¸¹ ¨ ¸ ¨© CD rCD ¸¹ ¨ ¸ ¨ ¸ ©0¹ ©0¹ © 0 ¹ © F ¹ § TBD · ¨ ¸ ¨ TCD ¸ ¨ Ax ¸ ¨ ¸ ¨ Ay ¸ ¨ A ¸ © z ¹

A

1.5 m

§ TBD · ¨ ¸ © TCD ¹ § Ax · ¨ ¸ ¨ Ay ¸ ¨A ¸ © z¹

0

§ 116.7 · ¨ ¸N © 116.7 ¹ § 66.7 · ¨ 0 ¸N ¨ ¸ © 100 ¹

Ans.

Ans.

Ans: T BD = 116.7 N, T CD = 116.7 N A x = 66.7 N, A y = 0 N, A z = 100 N 480

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*5–84. z

Both pulleys are fixed to the shaft and as the shaft turns with constant angular velocity, the power of pulley A is transmitted to pulley B. Determine the horizontal tension T in the belt on pulley B and the x, y, z components of reaction at the journal bearing C and thrust bearing D if u = 0°. The bearings are in proper alignment and exert only force reactions on the shaft.

200 mm 250 mm

50 N D u

300 mm C 80 mm A

x

150 mm B

y T

65 N

SOLUTION

80 N

Equations of Equilibrium: ©Mx = 0;

6510.082 - 8010.082 + T10.152 - 5010.152 = 0 Ans.

T = 58.0 N ©My = 0;

165 + 80210.452 - Cz 10.752 = 0 Ans.

Cz = 87.0 N ©Mz = 0;

©Fx = 0; ©Fy = 0;

150 + 58.0210.22 - Cy 10.752 = 0 Cy = 28.8 N

Ans.

Dx = 0

Ans.

Dy + 28.8 - 50 - 58.0 = 0 Ans.

Dy = 79.2 N ©Fz = 0;

Dz + 87.0 - 80 - 65 = 0 Ans.

Dz = 58.0 N

Ans: T = 58.0 N Cz = 87.0 N Cy = 28.8 N Dx = 0 Dy = 79.2 N Dz = 58.0 N 481

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5–85. The sign has a mass of 100 kg with center of mass at G. Determine the x, y, z components of reaction at the ball-andsocket joint A and the tension in wires BC and BD.

z 1m

D

2m

C 1m 2m A x

SOLUTION

B G

Equations of Equilibrium: Expressing the forces indicated on the free-body diagram, Fig. a, in Cartesian vector form, we have

1m 1m

FA = A xi + A yj + A zk W = {- 100(9.81)k} N = {- 981k} N FBD = FBDuBD = FBD ≥

FBC = FBCuBC = FBC ≥

(-2 - 0)i + (0 - 2)j + (1 - 0)k 2( - 2 - 0)2 + (0 - 2)2 + (1 - 0)2

(1 - 0)i + (0 - 2)j + (2 - 0)k 2(1 - 0)2 + (0 - 2)2 + (2 - 0)2

¥ = a-

2 2 1 FBDi - FBDj + FBDkb 3 3 3

1 2 2 ¥ = a FBCi - FBCj + FBCkb 3 3 3

Applying the forces equation of equilibrium, we have ©F = 0;

FA + FBD + FBC + W = 0

2 2 1 1 2 2 (A xi + A yj + A zk) + a - FBDi - FBDj + FBDk b + a FBCi - FBCj + FBCk b + ( - 981 k) = 0 3 3 3 3 3 3 a Ax -

2 1 2 2 1 2 FBD + FBC bi + aA y - FBD - FBC b j + aA z + FBD + FBC - 981 bk = 0 3 3 3 3 3 3

Equating i, j, and k components, we have Ax -

2 1 FBD + FBC = 0 3 3

(1)

Ay -

2 2 F - FBC = 0 3 BD 3

(2)

Az +

1 2 FBD + FBC - 981 = 0 3 3

(3)

In order to write the moment equation of equilibrium about point A, the position vectors rAG and rAB must be determined first. rAG = {1j} m rAB = {2j} m

482

y

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5–85. Continued

Thus, ©M A = 0; rAB * (FBC + FBD) + (rAG * W) = 0 2 2 2 2 1 1 (2j) * c a FBC - FBD b i - a FBC + FBD bj + a FBC + FBD bk d + (1j) * (- 981k) = 0 3 3 3 3 3 3 4 2 4 2 a FBC + FBD - 981 bi + a FBD - FBC b k = 0 3 3 3 3 Equating i, j, and k components we have 2 4 F + FBC - 981 = 0 3 BC 3

(4)

4 2 F - FBC = 0 3 BC 3

(5)

Ans: FBD = 294 N FBC = 589 N Ax = 0 Ay = 589 N Az = 490.5 N 483

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6–1.

300 N

Determine the force in each member of the truss, and state if the members are in tension or compression.

400 N D

C

2m

250 N

A

SOLUTION

B 2m

Method of Joints: Here, the support reactions A and C do not need to be determined. We will first analyze the equilibrium of joints D and B, and then proceed to analyze joint C.

200 N

Joint D: From the free-body diagram in Fig. a, we can write + ©F = 0; : x

400 - FDC = 0 FDC = 400 N (C)

+ c ©Fy = 0;

Ans.

FDA - 300 = 0 FDA = 300 N (C)

Ans.

Joint B: From the free-body diagram in Fig. b, we can write + ©F = 0; : x

250 - FBA = 0 FBA = 250 N (T)

+ c ©Fy = 0;

Ans.

FBC - 200 = 0 FBC = 200 N (T)

Ans.

Joint C: From the free-body diagram in Fig. c, we can write + c ©Fy = 0;

FCA sin 45° - 200 = 0 FCA = 282.84 N = 283 N (C)

+ ©F = 0; : x

Ans.

400 + 282.84 cos 45° - NC = 0 NC = 600 N

Note: The equilibrium analysis of joint A can be used to determine the components of support reaction at A.

Ans: Joint D, FDC = 400 N (C) FDA = 300 N (C) Joint B, FBA = 250 N (T) FBC = 200 N (T) Joint C, FCA = 283 N (C) 484

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6–2. Determine the force in each member of the truss and state if the members are in tension or compression. Set P1 = 20 kN, P2 = 10 kN.

P1 B

C

1.5 m P2 D 2m

Solution

A

Method of Joints. Start at joint C and then proceed to join D. Joint C. Fig. a + ΣFx = 0;     FCB = 0  S + c ΣFy = 0;   

FCD - 20 = 0  

Ans. Ans.

FCD = 20.0 kN (C) 

Joint D. Fig. b 3 + c ΣFy = 0;  FDB a b - 20.0 = 0  FDB = 33.33 kN (T) = 33.3 kN (T) Ans. 5 + ΣFx = 0;    10 + 33.33 a 4 b - FDA = 0 S 5 FDA = 36.67 kN (C) = 36.7 kN (C) Ans.

Ans: FCB = FCD = FDB = FDA = 485

0 20.0 kN (C) 33.3 kN (T) 36.7 kN (C)

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6–3. Determine the force in each member of the truss and state if the members are in tension or compression. Set P1 = 45 kN, P2 = 30 kN.

P1 B

C

1.5 m P2 D 2m

Solution

A

Method of Joints. Start at joint C and then proceed to joint D. Joint C. Fig. a + ΣFx = 0;   FCB = 0  S

Ans.

+ c ΣFy = 0;      FCD - 45 = 0    FCD = 45.0 kN (C) 

Ans.

Joint D. Fig. b 3 + c ΣFy = 0;   FDB a b - 45.0 = 0   FDB = 75.0 kN (T) 5

Ans.

+ ΣFx = 0;   30 + 75.0 a 4 b - FDA = 0   FDA = 90.0 kN (C) S 5

Ans.

Ans: FCB = FCD = FDB = FDA = 486

0 45.0 kN (C) 75.0 kN (T) 90.0 kN (C)

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*6–4. Determine the force in each member of the truss, and state if the members are in tension or compression. Set u = 0°.

D

3 kN

1.5 m A

SOLUTION

B

Support Reactions: Applying the equations of equilibrium to the free-body diagram of the entire truss,Fig.a, we have a + ©MA = 0;

2m

3 - Ax = 0 A x = 3 kN

+ c ©Fy = 0;

A y + 3.125 - 4 = 0 A y = 0.875 kN

Method of Joints: We will use the above result to analyze the equilibrium of joints C and A, and then proceed to analyze of joint B. Joint C: From the free-body diagram in Fig. b, we can write + c ©Fy = 0;

3 3.125 - FCD a b = 0 5 FCD = 5.208 kN = 5.21 kN (C)

+ ©F = 0; : x

Ans.

4 5.208 a b - FCB = 0 5 FCB = 4.167 kN = 4.17 kN (T)

Ans.

Joint A: From the free-body diagram in Fig. c, we can write + c ©Fy = 0;

3 0.875 - FAD a b = 0 5 FAD = 1.458 kN = 1.46 kN (C)

+ : ©Fx = 0;

Ans.

4 FAB - 3 - 1.458a b = 0 5 FAB = 4.167 kN = 4.17 kN (T)

Ans.

Joint B: From the free-body diagram in Fig. d, we can write + c ©Fy = 0;

FBD - 4 = 0 FBD = 4 kN (T)

+ ©F = 0; : x

4.167 - 4.167 = 0

Ans. (check!)

Note: The equilibrium analysis of joint D can be used to check the accuracy of the solution obtained above. Ans: FCD = FCB = FAD = FAB = FBD = 487

2m 4 kN

NC (2 + 2) - 4(2) - 3(1.5) = 0 NC = 3.125 kN

+ : ©Fx = 0;

C

5.21 kN (C) 4.17 kN (T) 1.46 kN (C) 4.17 kN (T) 4 kN (T)

u

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6–5. Determine the force in each member of the truss, and state if the members are in tension or compression. Set u = 30°.

D

3 kN

1.5 m A

SOLUTION

B

Support Reactions: From the free-body diagram of the truss, Fig. a, and applying the equations of equilibrium, we have a + ©MA = 0;

NC cos 30°(2 + 2) - 3(1.5) - 4(2) = 0 NC = 3.608 kN

+ : ©Fx = 0;

3 - 3.608 sin 30° - A x = 0 A x = 1.196 kN

+ c ©Fy = 0;

A y + 3.608 cos 30° - 4 = 0 A y = 0.875 kN

Method of Joints: We will use the above result to analyze the equilibrium of joints C and A, and then proceed to analyze of joint B. Joint C: From the free-body diagram in Fig. b, we can write + c ©Fy = 0;

3 3.608 cos 30° - FCD a b = 0 5 FCD = 5.208 kN = 5.21 kN (C)

+ ©F = 0; : x

Ans.

4 5.208 a b - 3.608 sin 30° - FCB = 0 5 FCB = 2.362 kN = 2.36 kN (T)

Ans.

Joint A: From the free-body diagram in Fig. c, we can write + c ©Fy = 0;

3 0.875 - FAD a b = 0 5 FAD = 1.458 kN = 1.46 kN (C)

+ : ©Fx = 0;

Ans.

4 FAB - 1.458 a b - 1.196 = 0 5 FAB = 2.362 kN = 2.36 kN (T)

Ans.

Joint B: From the free-body diagram in Fig. d, we can write + c ©Fy = 0;

FBD - 4 = 0 FBD = 4 kN (T)

+ : ©Fx = 0;

C

2.362 - 2.362 = 0

Ans. (check!)

Note: The equilibrium analysis of joint D can be used to check the accuracy of the solution obtained above. Ans: FCD = FCB = FAD = FAB = FBD = 488

5.21 kN (C) 2.36 kN (T) 1.46 kN (C) 2.36 kN (T) 4 kN (T)

2m

2m 4 kN

u

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6–6. The truss, used to support a balcony, is subjected to the loading shown. Approximate each joint as a pin and determine the force in each member. State whether the members are in tension or compression.

40 kN

60 kN

B

A

C 45°

45° 4m

SOLUTION

E D

+ c ΣFy = 0; +

S ΣFx = 0; Joint B: + c ΣFy = 0; +

S ΣFx = 0; Joint D: + c ΣFy = 0; +

S ΣFx = 0;

4m

4m

Joint A: FAD sin 45° - 60 = 0 FAD = 84.853 = 84.9 kN

Ans.

FAB - 84.853 cos 45° = 0 FAB = 60 kN1T 2

Ans.

FBD - 40 = 0 FBD = 40 kN1C 2

Ans. 60 kN

FBC - 60 = 0 FBC = 60 kN1T 2

Ans.

FDC sin 45° - 40 - 84.853 sin 45° = 0 FDC = 141.42 kN = 141 kN1T 2

Ans.

84.853 cos 45° + 141.42 cos 45° - FOE = 0 FDE = 160 kN1C 2

84.853 kN

40 kN

Ans. 60 kN

40 kN

Ans: Joint A, FAD = 84.9 kN FAB = 60 kN (T) Joint B, FBD = 40 kN (C) FBC = 60 kN (T) Joint D, FDC = 141 kN (T) FDE = 160 kN (C) 489

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6–7. Determine the force in each member of the truss and state if the members are in tension or compression.

4 kN 3m

3m B

3m

C

D 3m 5m

A F

SOLUTION

E

Support Reactions: a + ©MD = 0;

4162 + 5192 - Ey 132 = 0

+ c ©Fy = 0;

23.0 - 4 - 5 - D y = 0

+ ©F = 0 : x

5 kN

Ey = 23.0 kN Dy = 14.0 kN

Dx = 0

Method of Joints: Joint D: + c ©Fy = 0;

FDE ¢

5 234

≤ - 14.0 = 0

FDE = 16.33 kN 1C2 = 16.3 kN 1C2 + ©F = 0; : x

16.33 ¢

3 234

Ans.

≤ - FDC = 0

FDC = 8.40 kN 1T2

Ans.

Joint E: + ©F = 0; : x

FEA ¢

3 210

≤ - 16.33 ¢

3 234

≤ = 0

FEA = 8.854 kN 1C2 = 8.85 kN 1C2 + c ©Fy = 0;

23.0 - 16.33 ¢

5 234

≤ - 8.854 ¢

FEC = 6.20 kN 1C2

1 210

Ans.

≤ - FEC = 0 Ans.

Joint C: + c ©Fy = 0;

6.20 - FCF sin 45° = 0 FCF = 8.768 kN 1T2 = 8.77 kN 1T2

+ ©F = 0; : x

Ans.

8.40 - 8.768 cos 45° - FCB = 0 F CB = 2.20 kN T

Ans.

490

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6–7. Continued

Joint B: + ©F = 0; : x

2.20 - FBA cos 45° = 0 FBA = 3.111 kN 1T2 = 3.11 kN 1T2

+ c ©Fy = 0;

Ans.

FBF - 4 - 3.111 sin 45° = 0 FBF = 6.20 kN 1C2

Ans.

+ c ©Fy = 0;

8.768 sin 45° - 6.20 = 0

(Check!)

+ ©F = 0; : x

8.768 cos 45° - FFA = 0

Joint F:

FFA = 6.20 kN 1T2

Ans.

Ans: FDE = 16.3 kN (C) FDC = 8.40 kN (T) FEA = 8.85 kN (C) FEC = 6.20 kN (C) FCF = 8.77 kN (T) FCB = 2.20 kN (T) FBA = 3.11 kN (T) FBF = 6.20 kN (C) FFA = 6.20 kN (T) 491

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*6–8.

B

C

Determine the force in each member of the truss. State whether the members are in tension or compression. Set P = 8 kN. A

SOLUTION

60°

4m

Method of Joints: In this case, the support reactions are not required for determining the member forces. Joint D: + c ©Fy = 0;

FDC sin 60° - 8 = 0 FDC = 9.238 kN 1T2 = 9.24 kN 1T2

+ ©F = 0; : x

Ans.

FDE - 9.238 cos 60° = 0 FDE = 4.619 kN 1C2 = 4.62 kN 1C2

Ans.

Joint C: + c ©Fy = 0;

FCE sin 60° - 9.238 sin 60° = 0 FCE = 9.238 kN 1C2 = 9.24 kN 1C2

+ ©F = 0; : x

Ans.

219.238 cos 60°2 - FCB = 0 FCB = 9.238 kN 1T2 = 9.24 kN 1T2

Ans.

Joint B: + c ©Fy = 0;

FBE sin 60° - FBA sin 60° = 0 FBE = FBA = F

+ ©F = 0; : x

9.238 - 2F cos 60° = 0 F = 9.238 kN

Thus, FBE = 9.24 kN 1C2

FBA = 9.24 kN 1T2

Ans.

Joint E: + c ©Fy = 0;

Ey - 219.238 sin 60°2 = 0

+ ©F = 0; : x

FEA + 9.238 cos 60° - 9.238 cos 60° - 4.619 = 0

Ey = 16.0 kN

FEA = 4.62 kN 1C2

Ans.

Note: The support reactions Ax and Ay can be determinedd by analyzing Joint A using the results obtained above.

Ans: FDC FCE FBE FEA 492

= = = =

60°

E

9.24 kN 1T2, FDE = 4.62 kN 1C2 9.24 kN 1C2, FCB = 9.24 kN 1T2 9.24 kN 1C2, FBA = 9.24 kN 1T2 4.62 kN 1C2

D

4m P

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6–9.

B

C

If the maximum force that any member can support is 8 kN in tension and 6 kN in compression, determine the maximum force P that can be supported at joint D. A

SOLUTION

60°

4m

Method of Joints: In this case, the support reactions are not required for determining the member forces.

60°

E

D

4m P

Joint D: FDC = 1.1547P 1T2

+ c ©Fy = 0;

FDC sin 60° - P = 0

+ ©F = 0; : x

FDE - 1.1547P cos 60° = 0

FDE = 0.57735P 1C2

Joint C: + c ©Fy = 0;

FCE sin 60° - 1.1547P sin 60° = 0 FCE = 1.1547P 1C2 211.1547P cos 60°2 - FCB = 0

FCB = 1.1547P 1T2

+ c ©Fy = 0;

FBE sin 60° - FBA sin 60° = 0

FBE = FBA = F

+ ©F = 0; : x

1.1547P - 2F cos 60° = 0

+ ©F = 0; : x Joint B:

F = 1.1547P

Thus, FBE = 1.1547P 1C2

FBA = 1.1547P 1T2

Joint E: + ©F = 0; : x

FEA + 1.1547P cos 60° - 1.1547P cos 60° - 0.57735P = 0 FEA = 0.57735P 1C2

From the above analysis, the maximum compression and tension in the truss member is 1.1547P. For this case, compression controls which requires 1.1547P = 6 P = 5.20 kN

Ans.

Ans: P = 5.20 kN 493

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6–10.

L

B

Determine the force in each member of the truss in terms of the load P and state if the members are in tension or compression.

L

L

C

L

L

A

D

E

SOLUTION

L

L P

Entire truss: a + ©MA = 0;

-P(L) + Dy (2 L) = 0 Dy =

P 2

P - P + Ay = 0 2

+ c ©Fy = 0;

Ay = + ©F = 0; : x

P 2

Ax = 0

Joint D: + c ©Fy = 0;

-FCD sin 60° +

P = 0 2

FCD = 0.577 P (C) + ©F = 0; ; x

Ans.

FDB - 0.577P cos 60° = 0 FDB = 0.289 P(T)

Ans.

Joint C: + c ©Fy = 0;

0.577 P sin 60° - FCE sin 60° = 0 FCE = 0.577 P (T)

+ ©F = 0; : x

Ans.

FBC - 0.577 P cos 60° - 0.577P cos 60° = 0 FBC = 0.577 P (C)

Ans.

Due to symmetry: FBE = FCE = 0.577 P (T)

Ans.

FAB = FCD = 0.577 P (C)

Ans.

FAE = FDE = 0.577 P (T)

Ans. Ans: Joint D: FCD = 0.577 P (C) FDB = 0.289 P (T) Joint C: FCE = 0.577 P (T) FBC = 0.577 P (C) Due to symmetry: FBE = FCE = 0.577 P (T) FAB = FCD = 0.577 P (C) FAE = FDE = 0.577 P (T) 494

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6–11. Each member of the truss is uniform and has a weight W. Remove the external force P and determine the approximate force in each member due to the weight of the truss. State if the members are in tension or compression. Solve the problem by assuming the weight of each member can be represented as a vertical force, half of which is applied at each end of the member.

L

B

L

L

C

L

L

A

D

E L

L P

SOLUTION Entire truss: a + ©MA = 0;

-

3 L 3 3 W a b - 2 W(L) - W a L b - W(2 L) + Dy (2 L) = 0 2 2 2 2

Dy =

7 W 2

Joint D: + c ©Fy = 0;

7 W - W - FCD sin 60° = 0 2 FCD = 2.887W = 2.89 W (C)

+ ©F = 0; : x

Ans.

2.887W cos 60° - FDE = 0 FDE = 1.44 W (T)

Ans.

Joint C: + c ©Fy = 0;

2.887W sin 60° -

3 W - FCE sin 60° = 0 2

FCE = 1.1547W = 1.15 W (T) + ©F = 0; : x

Ans.

FBC - 1.1547W cos 60° - 2.887W cos 60° = 0 FBC = 2.02 W (C)

Ans.

Due to symmetry: FBE = FCE = 1.15 W (T)

Ans.

FAB = FCD = 2.89 W (C)

Ans.

FAE = FDE = 1.44 W (T)

Ans. Ans: Joint D: FCD = 2.89 W (C) FDE = 1.44 W (T) Joint C: FCE = 1.15 W (T) FBC = 2.02 W (C) Due to symmetry: FBE = FCE = 1.15 W (T) FAB = FCD = 2.89 W (C) FAE = FDE = 1.44 W (T) 495

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*6–12. Determine the force in each member of the truss and state if the members are in tension or compression. Set P1 = 3 kN, P2 = 6 kN.

E

6m D

A B

C

4m

4m P1

Solution Support Reactions. Referring to the FBD of the entire truss shown in Fig. a, a+ΣMA = 0;   ND(12) - 3(4) - 6(8) = 0   ND = 5.00 kN a+ΣMD = 0;   

6(4) + 3(8) - Ay(12) = 0   Ay = 4.00 kN

+ ΣFx = 0;   Ax = 0 S Method of Joints. We will carry out the analysis of joint equilibrium according to the sequence of joints A, D, B and C. Joint A. Fig. b + c ΣFy = 0;  4.00 - FAEa



1 22

b = 0

Ans.

FAE = 4 22 kN (C) = 5.66 kN (C)

+ ΣFx = 0;    FAB - 4 22 a 1 b = 0  FAB = 4.00 kN (T) S 22

496

Ans.

4m P2

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6–12.   Continued

Joint D. Fig. c + c ΣFy = 0;  5.00 - FDE a

1 22

b = 0  FDE = 5 22 kN (C) = 7.07 kN (C) Ans.

+ ΣFx = 0;    5 22 a 1 b - FDC = 0  FDC = 5.00 kN (T) S 22

Ans.

Joint B. Fig. d

+ c ΣFy = 0;  FBE a

3 210

b - 3 = 0  FBE = 210 kN (T) = 3.16 kN (T) Ans.

+ ΣFx = 0;  FBC + 210 a S

Joint C. Fig. e

+ c ΣFy = 0;  FCE a

3 210

1

210

b - 4.00 = 0  FBC = 3.00 kN (T)

Ans.

b - 6 = 0  FCE = 2210 kN (T) = 6.32 kN (T) Ans.

+ ΣFx = 0;    5.00 - 3.00 - a2210b a S

1

210

b = 0(Check!!)

Ans: FAE = 5.66 kN (C) FAB = 4.00 kN (T) FDE = 7.07 kN (C) FDC = 5.00 kN (T) FBE = 3.16 kN (T) FBC = 3.00 kN (T) FCE = 6.32 kN (T) 497

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6–13. Determine the force in each member of the truss and state if the members are in tension or compression. Set P1 = 6 kN, P2 = 9 kN.

E

6m D

A B

C

4m

4m P1

Solution Support Reactions. Referring to the FBD of the entire truss shown in Fig. a, a+ΣMA = 0;   ND(12) - 6(4) - 9(8) = 0   ND = 8.00 kN a+ΣMD = 0;   9(4) + 6(8) - Ay(12) = 0   Ay = 7.00 kN + ΣFx = 0;   Ax = 0 S Method of Joints. We will carry out the analysis of joint equilibrium according to the sequence of joints A, D, B and C. Joint A. Fig. a + c ΣFy = 0;  7.00 - FAE a

1 22

b = 0  FAE = 7 22 kN (C) = 9.90 kN (C)

+ ΣFx = 0;  FAB - 722 a 1 b = 0  FAB = 7.00 kN (T) S 22

Ans. Ans.

Joint D. Fig. c

+ c ΣFy = 0;  8.00 - FDE a

1 22

b = 0  FDE = 8 22 kN (C) = 11.3 kN (C)

+ ΣFx = 0;  822 a 1 b - FDC = 0  FDC = 8.00 kN (T) S 22

Ans. Ans.

498

4m P2

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6–13.   Continued

Joint B. Fig. d + c ΣFy = 0;    FBE a

3 210

b - 6 = 0  FBE = 2 210 kN (T) = 6.32 kN (T) Ans.

+ ΣFx = 0;  FBC - 7.00 + a2210b a 1 b = 0  FBC = 5.00 kN (T) S 210

Ans.

Joint C. Fig. e

+ c ΣFy = 0;    FCE a

3 210

b - 9 = 0  FCE = 3210 kN = 9.49 kN (T) Ans.

+ ΣFx = 0;  8.00 - 5.00 - a3210ba 1 b = 0  (Check!!) S 210

Ans: FAE = 9.90 kN (C) FAB = 7.00 kN (T) FDE = 11.3 kN (C) FDC = 8.00 kN (T) FBE = 6.32 kN (T) FBC = 5.00 kN (T) FCE = 9.49 kN (T) 499

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6–14. Determine the force in each member of the truss, and state if the members are in tension or compression.

600 N

D

4m

SOLUTION

900 N

E C

Method of Joints: We will begin by analyzing the equilibrium of joint D, and then proceed to analyze joints C and E. Joint D: From the free-body diagram in Fig. a, + : ©Fx = 0;

B

A

3 FDE a b - 600 = 0 5

6m

FDE = 1000 N = 1.00 kN (C) + c ©Fy = 0;

4m

Ans.

4 1000 a b - FDC = 0 5 FDC = 800 N (T)

Ans.

Joint C: From the free-body diagram in Fig. b, + ©F = 0; : x

FCE - 900 = 0 FCE = 900 N (C)

+ c ©Fy = 0;

Ans.

800 - FCB = 0 FCB = 800 N (T)

Ans.

Joint E: From the free-body diagram in Fig. c, R+ ©Fx ¿ = 0;

- 900 cos 36.87° + FEB sin 73.74° = 0 FEB = 750 N (T)

Q+ ©Fy ¿ = 0;

Ans.

FEA - 1000 - 900 sin 36.87° - 750 cos 73.74° = 0 FEA = 1750 N = 1.75 kN (C)

Ans.

Ans: FDE = FDC = FCE = FCB = FEB = FEA = 500

1.00 kN (C) 800 N (T) 900 N (C) 800 N (T) 750 N (T) 1.75 kN (C)

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6–15.

B

P

Determine the force in each member of the truss in terms of the load P, and indicate whether the members are in tension or compression.

d C

A D

F

d

SOLUTION Support Reactions: a + ©ME = 0;

3 P(2d) - Ay a d b = 0 2

+ c ©Fy = 0;

4 P - Ey = 0 3

+ ©F = 0; : x

Ex - P = 0

Ey =

E

4 Ay = P 3

d

4 P 3

Ex = P

Method of Joints: By inspection of joint C, members CB and CD are zero force members. Hence Ans. FCB = FCD = 0 Joint A: + c ©Fy = 0;

FAB ¢

1 23.25

≤ -

4 P = 0 3

FAB = 2.40P (C) = 2.40P (C) + ©F = 0; : x

FAF - 2.404P ¢

1.5 23.25

Ans.

≤ = 0

FAF = 2.00P (T)

Ans.

Joint B: + ©F = 0; : x

2.404P ¢

1.5 23.25

≤ - P - FBF ¢

0.5 21.25

≤ - FBD ¢

0.5 21.25

≤ = 0

1.00P - 0.4472FBF - 0.4472FBD = 0 + c ©Fy = 0;

2.404P ¢

1 23.25

≤ + FBD ¢

1 21.25

≤ - FBF ¢

(1) 1 21.25

1.333P + 0.8944FBD - 0.8944FBF = 0

≤ = 0 (2)

Solving Eqs. (1) and (2) yield, FBF = 1.863P (T) = 1.86P (T)

Ans.

FBD = 0.3727P (C) = 0.373P (C)

Ans.

Joint F: + c ©Fy = 0;

FFE + ©F = 0; : x

1 1 ≤ - FFE ¢ ≤ = 0 21.25 21.25 = 1.863P (T) = 1.86P (T)

1.863P ¢

FFD + 2 B 1.863P ¢

0.5 21.25

Ans.

≤ R - 2.00P = 0

FFD = 0.3333P (T) = 0.333P (T)

Ans.

Joint D: + c ©Fy = 0;

FDE ¢

1 21.25

≤ - 0.3727P ¢

1 21.25

≤ = 0

FDE = 0.3727P (C) = 0.373P (C) + ©F = 0; : y

2 B 0.3727P ¢

0.5 21.25

Ans.

≤ R - 0.3333P = 0 (Check!)

501

d/2

d/2

d

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*6–16. B

P

If the maximum force that any member can support is 4 kN in tension and 3 kN in compression, determine the maximum force P that can be applied at joint B. Take d = 1 m.

d

SOLUTION

C

A

Support Reactions:

D

F

a + ©ME = 0;

3 P(2d) - Ay a d b = 0 2

+ c ©Fy = 0;

4 P - Ey = 0 3

Ey =

+ ©F = 0; : x

Ex - P = 0

Ex = P

4 Ay = P 3

d

4 P 3

E d

d/2

d/2

d

Method of Joints: By inspection of joint C, members CB and CD are zero force members. Hence FCB = FCD = 0 Joint A: 1

+ c ©Fy = 0;

FAB ¢

+ ©F = 0; : x

FAF - 2.404P ¢

4 P = 0 3

≤ -

23.25

1.5 23.25

FAB = 2.404P (C)

≤ = 0

FAF = 2.00P (T)

Joint B: + ©F = 0; : x

2.404P ¢

1.5 23.25

≤ - P - FBF ¢

0.5 21.25

≤ - FBD ¢

0.5 21.25

≤ = 0

1.00P - 0.4472FBF - 0.4472FBD = 0 + c ©Fy = 0;

2.404P ¢

1 23.25

≤ + FBD ¢

1 21.25

≤ - FBF ¢

(1) 1 21.25

≤ = 0

1.333P + 0.8944FBD - 0.8944FBF = 0

(2)

Solving Eqs. (1) and (2) yield, FBF = 1.863P (T)

FBD = 0.3727P (C)

Joint F: + c ©Fy = 0;

1.863P ¢

1 21.25

≤ - FFE ¢

1 21.25

≤ = 0

FFE = 1.863P (T) + ©F = 0; : x

FFD + 2 B 1.863P ¢

0.5 21.25

≤ R - 2.00P = 0

FFD = 0.3333P (T) Joint D: + c ©Fy = 0;

FDE ¢

1 21.25

≤ - 0.3727P ¢

1 21.25

≤ = 0

FDE = 0.3727P (C) + ©F = 0; : y

2 B 0.3727P ¢

0.5 21.25

≤ R - 0.3333P = 0 (Check!)

From the above analysis, the maximum compression and tension in the truss members are 2.404P and 2.00P, respectively. For this case, compression controls which requires Ans:

2.404P = 3

P = 1.25 kN

P = 1.25 kN 502

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6–17. Determine the force in each member of the Pratt truss, and state if the members are in tension or compression.

J 2m

SOLUTION

2m

Joint A: + c ©Fy = 0;

2m

20 - FAL sin 45° = 0

I

L

H

A B D C E F 2m 2m 2m 2m 2m 2m

FAL = 28.28 kN (C) + ©F = 0; : x

K

FAB - 28.28 cos 45° = 0

10 kN

FAB = 20 kN (T)

20 kN

G

10 kN

Joint B: + ©F = 0; : x

FBC - 20 = 0 FBC = 20 kN (T)

+ c ©Fy = 0;

FBL = 0

Joint L: R + ©Fx = 0;

FLC = 0

+Q©Fy = 0;

28.28 - FLK = 0 FLK = 28.28 kN (C)

Joint C: + : ©Fx = 0;

FCD - 20 = 0 FCD = 20 kN (T)

+ c ©Fy = 0;

FCK - 10 = 0 FCK = 10 kN (T)

Joint K: R + ©Fx - 0;

10 sin 45° - FKD cos (45° - 26.57°) = 0 FKD = 7.454 kN (L)

+ Q©Fy = 0;

28.28 - 10 cos 45° + 7.454 sin (45° - 26.57°) - FKJ = 0 FKJ = 23.57 kN (C)

Joint J: + : ©Fx = 0;

23.57 sin 45° - FJI sin 45° = 0 FJI = 23.57 kN (L)

+ c ©Fy = 0;

2 (23.57 cos 45°) - FJD = 0 FJD = 33.3 kN (T)

Ans.

FAL = FGH = FLK = FHI = 28.3 kN (C)

Ans.

Due to Symmetry Ans: Ans. FJD = 33.3 kN (T) Ans. FAL = FGH = FLK = FHI = 28.3 kN (C) FAB = FGF = FBC = FFE = FCD = FED = 20 kN (T) Ans. FBL = FFH = FLC = FHE = 0 F = FEI = 10 kN (T) Ans. CK FKJ = FIJ = 23.6 kN (C) Ans. FKD = FID = 7.45 kN (C)

FAB = FGF = FBC = FFE = FCD = FED = 20 kN (T) FBL = FFH = FLC = FHE = 0 FCK = FEI = 10 kN (T) FKJ = FIJ = 23.6 kN (C) FKD = FID = 7.45 kN (C) 503

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6–18. Determine the force in each member of the truss and state if the members are in tension or compression. Set P1 = 9 kN, P2 = 15 kN.

E

F

P1

D

4m

A

Solution

3m

Support Reactions. Referring to the FBD of the entire truss shown in Fig. a

a+ΣMC = 0;  15(3) - 9(4) - Ay(6) = 0  Ay = 1.50 kN + ΣFx = 0;      9 - Ax = 0  Ax = 9.00 kN S Method of Joints: By inspecting joints D and F, we notice that members DE, DC and FA are zero force members. Thus Ans.

FDE = FDC = FFA = 0

We will perform the joint equilibrium analysis by following the sequence of joints C, B, A and F. Joint C. Fig. b 4 + c ΣFy = 0;  13.5 - FCE a b = 0   5 FCE = 16.875 kN (C) = 16.9 kN (C)

Ans.

+ ΣFx = 0;  16.875 a 3 b - FCB = 0  FCB = 10.125 kN (T) = 10.1 kN (T) Ans. S 5

Joint B. Fig. c

+ ΣFx = 0;  10.125 - FBA = 0  FBA = 10.125 kN (T) = 10.1 kN (T) Ans. S + c ΣFy = 0;  FBE - 15 = 0  FBE = 15.0 kN (T)

Ans.

504

3m P2

a+ΣMA = 0;  NC(6) - 15(3) - 9(4) = 0  NC = 13.5 kN



C

B

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6–18. Continued

Joint A. Fig. d 4 + c ΣFy = 0;  1.50 - FAE a b = 0  FAE = 1.875 kN (C) 5

Ans.

+ ΣFx = 0;  10.125 - 1.875a 3 b - 9.00 = 0(Check!!) S 5

Joint F. Fig. e

+ ΣFx = 0;  9 - FFE = 0   FFE = 9.00 kN (C) S

Ans.

Ans: FCE = FCB = FBA = FBE = FAE = FFE = 505

16.9 kN (C) 10.1 kN (T) 10.1 kN (T) 15.0 kN (T) 1.875 kN (C) 9.00 kN (C)

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6–19. Determine the force in each member of the truss and state if the members are in tension or compression. Set P1 = 30 kN, P2 = 15 kN.

E

F

P1

D

4m

A

C

B

Solution

3m

Support Reactions.

3m P2

a+ΣMA = 0;  NC(6) - 15(3) - 30(4) = 0  NC = 27.5 kN Method of Joints: By inspecting joints D and F, we notice that members DE, DC and FA are zero force members. Thus FDE = FDC = FFA = 0 Ans. We will perform the joint equilibrium analysis by following the sequence of joints C, B, F and E. Joint C. Fig. b 4 + c ΣFy = 0;  27.5 - FCEa b = 0  FCE = 34.375 kN (C) = 34.4 kN (C)Ans. 5 + ΣFx = 0;  34.375a 3 b - FCB = 0  FCB = 20.625 kN (T) = 20.6 kN (T)Ans. S 5

Joint B. Fig. c

+ ΣFx = 0;  20.625 - FBA = 0  FBA = 20.625 kN (T) = 20.6 kN (T)Ans. S + c ΣFy = 0;  FBE - 15 = 0  FBE = 15.0 kN (T)Ans.

506

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6–19.

Continued

Joint F. Fig. d + ΣFx = 0;  30 - FFE = 0  FFE = 30.0 kN (C)Ans. S Joint E. Fig. e 4 4 + c ΣFy = 0;  34.375a b - 15.0 - FEAa b = 0 5 5



Ans.

FEA = 15.625 kN (T) = 15.6 kN (T)

+ ΣFx = 0;  30.0 - 15.625a 3 b - 34.375a 3 b = 0 (Check!!) S 5 5

Ans: FDE = FDC = FFA = 0 FCE = 34.4 kN (C) FCB = 20.6 kN (T) FBA = 20.6 kN (T) FBE = 15.0 kN (T) FFE = 30.0 kN (C) FEA = 15.6 kN (T) 507

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*6–20. Determine the force in each member of the truss and state if the members are in tension or compression. Set P1 = 10 kN, P2 = 8 kN.

G

F

E

2m

B

A

C

1m P1

Solution Support Reactions. Not required. Method of Joints. We will perform the joint equilibrium according to the sequence of joints D, C, E, B and F. Joint D. Fig. a + c ΣFy = 0;    FDE a

2

b - 8 = 0  FDE = 4 25 kN (T) = 8.94 kN (T)Ans.

25

+ ΣFx = 0;  FDC - a4 25ba 1 b = 0  FDC = 4.00 kN (C) S 25

Ans.

Joint C. Fig. b

+ ΣFx = 0;  FCB - 4.00 = 0   FCB = 4.00 kN(C) S

Ans.

+ c ΣFy = 0;    FCE = 0 

Ans.

Joint E. Fig. c + c ΣFy = 0;  FEB a

1 22

b - a425ba

2 25

b =0

FEB = 8 22 kN (C) = 11.3 kN (C)

Ans.

FEF = 12.0 kN (T)

Ans.

+ ΣFx = 0;  a822ba 1 b + a425ba 1 b - FEF = 0  S 22 25



2m

508

D 1m P2

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6–20. Continued

Joint B. Fig. d + ΣFx = 0;  FBA - 4.00 - a8 22ba 1 b = 0  FBA = 12.0 kN (C) Ans. S 22

+ c ΣFy = 0;  FBF - 10 - a8 22ba

Joint F. Fig. e

+ c ΣFy = 0;  FFA a

2 25

1

22

b = 0  FBF = 18.0 kN (T)

Ans.

b - 18.0 = 0  FFA = 9 25 kN (C) = 20.1 kN (C) Ans.

+ ΣFx = 0;  12.0 + a9 25ba 1 b - FFG = 0  FFG = 21.0 kN (T) S 25

Ans.

Ans: FDE = 8.94 kN (T) FDC = 4.00 kN (C) FCB = 4.00 kN (C) FCE = 0 FEB = 11.3 kN (C) FEF = 12.0 kN (T) FBA = 12.0 kN (C) FBF = 18.0 kN (T) FFA = 9 25 kN (C) = 20.1 kN (C) FFG = 21.0 kN (T) 509

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6–21. Determine the force in each member of the truss and state if the members are in tension or compression. Set P1 = 8 kN, P2 = 12 kN.

G

F

E

2m

B

A

C

1m P1

Solution Support Reactions. Not required. Method of Joints. We will perform the joint equilibrium according to the sequence of joints D, C, E, B and F. Joint D. Fig. a + c ΣFy = 0;  FDE a

2 25

b - 12 = 0  FDE = 6 25 kN (T) = 13.4 kN (T)

+ ΣFx = 0;  FDC - a6 25ba 1 b = 0  FDC = 6.00 kN (C) S 25

Ans. Ans.

Joint C. Fig. b

+ ΣFx = 0;  FCB - 6.00 = 0   FCB = 6.00 kN (C) S

Ans.

+ c ΣFy = 0;  FCE = 0 

Ans.

Joint E. Fig. c + c ΣFy = 0;  FEBa





2m

1 22

b - a6 25ba

2 25

b =0

Ans.

FEB = 12 22 kN (C) = 17.0 kN (C)

+ ΣFx = 0;  a12 22ba 1 b + a6 25ba 1 b - FEF = 0 S 22 25

Ans.

FEF = 18.0 kN (T)

510

D 1m P2

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6–21. Continued

Joint B. Fig. d + ΣFx = 0;  FBA - 6.00 - a12 22ba 1 b = 0  FBA = 18.0 kN (C) Ans. S 22

+ c ΣFy = 0;  FBF - 8 - a12 22ba

Joint F. Fig. e

+ c ΣFy = 0;  FFA a

2 25

1

22

b = 0  FBF = 20.0 kN (T)

Ans.

b - 20.0 = 0  FFA = 10 25 kN (C) = 22.4 kN (C) Ans.

+ ΣFx = 0;  a10 25ba 1 b + 18.0 - FFG = 0  FFG = 28.0 kN(T) S 25

Ans.

Ans: FDE = 13.4 kN (T) FDC = 6.00 kN (C) FCB = 6.00 kN (C) FCE = 0 FEB = 17.0 kN (C) FEF = 18.0 kN (T) FBA = 18.0 kN (C) FBF = 20.0 kN (T) FFA = 22.4 kN (C) FFG = 28.0 kN (T) 511

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6–22.

Determine the force in each member of the double scissors truss in terms of the load P and state if the members are in tension or compression.

B

C

L/3

SOLUTION

L 2L b + Pa b - (Dy)(L) = 0 3 3 Dy = P

c + ©MA = 0; + c ©F y = 0;

E

A

Pa

L/3

Ay = P

F L/3

P

D L/3

P

Joint F: FFD - FFE - FFB a

+ ©F = 0; : x

1 22

b = 0

(1)

F FD - FFE = P + ©F = 0; : y

1

FFB a

22

b -P = 0

FFB = 22P = 1.41 P (T) Similarly, FEC = 22P Joint C:

2 25 +c ©Fy = 0;

2

FCA a

+ ©F = 0; : x

F CA

25

b - 22P a

FCA 1 25

1 22

1 22

b - FCD a

1 22

b = 0

FCD = P

- 22P

1 22

+ FCD

1 22

=0

FCA =

2 25 P = 1.4907P = 1.49P (C) 3

F CD =

22 P = 0.4714P = 0.471P (C) 3

FAE -

1 2 22 2 25 Pa Pa b b = 0 3 3 22 25

FAE =

5 P = 1.67 P (T) 3

Joint A: + ©F = 0; : x

Similarly, FFD=1.67 P (T) From Eq.(1), and Symmetry, FFE = 0.667 P (T)

Ans.

FFD = 1.67 P (T)

Ans.

FAB = 0.471 P (C)

Ans.

FAE = 1.67 P (T)

Ans.

FAC = 1.49 P (C)

Ans.

FBF = 1.41 P (T)

Ans.

FBD = 1.49 P (C)

Ans.

FEC = 1.41 P (T)

Ans.

FCD = 0.471 P (C)

Ans. 512

Ans: FFE = 0.667P (T) FFD = 1.67P (T) FAB = 0.471P (C) FAE = 1.67P (T) FAC = 1.49P (C) FBF = 1.41P (T) FBD = 1.49P (C) FEC = 1.41P (T) FCD = 0.471P (C)

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6–23. Determine the force in each member of the truss in terms of the load P and state if the members are in tension or compression.

E

d

D

SOLUTION

d/ 2

Support reactions: a + ©ME = 0;

3 Ax a d b - Pd = 0 2

+ ©F = 0; : x

2 P - Ex = 0 3

A

C

Ax = 0.6667P

d/ 2 P

Ex = 0.6667P d

+ c ©Fy = 0;

Ey - P = 0

B

d

Ey = P

Joint E: + ©F = 0; : x

FEC sin 33.69° - 0.6667P = 0 FEC = 1.202P = 1.20P (T)

+ c ©Fy = 0;

P - FED - 1.202P cos 33.69° = 0

FED = 0

Ans. Ans.

Joint A: + c ©Fy = 0; + ©F = 0; : x

FAB sin 26.57° - FAD sin 26.57° = 0 0.6667P - 2F cos 26.57° = 0

FAB = FAD = F

F = 0.3727P

FAB = FAD = F = 0.373P (C)

Ans.

Joint D: + ©F = 0; : x

0.3727P cos 26.57° - FDC cos 26.57° = 0 FDC = 0.3727P = 0.373P (C)

+ c ©Fy = 0;

Ans.

2(0.3727P sin 26.57°) - FDB = 0 FDB = 0.3333P = 0.333P (T)

Ans.

Joint B: + ©F = 0; : x

0.3727P cos 26.57° - FBC cos 26.57° = 0 FBC = 0.3727P = 0.373P (C)

+ c ©Fy = 0;

Ans.

0.3333P - 2(0.3727 sin 26.57°) = 0 (Check)

Ans: FEC = FED = FAB = FDC = FDB = FBC = 513

1.20P (T) 0 FAD = 0.373P (C) 0.373P (C) 0.333P (T) 0.373P (C)

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*6–24. The maximum allowable tensile force in the members of the truss is 1Ft2max = 5 kN, and the maximum allowable compressive force is 1Fc2max = 3 kN. Determine the maximum magnitude of the load P that can be applied to the truss. Take d = 2 m.

E

d

D

SOLUTION Support reactions:

d/ 2

a + ©ME = 0;

3 Ax a db - Pd = 0 2

+ ©F = 0; : x

2 P - Ex = 0 3

+ c ©Fy = 0; Joint E: + ©F = 0; : x

A

C

Ax = 0.6667P d/ 2

Ex = 0.6667P

Ey - P = 0

P d

Ey = P

B

FEC sin 33.69° - 0.6667P = 0 FEC = 1.202P = 1.20P (T)

+ c ©Fy = 0;

P - FED - 1.202P cos 33.69° = 0

FED = 0

Joint A: + c ©Fy = 0; + ©F = 0; : x

FAB sin 26.57° - FAD sin 26.57° = 0 0.6667P - 2F cos 26.57° = 0

FAB = FAD = F

F = 0.3727P

FAB = FAD = F = 0.373P (C) Joint D: + ©F = 0; : x

0.3727P cos 26.57° - FDC cos 26.57° = 0 FDC = 0.3727P = 0.373P (C)

+ c ©Fy = 0;

2(0.3727P sin 26.57°) - FDB = 0 FDB = 0.3333P = 0.333P (T)

Joint B: + ©F = 0; : x

0.3727P cos 26.57° - FBC cos 26.57° = 0 FBC = 0.3727P = 0.373P (C)

+ c ©Fy = 0;

0.3333P - 2(0.3727 sin 26.57°) = 0 (Check)

Maximum tension is in member EC. FEC = 1.202 P = 5 P = 4.16 kN Maximum compression is in members AB, AD, DC, and BC. F = 0.3727 P = 3 P = 8.05 kN Thus, the maximum allowable load is P = 4.16 kN

Ans. Ans: P = 4.16 kN 514

d

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6–25. Determine the force in each member of the truss in terms of the external loading and state if the members are in tension or compression. Take P = 2 kN.

P

P 2m

D

C

30 2m

2m

A

Solution

B 2m

Support Reactions. Not required. Method of Joints: We will start the joint equilibrium analysis at joint C, followed by joint D and finally joint A. Joint C. Fig. a + c ΣFy = 0;  FCB sin 60° - 2 = 0        FCB = 2.309 kN (C) = 2.31 kN (C)     Ans. + ΣFx = 0;  FCD - 2.309 cos 60° = 0  FCD = 1.1547 kN (C) = 1.15 kN (C) Ans. S Joint D. Fig. b + ΣFx = 0;  FDB cos 30° - FDA sin 30° - 1.1547 = 0 S + c ΣFy = 0;  FDA cos 30° - FDB sin 30° - 2 = 0

(1) (2)

Solving Eqs. (1) and (2) FDB = 4.00 kN (T)  FDA = 4.6188 kN (C) = 4.62 kN (C)

Ans.

Joint A. Fig. c + ΣFx = 0;  4.6188 sin 30° - FAB = 0  FAB = 2.3094 kN (C) = 2.31 kN (C)Ans. S + c ΣFy = 0;  - 4.6188 cos 30° + NA = 0  NA = 4.00 kN

Ans: FCB = 2.31 kN (C) FCD = 1.15 kN (C) FDB = 4.00 kN (T) FDA = 4.62 kN (C) FAB = 2.31 kN (C) 515

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6–26. The maximum allowable tensile force in the members of the truss is (Ft)max = 5 kN, and the maximum allowable compressive force is (Fc)max = 3 kN. Determine the maximum magnitude P of the two loads that can be applied to the truss.

P

P 2m

D

C

30 2m

2m

A

Solution

B 2m

Support Reactions. Not required. Method of Joints: We will start the joint equilibrium analysis at joint C, followed by joint D and finally joint A. Joint C. Fig. a + c ΣFy = 0;  FCB sin 60° - P = 0  FCB = 1.1547P (C) + ΣFx = 0;  FCD - 1.1547P cos 60° = 0  FCD = 0.5774P (C) S Joint D. Fig. b + ΣFx = 0;  FDB cos 30° - FDA sin 30° - 0.5774P = 0 S

(1)

+ c ΣFy = 0;  FDA cos 30° - FDB sin 30° - P = 0

(2)

Solving Eqs. (1) and (2), FDA = 2.3094P (C)  FDB = 2.00P (T) Joint C. Fig. c + ΣFx = 0;  2.3094 P sin 30° - FAB = 0  FAB = 1.1547P (C) S + c ΣFy = 0;  NA - 2.3094 p cos 30° = 0  NA = 2.00P By observation, members DA and DB are subjected to maximum compression and tension, respectively. Thus, they will reach the limit first. FDA = (FC)max;  2.3094P = 3  P = 1.299 kN = 1.30 kN(Control!) Ans. FDB = (FC)max;  2.00P = 5  P = 2.50 kN

Ans: P = 1.30 kN 516

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6–27. Determine the force developed in members FE, EB, and BC of the truss and state if these members are in tension or compression.

2m

1.5 m

2m

F

E

B

C

2m A

D

11 kN

Solution

22 kN

Support Reactions. Referring to the FBD of the entire truss shown in Fig. a, ND can be determined directly by writing the moment equation of equilibrium about point A. a+ΣMA = 0;  ND(5.5) - 11(2) - 22(3.5) = 0  ND = 18.0 kN Method of Sections. Referring to the FBD of the right portion of the truss ­sectioned through a–a shown in Fig. b, FBC and FFE can be determined directly by writing the moment equations of equilibrium about point E and B, respectively. a+ΣME = 0;  18.0(2) - FBC(2) = 0  FBC = 18.0 kN (T)

Ans.

a+ΣMB = 0;  18.0(3.5) - 22(1.5) - FFE(2) = 0  FFE = 15.0 kN (C) Ans. Also, FEB can be obtained directly by writing force equation of equilibrium along the y axis 4 + c ΣFy = 0;  FEB a b + 18.0 - 22 = 0  FEB = 5.00 kN (C) Ans. 5

Ans: FBC = 18.0 kN (T) FFE = 15.0 kN (C) FEB = 5.00 kN (C) 517

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*6–28. Determine the force in members EF, BE, BC and BF of the truss and state if these members are in tension or compression. Set P1 = 9 kN, P2 = 12 kN, and P3 = 6 kN.

E

F

P3

3m D

A B

C 3m

3m P1

3m P2

Solution Support Reactions. Referring to the FBD of the entire truss shown in Fig. a, ND can be determined directly by writing the moment equation of equilibrium about point A. a+ΣMA = 0;  ND(9) - 12(6) - 9(3) - 6(3) = 0  ND = 13.0 kN Method of sections. Referring to the FBD of the right portion of the truss ­sectioned through a–a shown in Fig. b, FEF and FBC can be determined directly by writing the moment equation of equilibrium about points B and E, respectively. a+ΣMB = 0;  13.0(6) - 12(3) - FEF(3) = 0  FEF = 14.0 kN (C)

Ans.

a+ΣME = 0;  13.0(3) - FBC (3) = 0  FBC = 13.0 kN (T)

Ans.

Also, FBE can be determined by writing the force equation of equilibrium along the y axis. + c ΣFy = 0;  13.0 - 12 - FBE a

1 22

b = 0  FBE = 22 kN (T) = 1.41 kN (T)Ans.

Method of Joints. Using the result of FBE, the equilibrium of joint B, Fig. c, requires + c ΣFy = 0;  FBF + a 22ba

1 22

b - 9 = 0  FBF = 8.00 kN (T)

Ans.

Ans: FEF = FBC = FBE = FBF = 518

14.0 kN (C) 13.0 kN (T) 1.41 kN (T) 8.00 kN (T)

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6–29. Determine the force in members BC, BE, and EF of the truss and state if these members are in tension or compression. Set P1 = 6 kN, P2 = 9 kN, and P3 = 12 kN.

E

F

P3

3m D

A B

C 3m

3m P1

3m P2

Solution Support Reactions. Referring to the FBD of the entire truss shown in Fig. a, ND can be determined directly by writing the moment equation of equilibrium about point A. a+ΣMA = 0;  ND(9) - 12(3) - 6(3) - 9(6) = 0  ND = 12.0 kN Method of Sections. Referring to the FBD of the right portion of the truss ­sectioned through a–a shown in Fig. b, FEF and FBC can be determined directly by writing the moment equation of equilibrium about points B and E, respectively. a+ΣMB = 0;  12.0(6) - 9(3) - FEF(3) = 0  FEF = 15.0 kN (C)

Ans.

a+ΣME = 0;  12.0(3) - FBC(3) = 0  FBC = 12.0 kN (T)

Ans.

Also, FBE can be obtained directly by writing the force equation of equilibrium along the y axis + c ΣFy = 0;  12.0 - 9 - FBE a



1 22

b = 0

Ans.

FBE = 322 kN = 4.24 kN (T)

Ans: FEF = 15.0 kN (C) FBC = 12.0 kN (T) FBE = 4.24 kN (T) 519

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6–30. Determine the force in members BC, HC, and HG. After the truss is sectioned use a single equation of equilibrium for the calculation of each force. State if these members are in tension or compression.

4 kN

4 kN

B

C

5 kN 3 kN

2 kN A

D

E

3m F

H

2m

G

5m

SOLUTION a+ ΣME = 0;

5m

5m

5m

- Ay(20) + 2(20) + 4(15) + 4(10) + 5(5) = 0 Ay = 8.25 kN

a+ ΣMH = 0;

-8.25(5) + 2(5) + FBC (3) = 0 Ans.

FBC = 10.4 kN (C) a+ ΣMC = 0;

- 8.25(10) + 2(10) + 4(5) + FHG = 9.155 = 9.16 kN (T)

a+ ΣMO′ = 0;

5 229

-2(2.5) + 8.25(2.5) - 4(7.5) +

FHG (5) = 0

3 234

Ans. FHC (12.5) = 0 Ans.

FHC = 2.24 kN (T)

Ans: FBC = 10.4 kN (C) FHG = 9.16 kN (T) FHC = 2.24 kN (T) 520

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6–31. Determine the force in members CD, CF, and CG and state if these members are in tension or compression.

4 kN

4 kN

B

C

5 kN 3 kN

2 kN A

D

E

3m F

H

2m

G

5m

SOLUTION + ΣFx = 0; S a+ ΣMA = 0;

5m

5m

5m

Ex = 0 -4(5) - 4(10) - 5(15) - 3(20) + Ey(20) = 0 Ey = 9.75 kN

a+ ΣMC = 0;

- 5(5) - 3(10) + 9.75(10) FFG = 9.155 kN (T)

a+ ΣMF = 0;

5 229

FFG(5) = 0

-3(5) + 9.75(5) - FCD(3) = 0 Ans.

FCD = 11.25 = 11.2 kN (C) a+ ΣMO′ = 0;

- 9.75(2.5) + 5(7.5) + 3(2.5) -

3 234

FCF (12.5) = 0 Ans.

FCF = 3.21 kN (T) Joint G: + ΣFx = 0; S + c ΣFy = 0;

FGH = 9.155 kN (T) 2 229

(9.155)(2) - FCG = 0 Ans.

FCG = 6.80 kN (C)

Ans: FCD = 11.2 kN (C) FCF = 3.21 kN (T) FCG = 6.80 kN (C) 521

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*6–32. Determine the force in members EF, CF, and BC, and state if the members are in tension or compression.

4 kN

E

1.5 m

D

2m

8 kN

F C

Solution

2m

Support Reactions. Not required. Method of Sections. FBC and FEF can be determined directly by writing the moment equations of equilibrium about points F and C, respectively, by referring to the FBD of the upper portion of the truss section through a–a shown in Fig. a. a+ΣMF = 0;  FBC(1.5) - 4(2) = 0   FBC = 5.333 kN (C) = 5.33 kN (C)

Ans.

a+ΣMC = 0;  FEF(1.5) - 4(2) = 0   FEF = 5.333 kN (T) = 5.33 kN (T)

Ans.

A

B

Also, by writing the force equation of equilibrium along the x axis, + ΣFx = 0;  4 - FCF = 0  FCF = 4.00 kN (T) S

Ans.

Ans: FBC = 5.33 kN (C) FEF = 5.33 kN (T) FCF = 4.00 kN (T) 522

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6–33. Determine the force in members AF, BF, and BC, and state if the members are in tension or compression.

4 kN

E

1.5 m

D

2m

8 kN

F C

Solution

2m

Support Reactions. Not required. Method of Sections: Referring to the FBD of the upper portion of the truss section through a–a shown in Fig. a, FAF and FBC can be determined directly by writing the moment equations of equilibrium about points B and F, respectively.

A

B

a+ΣMB = 0;  FAF(1.5) - 8(2) - 4(4) = 0

Ans.

FAF = 21.33 kN (T) = 21.3 kN (T)

a+ΣMF = 0;  FBC(1.5) - 4(2) = 0  

Ans.

FBC = 5.333 kN (C) = 5.33 kN (C)

Also, write the force equation of equilibrium along the x axis, we can obtain FBF directly. + ΣFx = 0;  4 + 8 - FBF a 3 b = 0  FBF = 20.0 kN (C) S 5

Ans.

Ans: FAF = 21.3 kN (T) FBC = 5.33 kN (C) FBF = 20.0 kN (C) 523

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6–34.

L

N

Determine the force in members CD and CM of the Baltimore bridge truss and state if the members are in tension or compression. Also, indicate all zero-force members.

O

K 2m

J

P

A

I B

C 2 kN

D

E

F

G

5 kN 3 kN

2m

H

2 kN

16 m, 8 @ 2 m

SOLUTION Support Reactions: a + ©MI = 0;

21122 + 5182 + 3162 + 2142 - Ay 1162 = 0 Ay = 5.625 kN

+ ©F = 0; : x

Ax = 0

Method of Joints: By inspection, members BN, NC, DO, OC, HJ LE and JG are zero force members.

Ans.

Method of Sections: a + ©MM = 0;

a + ©MA = 0;

FCD142 - 5.625142 = 0

FCD = 5.625 kN 1T2

Ans.

FCM = 2.00 kN T

Ans.

FCM 142 - 2142 = 0

Ans: FCD = 5.625 kN (T) FCM = 2.00 kN (T) 524

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6–35.

L

N

Determine the force in members EF, EP, and LK of the Baltimore bridge truss and state if the members are in tension or compression. Also, indicate all zero-force members.

O

K 2m

J

P

A

I B

C 2 kN

D

E

F

5 kN 3 kN

G

2m

H

2 kN

16 m, 8 @ 2 m

SOLUTION Support Reactions: a + ©MA = 0;

Iy 1162 - 21122 - 31102 - 5182 - 2142 = 0 Iy = 6.375 kN

Method of Joints: By inspection, members BN, NC, DO, OC, HJ LE and JG are zero force members.

Ans.

Method of Sections: a + ©MK = 0;

a + ©ME = 0;

+ c ©Fy = 0;

3122 + 6.375142 - FEF142 = 0

FEF = 7.875 = 7.88 kN 1T2

Ans.

FLK = 9.25 kN 1C2

Ans.

6.375182 - 2142 - 3122 - FLK 142 = 0

6.375 - 3 - 2 - FED sin 45° = 0 FED = 1.94 kN T

Ans.

Ans: FEF = 7.88 kN (T) FLK = 9.25 kN (C) FED = 1.94 kN (T) 525

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*6–36. The Howe truss is subjected to the loading shown. Determine the force in members GF, CD, and GC, and state if the members are in tension or compression.

5 kN G 5 kN 3m

5 kN F

H

2 kN

2 kN

A B

2m

C

2m

E

D

2m

2m

SOLUTION a + ©MA = 0;

Ey(8) - 2(8) - 5(6) - 5(4) - 5(2) = 0

a + ©MD = 0;

4 - FGF (1.5) - 2(2) + 9.5(2) = 0 5

Ey = 9.5kN

FGF = 12.5 kN (C) a + ©MG = 0;

Ans.

9.5(4) - 2(4) - 5(2) - FCD(3) = 0 FCD = 6.67 kN (T)

Ans.

Joint C: + c ©Fy = 0;

FGC = 0

Ans.

Ans: FGF = 12.5 kN (C) FCD = 6.67 kN (T) FGC = 0 526

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6–37. The Howe truss is subjected to the loading shown. Determine the force in members GH, BC, and BG of the truss and state if the members are in tension or compression.

5 kN G 5 kN 3m

5 kN F

H

2 kN

2 kN

A

SOLUTION a + ©MB = 0;

B

-7.5(2) + FGH sin 36.87°(2) = 0

2m

FGH = 12.5 kN (C) a + ©MA = 0;

2m

E

D

2m

2m

Ans.

- 5 (2) + FBG sin 56.31°(2) = 0 FBG = 6.01 kN (T)

a + ©MH = 0;

C

Ans.

- 7.5(4) + 5(2) + FBC(3) = 0 FBC = 6.67 kN (T)

Ans.

Ans: FGH = 12.5 kN (C) FBG = 6.01 kN (T) FBC = 6.67 kN (T) 527

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6–38. Determine the force in members KJ, NJ, ND, and CD of the K truss. Indicate if the members are in tension or compresion. Hint: Use sections aa and bb.

L

K

a b

J

I

H

1.5 m 1.5 m

M

N

B

C

A

1200 N 2m

O

a b

D

P G

F

E

1500 N

2m

1800 N 2m 2m

2m

2m

Solution Support Reactions: a+ ΣMG = 0; +

1.201102 + 1.50182 + 1.80162 - Ay 1122 = 0 Ay = 2.90 kN

S ΣFx = 0;

Az = 0 2 m 2 m

Method of Sections: From section a–a, FKJ and FCD can be obtained directly by summing moment about points C and K respectively. a+ ΣMC = 0; a+ ΣMK = 0;

FKJ 132 + 1.20122 - 2.90142 = 0 FKJ = 3.067 kN1C 2 = 3.07 kN1C 2

m

2 m

kN

kN kN

Ans.

FCD 132 + 1.20122 - 2.90142 = 0 FCD = 3.067 kN1T 2 = 3.07 kN1T 2

Ans.

3 m

From sec b–b, summing forces along x and y axes yields +

S ΣFx = 0;

+ c ΣFy = 0;

2 m

4 4 FND a b - FNJ a b + 3.067 - 3.067 = 0 5 5 FND = FNJ

kN

kN kN

[1]

3 3 2.90 - 1.20 - 1.50 - FND a b - FNJ a b = 0 5 5 FND = FNJ = 0.3333

3 m

[2]

Solving Eqs. [1] and [2] yields FND = 0.167 kN1T 2

2 m kN

2 m

FNJ = 0.167 kN1C 2

Ans.

kN

kN

2 m kN

kN

Ans: FKJ = 3.07 kN (T) FCD = 3.07 kN (T) FND = 0.167 kN (T) FNJ = 0.167 kN (C) 528

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L

6–39. Determine the force in members JI and DE of the K truss. Indicate if the members are in tension or compression.

K

a b

J

I

H

1.5 m 1.5 m

M

N

B

C

A

1200 N 2m

O

a b

D

P G

F

E

1500 N

1800 N 2m 2m

2m

2m

2m

Solution Support Reactions: a+ ΣMA = 0;

Gy 11202 - 1.80162 - 1.50142 - 1.20122 = 0 Gy = 1.60 kN

Method of Sections: a+ ΣME = 0; a+ ΣMI = 0;

1.60142 - FJI 132 = 0 FJI = 2.13 kN 1C 2

Ans.

1.60142 - FDE 132 = 0 FDE = 2.13 kN 1T 2

Ans.

3 m

4m kN

Ans: FJI = 2.13 kN (C) FDE = 2.13 kN (T) 529

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*6–40. Determine the force in members DC, HC, and HI of the truss, and state if the members are in tension or compression.

50 kN

40 kN 2m

2m

2m D

E

C

F G

SOLUTION

I

Support Reactions: Applying the moment equation of equilibrium about point A to the free - body diagram of the truss, Fig. a, a + ©MA = 0;

H

1.5 m 30 kN

1.5 m B 40 kN 1.5 m

A

40(1.5) + 30(3) + 40(2) - Fy(4) = 0 Fy = 57.5 kN

+ ©F = 0; : x

Ax - 30 - 40 = 0;

Ax = 70 kN

+ c ©Fy = 0;

57.5 - 40 - 50 + Ay = 0;

Ay = 32.5 kN

Method of Sections: Using the bottom portion of the free - body diagram, Fig. b. a + ©MC = 0;

70(3) - 32.5(2) - 40(1.5) - FHI(2) = 0 Ans.

FHI = 42.5 kN (T) a + ©MD = 0;

70(4.5) - 40(3) - 30(1.5) - FHC(1.5) = 0 Ans.

FHC = 100 kN (T) + c ©Fy = 0;

3 32.5 + 42.5 - FDC ( ) = 0 5 Ans.

FDC = 125 kN (C)

Ans: FHI = 42.5 kN (T) FHC = 100 kN (T) FDC = 125 kN (C) 530

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6–41. Determine the force in members ED, EH, and GH of the truss, and state if the members are in tension or compression.

50 kN

40 kN 2m

2m

2m D

E

C

F G

SOLUTION

I

Support Reactions: Applying the moment equation of equilibrium about point A to the free - body diagram of the truss, Fig. a, a + ©MA = 0;

H

1.5 m 30 kN

1.5 m B 40 kN 1.5 m

A

40(1.5) + 30(3) + 40(2) - Fy(4) = 0 Fy = 57.5 kN

+ ©F = 0; : x

Ax - 30 - 40 = 0;

Ax = 70 kN

+ c ©Fy = 0;

57.5 - 40 - 50 + Ay = 0;

Ay = 32.5 kN

Method of Sections: Using the left portion of the free - body diagram, Fig. b. a + ©ME = 0;

- 57.5(2) + FGH(1.5) = 0 Ans.

FGH = 76.7 kN (T) a + ©MH = 0;

- 57.5(4) + FED(1.5) + 40(2) = 0 Ans.

FED = 100 kN (C) + c ©Fy = 0;

3 57.5 - FEH ( ) - 40 = 0 5 Ans.

FEH = 29.2 kN (T)

Ans: FGH = 76.7 kN (T) FED = 100 kN (C) FEH = 29.2 kN (T) 531

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6–42. Determine the force in members JK, CJ, and CD of the truss, and state if the members are in tension or compression.

J

K 3m

I H

L

G

A B 2m

2m

4 kN

SOLUTION

C 2m

5 kN

D 2m

F

E 2m

8 kN

2m

6 kN

Method of Joints: Applying the equations of equilibrium to the free - body diagram of the truss, Fig. a, + ©F = 0; : x

Ax = 0

a + ©MG = 0

6(2) + 8(4) + 5(8) + 4(10) - Ay(12) = 0 Ay = 10.33 kN

Method of Sections: Using the left portion of the free - body diagram, Fig. a. a + ©MC = 0;

FJK(3) + 4(2) - 10.33(4) = 0 Ans.

FJK = 11.111 kN = 11.1 kN (C) a + ©MJ = 0;

FCD(3) + 5(2) + 4(4) - 10.33(6) = 0 Ans.

FCD = 12 kN (T) + c ©Fy = 0;

10.33 - 4 - 5 - FCJ sin 56.31° = 0 Ans.

FCJ = 1.602 kN = 1.60 kN (C)

Ans: FJK = 11.1 kN (C) FCD = 12 kN (T) FCJ = 1.60 kN (C) 532

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6–43. Determine the force in members HI, FI, and EF of the truss, and state if the members are in tension or compression.

J

K 3m

I H

L

G

A B 2m

2m

4 kN

SOLUTION

C 2m

5 kN

D 2m

F

E 2m

8 kN

2m

6 kN

Support Reactions: Applying the moment equation of equilibrium about point A to the free - body diagram of the truss, Fig. a, a + ©MA = 0;

NG(2) - 4(2) - 5(4) - 8(8) - 6(10) = 0 NG = 12.67 kN

Method of Sections: Using the right portion of the free - body diagram, Fig. b. a + ©MI = 0;

12.67(4) - 6(2) - FEF(3) = 0 Ans.

FEF = 12.89 kN = 12.9 kN (T) a + ©MG = 0;

- FFI sin 56.31°(2) + 6(2) = 0 Ans.

FFI = 7.211 kN = 7.21 kN (T) a + ©MF = 0;

3 12.67(2) - FHI a b(2) = 0 5 Ans.

FHI = 21.11 kN = 21.1 kN (C)

Ans: FEF = 12.9 kN (T) FFI = 7.21 kN (T) FHI = 21.1 kN (C) 533

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*6–44. Determine the force in members BC, CH, GH, and CG of the truss and state if the members are in tension or compression.

G H

2m

F E

A B 4m

C 4m

D 4m

4 kN 8 kN

3m

4m 5 kN

Solution Support Reactions. Referring to the FBD of the entire truss shown in Fig. a, Ay can be determined directly by writing the moment equation of equilibrium about point E. a+ΣME = 0;  5(4) + 8(8) + 4(12) - Ay(16) = 0  Ay = 8.25 kN + ΣFx = 0;  Ax = 0 S Method of Sections. Referring to the FBD of the left portion of the truss section through a–a shown in Fig. a, FBC, FGH and FCH can be determined directly by writing the moment equations of equilibrium about points H, C, and O, respectively, Ans.

a+ΣMH = 0;  FBC(3) - 8.25(4) = 0  FBC = 11.0 kN (T) a+ΣMC = 0;  FGHa

1 25

b (10) + (4)(4) - 8.25(8) = 0

Ans.

FGH = 525 kN (C) = 11.2 kN (C)

3 a+ΣMO = 0;  FCH a b(10) + (8.25)(2) - 4(6) = 0  FCH = 1.25 kN (C) 5

Ans.

Method of Joints. Using the result of FGH , equilibrium of joint G, Fig. c, requires + ΣFx = 0;  a525ba 2 b - FGF a 2 b = 0  S 25 25

+ c ΣFy = 0;  2a525ba

1

25

b - FCG = 0 

FGF = a525b kN (C)

FCG = 10.0 kN (T)

Ans.

Ans: FBC = 11.0 kN(T) FGH = 11.2 kN(C) FCH = 1.25 kN(C) FCG = 10.0 kN(T) 534

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6–45. 6 kN

Determine the force in members CD, CJ, and KJ and state if these members are in tension or compression.

6 kN

6 kN

J

6 kN

I

K

6 kN

L

3m

H G

A B

C

D

E

F

12 m, 6 @ 2 m

Solution Support Reactions. Referring to the FBD of the entire truss shown in Fig. a, Ay can be determined directly by writing the moment equation of equilibrium about point G. a+ΣMG = 0;  6(2) + 6(4) + 6(6) + 6(8) + 6(10) - Ay(12) = 0  Ay = 15.0 kN + ΣFx = 0;   Ax = 0   S Method of Sections. Referring to the FBD of the left portion of the truss sectioned through a - a shown in Fig. b, FCD, FCJ and FKJ can be determined directly by writing the moment equations of equilibrium about points J, A and C, respectively. a+ΣMJ = 0;  FCD(3) + 6(2) + 6(4) - 15.0(6) = 0  FCD = 18.0 kN (T)Ans. a+ΣMA = 0;  FCJ a

a+ΣMC = 0;  FKJ a



3 213 1 25

b(4) - 6(4) - 6(2) = 0

FCJ = 3213 kN (T) = 10.8 kN (T)

b(4) + 6(2) - 15.0 (4) = 0

FKJ = 1225 kN (C) = 26.8 kN (C)

Ans.

Ans.

Ans: FCD = 18.0 kN (T) FCJ = 10.8 kN (T) FKJ = 26.8 kN (T) 535

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6–46. Determine the force in members BE, EF, and CB, and state if the members are in tension or compression.

D 5 kN 5 kN

4m C

E

10 kN 4m

SOLUTION + ©F = 0; : x

B

5 + 10 - FBE cos 45° = 0 Ans.

FBE = 21.2 kN (T) a + ©ME = 0;

4m A

- 5 (4) + FCB (4) = 0 Ans.

FCB = 5 kN (T) a + ©MB = 0;

F

10 kN

G 4m

- 5 (8) - 10 (4) - 5 (4) + FEF (4) = 0 Ans.

FEF = 25 kN (C)

Ans: FBE = 21.2 kN (T) FCB = 5 kN (T) FEF = 25 kN (C) 536

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6–47. Determine the force in members BF, BG, and AB, and state if the members are in tension or compression.

D 5 kN 5 kN

4m C

E

10 kN 4m

SOLUTION

B

Joint F: + ©F = 0; : x

Ans.

FBF = 0

G 4m

5 + 10 + 10 - FBG cos 45° = 0 Ans.

FBG = 35.4 kN (C) a + ©MG = 0;

4m A

Section: + ©F = 0; : x

F

10 kN

FAB (4) - 10 (4) - 10 (8) - 5 (12) = 0 Ans.

FAB = 45 kN (T)

Ans: FBF = 0 FBG = 35.4 kN (C) FAB = 45 kN (T) 537

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*6–48. Determine the force in members BC, HC, and HG. State if these members are in tension or compression.

12 kN 6 kN 4 kN

9 kN

G

H

6 kN

J 2m

1m

1m

A

E B 1.5 m

Solution

C 1.5 m

D 1.5 m

1.5 m

Support Reactions. Referring to the FBD of the entire truss shown in Fig. a, NA can be determined directly by writing the moment equation of equilibrium about point E. a+ΣME = 0;  9(1.5) + 12(3) + 6(4.5) + 4(6) - NA(6) = 0  NA = 16.75 kN Method of Sections. Referring to the FBD of the left portion of the truss sectioned through a–a shown in Fig. b, FHG, FHC and FBC can be determined directly by ­writing the moment equations of equilibrium about points C, A, and H, respectively. a+ΣMC = 0;  FHG a

213

b (3) + 6(1.5) + 4(3) - 16.75(3) = 0

FHG = 4.875213 kN (C) = 17.6 kN (C)

a+ΣMA = 0;  FHC a

2

2 213

b (3) - 6(1.5) = 0 

Ans.

Ans.

FHC = 1.5213 kN (C) = 5.41 kN (C)

a+ΣMH = 0;  FBC(1) + 4(1.5) - 16.75(1.5) = 0 

Ans.

FBC = 19.125 kN (T) = 19.1 kN (T)

Ans: FHG = 17.6 kN (C) FHC = 5.41 kN (C) FBC = 19.1 kN (T) 538

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6–49. Determine the force in members CD, CJ, GJ, and CG and state if these members are in tension or compression.

12 kN 6 kN 4 kN

9 kN

G

H

6 kN

J 2m

1m

1m

A

E B 1.5 m

Solution

C 1.5 m

D 1.5 m

1.5 m

Support Reactions. Referring to the FBD of the entire truss shown in Fig. a, Ey can be determined directly by writing the moment equation of equilibrium about point A. a+ΣMA = 0;  Ey(6) - 6(6) - 9(4.5) - 12(3) - 6(1.5) = 0  Ey = 20.25 kN + ΣFx = 0;  Ex = 0 S Method of Sections. Referring to the FBD of the right portion of the truss sectioned through a–a shown in Fig. b, FGJ , FCJ and FCD can be determined directly by writing moment equations of equilibrium about point C, E and J, respectively. a+ΣMC = 0;  20.25(3) - 6(3) - 9(1.5) - FGJa



213

b(3) = 0

FGJ = 4.875 213 kN (C) = 17.6 kN (C)

a+ΣME = 0;  9(1.5) - FCJ a



2

2

213

b(3) = 0

Ans.

Ans.

FCJ = 2.25213 kN (C) = 8.11 kN (C)

a+ΣMJ = 0;  20.25(1.5) - 6(1.5) - FCD(1) = 0

Ans.

FCD = 21.375 kN (T) = 21.4 kN (T)

Method of Joints. Using the result of FGJ to consider the equilibrium of joint G, Fig. c, + ΣFx = 0;  FHG a 3 b - (4.875213)a 3 b = 0  FHG = 4.875213 kN (C) S 213 213

2 + c ΣFy = 0;  2 a4.875213ba b - 12 - FCG = 0  FCG = 7.50 kN (T) 213 Ans.

Ans: FGJ = 17.6 kN (C) FCJ = 8.11 kN (C) FCD = 21.4 kN (T) FCG = 7.50 kN (T) 539

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6–50. If the truss supports a force of F = 200 N, determine the force in each member and state if the members are in tension or compression.

z 200 mm 200 mm

D C 200 mm

200 mm

E B

x y

Solution

500 mm

Method of Joints: We will begin by analyzing the equilibrium of joint A, and then proceed to analyzing that of joint B.

A 300 mm

Joint A: From the free-body diagram in Fig. b, ΣFx = 0;

FAE a

ΣFy = 0;

FAB a

ΣFz = 0;

FAC a

0.2 20.54 0.3

20.34 0.5

20.54

b - FAC a b - FAE a b + FAE a

Solving Eqs. (1) through (3) yields

0.2 20.54 0.5

20.54 0.5

20.54

b = 0 112 b - FAC a b - FAB a

0.5

20.54 0.5

20.34

b = 0 122 b + 200 = 0 132 Ans.

FAE = FAC = 220.45 N = 200 N1T 2

Ans.

FAB = 583.10 N = 583 N1C 2

540

F

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6–50.

Continued

Joint B: From the free-body diagram in Fig. b, ΣFz = 0;

583.10 a FBD

0.5

b - FBD sin 45° = 0 20.34 = 707.11 N = 707 N1C 2

ΣFx = 0;

FBE cos 45° - FBC cos 45° = 0 FBE = FBC = F

ΣFy = 0;

707.11 cos 45° - 583.10 a

Ans.

0.3

b - 2F sin 45° = 0 20.34 F = 141.42 N

Ans.

Thus, Ans.

FBE = FBC = 141.42 N = 141 N1T 2

Ans: FAE = FAB = FBD = FBE = 541

FAC = 220 N(T) 583 N (C) 707 N (C) FBC = 141.4 N (T)

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6–51. z

Determine the force in each member of the space truss and state if the members are in tension or compression. The truss is supported by ball-and-socket joints at A, B, C, and D.

2m 2m

D 6 kN 4 kN

C G 4m

A B

Solution E

Support Reactions. Not required Method of Joints. Perform the joint equilibrium analysis at joint G first and then proceed to joint E.

y 4m

x

Joint G. Fig. a   ΣFx = 0;  FGDa

2

2

b - FGCa

b = 0  FGD = FGC = F 25 25 1 1   ΣFy = 0;  F a b + Fa b - 4 = 0  F = 225 kN 25 25 Thus,

FGC = 225 kN (T) = 4.47 kN (T)



FGD = 225 kN (C) = 4.47 kN (C)

  ΣFz = 0;  FGE - 6 = 0

Ans. Ans. Ans.

FGE = 6.00 kN (C)

Joint E. Fig. b

2   ΣFz = 0;  FEDa b - 6.00 = 0   FED = 9.00 kN (T)  3   ΣFx = 0;  FEAa

  ΣFy = 0;  FEAa

2

25 1

25

Solving Eqs. (1) and (2)

b - FEBa

Ans.

2 b - 9.00a b = 0 3 25

b + FEBa

2

(1)

1

1 b - 9.00a b = 0 3 25

(2)

Ans.

FEA = 325 kN (C) = 6.71 kN (C)  FEB = 0

Ans:  FGC = 4.47 kN (T)  FGD = 4.47 kN (C) FGE = 6.00 kN (C)  FED = 9.00 kN (T) FEA = 6.71 kN (C)  FEB = 0 542

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*6–52.

z

The space truss supports a force F = [300i + 400j - 500k] N. Determine the force in each member, and state if the members are in tension or compression.

F D

SOLUTION Method of Joints: In this case, there is no need to compute the support reactions. We 1.5 m will begin by analyzing the equilibrium of joint D, and then that of joints A and C.

©Fx = 0;

FDA a

©Fy = 0;

FDB a

©Fz = 0;

- FDA a

(1)

1 1 1 b - FDA a b - FDC a b + 400 = 0 3.5 3.5 110

(2)

3 3 3 b - FDC a b - FDB a b - 500 = 0 3.5 3.5 110

(3)

Solving Eqs. (1) through (3) yields FDB = - 895.98 N = 896 N (C)

Ans.

FDC = 554.17 N = 554 N (T)

Ans.

FDA = - 145.83 N = 146 N (C)

Ans.

Joint A: From the free-body diagram, Fig. b, FAB a

©Fx = 0;

1.5 1.5 b - 52.08a b - FAC = 0 145.83 a 3.5 2.5

2 1 b - 145.83a b = 0 2.5 3.5 FAB = 52.08 N = 52.1 N (T)

Ans.

FAC = 31.25 N (T) ©Fz = 0;

Ans.

3 b = 0 A z - 145.83 a 3.5 A z = 125 N

Ans.

Joint C: From the free-body diagram, Fig. c, 1.5 1.5 b - FCB a b = 0 3.5 2.5

©Fx = 0;

31.25 + 554.17 a

©Fy = 0;

2 1 b - 447.92a b + Cy = 0 554.17 a 3.5 2.5

FCB = 447.92 N = 448 N (C)

Ans.

Cy = 200 N ©Fz = 0;

554.17 a

B y

1.5 1.5 b - FDC a b + 300 = 0 3.5 3.5

©Fy = 0;

C A

Joint D: From the free-body diagram, Fig. a, we can write

3m

1.5 m

3 b - Cz = 0 3.5 Cz = 475 N

Note: The equilibrium analysis of joint B can be used to determine the components of support reaction of the ball and socket support at B.

543

1m x

1m

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6–53. The space truss supports a force F = {-400i + 500j + 600k} N. Determine the force in each member, and state if the members are in tension or compression.

F D

C

1.5 m

Method of Joints: In this case, there is no need to compute the support reactions. We will begin by analyzing the equilibrium of joint D, and then that of joints A and C.

©Fy = 0; ©Fz = 0;

FDA a

1.5 1.5 b - FDC a b - 400 = 0 3.5 3.5 1 1 1 b - FDA a b - FDC a b + 500 = 0 FDB - a 3.5 3.5 110 3 3 3 600 - FDA a b - FDC a b - FDB a b = 0 3.5 3.5 110

(1) (2) (3)

FDB = - 474.34 N = 474 N (C)

Ans.

FDC = 145.83 N = 146 N (T)

Ans.

FDA = 1079.17 N = 1.08 kN (T)

Ans.

Joint A: From the free-body diagram, Fig. b, 1079.17 a

1 2 b - FAB a b = 0 3.5 2.5 FAB = 385.42 N = 385 N (C)

©Fx = 0;

385.42 a

©Fz = 0;

1 b - Az = 0 1079.17 a 3.5

Ans.

1.5 1.5 b - 1079.17 a b + FAC = 0 2.5 3.5 FAC = 231.25 N = 231 N (C)

Ans.

A z = 925 N

Ans.

Joint C: From the free-body diagram, Fig. c, ©Fx = 0;

FCB a

1.5 1.5 b - 231.25 + 145.83 a b = 0 2.5 3.5 FCB = 281.25 N = 281 N (T)

Ans.

2 1 b + 145.83a b - Cy = 0 2.5 3.5

©Fy = 0;

281.25 a

©Fz = 0;

Cy = 266.67 N 3 b - Cz = 0 145.83 a 3.5 Cz = 125 N

Note: The equilibrium analysis of joint B can be used to determine the components of support reaction of the ball and socket support at B.

544

B y

x

Solving Eqs. (1) through (3) yields

©Fy = 0;

A 1m

Joint D: From the free-body diagram, Fig. a, we can write ©Fx = 0;

3m

1.5 m

SOLUTION

1m

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6–54. Determine the force in each member of the space truss and state if the members are in tension or compression. Hint: The support reaction at E acts along member EB. Why?

2m E

SOLUTION

B

5m

3m C

Method of Joints: In this case, the support reactions are not required for determining the member forces.

3m

Joint A: ©Fz = 0;

A

FAB ¢

5 229

y

D

≤ - 6 = 0 x

FAB = 6.462 kN 1T2 = 6.46 kN 1T2

4m

6 kN

Ans.

©Fx = 0;

3 3 FAC a b - FAD a b = 0 5 5

©Fy = 0;

2 4 4 FAC a b + FAD a b - 6.462 ¢ ≤ = 0 5 5 229

FAC = FAD

(1)

FAC + FAD = 3.00

(2)

Solving Eqs. (1) and (2) yields FAC = FAD = 1.50 kN 1C2

Ans.

Joint B: ©Fx = 0;

FBC ¢

©Fz = 0;

FBC ¢

3 238 5 238

≤ - FBD ¢ ≤ + FBD ¢

3 238 5 238

≤ = 0

FBC = FBD

≤ - 6.462 ¢

5 229

(1)

≤ = 0

FBC + FBD = 7.397

(2)

Solving Eqs. (1) and (2) yields FBC = FBD = 3.699 kN 1C2 = 3.70 kN 1C2 ©Fy = 0;

2 B 3.699 ¢

2 238

≤ R + 6.462 ¢

2 229

Ans.

≤ - FBE = 0

FBE = 4.80 kN 1T2

Ans.

Note: The support reactions at supports C and D can be determined by analyzing joints C and D, respectively using the results obtained above.

Ans: FAB = 6.46 kN (T), FAC = FAD = 1.50 kN (C), FBC = FBD = 3.70 kN (C), FBE = 4.80 kN (T) 545

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6–55.

z

Determine the force in members BE, BC, BF, and CE of the space truss, and state if the members are in tension or compression.

F

SOLUTION

D

1.5 m A

Method of Joints: In this case, there is no need to compute the support reactions.We will begin by analyzing the equilibrium of joint C, and then that of joints E and B. x

Joint C: From the free-body diagram, Fig. a, we can write ©Fz = 0;

©Fx = 0;

E

1.5 b - 600 = 0 FCE a 13.25 FCE = 721.11 N = 721 N (T) 1 b - FBC = 0 721.11 a 13.25 FBC = 400 N (C)

3m B

Ans.

1m

C 1m 600 N

y

900 N

Ans.

Joint E: From the free-body diagram, Fig, b, notice that FEF, FED, and FEC lie in the same plane (shown shaded), and FBE is the only force that acts outside of this plane. If the x⬘ axis is perpendicular to this plane and the force equation of equilibrium is written along this axis, we have ©Fx¿ = 0;

FBE = cos u = 0 FBE = 0

Ans.

Joint B: From the free-body diagram, Fig. c, ©Fz = 0;

FBF a

1.5 b - 900 = 0 3.5 FBF = 2100 N = 2.10 kN (T)

Ans.

Ans: FCE = 721 N (T), FBC = 400 N (C) FBE = 0, FBF = 2.10 kN (T) 546

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*6–56.

z

Determine the force in members AF, AB, AD, ED, FD, and BD of the space truss, and state if the members are in tension or compression.

F

SOLUTION

D

1.5 m

Support Reactions: In this case, it will be easier to compute the support reactions first. From the free-body diagram of the truss, Fig. a, and writing the equations of x equilibrium, we have ©Mx = 0; Fy(1.5) - 900(3) - 600(3) = 0

Fy = 3000 N

©My = 0; 900(2) - A z(2) = 0

A z = 900 N

©Mz = 0;

A y(2) - 3000(1) = 0

A y = 1500 N

©Fx = 0;

Ax = 0

©Fy = 0;

Dy + 1500 - 3000 = 0

Dy = 1500 N

©Fz = 0;

Dz + 900 - 900 - 600 = 0

Dz = 600 N

©Fz = 0;

1500 - FAB = 0

900 - FAF a

Ans.

1.5 b = 0 13.25

FAF = 1081.67 N = 1.08 kN (C) ©Fx = 0;

1081.67 a

Ans.

1 b - FAD = 0 13.25 FAD = 600 N (T)

Ans.

Joint C: From the free-body diagram of the joint in Fig. c, notice that FCE, FCB, and the 600-N force lie in the x–z plane (shown shaded). Thus, if we write the force equation of equilibrium along the y axis, we have ©Fy = 0;

FDC = 0

Joint D: From the free-body diagram, Fig. d, ©Fx = 0; ©Fy = 0; ©Fz = 0;

FBC - a

2 1 1 b + FFD a b + FFD a b + 600 = 0 3.5 113 13.25 3 3 b + FED a b + 1500 = 0 FBD a 3.5 113 1.5 1.5 b + FED a b + 600 = 0 FFD a 3.5 113

(1) (2) (3)

Solving Eqs. (1) through (3) yields FFD = 0

FED = - 1400N = 1.40 kN (C)

Ans.

FBD = - 360.56 N = 361 N (C)

Ans.

547

3m 1m

900 N

Joint A: From the free-body diagram, Fig. b, we can write

FAB = 1500 N = 1.50 kN (C)

A

B

Method of Joints: Using the above results, we will begin by analyzing the equilibrium of joint A, and then that of joints C and D.

©Fy = 0;

E

C 1m 600 N

y

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6–57. z

Determine the force in members EF, AF, and DF of the space truss and state if the members are in tension or compression. The truss is supported by short links at A, B, D, and E.

3 kN

4 kN E

2 kN F 3m

3m

C

D

Solution

3m

Support Reactions. Referring to the FBD of the entire truss shown in Fig. a,

x

  ΣFx = 0;  Dx - Bx - Ex + 2 = 0 

(1)

  ΣFy = 0;  Ay - 3 = 0

(2)

  ΣFz = 0;  Dz + Bz - 4 = 0 

(3)

  ΣMx = 0;  4(1.5) + 3(3 sin 60°) - Dz(3) = 0 

(4)

  ΣMy = 0;  2(3 sin 60°) + 4(5) - Dz (5) - Ex (3 sin 60°) = 0 

(5)

  ΣMz = 0;  Dx (3) + Ay (5) - Ex (1.5) + 2(1.5) - 3(5) = 0 

(6)

Solving Eqs. (1) to (6)

Ay = 3.00 kN   Dz = 4.5981 kN    Bz = -0.5981 kN



Ex = 0.8490 kN  Dx = -0.5755 kN  Bx = 0.5755 kN

Method of Joints. We will analyse the equilibrium of the joint at joint D first and then proceed to joint F. Joint D. Fig. b + c ΣFz = 0;  4.5981 - FDF sin 60° = 0   FDF = 5.3094 kN (C) = 5.31 kN (C)Ans.

548

B

A

5m

y

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6–57. Continued

Joint F. Fig. c   ΣFx = 0;  2 - FEF -

5 234

(7)

FCF = 0

  ΣFy = 0;  FAF cos 60° + 5.3094 cos 60° - 3 - FCF a   ΣFz = 0;  5.3094 sin 60° - 4 - FAF sin 60° - FCF a

Solving Eqs. (7), (8) and (9)

1.5 234

b = 0

3 sin 60° 234

b = 0

(8) (9)

Ans.

FCF = 0  FEF = 2.00 kN (T) FAF = 0.6906 kN (T) = 0.691 kN (T)

Ans.

Ans: FDF = 5.31 kN (C) FEF = 2.00 kN (T) FAF = 0.691 kN (T) 549

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6–58. The space truss is used to support the forces at joints B and D. Determine the force in each member and state if the members are in tension or compression.

C

2m B

12 kN

90 D

20 kN

3m F A 2.5 m

Solution Support Reactions. Not required Method of Joints. Analysis of joint equilibrium will be in the sequence of joints D, C, B, A and F. Joint D. Fig. a 4   ΣFx = 0;  20 - FDB a b = 0  FDB = 25.0 kN (T) 5

Ans.

3   ΣFy = 0;  25.0 a b - FDC = 0  FDC = 15.0 kN (T) 5

Ans.

  ΣFz = 0;  FDE - 12 = 0     FDE = 12.0 kN (C)

Ans.

Joint C. Fig. b   ΣFx = 0;  FCB = 0   ΣFy = 0;  FCE a

1 25

Ans. b + 15.0 = 0  FCE = - 1525 kN = 33.5 kN (C) Ans.

  ΣFz = 0;   - FCF - a - 1525ba

Joint B. Fig. c

  ΣFy = 0;  FBE a

2

25

b = 0  FCF = 30.0 kN (T)

3 b - 25.0 a b = 0 5 215.25

1.5

FBE = 10215.25 kN (T) = 39.1 kN (T)

4 2 2   ΣFx = 0;  25.0 a b - a10215.25b a b - FBF a b = 0 5 215.25 213 FBF = 0   ΣFz = 0;   - FBA - a10215.25b a

Ans.

3 215.25

b = 0

FBA = - 30.0 kN = 30.0 kN (C)

550

Ans.

Ans.

Ans.

1.5 m E

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6–58.

Continued

Joint A. Fig. d ΣFy = 0;

FAE = 0

Ans.

ΣFx = 0;

FAF = 0

Ans.

FFE = 0

Ans.

Joint F. Fig. e ΣFy = 0;

Ans: FDB = 25.0 kN (C) FDC = 15.0 kN (T) FDE = 12.0 kN (C) FCB = 0 FCE = 33.5 kN (C) FCF = 30.0 kN (T) FBE = 39.1 kN (T) FBF = 0 FBA = 30.0 kN (C) FAE = 0 FAF = 0 FFE = 0 551

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6–59. Determine the force in each member of the space truss and state if the members are in tension or compression. The The truss is supported by ball-and-socket joints at A, B, and E. Set F = 5800j6 N. Hint: The support reaction at E acts along member EC. Why?

z

D F

1m

A

2m C

5m

y

E

Solution

x

2m

B 1.5 m

Joint D: ΣFx = 0; ΣFy = 0; ΣFz = 0;

Joint C: ΣFx = 0;

1 5 1 F + FBD + FCD = 0 3 AD 231.25 27.25 2 1.5 1.5 - FAD + FBD FCD + 800 = 0 3 231.25 27.25 2 2 2 - FAD FBD + FCD = 0 3 231.25 27.25 -

FAD = 686 N1T 2

Ans.

FBD = 0

Ans.

FCD = 615.4 = 615 N1C 2

Ans.

FBC ΣFy = 0;

ΣFz = 0;

1

1615.42 = 0 27.25 = 229 N1T 2

FBC -

Ans.

1.5

1615.42 - FAC = 0 27.25 FAC = 343 N1T 2

Ans.

FEC

Ans.

2

1615.42 = 0 27.25 = 457 N1C 2

FBC -

Ans: FAD = FBD = FCD = FBC = FAC = FEC = 552

686 N (T) 0 615 N (C) 229 N (T) 343 N (T) 457 N (C)

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*6–60. Determine the force in each member of the space truss and state if the members are in tension or compression. The truss is supported by ball-and-socket joints at A, B, and E. Set F = 5 -200i + 400j6 N. Hint: The support reaction at E acts along member EC. Why?

z

D F

1m

2m

A

C 5m E

SOLUTION Joint D:

x

©Fx = 0;

1 5 1 - FAD + FBD + FCD - 200 = 0 3 231.25 27.25

©Fy = 0;

2 1.5 1.5 - FAD + FBD FCD + 400 = 0 3 231.25 27.25

©Fz = 0;

2 2 2 - FAD FBD + FCD = 0 3 231.25 27.25 FAD = 343 N (T)

Ans.

FBD = 186 N (T)

Ans.

FCD = 397.5 = 397 N (C)

Ans.

Joint C: ©Fx = 0;

FBC -

1 27.25

(397.5) = 0

FBC = 148 N (T) ©Fy = 0;

1.5 27.25

Ans.

(397.5) - FAC = 0

FAC = 221 N (T) ©Fz = 0;

FEC -

2 27.25

Ans.

(397.5) = 0

FEC = 295 N (C)

Ans.

553

2m

B 1.5 m

y

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6–61. Determine the force P required to hold the 50-kg mass in equilibrium.

C B

A

SOLUTION

P

Equations of Equilibrium: Applying the force equation of equilibrium along the y axis of each pulley. + c ©Fy = 0;

R - 3P = 0;

R = 3P

+ c ©Fy = 0;

T - 3R = 0;

T = 3R = 9P

+ c ©Fy = 0;

2P + 2R + 2T - 50(9.81) = 0

Ans.

Substituting Eqs.(1) and (2) into Eq.(3) and solving for P, 2P + 2(3P) + 2(9P) = 50(9.81) P = 18.9 N

Ans.

Ans: P = 18.9 N 554

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6–62. The principles of a differential chain block are indicated schematically in the figure. Determine the magnitude of force P needed to support the 800-N force. Also, find the distance x where the cable must be attached to bar AB so the bar remains horizontal. All pulleys have a radius of 60 mm.

x B

A 800 N

180 mm

240 mm

P

SOLUTION Equations of Equilibrium: From FBD(a), + c ©Fy = 0;

4P¿ - 800 = 0

P¿ = 200 N

From FBD(b), + c ©Fy = 0;

200 - 5P = 0

a + ©MA = 0;

2001x2 - 40.011202 - 40.012402

P = 40.0 N

Ans.

- 40.013602 - 40.014802 = 0 x = 240 mm

Ans.

Ans: P = 40.0 N x = 240 mm 555

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6–63. Determine the force P required to hold the 150-kg crate in equilibrium. B

SOLUTION

A

C

Equations of Equilibrium: Applying the force equation of equilibrium along the y axis of pulley A on the free - body diagram, Fig. a, + c ©Fy = 0;

2TA - 150(9.81) = 0

P

TA = 735.75 N

Using the above result and writing the force equation of equilibrium along the y¿ axis of pulley C on the free - body diagram in Fig. b, ©Fy¿ = 0;

735.75 - 2P = 0

Ans.

P = 367.88 N = 368 N

Ans: P = 368 N 556

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*6–64.

A H

Determine the force P needed to support the 20-kg mass using the Spanish Burton rig. Also, what are the reactions at the supporting hooks A, B, and C?

B G

P

C F

E

D

SOLUTION For pulley D: + c ©Fy = 0;

9P - 2019.812 = 0 P = 21.8 N

Ans.

At A,

RA = 2P = 43.6 N

Ans.

At B,

RB = 2P = 43.6 N

Ans.

At C,

RC = 6P = 131 N

Ans.

Ans: P RA RB RC 557

= = = =

21.8 N 43.6 N 43.6 N 131 N

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6–65. 12 kN

Determine the reactions at the supports A, C, and E of the compound beam.

3 kN/ m

A 3m

B

C 4m

2m

D

E 6m

3m

Solution Free Body Diagram. The compound beam is being dismembered into members AB, BD and DE of which their respective FBDs are shown in Fig. a, b and c. Equations of Equilibrium. Equilibrium of member DE will be considered first by referring to Fig. c. Ans.

a+ΣMD = 0;  NE(6) - 12(9) = 0  NE = 18.0 kN a+ΣME = 0;  Dy(6) - 12(3) = 0 

Dy = 6.00 kN

+ ΣFx = 0;  Dx = 0 S Next, member BD, Fig. b. a+ΣMC = 0;  6.00(2) + 3(6)(1) - By(4) = 0 

 By = 7.50 kN

a+ΣMB = 0;  NC(4) + 6.00(6) - 3(6)(3) = 0  NC = 4.50 kN

Ans.

+ ΣFx = 0;  Bx = 0 S Finally, member AB, Fig. a + ΣFx = 0;  Ax = 0 S + c ΣFy = 0;  Ay - 7 - 50 = 0 

Ans. Ans.

 Ay = 7.50 kN

a+ΣMA = 0;  MA - 7.50(3) = 0  MA = 22.5 kN # m

Ans.

Ans: NE = 18.0 kN NC = 4.50 kN Ax = 0 Ay = 7.50 kN MA = 22.5 kN # m 558

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6–66. 900 N/m

Determine the reactions at the supports at A, E, and B of the compound beam.

900 N/m B

A 3m

3m

E

D

C 4m

3m

3m

Solution Free Body Diagram. The solution will be very much simplified if one realizes that member CD is a two force member. Equation of Equilibrium. Consider the equilibrium of member BD, Fig. b + ΣFx = 0;  FCD = 0 S 1 a+ΣMB = 0;   (900)(6)(4) - NE (3) = 0  NE = 3600 N = 3.60 kN 2 1 a+ΣME = 0;   (900)(6)(1) - NB (3) = 0  NB = 900 N 2

Ans. Ans.

Then the equilibrium of member AC gives + ΣFx = 0;  Ax = 0 S

Ans.

1 (900)(6) = 0  Ay = 2700 N = 2.70 kN Ans. 2 1 a+ΣMA = 0;  MA - (900)(6)(3) = 0  MA = 8100 N # m = 8.10 kN # m Ans. 2 + c ΣFy = 0;  Ay -

Ans:  NE =  NB = Ax = Ay = MA = 559

3.60 kN 900 N 0 2.70 kN 8.10 kN # m

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6–67. Determine the resultant force at pins A, B, and C on the three-member frame.

2m 800 N C 200 N/ m 2m 60 A

B

Solution Free Body Diagram. The frame is being dismembered into members AC and BC of which their respective FBDs are shown in Fig. a and b. Equations of Equilibrium. Write the moment equation of equilibrium about point A for member AC, Fig. a and point B for member BC, Fig. b. a+ΣMA = 0;  Cy a

2 2 1 b + Cx (2) - 200 a ba b = 0 tan 60° sin 60° sin 60°

a+ΣMB = 0;  Cy(2) - Cx(2) + 800(2) = 0

(1) (2)

Solving Eqs. (1) and (2) Cy = -338.12 N  Cx = 461.88 N



The negative sign indicates that Cy acts in the sense opposite to that shown in the FBD write the force equation of equilibrium for member AC, Fig. a, 2 b sin 60° - 461.88 = 0   Ax = 61.88 N sin 60° 2 + c ΣFy = 0;  Ay + ( - 338.12) - 200 a b cos 60° = 0  Ay = 569.06 N sin 60° + ΣFx = 0;  Ax + 200 a S

Also, for member BC, Fig. b

+ ΣFx = 0;  Bx + 461.88 - 800 = 0  Bx = 338.12 N S + c ΣFy = 0;   - By - ( - 338.12) = 0   By = 338.12 N Thus,

FC = 2C 2x + C 2y = 2461.88 2 + ( - 338.12)2 = 572.41 N = 572 N

Ans.

FB = 2B2x + B2y = 2338.122 + 338.122 = 478.17 N = 478 N

Ans.

FA = 2A2x + A2y = 261.882 + 569.062 = 572.41 N = 572 N

Ans.

Ans: FC = 572 N FA = 572 N FB = 478 N 560

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*6–68.

600 N

750 N

Determine the reactions on the collar at A and the pin at C. The collar fits over a smooth rod, and rod AB is fixed connected to the collar.

1.25 m

SOLUTION Equations of Equilibrium: From the force equation of equilibrium of member AB, Fig. a, we can write + ©MA = 0;

MA - 750(1.25) - By(2.5) = 0

(1)

+ ©F = 0; : x

NA cos 45° - Bx = 0

(2)

+ c ©Fy = 0;

NA sin 45° - 750 - By = 0

(3)

From the free-body diagram of member BC in Fig. b, + ©MC = 0;

Bx(2 sin 30°) - By(2 cos 30°) + 600(1) = 0

(4)

+ ©F = 0; : x

Bx + 600 sin 30° - Cx = 0

(5)

+ c ©Fy = 0;

By - Cy - 600 cos 30° = 0

(6)

Solving Eqs. (2), (3), and (4) yields By = 1844.13

N = 1.84 kN

NA = 3668.66

N = 3.67 kN

1.25 m

A 45

Bx = 2594.13 N Ans.

Substituting the results of Bx and By into Eqs. (1), (5), and (6) yields MA = 5547.84 N # m = 5.55 kN # m

Ans.

Cx = 2894.13

N = 2.89 kN

Ans.

Cy = 1324.52

N = 1.32 kN

Ans.

561

C

30 B

1m

1m

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6–69. 0.1 m

Determine the greatest force P that can be applied to the frame if the largest force resultant acting at A can have a magnitude of 2 kN.

0.5 m

A 0.75 m

0.75 m P

SOLUTION a + ©MA = 0;

T(0.6) - P(1.5) = 0

+ ©F = 0; : x

Ax - T = 0

+ c ©Fy = 0;

Ay - P = 0

Thus, Ax = 2.5 P, Ay = P Require, 2 = 2(2.5P)2 + (P)2 P = 0.743 kN = 743 N

Ans.

Ans: P = 743 N 562

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6–70. Determine the horizontal and vertical components of force at pins B and C. The suspended cylinder has a mass of 75 kg.

0.3 m B

C 1.5 m

Solution

A

Free Body Diagram. The solution will be very much simplified if one realizes that member AB is a two force member. Also, the tension in the cable is equal to the weight of the cylinder and is constant throughout the cable. Equations of Equilibrium. Consider the equilibrium of member BC by referring to its FBD, Fig. a, 3 a+ΣMC = 0;  FAB a b(2) + 75(9.81)(0.3) - 75(9.81)(2.8) = 0 5

2m

0.5 m

FAB = 1532.81 N



a+ΣMB = 0;  Cy (2) + 75(9.81)(0.3) - 75(9.81)(0.8) = 0

Ans.

Cy = 183.94 N = 184 N

+ ΣFx = 0;  1532.81a 4 b - 75(9.81) - Cx = 0 S 5 Cx = 490.5 N

Ans.

Thus,

FB = FAB = 1532.81 N



Bx =

4 (1532.81) = 1226.25 N = 1.23 kN 5

Ans.



By =

3 (1532.81) = 919.69 N = 920 kN 5

Ans.

Ans: Cy = Cx = Bx = By = 563

184 N 490.5 N 1.23 kN 920 kN

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6–71. Determine the greatest force P that can be applied to the frame if the largest force resultant acting at A can have a magnitude of 5 kN.

0.1 m

0.5 m

A

0.8 m

0.8 m

P

Solution a+ ΣMA = 0; + S ΣFx = 0; + c ΣFy = 0; Thus,

T 10.82 - P 11.62 = 0 Ax - T = 0 Ay - P = 0

Ax = 2P,

Ay = P

Require,

0.8 m 2

5 = 212P 2 + 1P 2 P = 2.24 kN

2

1.6 m

Ans.

Ans: P = 2.24 kN 564

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*6–72. Determine the horizontal and vertical components of force at pins A and D.

D

2m 0.3 m C A

1.5 m

B

1.5 m E

Solution Free Body Diagram. The assembly will be dismembered into member AC, BD and pulley E. The solution will be very much simplified if one recognizes that member BD is a two force member. The FBD of pulley E and member AC are shown in Fig. a and b respectively.

12 kN

Equations of Equilibrium. Consider the equilibrium of pulley E, Fig. a, + c ΣFy = 0;  2T - 12 = 0  T = 6.00 kN Then, the equilibrium of member AC gives 4 a+ΣMA = 0;  FBD a b(1.5) + 6(0.3) - 6(3) - 6(3.3) = 0 5 FBD = 30.0 kN + ΣFx = 0;  Ax - 30.0 a 3 b - 6 = 0   S 5

Ans.

Ax = 24.0 kN

4 + c ΣFy = 0;  30.0 a b - 6 - 6 - Ay = 0  Ay = 12.0 kN 5

Ans.

Thus,

FA = 2A2x + A2y = 224.02 + 12.02 = 26.83 kN = 26.8 kN FB = FBD = 30.0 kN



Dx =

3 (30) = 18.0 kN  5

Ans.



Dy =

4 (30) = 24.0 kN  5

Ans.

Ans: Ax = Ay = Dx = Dy = 565

24.0 kN 12.0 kN 18.0 kN 24.0 kN

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6–73. Determine the horizontal and vertical components of force that pins A and B exert on the frame.

C 2 kN/m

4m

B

A

Solution Free Body Diagram. The frame will be dismembered into members BC and AC. The solution will be very much simplified if one recognizes that member AC is a two force member. The FBDs of member BC and pin A are shown in Figs. a and b, respectively.

3m

Equations of Equilibrium. Consider the equilibrium of member BC, Fig. a, 3 a+ΣMB = 0;  2(4)(2) - FAC a b(4) = 0  FAC = 6.6667 kN 5 a+ΣMC = 0;  Bx (4) - 2(4)(2) = 0  Bx = 4.00 kN

4 + c ΣFy = 0;   6.6667 a b - By = 0  By = 5.333 kN = 5.33 kN 5

Ans. Ans.

Then, the equilibrium of pin A gives

+ ΣFx = 0;  Ax - 6.6667 a 3 b = 0  Ax = 4.00 kN S 5

Ans.

4 + c ΣFy = 0;  Ay - 6.6667 a b = 0  Ay = 5.333 kN = 5.33 kN 5

Ans.

Ans: Bx = By = Ax = Ay = 566

4.00 kN 5.33 kN 4.00 kN 5.33 kN

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6–74. Determine the horizontal and vertical components of force which the pins at A and B exert on the frame.

2m 400 N/m C

D 1.5 m E

3m

3m

Solution

F

Free Body Diagram. The frame will be dismembered into members AD, EF, CD and BC. The solution will be very much simplified if one realizes that members CD and EF are two force member. Therefore, only the FBD of members AD and BC, Fig. a and b respectively, need to be drawn

1.5 m A

B

Equations of Equilibrium. Write the moment equations of equilibrium about point A for member AD, Fig. a, and point B for member BC, Fig. b. 4 a+ΣMA = 0;  FEF a b(3) - FCD (4.5) - 400 (4.5)(2.25) = 0 5

(1)

4 a+ΣMB = 0;   - FEF a b(1.5) + FCD (4.5) = 0 5

(2)

Solving Eqs. (1) and (2)

FEF = 3375 N  FCD = 900 N

Write the force equation of equilibrium for member AD, Fig. a, + ΣFx = 0;  Ax + 400(4.5) + 900 - 3375 a 4 b = 0  Ax = 0 S 5

3 + c ΣFy = 0;  3375 a b - Ay = 0  Ay = 2025 N = 2.025 kN 5

Ans. Ans.

Also, for member BC, Fig. b

+ ΣFx = 0;  3375a 4 b - 900 - Bx = 0  Bx = 1800 N = 1.80 kN S 5 + c ΣFy = 0;  

3 By - 3375a b = 0 5

    By = 2025 N = 2.025 kN

Ans. Ans.

Ans: Ax = Ay = Bx = By = 567

0 2.025 kN 1.80 kN 2.025 kN

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140 mm

6–75. If P = 75 N, determine the force F that the toggle clamp exerts on the wooden block.

85 mm 140 mm

50 mm

P

A D

50 mm

C

B

20 mm E

F

P

Solution Equations of Equilibrium: First, we will consider the free-body diagram of the upper handle in Fig. a. a+ ΣMA = 0; +

S ΣFx = 0; + c ΣFy = 0;

Bx 1502 - 7511402 = 0 Bx = 210 N 210 - Ax = 0 Ax = 210 N By - Ay - 75 = 0

(1)

Using the result for Bx and applying the moment equation of equilibrium about point C on the free-body diagram of the lower handle in Fig. b, a+ ΣMC = 0;

2101502 + 7511602 - By 1202 = 0 By = 1125 N

Substituting By = 1125 N into Eq. (1) yields Ay = 1050 N Writing the moment equation of equilibrium about point D on the free-body diagram of the clamp shown in Fig. c, a+ ΣMD = 0;

10501852 - 2101502 - F 11402 = 0 F = 562.5 N

Ans.

Ans: F = 562.5 N 568

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*6–76. If the wooden block exerts a force of F = 600 N on the toggle clamp, determine the force P applied to the handle.

140 mm

85 mm 140 mm

50 mm

P

A D

50 mm

C

B

20 mm E

F

P

Solution Equations of Equilibrium: First, we will consider the free-body diagram of the upper handle in Fig. a. a+ ΣMA = 0; +

S ΣFx = 0; + c ΣFy = 0;

Bx 1502 - P 11402 = 0 Bx = 2.8P 2.8P - Ax = 0 Ax = 2.8P Bx - Ay - P = 0

(1)

Using the result for Bx and applying the moment equation of equilibrium about point C on the free-body diagram of the lower handle in Fig. b, a+ ΣMC = 0;

P 11602 + 2.8P 1502 - By 1202 = 0 By = 15P

Substituting By = 15 P into Eq. (1) yields Ay = 14P Writing the moment equation of equilibrium about point D on the free-body diagram of the clamp shown in Fig. c, a+ ΣMD = 0;

14P 1852 - 2.8P 1502 - 60011402 = 0 P = 80 N

Ans.

Ans: P = 80 N 569

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6–77.

b

3b D

Show that the weight W1 of the counterweight at H required for equilibrium is W1 = (b>a)W, and so it is independent of the placement of the load W on the platform.

c 4 A

H

G

W E

B

SOLUTION

F

a

C

c

Equations of Equilibrium: First, we will consider the free-body diagram of member BE in Fig. a, W(x) - NB a 3b +

+ ©ME = 0;

NB =

+ c ©Fy = 0;

FEF +

3 cb = 0 4

Wx 3 a 3b + c b 4 Wx - W = 0 3 a 3b + c b 4

FEF = W § 1 -

x 3b +

3 c 4

¥

Using the result of NB and applying the moment equation of equilibrium about point A on the free-body diagram in Fig. b, we obtain + ©MA = 0;

Wx 1 a cb = 0 3 4 3b + c 4 Wx = 12b + 3c

FCD (c) -

NCD

Writing the moment equation of equilibrium about point G on the free-body diagram in Fig. c, we have + ©MG = 0;

Wx (4b) + W § 1 12b + 3c

W1 =

x 3b +

3 c 4

¥(b) - W1(a) = 0

b W a

Ans.

This result shows that the required weight W1 of the counterweight is independent of the position x of the load on the platform.

Ans: W1 = 570

b W a

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6–78.

The toggle clamp is subjected to a force F at the handle. Determine the vertical clamping force acting at E.

a/2

F

B A

a/2

C 60

D a/2

E

SOLUTION

1.5 a

1.5 a

Free Body Diagram: The solution for this problem will be simplified if one realizes that member CD is a two force member. Equations of Equilibrium: From FBD (a), a + ©MB = 0;

a a FCD cos 30°a b - FCD sin 30°a b - F12a2 = 0 2 2 FCD = 10.93F

+ ©F = 0; : x

Bx - 10.93 sin 30° = 0 Bx = 5.464F

From (b), a + ©MA = 0;

5.464F1a2 - FE 11.5a2 = 0 F E = 3.64F

Ans.

Ans: FE = 3.64 F 571

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6–79. The hoist supports the 125-kg engine. Determine the force the load creates in member DB and in member FB, which contains the hydraulic cylinder H.

1m

2m F

G

E

2m

SOLUTION

H

Free Body Diagrams: The solution for this problem will be simplified if one realizes that members FB and DB are two-force members.

1226.25(3) - FFB ¢

1938.87 ¢

3 210

B

A

3 210

≤ (2) = 0

2m

FFB = 1938.87 N = 1.94 kN + c ©Fy = 0;

1m C

Equations of Equilibrium: For FBD(a), a + ©ME = 0;

D

1m

Ans.

≤ - 1226.25 - Ey = 0

Ey = 613.125N + ©F = 0; : x

Ex - 1938.87 ¢

1 210

≤ = 0

Ex = 613.125 N From FBD (b), a + ©MC = 0;

613.125(3) - FBD sin 45°(1) = 0 FBD = 2601.27 N = 2.60 kN

Ans.

Ans: FFB = 1.94 kN FBD = 2.60 kN 572

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*6–80.

0.8 m

The scissors lift consists of two sets of cross members and two hydraulic cylinders, DE, symmetrically located on each side of the platform. The platform has a uniform mass of 60 kg, with a center of gravity at G1. The load of 85 kg, with center of gravity at G2, is centrally located between each side of the platform. Determine the force in each of the hydraulic cylinders for equilibrium. Rollers are located at B and D.

1.2 m

2m

G2 G1 A

B

C D

F 1.5 m

1m

1m E

1.5 m

SOLUTION Free Body Diagram: The solution for this problem will be simplified if one realizes that the hydraulic cyclinder DE is a two force member. Equations of Equilibrium: From FBD (a), a + ©MA = 0;

2NB 132 - 833.8510.82 - 588.6122 = 0 2NB = 614.76 N

+ ©F = 0; : x + c ©Fy = 0;

Ax = 0 2Ay + 614.76 - 833.85 - 588.6 = 0 2Ay = 807.69 N

From FBD (b), a + ©MD = 0;

807.69132 - 2Cy 11.52 - 2Cx 112 = 0 2Cx + 3Cy = 2423.07

(1)

From FBD (c), a + ©MF = 0;

2Cx 112 - 2Cy 11.52 - 614.76132 = 0 2Cx - 3Cy = 1844.28

(2)

Solving Eqs. (1) and (2) yields Cx = 1066.84 N

Cy = 96.465 N

From FBD (b), + ©F = 0; : x

211066.842 - 2FDE = 0 FDE = 1066.84 N = 1.07 kN

Ans.

Ans: FDE = 1.07 kN 573

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6–81. Determine the force in members FD and DB of the frame. Also, find the horizontal and vertical components of reaction the pin at C exerts on member ABC and member EDC.

E

G F 6 kN

2m

D 1m

B

A

C 2m

Solution

1m

Free Body Diagram. The assembly will be dismembered into members GFE, EDC, FD, BD and ABC. The solution will be very much simplified if one recognizes members FD and BD are two force members. The FBDs of members GFE, EDC and ABC are shown in Figs. a, b and c respectively. Equations of Equilibrium. First, consider the equilibrium of member GFE, Fig. a, 2 b(1) = 0  FFD = 925 kN = 20.1 kN 25 a+ΣMF = 0;  6(2) - Ey (1) = 0 Ey = 12.0 kN

a+ΣME = 0;  6(3) - FFD a

+ ΣFx = 0;  Ex - a925ba 1 b = 0 S 25

Ans.

Ex = 9.00 kN

Next, for member EDC, Fig. b,

a+ΣMC = 0;  9.00(3) - a925b a



1 25

b(1) - FBD a

1 22

b(1) = 0

FBD = 1822 kN = 25.5 kN

Ans.

a+ΣMD = 0;  9.00(2) - C′x (1) = 0  C′x = 18.0 kN



+ c ΣFy = 0;  12.0 + a1822ba

1 22

b - a925ba

2

25 C′y = 12.0 kN

Ans.

b - C′y = 0

Ans.

Finally, for member ABC, Fig. c a+ΣMA = 0;  C″y (3) - a1822b a

1 22

b(2) = 0  Cy″ = 12.0 kN

+ ΣFx = 0;  Cx″ - a1822ba 1 b = 0    Cx″ = 18.0 kN S 22

Ans. Ans.

Ans: FFD = 20.1 kN FBD = 25.5 kN Cx″ = 18.0 kN Cy″ = 12.0 kN 574

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6–82. Determine the force that the smooth 20-kg cylinder exerts on members AB and CDB. Also, what are the horizontal and vertical components of reaction at pin A?

D

C

1m A E 1.5 m

B 2m

Solution Free Body Diagram. The FBDs of the entire assembly, member CDB and the cylinder are shown in Figs. a, b and c, respectively. Equations of Equilibrium. First consider the equilibrium of the entire assembly, Fig. a, a+ΣMA = 0;  ND(1) - 20(9.81)(1.5) = 0  ND = 294.3 N + ΣFx = 0;  Ax - 294.3 = 0      Ax = 294.3 N = 294 N S

Ans.

+ c ΣFy = 0;  Ay - 20(9.81) = 0 

Ans.

Ay = 196.2 N = 196 N

Next, for member CDB, Fig. b a+ΣMB = 0;  294.3(1) - NC (2) = 0  NC = 147.15 N = 147 N

Ans.

Finally for the cylinder, Fig. c + c ΣFy = 0;  NE - 20(9.81) - 147.15 = 0  NE = 343.35 N = 343 N

Ans.

Ans: Ax = 294 N Ay = 196 N NC = 147 N NE = 343 N 575

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6–83.

20 N 60 mm

The pruner multiplies blade-cutting power with the compound leverage mechanism. If a 20-N force is applied to the handles, determine the cutting force generated at A. Assume that the contact surface at A is smooth.

150 mm F

E

B

10 mm A

C

D

45

25 mm 30 mm G

20 N

SOLUTION Equations of Equilibrium: Applying the moment equation of equilibrium about point C to the free-body diagram of handle CDG in Fig. a, we have + ©MC = 0;

20(150) - FDE sin 45°(25) = 0 FDE = 169.71 N

Using the result of FDE and applying the moment equation of equilibrium about point B on the free-body diagram of the cutter in Fig. b, we obtain + ©MB = 0;

169.71 sin 45°(55) + 169.71 cos 45°(10) - NA (60) = 0 FA = 130 N

Ans.

Ans: FA = 130 N 576

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*6–84. H

The symmetric coil tong supports the coil which has a mass of 800 kg and center of mass at G. Determine the horizontal and vertical components of force the linkage exerts on plate DEIJH at points D and E. The coil exerts only vertical reactions at K and L.

300 mm

D

J E

I

400 mm B 100 mm

45° A

C 30° 45°

F

30°

50 mm

100 mm

SOLUTION Free-Body Diagram: The solution for this problem will be simplified if one realizes that links BD and CF are two-force members.

K

G

L

Equations of Equilibrium : From FBD (a), 78481x2 - FK12x2 = 0

a + ©ML = 0;

FK = 3924 N

From FBD (b), a + ©MA = 0;

FBD cos 45°11002 + FBD sin 45°11002 - 39241502 = 0 FBD = 1387.34 N

+ ©F = 0; : x

A x - 1387.34 cos 45° = 0

+ c ©Fy = 0;

A y - 3924 - 1387.34 sin 45° = 0

A x = 981 N

A y = 4905 N From FBD (c), a + ©ME = 0;

4905 sin 45°17002 - 981 sin 45°17002 - FCF cos 15°13002 = 0 FCF = 6702.66 N

+ ©F = 0; : x

Ex - 981 - 6702.66 cos 30° = 0 Ex = 6785.67 N = 6.79 kN

+ c ©Fy = 0;

Ans.

Ey + 6702.66 sin 30° - 4905 = 0 Ey = 1553.67 N = 1.55 kN

Ans.

Dx = FBD cos 45° = 1387.34 cos 45° = 981 N

Ans.

Dy = FBD sin 45° = 1387.34 sin 45° = 981 N

Ans.

At point D,

Ans: Ex = Ey = Dx = Dy = 577

6.79 kN 1.55 kN 981 N 981 N

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6–85. Determine the force that the jaws J of the metal cutters exert on the smooth cable C if 100-N forces are applied to the handles. The jaws are pinned at E and A, and D and B. There is also a pin at F.

15

100 N

400 mm 15 J E C D B 30 mm 80 mm

20 mm A F

15 15 20 mm

400 mm

SOLUTION Free Body Diagram: The solution for this problem will be simplified if one realizes that member ED is a two force member.

15

100 N

Equations of Equilibrium: From FBD (b), + ©F = 0; : x

Ax = 0

From (a), a + ©MF = 0;

Ay sin 15°1202 + 100 sin 15°1202 - 100 cos 15°14002 = 0 Ay = 7364.10 N

From FBD (b), a + ©ME = 0;

7364.101802 - FC1302 = 0 Ans.

FC = 19637.60 N = 19.6 kN

Ans: FC = 19.6 kN 578

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6–86. A 300-kg counterweight, with center of mass at G, is mounted on the pitman crank AB of the oil-pumping unit. If a force of F = 5 kN is to be developed in the fixed cable attached to the end of the walking beam DEF, determine the torque M that must be supplied by the motor.

1.75 m

2.50 m

D 30⬚

E

M A

F

B G

30⬚ 0.5 m 0.65 m

F

Solution Equations of Equilibrium: Applying the moment equation of equilibrium about point E to the free-body diagram of the walking beam in Fig. a, a+ ΣME = 0;

FCD cos 30° 11.752 - 500012.52 = 0 FCD = 8247.86 N

Using the result of FCD and applying the moment equation of equilibrium about point A on the free-body diagram of the pitman crank in Fig. b, a+ ΣMA = 0;

8247.8610.652 - 30019.812 cos 30° 11.152 - M = 0 M = 2430.09 N # m = 2.43 kN # m

Ans.

Ans: M = 2.43 kN # m 579

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6–87. A 300-kg counterweight, with center of mass at G, is mounted on the pitman crank AB of the oil-pumping unit. If the motor supplies a torque of M = 2500 N # m, determine the force F developed in the fixed cable attached to the end of the walking beam DEF.

1.75 m

2.50 m

D 30⬚

E

M A

F

B G

30⬚ 0.5 m 0.65 m

F

Solution Equations of Equilibrium: Applying the moment equation of equilibrium about point A to the free-body diagram of the pitman crank in Fig. a, a+ ΣMA = 0;

FCD 10.652 - 30019.812 cos 30° 11.152 - 2500 = 0 FCD = 8355.41 N

Using the result of FCD and applying the moment equation of equilibrium about point E on the free-body diagram of the walking beam in Fig. b, a+ ΣME = 0;

8355.41 cos 30° 11.752 - F 12.52 = 0 F = 5065.20 N = 5.07 kN

Ans.

Ans: F = 5.07 kN 580

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*6–88. 160 mm

The double link grip is used to lift the beam. If the beam weighs 8 kN, determine the horizontal and vertical components of force acting on the pin at A and the horizontal and vertical components of force that the flange of the beam exerts on the jaw at B.

160 mm

D

45°

E

SOLUTION

C

A

140 mm

Free Body Diagram: The solution for this problem will be simplified if one realizes that members ED and CD are two force members.

300 mm B

F

Equations of Equilibrium: Using method of joint, [FBD (a)], + c ©Fy = 0;

8 - 2F sin 45° = 0

300 mm 300 mm

F = 5.657 kN

From FBD (b), + c ©Fy = 0;

2By - 8 = 0

By = 4.00 kN

Ans.

From FBD (c), a + ©MA = 0;

8 kN

Bx 13002 - 4.0013002 - 5.657 cos 45°11402

- 5.657 sin 45°11602 = 0

Bx = 8.00 kN + c ©Fy = 0;

Ans.

Ay + 5.657 sin 45° - 4.00 = 0 Ay = 0

+ ©F = 0; : x

Ans.

5.657 kN

8.00 + 5.657 cos 45° - Ax = 0

140 mm

Ax = 12.00 kN

Ans.

4.00

300 mm

300 mm

8 300 mm

300 mm

Ans: By Bx Ay Ax 581

= = = =

4.00 kN 8.00 kN 0 12.00 kN

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6–89. The constant moment of 50 N # m is applied to the crank shaft. Determine the compressive force P that is exerted on the piston for equilibrium as a function of u. Plot the results of P (ordinate) versus u (abscissa) for 0° … u … 90°.

B 0.45 m

u 0.2 m A

C P

50 N  m

SOLUTION Member AB: a + ©MA = 0;

FBC sin f(0.2 sin u) + FBC cos f(0.2 cos f)-50 = 0 FBC =

250 (sin f sin u + cos f cos u)

Piston: + ©F = 0; : x

FBC cos f - P = 0 P =

250 cos f (sin f sin u + cos f cos u)

(1)

d = 0.2 cos u = 0.45 sin f f = sin-1 a

cos u b 2.25

(2)

Select u, solve for f in Eq. (2), then P in Eq. (1). 250 cos csin-1 a P(u) =

=

cos u bd 2.25

sin u cos u cos u + cos csin-1 a b d # cos u 2.25 2.25 250 22.252 - cos2 u

Ans.

sin u cos u + 22.252 - cos2 u # cos u

Ans: P(u) = 582

250 22.252 - cos2 u

sin u cos u + 22.252 - cos2 u # cos u

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6–90. The tractor boom supports the uniform mass of 600 kg in the bucket which has a center of mass at G. Determine the force in each hydraulic cylinder AB and CD and the resultant force at pins E and F. The load is supported equally on each side of the tractor by a similar mechanism.

G

B

A

0.25 m E 0.3 m 0.1 m

C 1.5 m 1.25 m

0.2 m

F

SOLUTION

0.6 m

294310.12 - FAB10.252 = 0

a + ©ME = 0;

D

FAB = 1177.2 N = 1.18 kN + ©F = 0; : x

- Ex + 1177.2 = 0;

Ex = 1177.2 N

+ c ©Fy = 0;

Ey - 2943 = 0;

Ey = 2943 N

FE = 211177.22 + 129432 2

a + ©MF = 0;

2

Ans.

2943 N

= 3.17 kN

Ans.

294312.802 - FCD1cos 12.2°210.72 + FCD1sin 12.2°211.252 = 0 FCD = 19 618 N = 19.6 kN

+ ©F = 0; : x

Ans.

Fx - 19 618 sin 12.2° = 0 Fx = 4145.8 N

+ c ©Fy = 0;

0.4 m 0.3 m

2943 N

-Fy - 2943 + 19 618 cos 12.2° = 0 Fy = 16 232 N

FF = 214145.822 + 116 23222 = 16.8 kN

Ans.

Ans: FF = 16.8 kN 583

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6–91. The machine shown is used for forming metal plates. It consists of two toggles ABC and DEF, which are operated by the hydraulic cylinder H.The toggles push the movable bar G forward, pressing the plate p into the cavity. If the force which the plate exerts on the head is P = 12 kN, determine the force F in the hydraulic cylinder when u = 30°.

D

u  30 E

200 mm

F 200 mm

F P  12 kN

H G

F 200 mm A

B u  30

200 mm C

p

SOLUTION Member EF: a + ©ME = 0;

- Fy (0.2 cos 30°) + 6 (0.2 sin 30°) = 0 Fy = 3.464 kN

Joint E: + ©F = 0; : x

FDE cos 30° - 6 = 0 FDE = 6.928 kN

+ c ©Fy = 0;

F - 3.464 - 6.928 sin 30° = 0 Ans.

F = 6.93 kN

Ans: F = 6.93 kN 584

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*6–92. The nail cutter consists of the handle and the two cutting blades. Assuming the blades are pin connected at B and the surface at D is smooth, determine the normal force on the fingernail when a force of 5 N is applied to the handles as shown. The pin AC slides through a smooth hole at A and is attached to the bottom member at C.

6 mm

5N

6 mm 36 mm

A

B D C 5N

Solution F = 5N a = 6 mm b = 36 mm Handle: ΣMD = 0;

ΣFy = 0;

FA # a - F # b = 0 b FA = F # a ND - FA - F = 0 ND = FA + F

FA = 30 N

ND = 35 N

Top blade : ΣMB = 0;

ND # b - FN # 12a + b2 = 0 b FN = ND # # 2 a + b

FN = 26.25 N

Ans.

Ans: FN = 26.25 N 585

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6–93.

The pipe cutter is clamped around the pipe P. If the wheel at A exerts a normal force of FA = 80 N on the pipe, determine the normal forces of wheels B and C on the pipe. The three wheels each have a radius of 7 mm and the pipe has an outer radius of 10 mm.

C 10 mm B

A

10 mm

P

SOLUTION u = sin

-1

a

10 b = 36.03° 17

Equations of Equilibrium: + c ©Fy = 0;

NB sin 36.03° - NC sin 36.03° = 0 N B = NC

+ ©F = 0; : x

80 - NC cos 36.03° - NC cos 36.03° = 0 NB = NC = 49.5 N

Ans.

Ans: NB = NC = 49.5 N 586

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6–94. Determine the horizontal and vertical components of force which pin C exerts on member ABC. The 600-N load is applied to the pin.

2m

2m

E

D

3m A

C

B 1.5 m

Solution

600 N

Free Body Diagram. The solution will be very much simplified if one determines the support reactions first and then considers the equilibrium of two of its three members after they are dismembered. The FBDs of the entire assembly, member DBF and member ABC are shown in Figs. a, b and c, respectively.

F

300 N

Equations of Equilibrium. Consider the equilibrium of the entire assembly, Fig. a, a+ΣME = 0;  NA(3) - 300(4.5) - 600(4) = 0  NA = 1250 N Next, write the moment equation of equilibrium about point D for member DBF, Fig. b. a+ΣMD = 0;  Bx(1.5) - 300(3) = 0  Bx = 600 N Finally, consider the equilibrium of member ABC, Fig. c + ΣFx = 0;  1250 - 600 - Cx = 0  Cx = 650 N S

Ans. Ans.

a+ΣMB = 0;  Cy = 0

Ans: Cx = 650 N Cy = 0 587

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6–95. Determine the force created in the hydraulic cylinders EF and AD in order to hold the shovel in equilibrium. The shovel load has a mass of 1.25 Mg and a center of gravity at G. All joints are pin connected.

0.25 m

0.25 m

1.5 m

E C

30 H

D 10 60

A

F

2m

0.5 m G

SOLUTION Assembly FHG: a + ©MH = 0;

- 1250(9.81) (0.5) + FEF (1.5 sin 30°) = 0 Ans.

FEF = 8175 N = 8.18 kN (T) Assembly CEFHG: a + ©MC = 0;

FAD cos 40° (0.25) - 1250(9.81) (2 cos 10° + 0.5) = 0 Ans.

FAD = 158 130 N = 158 kN (C)

Ans: FEF = 8.18 kN (T) FAD = 158 kN (C) 588

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*6–96. 4 kN

Determine the horizontal and vertical components of force that the pins at A and B exert on the frame.

2 kN 2m

2m

2m

C B 3m 3 kN

Solution

1m

Free Body Diagram. The assembly will be dismembered into members BC, CD, CE and AD. The solution will be very much simplified if one recognizes member CE is a two force member. The FBDs of members CD, BC and AD are shown in Figs. a, b and c, respectively.

A E

D 3m

3m

Equations of Equilibrium. First, consider the equilibrium of member CD, Fig. a. a+ΣMC = 0;  3(3) - Dx (4) = 0  Dx = 2.25 kN a+ΣMD = 0;  Cx (4) - 3(1) = 0  Cx = 0.75 kN + c ΣFy = 0;  Cy - Dy = 0      Cy = Dy Next write the moment equation of equilibrium about point B for member BC, Fig. b and about point A for member AD, Fig. c, 4 a+ΣMB = 0;  2(2) + 4(4) + Dy (6) - FCE a b(6) = 0 5

(1)

4 a+ΣMA = 0;  FCE a b(3) - Dy(6) = 0 5

(2)

Solving Eqs. (1) and (2)

FCE = 8.3333 kN  Dy = 3.3333 kN

Finally, write the force equation of equilibrium for member BC, Fig. b, 4 + c ΣFy = 0;    By + 8.3333 a b - 3.3333 - 4 -2 = 0 5 By = 2.6667 kN = 2.67 kN

Ans.

+ ΣFx = 0;  Bx + 0.75 - 8.3333 a 3 b = 0  Bx = 4.25 kN S 5

Ans.

Also, for member AD, Fig. c.

4 + c ΣFy = 0;  Ay + 3.3333 - 8.3333 a b = 0  Ay = 3.3333 kN = 3.33 kN Ans. 5 + ΣFx = 0;  2.25 + 8.3333 a 3 b - Ax = 0  Ax = 7.25 kN S 5

Ans.

Ans: By = 2.67 kN Bx = 4.25 kN Ay = 3.33 kN Ax = 7.25 kN 589

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6–97. Operation of exhaust and intake valves in an automobile engine consists of the cam C, push rod DE, rocker arm EFG which is pinned at F, and a spring and valve, V. If the compression in the spring is 20 mm when the valve is open as shown, determine the normal force acting on the cam lobe at C. Assume the cam and bearings at H, I, and J are smooth. The spring has a stiffness of 300 N>m.

40 mm

25 mm

E G

F H

I

SOLUTION Fs = kx;

V

Fs = 300 (0.02) = 6 N

+ c ©Fy = 0;

J

- FG + 6 = 0 FG = 6 N

a + ©MF = 0;

- 6(40) + T(25) = 0

D

Ans.

T = 9.60 N

C

Ans: T = 9.60 N 590

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6–98. A man having a weight of 875 N attempts to hold himself using one of the two methods shown. Determine the total force he must exert on bar AB in each case and the normal reaction he exerts on the platform at C. Neglect the weight of the platform.

A

B A

B

C

C

(a)

(b)

SOLUTION (a) Bar: 437.5 N

+↑ΣFy = 0;

437.5 N

2(F/2) – 2(437.5) = 0 F = 875 N

Ans.

Man: +↑ΣFy = 0;

NC – 875 – 2(437.5) = 0 NC = 1750 N

875 N 437.5 N 437.5 N

Ans.

(b) Bar: +↑ΣFy = 0;

2(218.75) – 2(F/2) = 0 F = 437.5 N

Ans.

218.75 N

218.75 N

Man: +↑ΣFy = 0;

NC – 875 + 2(218.75) = 0 NC = 437.5 N

Ans.

875 N 218.75 N 218.75 N

Ans: a. F = 875 N, N C = 1750 N b. F = 437.5 N, N C = 437.5 N 591

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6–99. A man having a weight of 875 N attempts to hold himself using one of the two methods shown. Determine the total force he must exert on bar AB in each case and the normal reaction he exerts on the platform at C. The platform has a weight of 150 N.

A

B A

B

C

C

(a)

(b)

SOLUTION 512.5 N

512.5 N

512.5 N

512.5 N

150 N 875 N

(a) Bar: +↑ΣFy = 0;

2(F/2) – 512.5 – 512.5 = 0 F = 1025 N

Ans.

875 N 512.5 N 512.5 N

Man: +↑ΣFy = 0;

NC – 875 – 512.5 – 512.5 = 0 NC = 1900 N

Ans.

(b) Bar: 512.5 N

+↑ΣFy = 0;

2(F/2) – 256.25 – 256.25 = 0 F = 512.5 N

Ans.

256.25 N 256.25 N

Man: +↑ΣFy = 0;

NC – 875 + 256.25 + 256.25 = 0 NC = 362.5 N

256.25 N

256.25 N

Ans.

875 N 256.25 N 256.25 N

Ans: a. F = 1025 N, N C = 1900 N b. F = 512.5 N, N C = 362.5 N 592

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*6–100. If the 300-kg drum has a center of mass at point G, determine the horizontal and vertical components of force acting at pin A and the reactions on the smooth pads C and D. The grip at B on member DAB resists both horizontal and vertical components of force at the rim of the drum.

P

600 mm E

A 60 mm 60 mm

Equations of Equilibrium: From the free - body diagram of segment CAE in Fig. a,

100 mm

D

G

Ans.

Ax - 12 743.56 = 0 Ans.

Ax = 12 743.56 N = 12.7 kN + c ©Fy = 0;

B

300(9.81)(600 cos 30°) - NC(120) = 0 NC = 12 743.56 N = 12.7 kN

+ ©F = 0; : x

C

390 mm

SOLUTION a + ©MA = 0;

30

300(9.81) - Ay = 0 Ans.

Ay = 2943 N = 2.94 kN

Using the results for Ax and Ay obtained above and applying the moment equation of equilibrium about point B on the free - body diagram of segment BAD, Fig. b, a + ©MB = 0;

12 743.56(60) - 2943(100) - ND(450) = 0 Ans.

ND = 1045.14 N = 1.05 kN

Ans: NC = 12.7 kN Ax = 12.7 kN Ay = 2.94 kN ND = 1.05 kN 593

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6–101. The two-member frame is pin connected at E. The cable is attached to D, passes over the smooth peg at C, and supports the 500-N load. Determine the horizontal and vertical reactions at each pin.

1m

0.5 m 0.5 m

1m

A B

0.5 m E

C D

Solution Free Body Diagram. The frame will be dismembered into members BD and AC, of which their respective FBDs are shown in Figs. a and b.

500 N

Equations of Equilibrium. Write the moment equation of equilibrium about point B for member BD, Fig. a, and about point A for member AC, Fig. b, a+ΣMB = 0;  Ey(0.5) - Ex(0.5) + 500(2) = 0

(1)

a+ΣMA = 0;  Ey (0.5) + Ex (0.5) - 500(2) - 500(2) = 0

(2)

Solving Eqs. (1) and (2),

Ey = 1000 N = 1.00 kN  Ex = 3000 N = 3.00 kN

Ans.

Write the force equations of equilibrium for member BD, Fig. a. + ΣFx = 0;  Bx + 500 - 3000 = 0  Bx = 2500 N = 2.50 kN S

Ans.

+ c ΣFy = 0;  By - 1000 = 0    By = 1000 N = 1.00 kN

Ans.

Also, for member AC, Fig. b + ΣFx = 0;  3000 - 500 - Ax = 0  Ax = 2500 N = 2.50 kN S

Ans.

+ c ΣFy = 0;  1000 - 500 - Ay = 0  Ay = 500 N

Ans.

Ans: Ey = Ex = Bx = By = Ax = Ay = 594

1.00 kN 3.00 kN 2.50 kN 1.00 kN 2.50 kN 500 N

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6–102. If a clamping force of 300 N is required at A, determine the amount of force F that must be applied to the handle of the toggle clamp.

F

235 mm

70 mm

30 mm C 30 mm A

SOLUTION

30

275 mm

B E

D 30

Equations of Equilibrium: First, we will consider the free-body diagram of the clamp in Fig. a. Writing the moment equation of equilibrium about point D, a + ©MD = 0;

Cx (60) - 300(235) = 0 Cx = 1175 N

Subsequently, the free - body diagram of the handle in Fig. b will be considered. a + ©MC = 0;

FBE cos 30°(70) - FBE sin 30°(30) - F cos 30°(275 cos 30° + 70) -F sin 30°(275 sin 30°) = 0 45.62FBE - 335.62F = 0

+ ©F = 0; : x

(1)

1175 + F sin 30° - FBE sin 30° = 0 0.5FBE - 0.5F = 1175

(2)

Solving Eqs. (1) and (2) yields F = 369.69 N = 370 N

Ans.

FBE = 2719.69N

Ans: F = 370 N 595

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6–103. If a force of F = 350 N is applied to the handle of the toggle clamp, determine the resulting clamping force at A.

F

235 mm

70 mm

30 mm C 30 mm A

SOLUTION

30 B

275 mm

E

D 30

Equations of Equilibrium: First, we will consider the free-body diagram of the handle in Fig. a. a + ©MC = 0;

FBE cos 30°(70) - FBE sin 30°(30) - 350 cos 30°(275 cos 30° + 70) -350 sin 30°(275 sin 30°) = 0 FBE = 2574.81 N

+ ©F = 0; : x

Cx - 2574.81 sin 30° + 350 sin 30° = 0 Cx = 1112.41 N

Subsequently,, the free-body diagram of the clamp in Fig. b will be considered. Using the result of Cx and writing the moment equation of equilibrium about point D, a + ©MD = 0;

1112.41(60)-NA (235) = 0 Ans.

NA = 284.01 N = 284 N

Ans: NA = 284 N 596

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*6–104. The platform scale consists of a combination of third and first class levers so that the load on one lever becomes the effort that moves the next lever. Through this arrangement, a small weight can balance a massive object. If x = 450 mm, determine the required mass of the counterweight S required to balance a 90-kg load, L.

100 mm

250 mm

150 mm H

E

C

F

G

D

150 mm

S

350 mm B

A

x L

SOLUTION Equations of Equilibrium: Applying the moment equation of equilibrium about point A to the free - body diagram of member AB in Fig. a, a + ©MA = 0;

FBG (500) - 90(9.81)(150) = 0 FBG = 264.87 N

Using the result of FBG and writing the moment equation of equilibrium about point F on the free - body diagram of member EFG in Fig. b, a + ©MF = 0;

FED (250) - 264.87(150) = 0 FED = 158.922 N

Using the result of FED and writing the moment equation of equilibrium about point C on the free - body diagram of member CDI in Fig. c, a + ©MC = 0;

158.922(100) - mS(9.81)(950) = 0 mS = 1.705 kg = 1.71 kg

Ans.

Ans: mS = 1.71 kg 597

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6–105. The platform scale consists of a combination of third and first class levers so that the load on one lever becomes the effort that moves the next lever. Through this arrangement, a small weight can balance a massive object. If x = 450 mm and, the mass of the counterweight S is 2 kg, determine the mass of the load L required to maintain the balance.

100 mm

250 mm

150 mm H

E

C

F

G

D

150 mm

S

350 mm B

A

x L

SOLUTION Equations of Equilibrium: Applying the moment equation of equilibrium about point A to the free - body diagram of member AB in Fig. a, a + ©MA = 0;

FBG (500) - ML(9.81)(150) = 0 FBG = 2.943 lb

Using the result of FBG and writing the moment equation of equilibrium about point F on the free - body diagram of member EFG in Fig. b, a + ©MF = 0;

FED (250) - 2.943mL(150) = 0 FED = 1.7658mL

Using the result of FED and writing the moment equation of equilibrium about point C on the free - body diagram of member CDI in Fig. c, a + ©MC = 0;

1.7658mL(100) - 2(9.81)(950) = 0 mL = 105.56 kg = 106 kg

Ans.

Ans: mL = 106 kg 598

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10

6–106. The engine hoist is used to support the 200-kg engine. Determine the force acting in the hydraulic cylinder AB, the horizontal and vertical components of force at the pin C, and the reactions at the fixed connection D.

350 mm

1250 mm

C

G A

850 mm

B

SOLUTION

550 mm

Free-Body Diagram: The solution for this problem will be simplified if one realizes that member AB is a two force member. From the geometry,

D

lAB = 23502 + 8502 - 2(350)(850) cos 80° = 861.21 mm sin u sin 80° = 850 861.24

u = 76.41°

Equations of Equilibrium: From FBD (a), a + ©MC = 0;

1962(1.60) - FAB sin 76.41°(0.35) = 0 FAB = 9227.60 N = 9.23 kN

+ ©F = 0; : x

Ans.

Cx - 9227.60 cos 76.41° = 0 Cx = 2168.65 N = 2.17 kN

+ c ©Fy = 0;

Ans.

9227.60 sin 76.41° - 1962 - Cy = 0 Cy = 7007.14 N = 7.01 kN

Ans.

+ ©F = 0; : x

Dx = 0

Ans.

+ c ©Fy = 0;

Dy - 1962 = 0

From FBD (b),

Dy = 1962 N = 1.96 kN a + ©MD = 0;

Ans.

1962(1.60 - 1.40 sin 10°) - MD = 0 MD = 2662.22 N # m = 2.66 kN # m

Ans.

Ans: FAB = 9.23 kN, Cx = 2.17 kN, Cy = 7.01 kN Dx = 0, Dy = 1.96 kN, M D = 2.66 kN # m 599

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6–107. Determine force P on the cable if the spring is compressed 0.025 m when the mechanism is in the position shown. The spring has a stiffness of k = 6 kN>m.

E P

150 mm

D

30

k

200 mm C 200 mm A

200 mm

Solution

F B

800 mm

Free Body Diagram. The assembly will be dismembered into members ACF, CDE and BD. The solution will be very much simplified if one recognizes member BD is a two force member. Here, the spring force is FSP = kx = 600(0.025) = 150 N. The FBDs of members ACF and CDE are shown in Figs. a and b, respectively. Equations of Equilibrium. Write the moment equation of equilibrium about point A for member ACF, Fig. a, (1)

a+ΣMA = 0;  Cy (0.2) + Cx (0.2) - 150(1) = 0 Next, consider the equilibrium of member CDE

(2)

a+ΣMC = 0;  FBD sin 30°(0.2) - P (0.35) = 0 + ΣFx = 0;  Cx + P - FBD sin 30° = 0 S

(3)

+ c ΣFy = 0;  FBD cos 30° - Cy = 0

(4)

Solving Eqs. (1) to (4),

Cx = 148.77 N  Cy = 601.23 N  FBD = 694.24 N



P = 198.36 N = 198 N

Ans.

Ans: P = 198 N 600

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*6–108. The pillar crane is subjected to the load having a mass of 500 kg. Determine the force developed in the tie rod AB and the horizontal and vertical reactions at the pin support C when the boom is tied in the position shown.

2.4 m

B

10 20

1.8 m

A C

SOLUTION Given: M

500 kg

a

1.8 m

b

2.4 m

T1

10 °

T2

20° 2

g

9.81 m > s

M









g cos T 1  F AB cos T 2  FCB 2

M

g sin T 1  FAB sin T 2  FCB 2

§¨ FAB ·¸ ¨ FCB ¸ © ¹



Find FAB  FCB



§¨ Cx ·¸ ¨ Cy ¸ © ¹

b 2

0

2

a b a 2

2

 Mg

0

a b

§b· ¨ ¸ 2 2©a¹ a b FCB

§ FAB · ¨ ¸ ¨ Cx ¸ ¨ C ¸ © y ¹

§ 9.7 · ¨ 11.53 ¸ kN ¨ ¸ © 8.65 ¹

Ans.

Ans:

FAB Cx Cy 601

9.7 kN 11.53 kN 8.65 kN

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6–109. The spring has an unstretched length of 0.3 m. Determine the angle u for equilibrium if the uniform bars each have a mass of 20 kg.

C

k  150 N/ m

Solution

2m

u u

B

A

Free Body Diagram. The assembly is being dismembered into members AB and BC of which their respective FBD are shown in Fig. b and a. Here, the spring stretches x = 2(2 sin u) - 0.3 = 4 sin u - 0.3. Thus, FSP = kx = 150 (4 sin u - 0.3) = 600 sin u - 45. Equations of Equilibrium. Considered the equilibrium of member BC, Fig. a, (1)

a+ΣMC = 0;  By (2 cos u) - Bx (2 sin u) - 20(9.81) cos u = 0 Also, member AB, Fig. b a+ΣMA = 0;   - By (2 cos u) - Bx (2 sin u) - 20(9.81) cos u = 0 + c ΣFy = 0;  (600 sin u - 45) -20(9.81) - By = 0

(2) (3)

Solving Eq. (1) and (2) 9.81 cos u sin u Substitute the result of By = 0 into Eq. (3)



By = 0  Bx = -



600 sin u - 45 - 20(9.81) = 0



sin u = 0.402



u = 23.7°

Ans.

Ans: u = 23.7° 602

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6–110. The spring has an unstretched length of 0.3 m. Determine the mass m of each uniform bar if each angle u = 30° for equilibrium.

C

k  150 N/ m

Solution

2m

u u

B

A

Free Body Diagram. The assembly is being dismembered into members AB and BC of which their respective FBD are shown in Fig. b and a. Here, the spring stretches x = 2(2 sin 30°) - 0.3 = 1.7 m. Thus, FSP = kx = 150(1.7) = 255 N. Equations of Equilibrium. Consider the equilibrium of member BC, Fig. a, a+ΣMC = 0;  Bx (2 sin 30°) + By (2 cos 30°) - m(9.81) cos 30° = 0

(1)

Also, member AB, Fig. b a+ΣMA = 0;  Bx (2 sin 30°) - By (2 cos 30°) - m(9.81) cos 30° = 0 + c ΣFy = 0;  255 - m(9.81) - By = 0

(2) (3)

Solving Eqs. (1) and (2)

Bx = 8.4957 m  By = 0

Substitute the result of By = 0 into Eq. (3)

255 - m(9.81) = 0



m = 26.0 kg

Ans.

Ans: m = 26.0 kg 603

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6–111. Determine the required mass of the suspended cylinder if the tension in the chain wrapped around the freely turning gear is to be 2 kN. Also, what is the magnitude of the resultant force on pin A?

2 kN 45

B

30

2 kN 300 mm

A

SOLUTION a + ©MA = 0;

-4(2 cos 30°) + W cos 45°(2 cos 30°) + Wsin 45°(2 sin 30°) = 0 W = 3.586 kN m = 3.586(1000)/9.81 = 366 kg

+ ©F = 0; : x

Ans.

4 - 3.586 cos 45° - Ax = 0 Ax = 1.464 kN

+ c ©Fy = 0;

3.586 sin 45° - Ay = 0

300 mm

Ay = 2.536 kN FA = 2(1.464)2 + (2.536)2 = 2.93 kN

Ans.

Ans: m = 366 kg FA = 2.93 kN 604

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*6–112. The compound beam is fixed at A and supported by a rocker at B and C. There are hinges (pins) at D and E. Determine the reactions at the supports.

A

D

B

E C

6m

2m 2m 2m

6m

SOLUTION Equations of Equilibrium: From FBD(a), a + ©ME = 0;

Cy 162 = 0

+ c ©Fy = 0;

Ey - 0 = 0

+ ©F = 0; : x

Ex = 0

Cy = 0

Ans.

Ey = 0

From FBD(b), a + ©MD = 0;

By 142 - 15122 = 0 By = 7.50 kN

+ c ©Fy = 0;

Ans.

Dy + 7.50 - 15 = 0 Dy = 7.50 kN

+ ©F = 0; : x

Dx = 0

From FBD(c), a + ©MA = 0;

MA - 7.50162 = 0 MA = 45.0 kN # m

Ans.

+ c ©Fy = 0;

Ay - 7.5 = 0

Ans.

+ ©F = 0; : x

Ax = 0

Ay = 7.5 kN

Ans.

Ans: Cy = 0 By = 7.50 kN MA = 45.0 kN # m Ay = 7.5 kN Ax = 0 605

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16 kN

6–113. The compound beam is pin supported at B and supported by rockers at A and C. There is a hinge (pin) at D. Determine the reactions at the supports.

6 kN

7 kN

B

A 4m

4m

2m

60

D

4m

C 3m 4m

SOLUTION F1

7 kN

a

4m

F2

6 kN

b

2m

F3

16 kN

c

3m

T

60 °

Member DC : 6MD = 0;

F 1 sin  T ( a  c)  Cy a Cy

6F y = 0;

ac a

Dy  F 1 sin  T  Cy Dy

6F x = 0;

F 1 sin  T

Cy

1.52 kN

Dy

4.55 kN

Dx

3.5 kN

By

23.5 kN

Ans.

Ay

3.09 kN

Ans.

Bx

3.5 kN

Ans.

Ans.

0

F1 sin  T  Cy

Dx  F 1 cos  T Dx

0

0

F1 cos  T

Member ABD : 6MA = 0;

F 3 a  F2( 2 a  b)  Dy( 3 a  b)  By2 a By

6F y = 0;

2a

Ay  F 3  B y  F2  Dy Ay

6F x = 0;

F 3 a  F2( 2a  b)  Dy ( 3a  b)

Bx

0

Dy  F3  By  F 2

B x  F1 cos  T

0

0

F 1 cos  T

Ans: Cy = By = Ay = Bx = 606

1.52 kN 23.5 kN 3.09 kN 3.5 kN

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6–114. The skid steer loader has a mass of 1.18 Mg, and in the position shown the center of mass is at G1. If there is a 300-kg stone in the bucket, with center of mass at G2, determine the reactions of each pair of wheels A and B on the ground and the force in the hydraulic cylinder CD and at the pin E.There is a similar linkage on each side of the loader.

1.25 m

D G2

E

30

C G1 0.5 m

A

B 0.15 m

SOLUTION

1.5 m

0.75 m

Entire system: a + ©MA = 0;

300 (9.81)(1.5) -1.18 (103)(9.81)(0.6)+NB (0.75) = 0 NB = 3374.6 N = 3.37 kN

+ c ©Fy = 0;

Ans.

(Both wheels)

3374.6 - 300 (9.81) - 1.18(103)(9.81)+NA = 0 NA = 11.1 kN

Ans.

(Both wheels)

Upper member: a + ©ME = 0;

300(9.81)(2.75)-FCD sin 30° (1.25) = 0 FCD = 12 949 N = 12.9 kN F¿ CD =

+ ©F = 0; : x

FCD 12 949 = = 6.47 kN 2 2

Ans.

Ex - 12 949 cos 30° = 0 Ex = 11 214 N

+ c ©Fy = 0;

-Ey - 300(9.81) + 12 949 sin 30° = 0 Ey = 3532 N

FE = 2(11 214)2 + (3532)2 = 11 757 N Since there are two members, ¿ FE =

FE 11 757 = = 5.88 kN 2 2

Ans.

Ans: NA = 11.1 kN ( Both wheels ) F′CD = 6.47 kN F′E = 5.88 kN 607

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6–115. The handle of the sector press is fixed to gear G, which in turn is in mesh with the sector gear C. Note that AB is pinned at its ends to gear C and the underside of the table EF, which is allowed to move vertically due to the smooth guides at E and F. If the gears exert tangential forces between them, determine the compressive force developed on the cylinder S when a vertical force of 40 N is applied to the handle of the press.

S E

F A

40 N 0.5 m G D

1.2 m

0.2 m

H

B 0.35 m

C 0.65 m

SOLUTION F

40 N

a

0.5 m

b

0.2 m

c

1.2 m

d

0.35 m

e

0.65 m

Member GD: 6MG = 0;

F a  F CG b F CG

F

0

a

F CG

b

100 N

Sector gear : 6MH = 0; F CG( d  e)  FAB§ ¨

c 2

©

F AB

FCG ( d  e)

c d

·d

0

2¸

¹

§ c2  d2 · ¨ ¸ F © c d ¹ AB

297.62 N

Table: 6F y = 0;

F AB§

c · ¨ 2 2 ¸  Fs © c d ¹

Fs

FAB

c § · ¨ 2 2¸ © c d ¹

0

Fs

286 N

Ans.

Ans: FS = 286 N 608

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*6–116. The structure is subjected to the loading shown. Member AD is supported by a cable AB and roller at C and fits through a smooth circular hole at D. Member ED is supported by a roller at D and a pole that fits in a smooth snug circular hole at E. Determine the x, y, z components of reaction at E and the tension in cable AB.

z B E 0.8 m D

SOLUTION ©My = 0;

4 - FAB (0.6) + 2.5(0.3) = 0 5

0.3 m 0.4 m

FAB = 1.5625 = 1.56 kN ©Fz = 0;

C

0.5 m

x

Ans.

A 0.3 m

{ 2.5k} kN

4 (1.5625) - 2.5 + Dz = 0 5 Dz = 1.25 kN

©Fy = 0;

Dy = 0

©Fx = 0;

Dx + Cx -

©Mx = 0;

MDx +

3 (1.5625) = 0 5

(1)

4 (1.5625)(0.4) - 2.5(0.4) = 0 5

MDx = 0.5 kN # m 3 (1.5625)(0.4) - Cx (0.4) = 0 5

(2)

©Mz = 0;

MDz +

©Fz = 0;

Dz¿ = 1.25 kN

©Mx = 0;

MEx = 0.5 kN # m

Ans.

©My = 0;

MEy = 0

Ans.

©Fy = 0;

Ey = 0

Ans.

©Mz = 0;

E x (0.5) - MDz = 0

(3)

Solving Eqs. (1), (2) and (3): Cx = 0.938 kN MDz = 0 Ans.

Ex = 0

Ans: MEx = 0.5 kN # m MEy = 0 Ey = 0 Ex = 0 609

y

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6–117. The structure is subjected to the loadings shown. Member AB is supported by a ball-and-socket at A and smooth collar at B. Member CD is supported by a pin at C. Determine the x, y, z components of reaction at A and C.

z 250 N

60 60

45

D A

SOLUTION

4m 800 N  m

From FBD (a) x

©My = 0;

MBy = 0

©Mx = 0;

- MBx + 800 = 0

©Mz = 0;

By (3) - Bx (2) = 0

©Fz = 0;

Az = 0

©Fx = 0;

-Ax + Bx = 0

(2)

©Fy = 0;

- Ay + By = 0

(3)

MBx = 800 N # m

2m 1.5 m

B

3m

C

(1) Ans.

From FBD (b) ©Mg = 0;

By(1.5) + 800- 250 cos 45°(5.5) = 0

From Eq.(1)

114.85(3) - Bx(2) = 0

From Eq.(2)

Ax = 172 N

Ans.

From Eq.(3)

Ay = 115 N

Ans.

©Fx = 0;

Cx + 250 cos 60° - 172.27 = 0

Cx = 47.3 N

Ans.

©Fy = 0;

250 cos 45° - 114.85 - Cy = 0

Cy = 61.9N

Ans.

©Fz = 0;

250 cos 60° - Cz = 0

©My = 0;

MCy - 172.27(1.5) + 250 cos 60°(5.5) = 0

©Mz = 0;

By = 114.85 N

Bx = 172.27 N

Ans.

Cz = 125 N

MCy = -429 N # m

Ans.

MCz = 0

Ans.

Negative sign indicates that MCy acts in the opposite sense to that shown on FBD.

Ans: Az = 0 Ax = 172 N Ay = 115 N Cx = 47.3 N Cy = 61.9 N Cz = 125 N MCy = -429 N # m MCz = 0 610

y

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6–118. The four-member “A” frame is supported at A and E by smooth collars and at G by a pin. All the other joints are ball-and-sockets. If the pin at G will fail when the resultant force there is 800 N, determine the largest vertical force P that can be supported by the frame. Also, what are the x, y, z force components which member BD exerts on members EDC and ABC? The collars at A and E and the pin at G only exert force components on the frame.

z 300 mm 300 mm

E 600 mm D

x

600 mm

A B

600 mm

F C

SOLUTION

G

GF is a two - force member, so the 800 - N force acts along the axis of GF. Using FBD (a), ©Mx = 0;

- P(1.2) + 800 sin 45°(0.6) = 0 P = 283 N

©Mz = 0;

- Ay (0.3) + Ey (0.3) = 0

©Fy = 0;

- Ay - Ey + 800 sin 45° = 0

y

P

Pk

Ans.

Ay = Ey = 283 N ©Mx = 0;

Az (0.6) + Ez (0.6)- 283((0.6) = 0

©My = 0;

Az(0.3)- Ez (0.3) = 0 Az = Ez = 118 N

Using FBD (b), ©Fy = 0;

- By - Dy + 800 sin 45° = 0

©Mz = 0;

Dy(0.3) - By (0.3) = 0 By = Dy = 283 N

©Fz = 0;

- Bz - Dz + 800 cos 45° = 0

©My = 0;

- Dz (0.3) + Bz (0.3) = 0

Ans.

Bz = Dz = 283 N ©Fx = 0;

Ans.

- Bx + Dx = 0

Using FBD (c), ©Mz = 0;

- By (0.6) + 283(0.15) - 283(0.3) = 0 Bx = Dx = 42.5 N

Ans.

Ans: P = 283 N Bx = Dx = 42.5 N By = Dy = 283 N Bz = Dz = 283 N 611

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7–1. Determine the internal normal force, shear force, and moment at point C in the cantilever beam.

w0

A

B

C L –– 2

L –– 2

Solution The intensity of the triangular distributed loading at C can be computed using the similar triangles shown in Fig. a, wC w0 = or wC = w0 >2 L>2 L

With reference to Fig. b, +

S ΣFx = 0;

NC = 0

+ c ΣFy = 0;

VC - a

a+ ΣMC = 0;

Ans.

w0 L 1 w0 L ba b - a ba b = 0 2 2 2 2 2

- MC - a

VC =

w0 L L 1 w0 L L ba ba b - a ba ba b = 0 2 2 4 2 2 2 3

3w0L 8

MC = -

5 w L2 48 0

Ans. Ans.

The negative sign indicates that MC acts in the opposite sense to that shown on the free-body diagram.

Ans: NC = 0 3w 0L VC = 8 5 M C = - w 0L 2 48 612

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7–2. Determine the internal normal force, shear force, and moment at point C in the simply supported beam. Point C is located just to the right of the 2.5-kN # m couple moment.

10 kN/m B

A C

30

2.5 kN m 2m

2m

Solution Writing the moment equation of equilibrium about point A with reference to Fig. a,

哭

+ ΣMA = 0;

FB cos 30° (4) – 10 (4) (2) – 2.5 = 0

FB = 23.816 kN

Using the result of FB and referring to Fig. b, + → ΣFx

= 0;

+↑ΣFy = 0;

哭

+ ΣMC = 0;

–NC – 23.816 sin 30° = 0

NC = –11.908 kN

Ans.

VC + 23.816 cos 30° – 10 (2) = 0

VC = –0.625 kN

Ans.

23.816 cos 30° (2) – 10 (2) (1) – MC = 0

MC = 21.25 kN · m

Ans.

The negative sign indicates that NC and VC act in the opposite sense to that shown on the free – body diagram. 10 (4) kN

10 (2) kN

2.5 kN · m 2m

1m

2m

1m FB = 3608.44 kN

Ans: N C = -11.908 kN V C = - 0.625 kN M C = 21.25 kN # m 613

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7–3. Determine the internal normal force, shear force, and moment at point C in the double-overhang beam.

3 kN/m

A 1.5 m

B

C 1.5 m

1.5 m

1.5 m

Solution The intensity of the triangular distributed loading at C can be computed using the similar triangles shown in Fig. b, wC 3 = or wC = 2 kN>m 3 4.5 With reference to Fig. a, a+ ΣMB = 0; +

S ΣFx = 0;

1 1 (3)(4.5)(1.5) - (3)(1.5)(0.5) - Ay(3) = 0 2 2 Ax = 0

Ay = 3 kN

Using the results of Ax and Ay and referring to Fig. c, +

S ΣFx = 0;

NC = 0

+ c ΣFy = 0;

3 -

a+ ΣMC = 0;

Ans.

1 (2)(3) - VC = 0 2

MC +

Ans.

VC = 0

1 (2)(3)(1) - 3(1.5) = 0 2

MC = 1.5 kN # m

Ans.

Ans: NC = 0 VC = 0 M C = 1.5 kN # m 614

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B

*7–4 . The beam has a weight w per unit length. Determine the internal normal force, shear force, and moment at point C due to its weight.

L –– 2

L –– 2

C

u

Solution

A

With reference to Fig. a, a+ ΣMA = 0;

L Bx(L sin u) - wL cos ua b = 0 2

Bx =

wL cos u 2 sin u

Using this result and referring to Fig. b, +

S ΣFx′ = 0; + c ΣFy′ = 0; a+ ΣMC = 0;

- NC -

wL cos u L (cos u) - wa b sin u = 0 2 sin u 2

L wL cos u VC - wa b cos u + sin u = 0 2 2 sin u

NC = -

VC = 0

wL csc u 2

Ans. Ans.

L L wL cos u L a sin u b - wa b cos ua b - MC = 0 2 sin u 2 2 4 MC =

wL2 cos u 8

Ans.

The negative sign indicates that NC acts in the opposite sense to that shown on the freebody diagram

Ans: NC = -

wL csc u 2

VC = 0 wL2 MC = cos u 8 615

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7–5. Two beams are attached to the column such that structural connections transmit the loads shown. Determine the internal normal force, shear force, and moment acting in the column at a section passing horizontally through point A.

30 mm

40 mm 250 mm

16 kN 6 kN 23 kN 185 mm 6 kN

Solution

A

+ ΣFx = 0;     6 - 6 - VA = 0 S Ans.

VA = 0 + c ΣFy = 0;      - NA - 16 - 23 = 0

Ans.

NA = - 39 kN a + ΣMA = 0;

125 mm

- MA + 16(0.155) - 23(0.165) - 6(0.185) = 0

MA = - 2.42 kN # m

Ans.

Ans: V A = 0, N A = -39 kN M A = -2.425 kN # m 616

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7–6 . Determine the distance a in terms of the beam’s length L between the symmetrically placed supports A and B so that the internal moment at the center of the beam is zero.

w0

w0

A

B a –– 2

a –– 2 L

Solution In this problem, it is required that the internal moment at point C be equal to zero. With reference to Fig. a, a+ ΣMA = 0;

By(a) By =

1 L L a 1 L L a w a b c a + a - b d + w0 a ba - b = 0 2 0 2 3 2 2 2 3 2

1 wL 4 0

Using this result and referring to Fig. b, a+ ΣMC = 0;

1 a 1 L L w La b - w0 a ba b = 0 4 0 2 2 2 3 a =

2 L 3

Ans.

Ans: a= 617

L 3

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7–7. Determine the internal normal force, shear force, and moment at points C and D in the simply supported beam. Point D is located just to the left of the 5-kN force.

5 kN 3 kN/m

A

B C 1.5 m

Solution

1.5 m

D 3m

The intensity of the triangular distributed loading at C can be computed using the similar triangles shown in Fig. b, wC 3 = or wC = 1.5 kN>m 1.5 3 With reference to Fig. a, a+ ΣMA = 0;

By(6) - 5(3) -

1 (3)(3)(1) = 0 2

By = 3.25 kN

Using this result and referring to Fig. c, +

S ΣFx = 0;

NC = 0

+ c ΣFy = 0;

VC + 3.25 -

a+ ΣMC = 0;

3.25(4.5) -

Ans. 1 (1.5)(1.5) - 5 = 0 2

Ans.

VC = 2.875 kN

1 (1.5)(1.5)(0.5) - 5(1.5) - MC = 0 2

MC = 6.56 kN # m

Ans.

Also, referring to Fig. d, +

S ΣFx = 0;

ND = 0

+ c ΣFy = 0;

VD + 3.25 - 5 = 0

a+ ΣMD = 0;

3.25(3) - MD

Ans. Ans.

VD = 1.75 kN

MD = 9.75 kN # m

Ans.

Ans: NC = 0 V C = 2.875 kN M C = 6.56 kN # m ND = 0 V D = 1.75 kN M D = 9.75 kN # m 618

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*7–8.

Determine the distance a as a fraction of the beam’s length L for locating the roller support so that the moment in the beam at B is zero.

P

P

C

SOLUTION a + ©MA = 0;

a

-P a Cy =

a + ©M = 0;

B

A

L/3 L

2L - a b + Cy1L - a2 + Pa = 0 3 2P A L3 - a B L - a

M =

2P A L3 - a B L - a

2PL a a =

a

L b = 0 3

L - ab = 0 3

L 3

Ans.

Ans: L a = 3 619

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B

7–9. Determine the normal force, shear force, and moment at a section passing through point C. Take P = 8 kN.

0.1 m

0.5 m C

0.75 m

0.75 m

P

SOLUTION a + ©MA = 0;

0.75 m

A

-T(0.6) + 8(2.25) = 0 T = 30 kN

+ ©F = 0; : x

Ax = 30 kN

+ c ©Fy = 0;

Ay = 8 kN

+ ©F = 0; : x

-NC - 30 = 0 NC = - 30 kN

+ c ©Fy = 0;

Ans.

VC + 8 = 0 VC = - 8 kN

a + ©MC = 0;

Ans.

- MC + 8(0.75) = 0 MC = 6 kN # m

Ans.

Ans: N C = -30 kN V C = - 8 kN M C = 6 kN # m 620

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7–10.

B

The cable will fail when subjected to a tension of 2 kN. Determine the largest vertical load P the frame will support and calculate the internal normal force, shear force, and moment at a section passing through point C for this loading.

0.1 m

C 0.75 m

+ ©F = 0; : x

Ax = 2 kN

+ c ©Fy = 0;

Ay = 0.533 kN

+ ©F = 0; : x

-NC - 2 = 0

Ans.

NC = - 2 kN

Ans.

VC - 0.533 = 0 VC = -0.533 kN

a + ©MC = 0;

0.75 m

-2(0.6) + P(2.25) = 0 P = 0.533 kN

+ c ©Fy = 0;

0.75 m

A

P

SOLUTION a + ©MA = 0;

0.5 m

Ans.

- MC + 0.533(0.75) = 0 MC = 0.400 kN # m

Ans.

Ans: P = 0.533 kN N C = -2 kN VC = -0.533 kN M C = 0.400 kN # m 621

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7–11. C

Determine the internal normal force, shear force, and moment at points E and F in the beam.

SOLUTION

A

E

D

45

F

B

With reference to Fig. a, a + ©MA = 0; + ©Fx = 0; :

T(6) + T sin 45°(3) - 300(6)(3) = 0

T = 664.92 N

664.92 cos 45° - A x = 0

A x = 470.17 N

+ c ©Fy = 0;

A y + 664.92 sin 45° + 664.92 - 300(6) = 0

A y = 664.92 N

300 N/m 1.5 m

1.5 m

1.5 m

1.5 m

Use these result and referring to Fig. b, + ©Fx = 0; :

NE - 470.17 = 0 NE = 470 N

+ c ©Fy = 0;

Ans.

664.92 - 300(1.5) - VE = 0 VE = 215 N

a + ©ME = 0;

Ans.

ME + 300(1.5)(0.75) - 664.92(1.5) = 0

ME = 660 N # m

Ans.

Also, by referring to Fig. c, + ©Fx = 0; :

NF = 0

+ c ©Fy = 0;

VF + 664.92 - 300 = 0

Ans.

VF = - 215 N a + ©MF = 0;

Ans.

664.92(1.5) - 300(1.5)(0.75) - MF = 0

MF = 660 N # m

Ans.

The negative sign indicates that VF acts in the opposite sense to that shown on the free-body diagram.

Ans: N E = 470 N, V E = 215 N M E = 660 N # m, N F = 0 V F = -215 N, M F = 660 N # m 622

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*7–12. Determine the distance a between the bearings in terms of the shaft’s length L so that the moment in the symmetric shaft is zero at its center.

w

a L

SOLUTION Due to symmetry, Ay = By + c ©Fy = 0;

a + ©M = 0;

Ay + By -

w(L - a) w(L - a) - wa = 0 4 4

Ay = By =

w (L + a) 4

-M -

w(La) a L a w a wa a a b a + - b + (L + a) a b = 0 2 4 4 2 6 6 4 2

Since M = 0; 3a2 + (L - a)(L + 2a) - 3a (L + a) = 0 2a2 + 2a L - L2 = 0 Ans.

a = 0.366 L

Ans: a = 0.366 L 623

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7–13.

w0

Determine the distance a between the supports in terms of the shaft’s length L so that the bending moment in the symmetric shaft is zero at the shaft’s center. The intensity of the distributed load at the center of the shaft is w 0. The supports are journal bearings. a L

SOLUTION Support reactions: FBD(a) Moments Function: a + ©M = 0;

0 +

1 L 1 L a 1 (w )a b a b a b - w0 L a b = 0 2 0 2 3 2 4 2 a =

L 3

Ans.

Ans: a= 624

L 3

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7–14.

600 N/m

Determine the internal normal force, shear force, and moment at point D in the beam.

900 N⭈m A

B

D 1m

1m

1m 4 5 3

C

SOLUTION Writing the equations of equilibrium with reference to Fig. a, we have a + ©MA = 0;

4 FBC ¢ ≤ (2) - 600(3)(1.5) - 900 = 0 5

FBC = 2250 N

a + ©MB = 0;

600(3)(0.5) - 900 - A y(2) = 0

Ay = 0

+ ©Fx = 0; :

2 A x - 2250 ¢ ≤ = 0 5

A x = 1350 N

Using these results and referring to Fig. b, we have + ©Fx = 0; : + c ©Fy = 0;

ND + 1350 = 0

ND = - 1350 N = - 1.35 kN

Ans.

-VD - 600(1) = 0

VD = - 600 N

Ans.

a + ©MD = 0;

MD + 600(1)(0.5) = 0

MD =

Ans.

- 300 N # m

The negative sign indicates that ND, VD, and MD act in the opposite sense to that shown on the free-body diagram.

Ans: N D = - 1350 N = - 1.35 kN V D = -600 N M D = - 300 N # m 625

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7–15. Determine the internal normal force, shear force, and moment at point C.

6 kN/ m

B

A 3m

C

3m

Solution Support Reactions. Referring to the FBD of the entire beam shown in Fig. a,   a+ ΣMA = 0;  By(6) + ΣFx = 0;    S

1 (6)(6)(2) = 0  By = 6.00 kN 2 Bx = 0

Internal Loadings. Referring to the FBD of right segment of the beam sectioned through C, Fig. b + ΣFx = 0;    S

Ans.

NC = 0

1    + c ΣFy = 0; VC + 6.00 - (3)(3) = 0 VC = -1.50 kN 2 1   a+ ΣMC = 0;  6.00(3) - (3)(3)(1) - MC = 0  MC = 13.5 kN # m 2

Ans. Ans.

Ans: NC = 0 VC = - 1.50 kN MC = 13.5 kN # m 626

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*7–16. Determine the internal normal force, shear force, and moment at point C of the beam.

400 N/m 200 N/m A

B

C 3m

3m

SOLUTION Beam: a + ©MB = 0;

600 (2) + 1200 (3) - Ay (6) = 0 Ay = 800 N

+ ©F = 0; : x

Ax = 0

Segment AC: + ©F = 0; : x

NC = 0

+ c ©Fy = 0;

800 - 600 - 150 - VC = 0

Ans.

Ans.

VC = 50 N a + ©MC = 0;

- 800 (3) + 600 (1.5) + 150 (1) + MC = 0 MC = 1350 N # m = 1.35 kN # m

Ans.

Ans: NC = 0 V = 50 N

MC = 1.35 kN # m 627

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7–17.

8 kN 0.4 m 0.4 m 30⬚

Determine the internal normal force, shear force, and moment at points A and B in the column.

SOLUTION

A

Applying the equation of equilibrium to Fig. a gives

6 kN

0.9 m

3 kN

+ ©F = 0; : x

VA - 6 sin 30° = 0

VA = 3 kN

Ans.

+ c ©Fy = 0;

NA - 6 cos 30° - 8 = 0

NA = 13.2 kN

Ans.

a + ©MA = 0;

8(0.4) + 6 sin 30°(0.9) - 6 cos 30°(0.4) - MA = 0

MA = 3.82 kN # m

1.5 m 2m

Ans.

B

and to Fig. b, + ©F = 0; : x

VB - 6 sin 30° = 0

VB = 3 kN

Ans.

+ c ©Fy = 0;

NB - 3 - 8 - 6 cos 30° = 0

NB = 16.2 kN

Ans.

a + ©MB = 0;

3(1.5) + 8(0.4) + 6 sin 30°(2.9) - 6 cos 30°(0.4) - MB = 0

MB = 14.3 kN # m

Ans.

Ans:

V A = 3 kN, N A = 13.2 kN, M A = 3.82 kN # m V B = 3 kN, N B = 16.2 kN, M B = 14.3 kN # m 628

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7–18. Determine the internal normal force, shear force, and moment at point C.

0.2 m

400 N 1m

A

B

C 1.5 m 3m

SOLUTION

2m

Beam: + ©F = 0; : x

- Ax + 400 = 0 Ax = 400 N

a + ©MB = 0;

A y (5) -400(1.2) = 0 Ay = 96 N

Segment AC: + ©F = 0; : x

NC - 400 = 0 Ans.

NC = 400 N + c ©Fy = 0;

- 96 - VC = 0 Ans.

VC = - 96 N a + ©MC = 0;

MC + 96 (1.5) = 0 MC = -144 N # m

Ans.

Ans: NC = 400 N VC = - 96 N MC = -144 N # m 629

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7–19. Determine the internal normal force, shear force, and moment at points E and F of the compound beam. Point E is located just to the left of 800 N force.

800 N

1200 N 400 N/ m

5

4 3

A E 1.5 m

1m

C

B 2m

1m

D

F 1.5 m

1.5 m

Solution Support Reactions. Referring to the FBD of member BC shown in Fig. a, 4   a+ ΣMB = 0;  Cy(3) - 1200 a b(2) = 0  Cy = 640 N 5 4   a+ ΣMC = 0;  1200 a b(1) - By(3) = 0  By = 320 N 5 3 + ΣFx = 0;    S 1200 a b - Bx = 0 Bx = 720 N 5 Internal Loadings. Referring to the right segment of member AB sectioned through E, Fig. b + ΣFx = 0;    S

720 - NE = 0

NE = 720 N

Ans.

   + c ΣFy = 0;

VE - 800 - 320 = 0

VE = 1120 N = 1.12 kN

Ans.

  a+ ΣME = 0;     - ME - 320(1) = 0     ME = -320 N # m

Ans.

Referring to the left segment of member CD sectioned through F, Fig. c, + ΣFx = 0;     S

 

Ans.

  NF = 0

    + c ΣFy = 0;  -VF - 640 - 400(1.5) = 0  VF = -1240 N = - 1.24 kN Ans.   a+ ΣMF = 0;  MF + 400(1.5)(0.75) + 640(1.5) = 0

MF = -1410 N # m = - 1.41 kN # m

Ans.

Ans: NE = 720 N VE = 1.12 kN ME = - 320 N # m NF = 0 VF = - 1.24 kN MF = -1.41 kN # m 630

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*7–20. Determine the internal normal force, shear force, and moment at points D and E in the overhang beam. Point D is located just to the left of the roller support at B, where the couple moment acts.

2 kN/m

6 kN m

C A

D 3m

B

E

1.5 m

1.5 m

3

5 4

5 kN

SOLUTION The intensity of the triangular distributed load at E can be found using the similar triangles in Fig. b. With reference to Fig. a, a + ©MA = 0;

3 1 By (3) - 2(3)(1.5)-6 - (2)(3)(4)-5 a b(6) = 0 2 5

By = 15 kN Using this result and referring to Fig. c, 4 + ©F = 0; 5 a b -ND = 0 : x 5 3 1 (2)(3) - 5a b = 0 2 5

+ c ©Fy = 0;

VD + 15 -

a + ©MD = 0;

3 1 - MD - 6 - (2)(3)(1)-5 a b (3) = 0 2 5

ND = 4 kN

Ans.

VD = -9 kN

Ans.

MD = -18 kN # m

Ans.

NE = 4 kN

Ans.

VE = 3.75 kN

Ans.

ME = -4.875 kN # m

Ans.

Also, by referring to Fig. d, we can write + ©F = 0; : x

4 5 a b -NE = 0 5

+ c ©Fy = 0;

VE -

a + ©ME = 0;

- ME -

3 1 (1)(1.5)- 5a b = 0 2 5 3 1 (1)(1.5)(0.5)-5 a b (1.5) = 0 2 5

The negative sign indicates that VD, MD, and ME act in the opposite sense to that shown on the free-body diagram.

Ans: ND = 4 kN VD = - 9 kN MD = - 18 kN # m NE = 4 kN  VE = 3.75 kN ME = -4.875 kN # m 631

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7–21.

Determine the internal normal force, shear force, and bending moment at point C.

40 kN 8 kN/m 60° A 3m

C

B 3m

3m 0.3 m

SOLUTION Free body Diagram: The support reactions at A need not be computed. Internal Forces: Applying equations of equilibrium to segment BC, we have + ©F = 0; : x + c ©Fy = 0;

-40 cos 60° - NC = 0

Ans.

NC = -20.0 kN

VC - 24.0 - 12.0 - 40 sin 60° = 0 Ans.

VC = 70.6 kN a + ©MC = 0;

-24.011.52 - 12.0142 - 40 sin 60°16.32 - MC = 0 MC = - 302 kN # m

Ans.

Ans: NC = -20.0 kN VC = 70.6 kN MC = - 302 kN # m 632

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7–22. Determine the ratio of ab for which the shear force will be zero at the midpoint C of the beam.

w

A a

SOLUTION a + ©MB = 0;

-

A

C b/2

C

B b/2

B

a

2 w (2a + b)c (2a + b) - (a + b) d + Ay (b) = 0 2 3

Ay =

w (2a + b)(a - b) 6b

+ ©F = 0; : x

Ax = 0

+ c ©Fy = 0;

-

w w b (2a + b)(a - b) a a + b - VC = 0 6b 4 2

-

1 1 1 (2a + b)(a - b) = (2a + b) a b 6b 4 2

-

1 1 (a - b) = 6b 8

Since V C = 0,

-a+b =

3 b 4

a 1 = b 4

Ans.

Ans: a 1 = b 4 633

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7–23. Determine the internal normal force, shear force, and moment at points D and E in the compound beam. Point E is located just to the left of the 10-kN concentrated load. Assume the support at A is fixed and the connection at B is a pin.

10 kN

2 kN/m B

C A

D

1.5 m

1.5 m

E 1.5 m

1.5 m

Solution With reference to Fig. b, +

S ΣFx = 0; a+ ΣMB = 0; a+ ΣMC = 0;

Bx = 0 Cy(3) - 10(1.5) = 0 10(1.5) - By(3) = 0

Cy = 5 kN By = 5 kN

Using these results and referring to Fig. c, +

S ΣFx = 0;

ND = 0

+ c ΣFy = 0;

VD - 2(1.5) - 5 = 0

a+ ΣMD = 0;

Ans. Ans.

VD = 8 kN

- MD - 2(1.5)(0.75) - 5(1.5) = 0

MD = - 9.75 kN

Ans.

Also, by referring to Fig. d, +

S ΣFx = 0;

NE = 0

+ c ΣFy = 0;

VE - 10 + 5 = 0

a+ ΣME = 0;

5(1.5) - ME = 0

Ans. Ans.

VE = 5 kN

ME = 7.5 kN # m

Ans.

The negative sign indicates that MP acts in the opposite sense to that shown in the free-body diagram.

Ans: N D = 0, V D = 8 kN, M D = - 9.75 kN, N E = 0, V E = 5 kN, M E = 7.5 kN # m 634

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*7–24.

15 kN

3 kN/m

25 kN⭈m

Determine the internal normal force, shear force, and moment at points E and F in the compound beam. Point F is located just to the left of the 15-kN force and 25-kN # m couple moment.

A E 2.25 m

B 2.25 m 1.5 m

C

D

F 2m

2m

SOLUTION With reference to Fig. b, we have + ©F = 0; : x

Cx = 0

a + ©MC = 0;

Dy(4) - 15(2) - 25 = 0

Dy = 13.75 kN

a + ©MD = 0;

15(2) - 25 - Cy(4) = 0

Cy = 1.25 kN

Using these results and referring to Fig. a, we have + ©F = 0; : x

Ax = 0

a + ©MB = 0;

3(6)(1.5) - 1.25(1.5) - A y(4.5) = 0

A y = 5.583 kN

With these results and referring to Fig. c, + ©F = 0; : x

NE = 0

+ c ©Fy = 0;

5.583 - 3(2.25) - VE = 0

a + ©ME = 0;

ME + 3(2.25) - 3(2.25)(8.125) = 0

Ans. VE = - 1.17 kN

ME = 4.97 kN # m

Ans.

Ans.

Also, using the result of Dy referring to Fig. d, we have + ©F = 0; : x

NF = 0

+ c ©Fy = 0;

VF - 15 + 13.75 = 0

a + ©MF = 0;

13.75(2) - 25 - MF = 0

Ans. VF = 1.25 kN

MF = 2.5 kN # m

Ans. Ans.

The negative sign indicates that VE acts in the opposite sense to that shown in the free-body diagram.

Ans: NE = 0 VE = - 1.17 kN ME = 4.97 kN # m NF = 0 VF = 1.25 kN MF = 2.5 kN # m

635

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7–25.

Determine the internal normal force, shear force, and the moment at points C and D.

A 2m

C 2 kN/m

6m 45˚

B

D

SOLUTION

3m

3m

Support Reactions: FBD (a). a + ©MA = 0;

By 16 + 6 cos 45°2 - 12.013 + 6 cos 45°2 = 0 By = 8.485 kN

+ c ©Fy = 0; + ©F = 0 : x

A y + 8.485 - 12.0 = 0

A y = 3.515 kN

Ax = 0

Internal Forces: Applying the equations of equilibrium to segment AC [FBD (b)], we have Q+ ©Fx¿ = 0;

3.515 cos 45° - VC = 0

VC = 2.49 kN

Ans.

a+ ©Fy¿ = 0;

3.515 sin 45° - NC = 0

NC = 2.49 kN

Ans.

a + ©MC = 0;

MC - 3.515 cos 45°122 = 0 MC = 4.97 kN # m

Ans.

Applying the equations of equilibrium to segment BD [FBD (c)], we have + ©F = 0; : x + c ©Fy = 0; a + ©MD = 0;

ND = 0 VD + 8.485 - 6.00 = 0

Ans. VD = -2.49 kN

Ans.

8.485132 - 611.52 - MD = 0 MD = 16.5 kN # m

Ans.

Ans: VC = 2.49 kN NC = 2.49 kN MC = 4.97 kN # m ND = 0 VD = -2.49 kN MD = 16.5 kN # m 636

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7–26. Determine the internal normal force, shear force, and moment at points E and D of the compound beam.

800 N 200 N m A

D

2m

2m

B 4m

2m

E

C

2m

SOLUTION Given: M

200 N˜ m

c

4 m

F

800 N

d

2 m

a

2 m

e

2 m

b

2 m

Segment BC : M  Cy( d  e) B y  Cy

0

0

Cy

M de

By

Cy

Segment EC : NE

NE

0

V E  Cy

0N

Cy

VE

0

ME  M  Cy e

ME

0

NE

0.00

Ans.

VE

50.00 N

Ans.

Cy e  M ME

100.00 N˜ m Ans.

Segment DB : ND

0

VD  F  By

0

ND

0N

ND

0.00

Ans.

VD

F  By

VD

750.00 N

Ans.

MD  F b  By( b  c) MD

F b  B y( b  c)

0

MD

1300 N˜ m

Ans.

Ans: N E = 0, V E = - 50 N, M E = -100 N # m N D = 0, V D = 750 N, M D = -1300 N # m 637

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7–27. Determine the internal normal force, shear force, and moment at points C and D in the simply supported beam. Point D is located just to the left of the 10-kN concentrated load.

10 kN

6 kN/m

A

D

C 1.5 m

1.5 m

1.5 m

B

1.5 m

SOLUTION The intensity of the triangular distributed loading at C can be computed using the similar triangles shown in Fig. b, wC 6 = or wC = 3 kN>m 1.5 3 With reference to Fig. a, a + ©MA = 0; a + ©MB = 0; + ©F = 0 : x

1 By(6)-10(4.5) - (6)(3)(1) = 0 2 1 (6)(3)(5) + 10(1.5) - Ay(6) = 0 2

By = 9 kN Ay = 10 kN

Ax = 0

Using these results and referring to Fig. c, + ©F = 0; : x

NC = 0

+ c ©Fy = 0;

10 -

a + ©MC = 0;

MC + 3(1.5)(0.75) +

Ans.

1 (3)(1.5)-3(1.5) -VC = 0 2

VC = 3.25 kN

1 (3)(1.5)(1) - 10(1.5) = 0 2

Ans.

MC = 9.375 kN # m Ans.

Also, by referring to Fig. d, + ©F = 0; : x

ND = 0

+ c ©Fy = 0;

VD + 9 -10 = 0

VD = 1 kN

Ans.

a + ©MD = 0;

9(1.5)-MD = 0

MD = 13.5 kN # m

Ans.

Ans.

Ans: NC = 0 VC = 3.25 kN MC = 9.375 kN # m ND = 0 VD = 1 kN MD = 13.5 kN # m 638

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*7–28. Determine the internal normal force, shear force, and moment at point C.

B

E

200 N 1m

C A

D 1m

1m

2m

800 N  m

Solution Support Reactions. Referring to the FBD of the entire assembly shown in Fig. a,   a+ ΣMA = 0;  FBE (1) - 200(4) - 800 = 0     FBE = 1600 N + ΣFx = 0;    S

Ax - 1600 = 0

Ax = 1600 N

   + c ΣFy = 0;

Ay - 200 = 0

Ay = 200 N

Internal Loading. Referring to the FBD of the left segment of the assembly sectioned through C, Fig. b, + ΣFx = 0;      1600 + NC = 0    S

NC = -1600 N = - 1.60 kN

Ans.

   + c ΣFy = 0;     200 - VC = 0

VC = 200 N

Ans.

  a+ ΣMA = 0;  MC - 200(1) = 0    MC = 200 N # m

Ans.

Ans: NC = -1.60 kN VC = 200 N MC = 200 N # m 639

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7–29. Determine the internal normal force, shear force, and moment at point D of the two-member frame.

1.5 m

1.5 m B D

1.5 kN/m

1.5 m

E

SOLUTION Member BC: a + ©MC = 0;

1.5 m

4.5 (1.5) - Bx (3) = 0

+ ©F = 0; : x

A

2 kN/m

Bx = 2.25 kN

C

2.25 + Cx - 4.5 = 0 Cx = 2.25 kN

Member AB: a + ©MA = 0;

2.25 (3) - 3 (1) - By (3) = 0 By = 1.25 kN

Segment DB: + ©F = 0; : x

- ND - 2.25 = 0 Ans.

ND = - 2.25 kN + c ©Fy = 0;

VD - 1.25 = 0 Ans.

VD = 1.25 kN a + ©MD = 0;

- MD - 1.25 (1.5) = 0 MD = - 1.88 kN # m

Ans.

Ans: ND = - 2.25 kN VD = 1.25 kN -1.88 kN # m 640

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7–30. Determine the internal normal force, shear force, and moment at point E.

1.5 m

1.5 m B D

1.5 kN/m

1.5 m

E

SOLUTION Member BC: a + ©MC = 0;

1.5 m

4.5 (1.5) - Bx (3) = 0

+ ©F = 0; : x

A

2 kN/m

Bx = 2.25 kN

C

2.25 + Cx - 4.5 = 0 Cx = 2.25 kN

Member AB: a + ©MA = 0;

2.25 (3) - 3 (1) - By (3) = 0 By = 1.25 kN

Segment BE: + c ©Fy = 0;

1.25 - NE = 0 Ans.

NE = 1.25 kN + ©F = 0; : x

VE + 2.25 - 2.25 = 0 Ans.

VE = 0 + ©Mg = 0;

Mg - 2.25 (0.75) = 0 Mg = 1.6875 kN # m = 1.69 kN # m

Ans.

Ans: NE = 1.25 kN VE = 0 MB = 1.69 kN # m 641

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7–31. Determine the internal normal force, shear force, and moment at point D.

6 kN E 3m B 1m D 3m

Solution

A

C

Support Reactions. Notice that member BC is a two force member. Referring to the FBD of member ABE shown in Fig. a, 3   a+ ΣMA = 0;  FBC a b(4) - 6(7) = 0  FBC = 17.5 kN 5 3 + ΣFx = 0; Ax = 4.50 kN    S Ax - 17.5 a b + 6 = 0 5 4    + c ΣFy = 0; Ay - 17.5 a b = 0 Ay = 14.0 kN 5 Internal Loadings. Referring to the FBD of the lower segment of member ABE sectioned through D, Fig. b, + ΣFx = 0;    S

4.50 + VD = 0

VD = -4.50 kN

Ans.

   + c ΣFy = 0;

ND + 14.0 = 0

ND = -14.0 kN

Ans.

  a+ ΣMD = 0;

MD + 4.50(3) = 0

MD = -13.5 kN # m

3m

Ans.

Ans: VD = - 4.50 kN ND = -14.0 kN MD = - 13.5 kN # m 642

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*7–32. Determine the internal normal force, shear force, and moment acting at points D and E of the frame.

B A

E

2m

1.5 m

D C 4m

900 N . m 600 N

Solution Support Reactions. Notice that member AB is a two force member. Referring to the FBD of member BC,   a+ ΣMC = 0;  FAB (1.5) - 900 - 600(4) = 0  FAB = 2200 N + ΣFx = 0;    S

Cx - 2200 = 0

Cx = 2200 N

   + c ΣFy = 0;

Cy - 600 = 0

Cy = 600 N

Internal Loadings. Referring to the left segment of member AB sectioned through E, Fig. b, + ΣFx = 0;    S

NE - 2200 = 0

NE = 2200 N = 2.20 kN Ans.

   + c ΣFy = 0;

VE = 0

Ans.

  a+ ΣME = 0;

ME = 0

Ans.

Referring to the left segment of member BC sectioned through D, Fig. c + ΣFx = 0;    S

ND + 2200 = 0      ND = -2200 N = -2.20 kN

Ans.

   + c ΣFy = 0;

600 - VD = 0       VD = 600 N

Ans.

  a+ ΣMD = 0;  MD - 600(2) = 0    MD = 1200 N # m = 1.20 kN # m Ans.

Ans: NE = 2.20 kN VE = 0 ME = 0 ND = -2.20 kN VD = 600 N MD = 1.20 kN # m 643

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w

7–33. Determine the normal force, shear force, and moment at a section passing through point D. Take w = 150 N>m.

B

A D 4m

3m

4m C

SOLUTION a + ©MA = 0;

4m

3 -150182142 + FBC182 = 0 5 FBC = 1000 N

+ ©F = 0; : x

Ax -

4 110002 = 0 5

A x = 800 N + c ©Fy = 0;

A y - 150182 +

3 110002 = 0 5

A y = 600 N + ©F = 0; : x

ND = - 800 N

+ c ©Fy = 0;

600 - 150142 - VD = 0

Ans.

VD = 0 a + ©MD = 0;

Ans.

-600142 + 150142122 + MD = 0 MD = 1200 N # m = 1.20 kN # m

Ans.

Ans: N D = - 800 N VD = 0 M D = 1.20 kN # m 644

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w

7–34. The beam AB will fail if the maximum internal moment at D reaches 800 N # m or the normal force in member BC becomes 1500 N. Determine the largest load w it can support.

B

A D 4m

3m

4m C

4m

SOLUTION Assume maximum moment occurs at D; a + ©MD = 0;

MD - 4w(2) = 0 800 = 4w(2) w = 100 N/m

a + ©MA = 0;

(O.K.!)

- 800(4) + FBC (0.6)(8) = 0

Ans.

FBC = 666.7 N 6 1500 N w = 100 N/m

Ans: w = 100 N/m 645

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7–35. The distributed loading w = w0 sin u, measured per unit length, acts on the curved rod. Determine the internal normal force, shear force, and moment in the rod at u = 45°.

w = w0 sin θ

r

θ

SOLUTION w = w0 sin u Resultants of distributed loading: u

FRx =

u

w0 sin u(r du) cos u = rw0

L0 u

FRy =

L0

sin u cos u du =

1 r w0 sin2 u 2

u

w0 sin u(r du) sin u = rw0

Q+ ©Fx = 0;

L0

l 1 sin2 u du = rw0 c u - sin 2u d 2 4 L0

- V + FRx cos 45° + FRy sin 45° = 0 1 p 1 1 - sin 90°b sin 45° V = a r w0 sin2 45° b cos 45° + w0 a 2 2 4 4 Ans.

V = 0.278 w0r +a©Fy

= 0;

-N - FRy cos 45° + FRx sin 45° = 0 1 p 1 1 N = - r w0 c a b - sin 90° d cos 45° + a r w0 sin2 45°b sin 45° 2 4 4 2 Ans.

N = 0.0759 w0 r a + ©MO = 0;

M - (0.0759 r w0)(r) = 0 M = 0.0759 w0 r2

Ans.

Ans: V = 0.278 w0 r N = 0.0759 w0 r M = 0.0759 w0 r 2 646

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*7–36. w = w0 sin θ

Solve Prob. 7–35 for u = 120°.

r

θ

SOLUTION Resultants of distributed load: L0

u

L0

u

FRy = FRx =

1 2

FRx =

L0

w0 sin u(r du) cos u = rw0

u

sin u cos u =

1 rw sin2 u 2 0

u

w0 sin u(r du) sin u = rw0

r w0 sin2 120° = 0.375 r w0

u 1 1 sin2 u du = rw0 c u sin 2u drw0 (sin u) 2 = r w0 (sin u) 2 4 L0 0

1 120° 1 b - sin 240° d = 1.2637 r w0 FRy = r w0 c (p)a 2 180° 4 +

b©Fx¿ = 0;

N + 0.375 rw0 cos 30° + 1.2637 r w0 sin 30° = 0 Ans.

N = - 0.957 r w0 +a©Fy¿

= 0;

- V + 0.375 rw0 sin 30° - 1.2637 r w0 cos 30° = 0 Ans.

V = - 0.907 rw0 a + ©MO = 0;

- M - 0.957 r w0 (r) = 0 M = - 0.957 r2w0

Ans.

Ans: N = -0.957 r w0 V = - 0.907 rw0 M = - 0.957 r 2w0 647

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7–37. Determine the internal normal force, shear force, and moment at point D of the two-member frame.

250 N/m B

A

D 2m

1.5 m C

SOLUTION

E

300 N/m

4m

Member AB: a + ©MA = 0;

By (4) - 1000 (2) = 0 By = 500 N

Member BC: a + ©MC = 0;

- 500 (4) + 225 (0.5) + Bx (1.5) = 0 Bx = 1258.33 N

Segment DB: + ©F = 0; : x

-ND + 1258.33 = 0 Ans.

ND = 1.26 kN + c ©Fy = 0;

VD - 500 + 500 = 0 Ans.

VD = 0 a + ©MD = 0;

- MD + 500 (1) = 0 MD = 500 N # m

Ans.

Ans: ND = 1.26 kN VD = 0 MD = 500 N # m 648

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7–38. Determine the internal normal force, shear force, and moment at point E of the two-member frame.

250 N/m B

A

D 2m

1.5 m C

SOLUTION

E

300 N/m

4m

Member AB: a + ©MA = 0;

By (4) - 1000 (2) = 0 By = 500 N

Member BC: a + ©MC = 0;

- 500 (4) + 225 (0.5) + Bx (1.5) = 0 Bx = 1258.33 N

Segment EB: + ©F = 0; : x

-NE - 1258.33 - 225 = 0 Ans.

NE = - 1.48 kN + c ©Fy = 0;

VE - 500 = 0 Ans.

VE = 500 N a + ©ME = 0;

- ME + 225 (0.5) + 1258.33 (1.5) - 500 (2) = 0 ME = 1000 N # m

Ans.

Ans: NE = - 1.48 kN VE = 500 N ME = 1000 N # m 649

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7–39. The strongback or lifting beam is used for materials handling. If the suspended load has a weight of 2 kN and a center of gravity of G, determine the placement d of the padeyes on the top of the beam so that there is no moment developed within the length AB of the beam. The lifting bridle has two legs that are positioned at 45°, as shown.

3m

d

A

3m

45°

45°

0.2 m 0.2 m

SOLUTION

E

B

d

F

Support Reactions: From FBD (a), a + ©ME = 0;

FF162 - 2132 = 0

FE = 1.00 kN

+ c ©Fy = 0;

FF + 1.00 - 2 = 0

FF = 1.00 kN

G

From FBD (b), + ©F = 0; : x + c ©Fy = 0;

FAC cos 45° - FBC cos 45° = 0

FAC = FBC = F

2F sin 45° - 1.00 - 1.00 = 0 FAC = FBC = F = 1.414 kN

Internal Forces: This problem requires MH = 0. Summing moments about point H of segment EH[FBD (c)], we have a + ©MH = 0;

1.001d + x2 - 1.414 sin 45°1x2 - 1.414 cos 45°10.22 = 0 Ans.

d = 0.200 m

Ans: d = 0.200 m 650

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*7–40. Determine the internal normal force, shear force, and moment acting at points B and C on the curved rod.

A B 45 0.5 m

30 C

3

Solution

4

Support Reactions. Not required

5

200 N

Internal Loadings. Referring to the FBD of bottom segment of the curved rod sectioned through C, Fig. a + Q ΣFx = 0; NC - 200 sin (36.87° + 30°) = 0  NC = 183.92 N = 184 N

Ans.

+ aΣFy = 0; -VC - 200 cos (36.87° + 30° ) VC = -78.56 N = -78.6 N

Ans.

4 3 a+ ΣMC = 0;  200 a b(0.5 sin 30°) - 200 a b[0.5(1 - cos 30°)] + MC = 0 5 5



MC = -31.96 N # m = - 32.0 N # m

Ans.

Referring to the FBD of bottom segment of the curved rod sectioned through B, Fig. b a + ΣFx = 0;    NB - 200 sin (45° - 36.87°) = 0       NB = 28.28 N = 28.3 N Ans. + bΣFy = 0;    -VB + 200 cos (45° - 36.87°) = 0  VB = 197.99 N = 198 N Ans. 4 3 a+ ΣMB = 0;  MB + 200 a b(0.5 sin 45°) - 200 a b[0.5(1 + cos 45°] = 0 5 5

MB = 45.86 N # m = 45.9 N # m

Ans.

Ans: NC = 184 N VC = - 78.6 N MC = -32.0 N # m NB = 28.3 N VB = 198 N MB = 45.9 N # m 651

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7–41. Determine the x, y, z components of internal loading in the rod at point D. F = {7i - 12j - 5k} kN.

z

0.75 m

C 3 kN m

Solution

D

A

F B

Given: M

3 kN˜ m

F

§ 7 · ¨ 12 ¸ kN ¨ ¸ © 5 ¹

a

0.75 m

b

0.2 m

c

0.2 m

d

0.6 m

e

1 m

x

0.2 m

1m

0.2 m 0.6 m y

Guesses Cx

1 N

Cy

1 N

Bx

1 N

Bz

1 N

Ay

1 N

Az

1 N

Given

§¨ 0 ·¸ §¨ Bx ·¸ §¨ Cx ·¸ ¨ Ay ¸  ¨ 0 ¸  ¨ Cy ¸  F ¨ Az ¸ ¨ Bz ¸ ¨ 0 ¸ © ¹ © ¹ © ¹

0

§ e · §¨ 0 ·¸ § 0 · §¨ Bx ·¸ § 0 · §¨ Cx ·¸ § 0 § 0 · · ¨ ¨ ¨ ¸ ¸ ¸ ¨ b  c  d ¸ u ¨ Ay ¸  b  c u ¨ 0 ¸  0 u ¨ C ¸  b  c  d u F  ¨ 0 ¸ y ¨ ¸ ¨ ¨ ¸ ¸ ¸ ¨ ¸ ¨ ¸ ¨ ¨ ¸ ¨ © M ¹ ¹ © 0 ¹ © Az ¹ © 0 ¹ © Bz ¹ © a ¹ © 0 ¸¹ © 0

652

0

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§¨ Ay ·¸ ¨ Az ¸ ¨ ¸ ¨ Bx ¸ ¨ Bz ¸ ¨ ¸ ¨ Cx ¸ ¨ Cy ¸ © ¹

§¨ Ay ·¸ ¨ Az ¸ ¨ ¸ ¨ Bx ¸ ¨ Bz ¸ ¨ ¸ ¨ Cx ¸ ¨ Cy ¸ © ¹

Find  Ay  Az  Bx  B z  Cx  Cy

§ 53.60 · ¨ 87.00 ¸ ¨ ¸ ¨ 109.00 ¸ kN ¨ 82.00 ¸ ¨ 116.00 ¸ ¨ ¸ © 65.60 ¹

Guesses V Dx

1 N

NDy

1 N

V Dz

1N

MDx

1 N˜ m

MDz

1 N˜ m

MDy

1 N˜ m

Given

§¨ Cx ¸· §¨ VDx ¸· ¨ Cy ¸  ¨ NDy ¸ ¨ 0 ¸ ¨V ¸ © ¹ © Dz ¹

0

§ 0 · §¨ Cx ·¸ §¨ MDx ¸· § 0 · ¨ b ¸ u ¨ C ¸  ¨ MDy ¸  ¨ 0 ¸ ¨ ¸ ¨ ¸ ¨ y¸ ¨ © a ¹ © 0 ¹ © MDz ¸¹ © M ¹ §¨ VDx ¸· ¨ NDy ¸ ¨ ¸ ¨ VDz ¸ ¨ MDx ¸ ¨ ¸ ¨ MDy ¸ ¨ MDz ¸ © ¹

0

Find  VDx  NDy  VDz  MDx  MDy  MDz

§ VDx · ¨ ¸ ¨ NDy ¸ ¨V ¸ © Dz ¹

§ 116.00 · ¨ 65.60 ¸ kN ¨ ¸ © 0.00 ¹

Ans.

§ MDx · ¨ ¸ ¨ MDy ¸ ¨M ¸ © Dz ¹

§ 49.20 · ¨ 87.00 ¸ kN˜ m ¨ ¸ © 26.20 ¹

Ans.

Ans: V Dx = 116.00 kN, N Dy = - 65.60 kN, V Dz = 0.00, M Dx = 49.20, M Dy = 87.00, M Dz = 26.20 653

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7–42. Determine the x, y, z components of internal loading at a section passing through point C in the pipe assembly. Neglect the weight of the pipe. Take F1 = 5 -80i + 200j - 300k6kN and F2 = 5250i - 150j - 200k6 kN.

z

F2

C 1.5 m

x

2m y

F1 3m

Solution ΣFR = 0;

ΣMR = 0; i 3 MC + 3 - 80

FC + F1 + F2 = 0 FC = 5 -170i - 50j + 500k6kN Cx = -170 kN

Ans.

Cy = - 50 kN

Ans.

Cz = 500 kN

Ans.

MC + rC1 * F1 + rC2 * F2 = 0 j 2 200

k i 3 3 0 + 0 - 300 250

MC = 5 1000i - 900j - 260k6 kN # m

j 2 - 150

k 0 3 = 0 - 200

2m

3m

MCx = 1 MN # m

Ans.

MCy = 900 kN # m

Ans.

MCz = -260 kN # m

Ans.

.

Ans: Cx = -170 kN Cy = -50 kN Cz = 500 kN MCx = 1 MN # m MCy = 900 kN # m MCz = -260 kN # m 654

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7–43. Determine the x, y, z components of internal loading at a section passing through point B in the pipe assembly. Neglect the weight of the pipe. Take F1 =  5 200i - 100j - 400k6 N and F2 = 5 300i - 500k 6 N.

z

F2

A B 1m

y

1m x 1.5 m

Solution

F1

Internal Loadings. Referring to the FBD of the free end segment of the pipe assembly sectioned through B, Fig. a,   ΣFx = 0;

Nx + 300 + 200 = 0

Nx = -500 N

Ans.

  ΣFy = 0;

Vy - 100 = 0

Vy = 100 N

Ans.

  ΣFz = 0;

Vz - 500 - 400 = 0

Vz = 900 N

Ans.

  ΣMx = 0;

Mx - 400(1.5) = 0

Mx =

Ans.

  ΣMy = 0;

My + 500(1) + 400(1) = 0

My =

  ΣMz = 0

Mz - 200(1.5) - 100(1) = 0

Mz =

600 N # m -900 N # m 400 N # m

Ans. Ans.

The negative signs indicate that Nx and My act in the opposite sense to those shown in FBD.

Ans: Nx = -500 N Vy = 100 N Vz = 900 N Mx = 600 N # m My = - 900 N # m Mz = 400 N # m 655

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*7–44. Determine the x, y, z components of internal loading at a section passing through point B in the pipe assembly. Neglect the weight of the pipe. Take F1 = 5100i - 200j - 300k 6  N and F2 = 5 100i + 500j 6 N.

z

F2

A B 1m

y

1m x 1.5 m

Solution

F1

Internal Loadings. Referring to the FBD of the free end segment of the pipe assembly sectioned through B, Fig. a   ΣFx = 0;

Nx + 100 + 100 = 0

Nx = -200 N

Ans.

  ΣFy = 0;

Vy + 500 - 200 = 0

Vy = - 300 N

Ans.

  ΣFz = 0;

Vz - 300 = 0

Vz = 300 N

Ans.

  ΣMx = 0;

Mx - 300(1.5) = 0

  ΣMy = 0;

My + 300(1) = 0

  ΣMz = 0;

Mz + 500(1) - 100(1.5) - 200(1) = 0



Mx = 450 N # m My = - 300 N # m

Ans. Ans.

Mz = - 150 N # m

Ans.

The negative signs indicates that Nx, Vy, My and Mz act in the senses opposite to those shown in FBD.

Ans: Nx = -200 N Vy = -300 N Vz = 300 N Mx = 450 N # m My = - 300 N # m Mz = - 150 N # m 656

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7–45. Draw the shear and moment diagrams for the beam.

w A

C

B a

a

SOLUTION 0 … x 6 a + c ©Fy = 0;

- V - wx = 0 Ans.

V = - wx a + ©M = 0;

x M + wx a b = 0 2 M = -

w 2 x 2

Ans.

a 6 x … 2a + c ©Fy = 0;

- V + 2 wa - wx = 0 Ans.

V = w (2a - x) a + ©M = 0;

x M + wx a b - 2 wa (x - a) = 0 2 M = 2 wa x - 2 wa2 -

w 2 x 2

Ans.

Ans:

w 2 0 … x 6 a: V = - wx, M = - 2 x a 6 x … 2a: V = w(2a - x) w M = 2wax - 2wa2 - x2 2 657

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7–46.

Draw the shear and moment diagrams for the beam (a) in terms of the parameters shown; (b) set M0 = 500 N # m, L = 8 m.

M0

L/3

M0

L/3

L/ 3

SOLUTION (a) For 0 … x …

L 3

+ c ©Fy = 0;

V = 0

Ans.

a + ©M = 0;

M = 0

Ans.

+ c ©Fy = 0;

V = 0

Ans.

a + ©M = 0;

M = M0

Ans.

V = 0

Ans.

M = 0

Ans.

For

For

L 2L 6 x 6 3 3

2L 6 x … L 3

+ c ©Fy = 0; a + ©M = 0; (b)

Set M0 = 500 N # m , L = 8 m For 0 … x 6

8 m 3

+ c ©Fy = 0; c + ©M = 0; For

V = 0

Ans.

M = 0

Ans.

8 16 m 6 x 6 m 3 3

+ c ©F y = 0;

V = 0

Ans.

c + ©M = 0;

M = 500 N # m

Ans.

For

Ans: L : V = 0, M = 0 3 L 2L 6 x 6 : V = 0, M = M0 3 3 2L 6 x … L: V = 0, M = 0 3 8 0 … x 6 m: V = 0, M = 0 3 8 16 m 6 x 6 m: V = 0, M = 500 N # m 3 3 16 m 6 x … 8 m: V = 0, M = 0 3

0 … x 6

16 m 6 x … 8m 3

+ c ©Fy = 0;

V = 0

Ans.

c + ©M = 0;

M = 0

Ans.

658

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7–47.

If L = 9 m, the beam will fail when the maximum shear force is Vmax = 5 kN or the maximum bending moment is Mmax = 2 kN # m. Determine the magnitude M0 of the largest couple moments it will support.

M0

L/3

M0

L/ 3

L/ 3

SOLUTION See solution to Prob. 7–48 a. Mmax = M0 = 2 kN # m

Ans.

Ans: Mmax = 2 kN # m 659

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P

*7–48. Draw the shear and moment diagrams for the overhang beam.

SOLUTION Since the loading discontinues at B, the shear stress and moment equation must be written for regions 0 … x 6 b and b 6 x … a + b of the beam. The free-body diagram of the beam’s segment sectioned through an arbitrary point in these two regions are shown in Figs. b and c. Region 0 … x 6 b, Fig. b Pa - V = 0 b

V = -

Pa b

(1)

Pa x = 0 b

M = -

Pa x b

(2)

+ c ©Fy = 0;

-

a + ©M = 0;

M +

Region b 6 x … a + b, Fig. c ©Fy = 0;

V - P = 0

V = P

(3)

a + ©M = 0;

-M - P(a + b - x) = 0

M = - P(a + b - x)

(4)

The shear diagram in Fig. d is plotted using Eqs. (1) and (3), while the moment diagram shown in Fig. e is plotted using Eqs. (2) and (4). The value of moment at B is evaluated using either Eqs. (2) or (4) by substituting x = b; i.e., M ƒ x=b = -

C

A

Pa (b) = - Pa or M ƒ x = b = - P(a + b - b) = - Pa b

660

B b

a

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7–49.

8 kN/m

Draw the shear and moment diagrams for the overhang beam. C

A B 4m

SOLUTION 0 … x 6 5 m: + c ©Fy = 0;

2.5 - 2x - V = 0 V = 2.5 - 2x

a + ©M = 0;

1 M + 2x a x b - 2.5x = 0 2 M = 2.5x - x2

5 … x 6 10 m: + c ©Fy = 0;

2.5 - 10 - V = 0 V = - 7.5

a + ©M = 0;

M + 101x - 2.52 - 2.5x = 0 M = - 7.5x - 25

661

2m

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7–50. Draw the shear and moment diagrams for the beam.

1.5 kN/ m A

B 2m 4m

SOLUTION 0 … x … 2 m: + c ©Fy = 0;

0.75 - V = 0 Ans.

V = 0.75 kN a + ©M = 0;

M - 0.75 x = 0 M = 0.75 x kN # m

Ans.

2 m 6 x 6 4 m: + c ©Fy = 0;

0.75 - 1.5 (x - 2) - V = 0 Ans.

V = 3.75 - 1.5 x kN a + ©M = 0;

M +

1.5 (x - 2)2 - 0.75 x = 0 2

M = - 0.75 x2 + 3.75 x - 3 kN # m

Ans.

Ans: V = 0.75 kN M = 0.75 x kN # m V = 3.75 - 1.5 x kN M = -0.75x 2 + 3.75x - 3 kN # m 662

C

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7–51. Draw the shear and moment diagrams for the beam (a) in terms of the parameters shown; (b) set P = 30 kN, a = 2 m, b = 4 m.

P

A

B a

SOLUTION

663

b

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7–51. Continued

The support reactions are indicated on the FBD of the beam, Fig. a. Since the loading discontinues, the shear and moment functions must be written for regions 0 … x 6 2 m and 2 m 6 x … 6 m of the beam. The FBD of the beam’s segment sectioned through an arbitrary point in these two regions are shown in Fig. b and c. For region 0 … x 6 2 m, Fig. b, + c ΣFy = 0;

20.0 - V = 0

a+ ΣM = 0;

M - 20.0x = 0

(1)

V = 20.0 kN

M = {20.0x} kN # m

(2)

For region 2 m 6 x … 6 m, Fig. c, + c ΣFy = 0;

V + 10.0 = 0

a+ ΣM = 0;

10.0(6 - x) - M = 0

(3)

V = - 10.0 kN

M = {60.0 - 10.0x} kN # m

(4)

The shear diagram shown in Fig. d is plotted using Eqs. (1) and (3), while the moment diagram shown in Fig. e is plotted using Eqs. (2) and (4). The value of moment at x = 2 m can be evaluated using either Eqs. (2) or (4).

664

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*7–52. w

Draw the shear and moment diagrams for the compound beam. The beam is pin-connected at E and F.

A

L

SOLUTION Support Reactions: From FBD (b), a + ©ME = 0; + c ©Fy = 0;

Fy a

wL L L b a b = 0 3 3 6

Ey +

wL wL = 0 6 3

Fy = Ey =

wL 6

wL 6

From FBD (a), a + ©MC = 0; From FBD (c), a + ©MB = 0; + c ©Fy = 0;

Dy 1L2 +

wL L 4wL L a b a b = 0 6 3 3 3

4wL L wL L a b a b - A y 1L2 = 0 3 3 6 3 By +

4wL wL 7wL = 0 18 3 6

Dy =

7wL 18

Ay =

7wL 18

By =

10wL 9

Shear and Moment Functions: For 0 ◊ xm = 77.8 kN>m

Ans.

Ans: w0 = 77.8 kN>m 729

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7–106.

B

The cable AB is subjected to a uniform loading of 200 N/m. If the weight of the cable is neglected and the slope angles at points A and B are 30° and 60°, respectively, determine the curve that defines the cable shape and the maximum tension developed in the cable.

60°

y

SOLUTION y =

1 ¢ 200 dx ≤ dx FH L L

A

30° x

1 1100x2 + C1x + C22 y = FH 200 N/m

dy 1 1200x + C12 = dx FH At x = 0,

y = 0;

At x = 0,

dy = tan 30°; dx y =

15 m

C2 = 0 C1 = FH tan 30°

1 1100x2 + FH tan 30°x2 FH dy = tan 60°; dx

At x = 15 m,

FH = 2598 N

y = 138.5x2 + 577x2110-32 m

Ans.

umax = 60° Tmax =

FH 2598 = = 5196 N cos umax cos 60° Tmax = 5.20 kN

Ans.

Ans: y = (38.5x 2 + 577x)(10 - 3) m T max = 5.20 kN 730

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7–107.

y B

10°

Determine the maximum tension developed in the cable if it is subjected to a uniform load of 600 N/m.

A

SOLUTION

20 m x

600 N/m

The Equation of The Cable: y = =

100 m

1 1 w1x2dx2 dx FH L 1

1 w0 2 ¢ x + C1x + C2 ≤ FH 2

(1)

dy 1 = 1w x + C12 dx FH 0

(2)

Boundary Conditions: y = 0 at x = 0, then from Eq. (1) 0 =

1 1C 2 FH 2

C2 = 0

dy 1 = tan 10° at x = 0, then from Eq. (2) tan 10° = 1C 2 dx FH 1 y =

Thus,

C1 = FH tan 10°

w0 2 x + tan 10°x 2FH

(3)

y = 20 m at x = 100 m, then from Eq. (3) 20 =

600 110022 + tan 10°11002 2FH

FH = 1 267 265.47 N

and dy w0 = x + tan 10° dx FH =

600 x + tan 10° 1 267 265.47

= 0.4735110-32x + tan 10° u = umax at x = 100 m and the maximum tension occurs when u = umax . tan umax =

dy = 0.4735110-3211002 + tan 10° ` dx x = 100 m umax = 12.61°

The maximum tension in the cable is Tmax =

FH 1 267 265.47 = = 1 298 579.01 N = 1.30 MN cos umax cos 12.61°

Ans.

Ans: T max = 1.30 MN 731

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*7–108. Determine the maximum uniform distributed loading w0 N/m that the cable can support if it is capable of sustaining a maximum tension of 60 kN.

60 m 7m

w0

Solution The Equation of The Cable: y = =

1 ( w(x)dx)dx FH L 1

1 w0 2 a x + C1x + C2 b FH 2

[1]

dy 1 = (w x + C1) dx FH 0

[2]

Boundary Conditions: y = 0 at x = 0, then from Eq.[1] 0 =

1 (C )C = 0 FH 2 2

dy 1 = 0 at x = 0, then from Eq.[2] 0 = (C )C = 0 dx FH 1 1 Thus, w0 2 x 2FH

[3]

dy w0 = x dx FH

[4]

y =

y = 7 m at x = 30 m, then from Eq.[3] 7 =

w0 450 (302)FH = w 2FH 7 0

u = u max at x = 30 m and the maximum tension occurs when . u = u max . From Eq.[4] tan u max =

dy w0 ` = x = 0.01556(30) = 0.4667 dx x = 30 m 450 w0 7 u max = 25.02°

The maximum tension in the cable is T max =

FH cos u max

450 w 7 0 60 = cos 25.02° Ans.

w0 = 0.846 kN>m

Ans: w0 = 0.846 kN>m 732

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7–109. If the pipe has a mass per unit length of 1500 kg> m, determine the maximum tension developed in the cable.

30 m A

3m

B

SOLUTION As shown in Fig. a, the origin of the x, y coordinate system is set at the lowest point of the cable. Here, w(x) = w0 = 1500(9.81) = 14.715(103) N>m. Using Eq. 7–12, we can write y =

=

1 a w0dxb dx FH L L 1 14.715(103) 2 x + c1x + c2 b a FH 2

dy = 0 at x = 0, results in c1= 0. dx Applying the boundary condition y = 0 at x = 0 results in c2 = 0. Thus,

Applying the boundary condition

y =

7.3575(103) 2 x FH

Applying the boundary condition y = 3 m at x = 15 m, we have 3 =

7.3575(103) (15)2 FH

FH = 551.81(103) N

Substituting this result into Eq. (1), we have dy = 0.02667x dx The maximum tension occurs at either points at A or B where the cable has the greatest angle with the horizontal. Here, umax = tan - 1 a

dy ` b = tan-1 [0.02667(15)] = 21.80° dx 15 m

Thus, Tmax =

551.8(103) FH = = 594.32(103) N = 594 kN cos umax cos 21.80°

Ans: Tmax = 594 kN 733

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7–110. If the pipe has a mass per unit length of 1500 kg> m, determine the minimum tension developed in the cable.

30 m A

3m

B

SOLUTION As shown in Fig. a, the origin of the x, y coordinate system is set at the lowest point of the cable. Here, w(x) = w0 = 1500(9.81) = 14.715(103) N>m. Using Eq. 7–12, we can write y =

=

1 a w0dxb dx FH L L 3 1 14.715(10 ) 2 x + c1x + c2 b a FH 2

dy = 0 at x = 0, results in c1= 0. dx Applying the boundary condition y = 0 at x = 0 results in c2 = 0. Thus, Applying the boundary condition

y =

7.3575(103) 2 x FH

Applying the boundary condition y = 3 m at x = 15 m, we have 3 =

7.3575(103) (15)2 FH

FH = 551.81(103) N

Substituting this result into Eq. (1), we have dy = 0.02667x dx The minimum tension occurs at the lowest point of the cable, where u = 0°. Thus, Tmin = FH = 551.81(103) N = 552 kN

Ans: Tmin = 552 kN 734

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7–111. The cable supports the uniform distributed load of w0 = 12 kN/m. Determine the tension in the cable at each support A and B.

B A

4.5 m

3m

Solution w0

Use the equations of Example 7.12. w0

y=

4.5 =

7.5 m

x2

2 FH 12

2 FH

x2

12 (7.5 – x)2 2 FH

3=

12 12 x2 = (7.5 – x)2 2 (4.5) 2 (3)

4.5 m 3m

x2 = 1.5 (56.26 – 15x + x2)

7.5 – x

2

0.5x – 22.5x + 84.375 = 0 Choose root < 7.5 m x = 4.1288 m FH =

w0 2y

x2 =

12 (4.1288)2 = 22.729 kN 2 (4.5)

x2 =

12 x2 2 (22.729)

At B : y=

w0 2 FH

dy = tan ␪B = 0.52796 x | x = 4.1288 = 2.180 dx

␪B = 65.36° TB =

FH cos ␪ B

=

22.729 = 54.52 kN cos 65.36°

Ans.

At A : y= dy dx

w0 2 FH

x2 =

12 x2 2 (22.729)

= tan ␪A = 0.52796 x | x = (7.5 – 4.1288) = 1.780

␪A = 60.67° TA =

FH cos ␪ A

=

22.729 = 46.40 kN cos 60.67°

Ans.

Ans: T B = 54.52 kN T A = 46.40 kN 735

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*7–112. The cable will break when the maximum tension reaches Tmax = 10 kN. Determine the minimum sag h if it supports the uniform distributed load of w = 600 N>m.

25 m h

600 N/m

SOLUTION The Equation of The Cable: y = =

1 ( w(x)dx)dx FH L L 1 w0 2 ¢ x + C1x + C2 ≤ FH 2

[1]

dy 1 (w x + C1) = dx FH 0

[2]

Boundary Conditions: y = 0 at x = 0, then from Eq.[1] 0 =

1 (C ) FH 2

dy 1 = 0 at x = 0, then from Eq.[2] 0 = (C ) dx FH 1

C2 = 0

C1 = 0

Thus, w0 2 x 2FH

[3]

dy w0 x = dx FH

[4]

y =

y = h, at x = 12.5 m, then from Eq.[3] h =

w0 (12.52) 2FH

FH =

78.125 w0 h

u = umax at x = 12.5 m and the maximum tension occurs when u = umax. From Eq.[4] tan umax =

Thus, cos umax =

dy 2 = dx x - 12.5m

w0 x 18.125 h w0

= 0.0128h(12.5) = 0.160h

1 20.0256h2 + 1

The maximum tension in the cable is Tmax =

10 =

FH cos umax

18.125 h (0.6)

1 20.0256h2 + 1 Ans.

h = 7.09 m

736

Ans: h = 7.09 m

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7–113. If the slope of the cable at support A is zero, determine the deflection curve y = f(x) of the cable and the maximum tension developed in the cable.

12 m

B 4.5 m

SOLUTION

A

x

Using Eq. 7–12, y =

1 a w(x)dxb dx FH L L

y =

1 p a 4 cos * dx bdx FH L L 24

y =

1 24 p c 4(103) d sin x + C1 FH L p 24

y = -

4 kN/m

p w ⫽ 4 cos –– x 24

p 24 96(103) cos x d + C1x + C2 c p pFH 24

dy = 0 at x = 0 results in C1 = 0. dx Applying the boundary condition y = 0 at x = 0, we have Applying the boundary condition

0 = -

24 96(103) c cos 0° d + C2 p pFH

2304(103)

C2 =

p2FH

Thus, y =

2304(103) p c 1 - cos xd 24 p2FH

Applying the boundary condition y = 4.5 m at x = 12 m, we have 4.5 =

2304(103) p c 1 - cos (12) d 2 24 p FH

FH = 51.876(103) N Substituting this result into Eqs. (1) and (2), we obtain 96(103) dy p sin x = dx 24 p(51.876)(103) p = 0.5890 sin x 24 and y =

2304(103) p 2(51.876)(103)

= 4.5a 1 - cos

c 1 - cos

p xd 24

p xb m 24

Ans.

The maximum tension occurs at point B where the cable makes the greatest angle with the horizontal. Here, umax = tan-1 a

dy p b = tan-1 c 0.5890 sina (12) b d = 30.50° ` dx x = 12 m 24

Thus,

Ans: y = 4.5a1 - cos

3

51.876(10 ) FH = = 60.207(103) N = 60.2 kN Tmax = cos u cos 30.50° max 737

Ans.

T max = 60.2 kN

p xb m 24

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7–114. If the horizontal towing force is T = 20 kN and the chain has a mass per unit length of 15 kg>m, determine the maximum sag h. Neglect the buoyancy effect of the water on the chain. The boats are stationary.

40 m T

h

T

Solution As shown in Fig. a, the origin of the x, y coordinates system is set at the lowest point of the chain. Here, FH = T = 201103 2N and w 1s 2 = 1519.812N/m = 147.15 N/m. d 2y dx Set u =

2

=

dy 2 dy 2 147.15 1 + a b = 7.3575(10 - 3) 1 + a b 3 A dx A dx 20(10 )

dy d 2y du , then = . Thus, dx dx dx2 du

Integrating,

21 + u2

= 7.3575(10 - 3)dx

ln 1 u + 21 + u2 2 = 7.3575(10 - 3)x + C1

Applying the boundary condition u =

dy = 0 at x = 0 results in C1 = 0. Thus, dx

ln 1 u + 21 + u2 2 = 7.3575(10 - 3)x u + 21 + u2 = e 7.3575(10

dy e 7.3575(10 = u = dx Since sin h x =

-3

)x

-3

)x

- e - 7.3575(10 2

-3

)x

ex - e - x , then 2 dy = sin h 7.3575(10 - 3)x dx

Integrating, y = 135.92 cos h 7.3575(10 - 3)x + C2 Applying the boundary equation y = 0 at x = 0 results in C2 = –135.92. Thus, y = 135.923 cos h 7.3575(10 - 3)x - 14 Applying the boundary equation y = h at x = 20 m, h = 135.923cos h 7.3575(10 - 3) 1202 - 14 = 1.47 m

Ans.

Ans: h = 1.47 m 738

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7–115. The 80-m-long chain is fixed at its ends and hoisted at its midpoint B using a crane. If the chain has a weight of 0.5  kN/m, determine the minimum height h of the hook in order to lift the chain completely off the ground. What is the horizontal force at pin A or C when the chain is in this position? Hint: When h is a minimum, the slope at A and C is zero.

B h A

C 60 m

Solution Given: L = 80 m d = 60 m w = 0.5 kN/m Guesses Given a

FH = 10 kN FH # acos h a w # d b - 1b h = w FH 2

h b = Find 1h, FH 2 FH

FA = FH

h = 1m FH L # sin h a w # d b = 2 w FH 2

FC = FH

a

FA 11.1 b = a b kN FC 11.1 h = 23.5 m

Ans. Ans.

Ans: FA = 11.1 kN FC = 11.1 kN h = 23. 5 m 739

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*7–116 .

The cable has a mass of 0.5 kg>m, and is 25 m long. Determine the vertical and horizontal components of force it exerts on the top of the tower.

A

SOLUTION

B

ds

x =

L

b1 +

30

1

2 1 (w0 ds)2 r F2H

15 m

Performing the integration yields: x =

FH 1 b sinh - 1 c (4.905s + C1) d + C2 r 4.905 FH

(1)

From Eq. 7– 14 dy dx dy dx

=

1 w ds FH L 0

=

1 (4.905s + C1) FH

At s = 0;

dy = tan 30°. Hence C1 = FH tan 30° dx dy dx

=

4.905s + tan 30° FH

(2)

Applying boundary conditions at x = 0; s = 0 to Eq.(1) and using the result C1 = FH tan 30° yields C2 = - sinh - 1(tan 30°). Hence x =

FH 1 b sinh - 1 c (4.905s+FH tan 30°) d - sinh - 1(tan 30°)r 4.905 FH

(3)

At x = 15 m; s = 25 m. From Eq.(3) 15 =

FH 1 b sinh - 1 c (4.905(25) + FH tan 30°) R - sinh - 1(tan 30°) r 4.905 FH

By trial and error FH = 73.94 N At point A, s = 25 m From Eq.(2) tan uA =

dy 4.905(25) 2 + tan 30° = dx s = 25 m 73.94

uA = 65.90°

(Fv )A = FH tan uA = 73.94 tan 65.90° = 165 N

Ans.

(FH)A = FH = 73.9 N

Ans.

Ans: (Fv) A = 165 N (FH)A = 73.9 N 740

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7–117. A uniform cord is suspended between two points having the same elevation. Determine the sag-to-span ratio so that the maximum tension in the cord equals the cord’s total weight.

SOLUTION From Example 7.13.

At x =

s =

w0 FH sinh ¢ x≤ w0 FH

y =

FH w0 B cosh ¢ x ≤ - 1 R w0 FH

L , 2 dy w0L = tan umax = sinh ¢ ≤ ` dx max 2FH 1

cos umax =

cosh a

Tmax =

w0L b 2FH

FH cos umax

w012s2 = FH cosh ¢ 2FH sinh ¢ tanh ¢

w0L ≤ 2FH

w0L w0L ≤ = FH cosh ¢ ≤ 2FH 2FH

w0L 1 ≤ = 2FH 2

w0L = tanh-110.52 = 0.5493 2FH when x =

L ,y = h 2 h =

w0 FH B cosh ¢ x ≤ - 1 R w0 FH

h =

FH e w0

1

C

1 - tanh2 a

w0L b 2FH

- 1 u = 0.1547 ¢

FH ≤ w0

0.1547 L = 0.5493 2h h = 0.141 L

Ans.

Ans: h = 0.141 L

741

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7–118. A 50-m cable is suspended between two points a distance of 15 m apart and at the same elevation. If the minimum tension in the cables is 200 kN, determine the total weight of the cable and the maximum tension developed in the cable.

Solution Given: L = 50 m

d = 15 m

Tmin = 200 kN

Tmin = FH

FH = Tmin

FH = 200 kN

From Example 7.13: s =

w0 x FH sin h a b w0 FH

Guess

w0 = 1 kN>m w0 d FH L Given = sin h a b 2 w0 FH 2 Total weight = w0 L

w0 L tan1umax 2 = FH 2

umax

w0 = Find (w0) Total weight = 4.00 MN

L w0 a b 2 § = a tan FH £

w0 = 79.93 kN>m Ans.

umax = 84.284°

Then, Tmax =

FH cos 1umax 2

Ans.

Tmax = 2.01 MN

Ans: Total weight = 4.00 MN Tmax = 2.01 MN 742

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7–119. Show that the deflection curve of the cable discussed in Example 7.13 reduces to Eq. 4 in Example 7.12 when the hyperbolic cosine function is expanded in terms of a series and only the first two terms are retained. (The answer indicates that the catenary may be replaced by a parabola in the analysis of problems in which the sag is small. In this case, the cable weight is assumed to be uniformly distributed along the horizontal.)

SOLUTION cosh x = 1 +

x2 + Á 2!

Substituting into y =

FH w0 B cosh ¢ x ≤ - 1 R w0 FH

=

FH w 20x2 + Á - 1R B1 + w0 2F 2H

=

w0x2 2FH

Using Eq. (3) in Example 7.12, FH = We get

y =

w0L2 8h

4h 2 x L2

QED

743

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*7–120. A cable has a weight of 30 N/m and is supported at points that are 25 m apart and at the same elevation. If it has a length of 26 m, determine the sag.

Solution Given: g = 30 N>m

d = 25 m

L = 26 m

Guess FH = 1000 N g d FH L Given - c a bdd = 0 sin hc g 2 FH 2 FH 1 g h = acos h a db - 1b g 2 FH

FH = Find1FH 2

FH = 770 N Ans.

h = 3.104 m

Ans: h = 3.104 m 744

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7–121. A wire has a weight of 2 N/m. If it can span 10 m and has a sag of 1.2 m, determine the length of the wire. The ends of the wire are supported from the same elevation.

Solution Given : g = 2 N>m

d = 10 m

h = 1.2 m

From Eq. (5) of Example 7.13 : g d 2 b FH £ 2 FH § h = g 2 a

FH =

1 d2 g 8 h

FH = 20.83 N

From Eq. (3) of Example 7.13 : g d FH L = sin h c a bd g 2 FH 2

L = 2

FH 1 d sin h a g b g 2 FH

L = 10.39 m

Ans.

Ans: L = 10.39 m 745

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7–122. The 10 kg m cable is suspended between the supports A and B. If the cable can sustain a maximum tension of 1.5 kN and the maximum sag is 3 m, determine the maximum distance L between the supports

L A

3m

B

SOLUTION The origin of the x, y coordinate system is set at the lowest point of the cable. Here w0 = 10(9.81) N>m = 98.1 N>m. Using Eq. (4) of Example 7.13, y =

FH w0 Bcosh ¢ x ≤ - 1 R w0 FH

y =

FH 98.1x Bcosh ¢ ≤ - 1R 98.1 FH

Applying the boundary equation y = 3 m at x =

3 =

L , we have 2

FH 49.05L Bcosh ¢ ≤ - 1R 98.1 FH

The maximum tension occurs at either points A or B where the cable makes the greatest angle with the horizontal. From Eq. (1), tan umax = sinh ¢

49.05L ≤ FH

By referring to the geometry shown in Fig. b, we have 1

cos umax = A

1 + sinh2 49.05L ¢ F ≤ H

=

1 cosh ¢

49.05L ≤ FH

Thus, Tmax =

FH cos umax

1500 = FH cosh ¢

49.05L ≤ FH

(3)

Solving Eqs. (2) and (3) yields Ans.

L = 16.8 m FH = 1205.7 N

Ans: L = 16.8 m 746

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7–123. A cable having a weight per unit length of 0.1 kN/m is suspended between supports A and B. Determine the equation of the catenary curve of the cable and the cable’s length.

50 m A

SOLUTION As shown in Fig. a, the origin of the x, y coordinate system is set at the lowest point of the cable. Here, w(s) = 0.1 kN/m. ⎛ dy ⎞ 1+ ⎜ ⎟ ⎝ dx ⎠

d2y 0.1 = = dx 2 FH If we set u =

2

dy du d2y . Substituting these two values into the equation, , then = dx dx dx 2

du 1+u

= 2

0.1 FH

dx

Integrating, ln (u + 1 + u 2 ) =

0.1 FH

x + C1

Applying the boundary condition u = ln(u + 1 + u 2 ) =

0.1 FH

dy dx

= 0 at x = 0 results in C1 = 0. Thus,

x

0.1 x

u + 1 + u 2 = e FH 0.1

dy dx

=u=

Since sinh x =

x

e FH – e

– 0.1 x FH

2

ex – e–x , then 2

dy 0.1 x = sinh dx FH Applying the boundary equation

(1) dy = tan 30° at x = 25 m, dx

⎡ 0.1 ⎤ tan 30° = sinh ⎢ (25)⎥ ⎢⎣ FH ⎥⎦ FH = 4.5512 kN Substituting this result into Eq. (1), dy = sinh[(0.0219722)x] dx

(2)

747

30

30

B

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7–123 (Continued) Integrating, y = 45.512 cosh[(0.0219722)x] + C2 Applying the boundary equation y = 0 at x = 0 results in C2 = – 45.512. Thus, y = 45.512 {cosh[(0.0219722)x] – 1} m

Ans.

If we write the force equation of equilibrium along the x and y axes by referring to the free – body diagram shown in Fig. b, we have + → ΣFx

= 0;

+↑ΣFy = 0;

T cos ␪ – 4.5512 = 0 T sin ␪ – 0.1s = 0

Eliminating T, dy = tan ␪ = 0.0219722 dx

(3)

Equating Eqs. (2) and (3), (0.0219722)s = sinh[(0.0219722)x] s = 45.512 [(0.0219722)x] m Thus, the length of the cable is L = 2{45.512 sinh[(0.0219722) (25)]} = 52.553 m

25 m

Ans.

25 m

FH = 4.5512 kN

Ans: y = 45.512 {cosh[(0.0219722)x] - 1} m L = 52.553 m 748

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8–1. P 2 P 2

Determine the maximum force P the connection can support so that no slipping occurs between the plates. There are four bolts used for the connection and each is tightened so that it is subjected to a tension of 4 kN. The coefficient of static friction between the plates is ms = 0.4.

P

SOLUTION Free-Body Diagram: The normal reaction acting on the contacting surface is equal to the sum total tension of the bolts. Thus, N = 4(4) kN = 16 kN. When the plate is on the verge of slipping, the magnitude of the friction force acting on each contact surface can be computed using the friction formula F = msN = 0.4(16) kN. As indicated on the free-body diagram of the upper plate, F acts to the right since the plate has a tendency to move to the left. Equations of Equilibrium: + ©Fx = 0; :

0.4(16) -

P = 0 2

Ans.

p = 12.8 kN

Ans: P = 12.8 kN 749

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8–2. If the coefficient of static friction at A is m s= 0.4 and the collar at B is smooth so it only exerts a horizontal force on the pipe, determine the minimum distance x so that the bracket can support the cylinder of any mass without slipping. Neglect the mass of the bracket.

100 mm

B

x

C

200 mm A

Solution Free-Body Diagram. The weight of cylinder tends to cause the bracket to slide downward. Thus, the frictional force FA must act upwards as indicated in the freebody diagram shown in Fig. a. Here the bracket is required to be on the verge of slipping so that FA = msNA = 0.4NA. Equations of Equilibrium. + c ΣFy = 0; a+ ΣMB = 0;

0.4NA - mg = 0

NA = 2.5 mg

2.5 mg10.22 + 0.412.5 mg2 10.12 - m(g2 1x + 0.12 = 0 x = 0.5 m

Ans.

Note. Since x is independent of the mass of the cylinder, the bracket will not slip regardless of the mass of the cylinder provided x 7 0.5 m.

Ans: x = 0.5 m 750

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8–3.

10 kN

The mine car and its contents have a total mass of 6 Mg and a center of gravity at G. If the coefficient of static friction between the wheels and the tracks is ms = 0.4 when the wheels are locked, find the normal force acting on the front wheels at B and the rear wheels at A when the brakes at both A and B are locked. Does the car move?

0.9 m

G

B

A

0.15 m

0.6 m 1.5 m

SOLUTION Equations of Equilibrium: The normal reactions acting on the wheels at (A and B) are independent as to whether the wheels are locked or not. Hence, the normal reactions acting on the wheels are the same for both cases. a + ©MB = 0;

NA 11.52 + 1011.052 - 58.8610.62 = 0 Ans.

NA = 16.544 kN = 16.5 kN + c ©Fy = 0;

NB + 16.544 - 58.86 = 0 Ans.

NB = 42.316 kN = 42.3 kN

When both wheels at A and B are locked, then 1FA2max = msNA = 0.4116.5442 = 6.6176 kN and 1FB2max = msNB = 0.4142.3162 = 16.9264 kN. Since 1FA2max + FB max = 23.544 kN 7 10 kN, the wheels do not slip. Thus, the mine car does not move. Ans.

Ans: N A = 16.5 kN, N B = 42.3 kN, It does not move. 751

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*8–4. The fork lift has a weight of 12 kN and a center of gravity at G. If the rear wheels are powered, whereas the front wheels are free to roll, determine the maximum number of 150-kg crates the fork lift can push forward. The coefficient of static friction between the wheels and the ground is ms = 0.4, and between each crate and the ground is ms = 0.35.

G 0.75 m A

B 1.05 m

0.375 m

Solution Given: W1 = 12 kN W2 = 150 # (9.81) N ms = 0.4 m′s = 0.35 a = 0.75 m b = 0.375 m c = 1.05 m

Fork lift: ΣMB = 0;

ΣFx = 0;

W1 # c - NA # (b + c) = 0 c NA: = W1 # NA = 8.8 kN b + c ms # NA - P = 0 P: = ms # NA

P = 3.54 kN

Nc - W2 = 0 Nc: = W2

Nc = 1.471 kN

Crate: ΣFy = 0; ΣFx = 0;

P′ - ms # Nc = 0 P′: = m′s # Nc

P′ = 0.515 kN

Thus n =

P P′

n = 6.87

n = floor(n)

Ans.

n = 6

Ans:

n 752

6

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8–5. The pipe of weight W is to be pulled up the inclined plane of slope a using a force P. If P acts at an angle f, show that for slipping P = W sin(a + u)>cos(f - u), where u is the angle of static friction; u = tan - 1 ms.

P

φ

α

SOLUTION + a©Fy¿ = 0;

N + P sin f - W cos a = 0

+ Q©Fx¿ = 0;

P cos f - W sin a - tan u(W cos a - P sin f) = 0 P = =

N = W cos a - P sin f

W(sin a + tan u cos a) cos f + tan u sin f W sin(a + u) W(cos u sin a + sin u cos a) = cos f cos u + sin f sin u cos(f - u)

753

Q.E.D.

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8–6. Determine the angle f at which the applied force P should act on the pipe so that the magnitude of P is as small as possible for pulling the pipe up the incline. What is the corresponding value of P? The pipe weighs W and the slope a is known. Express the answer in terms of the angle of kinetic friction, u = tan - 1 mk.

P

φ

α

SOLUTION + a©Fy¿ = 0;

N + P sin f - W cos a = 0

+Q©Fx¿ = 0;

P cos f - W sin a - tan u (W cos a - P sin f) = 0

P =

N = W cos a - P sin f

W(sin a + tan u cos a) cos f + tan u sin f

=

W(cos u sin a + sin u cos a) cos f cos u + sin f sin u

=

W sin (a + u) cos (f - u)

W sin (a + u) sin (f - u) dP = 0 = df cos2(f - u) W sin (a + u) sin (f - u) = 0

W sin (a + u) = 0

sin (f - u) = 0

f = u

P =

f - u = 0

Ans.

W sin (a + u) = W sin (a + u) cos (u - u)

Ans.

Ans: f = u P = W sin (a + u) 754

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8–7. The automobile has a mass of 2 Mg and center of mass at G. Determine the towing force F required to move the car if the back brakes are locked, and the front wheels are free to roll. Take ms = 0.3.

F G

30

0.6 m

C

0.3 m

0.75 m

A 1m

B

1.50 m

Solution Equations of Equilibrium. Referring to the FBD of the car shown in Fig. a, + ΣFx = 0;  FB - F cos 30° = 0  S

(1)

    + c ΣFy = 0;  NA + NB + F sin 30° - 2000(9.81) = 0

(2)

  a + ΣMA = 0;  F cos 30°(0.3) - F sin 30°(0.75) + (3)

NB (2.5) - 2000(9.81)(1) = 0



Friction. It is required that the rear wheels are on the verge to slip. Thus

(4)

FB = ms NB = 0.3 NB

Solving Eqs. (1) to (4),

Ans.

F = 2,762.72 N = 2.76 kN NB = 7975.30 N  NA = 10, 263.34 N  FB = 2392.59 N

Ans: F = 2.76 kN 755

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*8–8. The automobile has a mass of 2 Mg and center of mass at G. Determine the towing force F required to move the car. Both the front and rear brakes are locked. Take ms = 0.3.

F G

30

0.6 m

C

0.3 m

0.75 m

A 1m

B

1.50 m

Solution Equations of Equilibrium. Referring to the FBD of the car shown in Fig. a, + ΣFx = 0;  FA + FB - F cos 30° = 0  S

(1)

 + c ΣFy = 0;  F sin 30° + NA + NB - 2000(9.81) = 0

(2)

  a + ΣMA = 0;  F cos 30°(0.3) - F sin 30°(0.75) + NB (2.5) - 200(9.81)(1) = 0



(3)

Friction. It is required that both the front and rear wheels are on the verge to slip. Thus

FA = ms NA = 0.3 NA

(4)



FB = ms NB = 0.3 NB

(5)

Solving Eqs. (1) to (5),

Ans.

F = 5793.16 N = 5.79 kN NB = 8114.93 N  NA = 8608.49 N  FA = 2582.55 N  FB = 2434.48 N

Ans: F = 5.79 kN 756

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8–9. The block brake consists of a pin-connected lever and friction block at B. The coefficient of static friction between the wheel and the lever is ms = 0.3, and a torque of 5 N # m is applied to the wheel. Determine if the brake can hold the wheel stationary when the force applied to the lever is (a) P = 30 N, (b) P = 70 N.

5N m

150 mm 50 mm

O P

A B 200 mm

400 mm

SOLUTION To hold lever: a + ©MO = 0;

FB (0.15) - 5 = 0;

FB = 33.333 N

Require NB =

33.333 N = 111.1 N 0.3

Lever, a + ©MA = 0;

PReqd. (0.6) - 111.1(0.2) - 33.333(0.05) = 0

PReqd. = 39.8 N a) P = 30 N 6 39.8 N

No

Ans.

b) P = 70 N 7 39.8 N

Yes

Ans.

Ans: No Yes 757

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8–10. The block brake consists of a pin-connected lever and friction block at B. The coefficient of static friction between the wheel and the lever is ms = 0.3, and a torque of 5 N # m is applied to the wheel. Determine if the brake can hold the wheel stationary when the force applied to the lever is (a) P = 30 N, (b) P = 70 N.

5N m

150 mm 50 mm

O P

A B 200 mm

400 mm

SOLUTION To hold lever: a + ©MO = 0;

- FB(0.15) + 5 = 0;

FB = 33.333 N

Require NB =

33.333 N = 111.1 N 0.3

Lever, a + ©MA = 0;

PReqd. (0.6) - 111.1(0.2) + 33.333(0.05) = 0

PReqd. = 34.26 N a) P = 30 N 6 34.26 N

No

Ans.

b) P = 70 N 7 34.26 N

Yes

Ans.

Ans: No Yes 758

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8–11. 0.75m

The car has a mass of 1.6 Mg and center of mass at G. If the coefficient of static friction between the shoulder of the road and the tires is ms = 0.4, determine the greatest slope u the shoulder can have without causing the car to slip or tip over if the car travels along the shoulder at constant velocity.

G B 1. 5 m A

θ

SOLUTION Tipping: a + ©MA = 0;

- W cos u10.752 + W sin u10.752 = 0

0.75m

tan u = 1 0.75m

u = 45° Slipping: Q + ©Fx = 0;

0.4 N - W sin u = 0

a + ©Fy = 0;

N - W cos u = 0 tan u = 0.4 u = 21.8°

Ans. (Car slips before it tips.)

Ans: u = 21.8 759

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*8–12. If a torque of M = 300 N # m is applied to the flywheel, determine the force that must be developed in the hydraulic cylinder CD to prevent the flywheel from rotating. The coefficient of static friction between the friction pad at B and the flywheel is ms = 0.4.

D

0.6 m 30

1m B

C M

0.3 m

60 mm 300 N m

A

O

SOLUTION Free-BodyDiagram: First we will consider the equilibrium of the flywheel using the free-body diagram shown in Fig. a. Here, the frictional force FB must act to the left to produce the counterclockwise moment opposing the impending clockwise rotational motion caused by the 300 N # m couple moment. Since the wheel is required to be on the verge of slipping, then FB = msNB = 0.4 NB. Subsequently, the free-body diagram of member ABC shown in Fig. b will be used to determine FCD. Equations of Equilibrium: We have a + ©MO = 0;

0.4 NB(0.3) - 300 = 0

NB = 2500 N

Using this result, a + ©MA = 0;

FCD sin 30°(1.6) + 0.4(2500)(0.06) - 2500(1) = 0 Ans.

FCD = 3050 N = 3.05 kN

Ans: FCD = 3.05 kN 760

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8–13. The block brake is used to stop the wheel from rotating when the wheel is subjected to a couple moment M0. If the coefficient of static friction between the wheel and the block is ms, determine the smallest force P that should be applied.

P a b C c

M0

O

r

SOLUTION a + ©MC = 0;

Pa - Nb - ms Nc = 0 N =

c + ©MO = 0;

Pa (b + ms c)

ms Nr - M0 = 0 ms P a P =

a br = M0 b + ms c

M0 (b + ms c) ms ra

Ans.

Ans: P = 761

M0 (b + ms c) msra

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8–14. The pipe is hoisted using the tongs. If the coefficient of static friction at A and B is ms, determine the smallest dimension b so that any pipe of inner diameter d can be lifted.

b

b

C h A

B d

Solution Require: W … msNB 2

FB =

a+ ΣMC = 0; NB = Thus,

W d W a b - NA(h) + b a b = 0 2 2 2

W d ab - b 2h 2

ms W W d … ab - b 2 2h 2 h … (b -

d )m 2 s

b Ú

h d + ms 2

b =

h d + ms 2

Ans.

Ans: b =

762

h d + ms 2

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8–15. If the coefficient of static friction at contacting surface between blocks A and B is ms, and that between block B and bottom is 2 ms, determine the inclination u at which the identical blocks, each of weight W, begin to slide.

A B u

Solution Free-Body Diagram. Here, we will assume that the impending motion of the upper block is down the plane while the impending motion of the lower block is up the plane. Thus, the frictional force F acting on the upper block acts up the plane while the friction forces F and F ¿ acting on the lower block act down the plane as indicated on the free-body diagram of the upper and lower blocks shown in Figs. a and b, respectively. Since both block are required to be on the verge of slipping, then F = msN and F′ = msN′. Equations of Equilibrium. Referring to Fig. a, + c ΣFy′ = 0; + c ΣFx′ = 0;

N - W cos u = 0 T + ms 1W cos u 2 - W sin u = 0

N = W cos u T = W sin u - msW cos u

Using these results and referring to Fig. b,

+ c ΣFy ′ = 0; N′ - W cos u - W cos u = 0

N′ = 2W cos u

+

S ΣFx ′ = 0; 21W sin u - msW cos u 2 - msW cos u - 2ms 12W cos u 2 - W sin u = 0 sin u - 7mscos u = 0 u = tan

-1

Ans.

7ms

Since the analysis yields a positive u, the above assumption is correct.

Fᦡ= 2μsNᦡ

Ans: u = tan - 1 7ms 763

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*8–16. The uniform hoop of weight W is suspended from the peg at A and a horizontal force P is slowly applied at B. If the hoop begins to slip at A when the angle is u = 30, determine the coefficient of static friction between the hoop and the peg.

A u r

P

B

Solution Given:

T 6F x = 0;

30 

P NA cos  T  P  NA sin  T P

6F y = 0;

P

cos  T  sin  T NA

P NA sin  T  W  NA cos  T W

0

P

0

sin  T  cos  T NA

60$ = 0; W r sin  T  P  r  r cos  T W sin  T

P P

0

P 1  cos  T

sin  T  cos  T sin  T sin  T

1  cos  T

P

 sin  T  P 0.27

cos  T  1  cos  T Ans.

Ans:

P 764

0.27

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8–17. The uniform hoop of weight W is suspended from the peg at A and a horizontal force P is slowly applied at B. If the coefficient of static friction between the hoop and peg is ms, determine if it is possible for the angle u = 30 before the hoop begins to slip.

A u r

B

P

Solution Given:

Ps T 6F x = 0;

0.2 30

P NA cos  T  P  NA sin  T P

6F y = 0;

P

cos  T  sin  T NA

P NA sin  T  W  NA cos  T W

P

0

0

sin  T  cos  T NA

60$ = 0; W r sin  T  P  r  r cos  T W sin  T

P

P If P s

P  1  cos  T

sin  T  cos  T sin  T

sin  T

1  cos  T

0.20 < P

0

P

 sin  T  P

cos  T  1  cos  T

0.27

0.27 then it is not possible to reach T

30.00 .

Ans.

Ans:

It is not possible since P 765

0.20 < 0.27

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8–18. The spool of wire having a mass of 150 kg rests on the ground at A and against the wall at B. Determine the force P required to begin pulling the wire horizontally off the spool. The coefficient of static friction between the spool and its points of contact is ms = 0.25.

P 250 mm

B

450 mm

A

Solution Given: M

150 kg

Ps

0.25

a

0.45 m

b

0.25 m

Initial guesses:

P

FA

100 N

NA

10 N

20 N

NB

30 N

FB

10 N

1.14 kN

Ans.

Given 6F y = 0;

NA  FB  M g

6F x = 0;

F A  NB  P

6MB = 0;

P b  M g a  NA a  FA a FA

§P · ¨ ¸ ¨ FA ¸ ¨ FB ¸ ¨ ¸ ¨ NA ¸ ¨N ¸ © B¹

P s NA

0 0

FB

Find  P  F A  FB  NA  NB

0

P s NB

§ FA · ¨ ¸ ¨ NA ¸ ¨ FB ¸ ¨ ¸ © NB ¹

§ 0.28 · ¨ ¸ ¨ 1.12 ¸ kN ¨ 0.36 ¸ ¨ ¸ © 1.42 ¹

P

Ans:

P 766

1.14 kN

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8–19. The spool of wire having a mass of 150 kg rests on the ground at A and against the wall at B. Determine the forces acting on the spool at A and B if P = 800 N. The coefficient of static friction between the spool and the ground at point A is ms = 0.35. The wall at B is smooth.

P 250 mm

B

450 mm

A

Solution

Assume no slipping

Given: P

800 N

a

0.45 m

M

150 kg

b

0.25 m

Ps

0.35

Initial guesses : F A

NA

10N

10N

NB

10N

F Amax

10N

Given 6F x = 0;

F A  NB  P

6F y = 0;

NA  M g

6M0 = 0;

P b  F A a F Amax

§ FA · ¨ ¸ ¨ FAmax ¸ ¨ NA ¸ ¨ ¸ © NB ¹ § NA · ¨ ¸ © FA ¹

0

0 0

P s NA

§ FA · ¨ ¸ © FAmax ¹

Find  FA  F Amax  NA  NB

§ 444 · ¨ ¸N © 515 ¹

If FA 444 N < F Amax 515 N then our no-slip assumption is good.

§ 1.47 · ¨ ¸ kN © 0.44 ¹

NB

1.24 kN

Ans.

Ans.

Ans: If FA = 444 N 6 FAmax = 515 N then our no-slip assumption is good. FA = 0.44 kN, N A = 1.47 kN, N B = 1.24 kN 767

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*8–20. The ring has a mass of 0.5 kg and is resting on the surface of the table. In an effort to move the ring a normal force P from the finger is exerted on it. If this force is directed towards the ring’s center O as shown, determine its magnitude when the ring is on the verge of slipping at A. The coefficient of static friction at A is mA = 0.2 and at B, mB = 0.3.

P

B 60

O

75 mm A

Solution

FA = FB



P cos 60° - FB cos 30° - FA = 0



NA - 0.5(9.81) - P sin 60° - FB sin 30° = 0



FA = 0.2 NA



NA = 19.34 N



FA = FB = 3.868 N



P = 14.4 N



(FB)max = 0.3(14.44) = 4.33 N 7 3.868 N(O.K!)

Ans.

Ans: P = 14.4 N 768

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8–21. A roofer, having a mass of 70 kg, walks slowly in an upright position down along the surface of a dome that has a radius of curvature of r = 20 m. If the coefficient of static friction between his shoes and the dome is ms = 0.7, determine the angle u at which he first begins to slip.

u

20 m 60

Solution + c ΣFy′ = 0; N - 7019.812 cos u = 0 +

S ΣFx′ = 0;

(1)

7019.812 sin u - 0.7N = 0

(2)

Solving Eqs. (1) and (2) yields: Ans.

u = 35.0° N = 562.6 N

Ans: u = 35.5 769

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8–22. The block brake is used to stop the wheel from rotating when the wheel is subjected to a couple moment M0 = 360 N # m. If the coefficient of static friction between the wheel and the block is ms = 0.6, determine the smallest force P that should be applied.

P

1m 0.4 m C B M0

0.05 m

O 0.3 m

Solution Equations of Equilibrium. Referring to the FBD of the lever arm shown in Fig. a, a + ΣMC = 0;  P(1) + FB (0.05) - NB (0.4) = 0

(1)

Also, the FBD of the wheel, Fig. b, a + ΣMO = 0;  FB (0.3) - 360 = 0  FB = 1200 N Friction. It is required that the wheel is on the verge to rotate thus slip at B. Then

FB = ms NB;  1200 = 0.6 NB  NB = 2000 N

Substitute the result of FB and NB into Eq. (1)

P(1) + 1200(0.05) - 2000(0.4) = 0 Ans.

P = 740 N

Ans: P = 740 N 770

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8–23. Solve Prob. 8–22 if the couple moment M0 is applied counterclockwise.

P

1m 0.4 m C B M0

0.05 m

O 0.3 m

Solution Equations of Equilibrium. Referring to the FBD of the lever arm shown in Fig. a, a+ ΣMC = 0;  P(1) - FB(0.05) - NB(0.4) = 0

(1)

Also, the FBD of the wheel, Fig. b a+ ΣMO = 0;  360 - FB(0.3) = 0  FB = 1200 N Friction. It is required that the wheel is on the verge to rotate thus slip at B. Then FB = ms NB;  1200 = 0.6 NB  NB = 2000 N Substituting the result of FB and NB into Eq. (1),

P(1) - 1200(0.05) - 2000(0.4) = 0 Ans.

P = 860 N

Ans: P = 860 N 771

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*8–24. The 45-kg disk rests on the surface for which the coefficient of static friction is mA = 0.2. Determine the largest couple moment M that can be applied to the bar without causing motion.

300 mm

C

M 400 mm

B 125 mm A

Solution a+ ΣM0 = 0; +

FA = By = 0.2NA

S ΣFx = 0;

Bx - 0.2NA = 0

+ c ΣFy = 0;

NA - By - 45(9.81) = 0 NA = 551.8 N Bx = 110.4 N By = 110.4 N

+

S ΣMc = 0;

-110.4(0.3) - 110.4(0.4) + M = 0 M = 77.3 N # m

Ans.

Ans:

M = 77.3 N # m 772

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8–25. The 45-kg disk rests on the surface for which the coefficient of static friction is mA = 0.15. If M = 50 N # m, determine the friction force at A.

300 mm

C

M 400 mm

B

Solution

125 mm

Bar: a+ ΣMc = 0; +

A

-By(0.3) - Bx(0.4) + 50 = 0

S ΣFx = 0;

Bx = Cx

+ c ΣFy = 0;

By = Cy

Disk: +

S ΣFx = 0; + c ΣFy = 0; a+ ΣMo = 0;

Bx = FA NA - By - 45(9.81) = 0 By 10.1252 - FA 10.1252 = 0 NA = 512.9 N FA = 71.4 N

Ans.

1FA 2 max = 0.151512.92 = 76.93 N > 71.43 N No motion of disk.

Ans: FA = 71.4 N 773

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8–26. The beam is supported by a pin at A and a roller at B which has negligible weight and a radius of 15 mm. If the coefficient of static friction is mB = mC = 0.3, determine the largest angle u of the incline so that the roller does not slip for any force P applied to the beam.

P 2m

2m B

A C

u

Solution a+ ΣMO = 0;  FB (15) - FC (15) = 0

(1)

+ ΣFx = 0;   - FB - FC cos u + NC sin u = 0 S

(2)

   + c ΣFy = 0;  NC cos u + FC sin u - NB = 0

(3)

Assume slipping at C so that FC = 0.3 NC



Then from Eqs. (1) and (2),

FB = FC



- 0.3 NC - 0.3 NC cos u + NC sin u = 0



( - 0.3 - 0.3 cos u + sin u ) NC = 0

(4)

The term in parentheses is zero when

Ans.

u = 33.4°

From Eq. (3),    NC (cos 33.4° + 0.3 sin 33.4°) = NB

NC = NB

Since Eq. (4) is satisfied for any value of NC, any value of P can act on the beam. Also, the roller is a “two-force member.”

2(90° - f) + u = 180°



f =



f = tan - 1 a

thus 

 

u 2 mN b = tan - 1 (0.3) = 16.7° N

Ans.

u = 2(16.7°) = 33.4°

Ans: u = 33.4° 774

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8–27. The friction pawl is pinned at A and rests against the wheel at B. It allows freedom of movement when the wheel is rotating counterclockwise about C. Clockwise rotation is prevented due to friction of the pawl which tends to bind the wheel. If 1ms2B = 0.6, determine the design angle u which will prevent clockwise motion for any value of applied moment M. Hint: Neglect the weight of the pawl so that it becomes a two-force member.

A

θ B

20°

M C

SOLUTION Friction: When the wheel is on the verge of rotating, slipping would have to occur. Hence, FB = mNB = 0.6NB . From the force diagram (FAB is the force developed in the two force member AB) tan120° + u2 =

0.6NB = 0.6 NB Ans.

u = 11.0°

Ans: u = 11.0° 775

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*8–28. The tongs are used to lift the 150-kg crate, whose center of mass is at G. Determine the least coefficient of static friction at the pivot blocks so that the crate can be lifted.

P 275 mm E 500 mm

C

30

F H

D

500 mm

SOLUTION Free - Body Diagram. Since the crate is suspended from the tongs, P must be equal to the weight of the crate; i.e., P = 150(9.81)N as indicated on the free - body diagram of joint H shown in Fig.a. Since the crate is required to be on the verge of slipping downward, FA and FB must act upward so that FA = msNA and FB = msNB as indicated on the free - body diagram of the crate shown in Fig. c.

A

G

B

300 mm

Equations of Equilibrium. Referring to Fig. a, + ©F = 0; : x

FHE cos 30° - FHF cos 30° = 0

FHE = FHF = F

+ c ©Fy = 0;

150(9.81) - 2F sin 30° = 0

F = 1471.5 N

Referring to Fig. b, a + ©MC = 0;

1471.5 cos 30°(0.5) + 1471.5 sin 30°(0.275) - NA (0.5) - msNA (0.3) = 0 (1)

0.5NA + 0.3msNA = 839.51 Due to the symmetry of the system and loading, NB = NA. Referring to Fig. c, + c ©Fy = 0;

(2)

2msNA - 150(9.81) = 0

Solving Eqs. (1) and (2), yields NA = 1237.57 N Ans.

ms = 0.595

Ans: ms = 0.595 776

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8–29. A man attempts to support a stack of books horizontally by applying a compressive force of F = 120 N to the ends of the stack with his hands. If each book has a mass of 0.95 kg, determine the greatest number of books that can be supported in the stack. The coefficient of static friction between the man’s hands and a book is (ms)h = 0.6 and between any two books (ms)b = 0.4.

F

120 N

F

120 N

SOLUTION Equations of Equilibrium and Friction: Let n¿ be the number of books that are on the verge of sliding together between the two books at the edge. Thus, Fb = (ms)b N = 0.4(120) = 48.0 N. From FBD (a), + c ©Fy = 0;

2(48.0) - n¿(0.95)(9.81) = 0

n¿ = 10.30

Let n be the number of books are on the verge of sliding together in the stack between the hands. Thus, Fk = (ms)k N = 0.6(120) = 72.0 N. From FBD (b), + c ©Fy = 0;

2(72.0) - n(0.95)(9.81) = 0

n = 15.45

Thus, the maximum number of books can be supported in stack is Ans.

n = 10 + 2 = 12

Ans: n = 12 777

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8–30. Determine the magnitude of force P needed to start towing the 40-kg crate. Also determine the location of the resultant normal force acting on the crate, measured from point A. Take ms = 0.3.

P 5 4 3

400 mm

200 mm 800 mm

A

SOLUTION Given: M

40 kg

c

200 mm

Ps

0.3

d

3

e

4

a

400 mm

b

800 mm

Initial guesses:

NC

P

200 N

50 N

Given 6F x = 0;

§ d ·P  P N s C ¨ 2 2¸ © d e ¹

6F y = 0;

NC  M g 

eP 2

0

0

2

d e

§ NC · ¨ ¸ ©P ¹ NC 6MO = 0;

Find  NC  P P

280.2 N

Ans.

140 N

§ e P ·§ b · ¸  N1 x  ¨ ¨ ¸ 2 2 ¸© 2 ¹ © 2¹ © d e ¹

P s NC§¨

x

a·

1 P s NC a 2

Thus, the distance from A is

2

0

2

d e ePb 2

2

NC

d e

A

x

b 2

A

x

123.51 mm

523.51 mm

Ans.

Ans: N C = 280.2 N P = 140 N A = 523.5 mm 778

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8–31. Determine the friction force on the 40-kg crate, and the resultant normal force and its position x, measured from point A, if the force is P = 300 N. Take ms = 0.5 and mk = 0.2.

P 5 4 3

400 mm

200 mm 800 mm

SOLUTION

A

Given: M

40 kg

Ps

0.5

a

400 mm

Pk

0.2

b

800 mm

d

3

c

200 mm

e

4

P

300 N

Initial guesses:

FC

25 N

NC

100 N

Given 6F x = 0;

6F y = 0;

P§

d · ¨ 2 2 ¸  FC © d e ¹

0

NC  Mg  P§

e · ¨ 2 2¸ © d e ¹

§ FC · ¨ ¸ © NC ¹

Find  F C  NC

0

F Cmax

P s NC

FC

P k NC

Since FC 180.00 N > F Cmax 76.13 N then the crate slips

6MO = 0;

NC x  P§

e § d · · ¨ 2 2 ¸ a  P¨ 2 2 ¸ c © d e ¹ © d e ¹

x

0.39 m

§ 30.5 · ¨ ¸N © 152.3 ¹

Ans.

0

e a  d c · ¨ 2 2¸ © NC d  e ¹

P§

b 2 Then the block does not tip. Since x

§ FC · ¨ ¸ © NC ¹


0 then W - P > 0  W > P

Also, P > 0. Thus

0 6 P 6 W



Choosing the smaller value of ms,

ms =

(P + W) - 2(W + 7P)(W - P) 2(2P - W)

for 0 6 P 6 W and P ≠

W  Ans. 2

1 1 W and P ≠ W, are continuous. 2 2 Note: Choosing the larger value of ms in the quadratic solution leads to NA, FA < 0, which is nonphysical. Also, (ms)max = 1. For ms > 1, the hoop will tend to climb the wall rather than rotate in place. The two solutions, for P =

Ans:

1 1 W , ms = 3 2 1 If P ≠ W 2 If P =

ms =

(P + W) - 2(W + 7P)(W - P)

for 0 6 P 6 W 783

2(2P - W)

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8–35. Determine the maximum horizontal force P that can be applied to the 12 kg hoop without causing it to rotate. The coefficient of static friction between the hoop and the surfaces A and B is ms = 0.2. Take r = 300 mm.

P

r

B B

A

Solution Equations of Equilibrium. Referring to the FBD of the hoop shown in Fig. a, +

S ΣFx = 0;

P + FA - NB = 0

(1)

+ c ΣFy = 0;

NA + FB - 12(9.81) = 0

(2)

a+ ΣMA = 0;

FB(0.3) + NB(0.3) - P(0.6) = 0

(3)

Friction. Assuming that the hoop is on the verge to rotate due to the slipping occur at A and B. Then FA = ms NA = 0.2 NA

(4)

FB = ms NB = 0.2 NB

(5)

Solving Eq. (1) to (5) NB = 53.51 N NA = 107.02 N FA = 21.40 N FB = 10.70 N Ans.

P = 32.11 N = 32.1 N

Since NA is positive, the hoop will be in contact with the floor. Thus, the assumption was correct.

Ans: P = 32.1 N 784

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*8–36. If u = 30° determine the minimum coefficient of static friction at A and B so that equilibrium of the supporting frame is maintained regardless of the mass of the cylinder C. Neglect the mass of the rods.

C

u

L

A

u

L

B

SOLUTION Free-Body Diagram: Due to the symmetrical loading and system, ends A and B of the rod will slip simultaneously. Since end B tends to move to the right, the friction force FB must act to the left as indicated on the free-body diagram shown in Fig. a. Equations of Equilibrium: We have + ©Fx = 0; :

FBC sin 30° - FB = 0

FB = 0.5FBC

+ c ©Fy = 0;

NB - FBC cos 30° = 0

NB = 0.8660 FBC

Therefore, to prevent slipping the coefficient of static friction ends A and B must be at least ms =

FB 0 .5FBC = = 0 .577 NB 0 .8660FBC

Ans.

Ans: ms = 0.577 785

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8–37. If the coefficient of static friction at A and B is ms = 0.6, determine the maximum angle u so that the frame remains in equilibrium, regardless of the mass of the cylinder. Neglect the mass of the rods.

C

u

L

A

u

L

B

SOLUTION Free-Body Diagram: Due to the symmetrical loading and system, ends A and B of the rod will slip simultaneously. Since end B is on the verge of sliding to the right, the friction force FB must act to the left such that FB = msNB = 0.6NB as indicated on the free-body diagram shown in Fig. a. Equations of Equilibrium: We have + c ©Fy = 0;

NB - FBC cos u = 0

+ ©Fx = 0; :

FBC sin u - 0.6(FBC cos u) = 0

NB = FBC cos u

tan u = 0 .6 Ans.

u = 31 .0°

Ans: u = 31.0° 786

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8–38. The 100-kg disk rests on a surface for which mB = 0.2. Determine the smallest vertical force P that can be applied tangentially to the disk which will cause motion to impend.

P

A 0.5 m B

Solution Equations of Equilibrium. Referring to the FBD of the disk shown in Fig. a,  

 + c ΣFy = 0;    NB - P - 100(9.81) = 0

(1)

a + ΣMA = 0;  P(0.5) - FB(1) = 0

(2)

Friction. It is required that slipping impends at B.Thus,

(3)

FB = mB NB = 0.2 NB

Solving Eqs. (1), (2) and (3)

P = 654 N



NB = 1635 N  FB = 327 N

Ans.

Ans: P = 654 N 787

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8–39. The coefficients of static and kinetic friction between the drum and brake bar are ms = 0.4 and mk = 0.3, respectively. If M = 50 N # m and P = 85 N determine the horizontal and vertical components of reaction at the pin O. Neglect the weight and thickness of the brake. The drum has a mass of 25 kg.

300 mm

700 mm

B O

125 mm 500 mm

M P

A

SOLUTION Equations of Equilibrium: From FBD (b), a + ©MO = 0

50 - FB (0.125) = 0

FB = 400 N

From FBD (a), a + ©MA = 0;

85(1.00) + 400(0.5) - NB (0.7) = 0 NB = 407.14 N

Friction: Since FB 7 (FB)max = msNB = 0.4(407.14) = 162.86 N, the drum slips at point B and rotates. Therefore, the coefficient of kinetic friction should be used. Thus, FB = mkNB = 0.3NB. Equations of Equilibrium: From FBD (b), a + ©MA = 0;

85(1.00) + 0.3NB (0.5) - NB (0.7) = 0 NB = 154.54 N

From FBD (b), + c ©Fy = 0;

Oy - 245.25 - 154.54 = 0

+ ©F = 0; : x

0.3(154.54) - Ox = 0

Ans.

Oy = 400 N

Ans.

Ox = 46.4 N

Ans: Oy = 400 N

Ox = 46.4 N

788

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*8–40. The coefficient of static friction between the drum and brake bar is ms = 0.4. If the moment M = 35 N # m, determine the smallest force P that needs to be applied to the brake bar in order to prevent the drum from rotating. Also determine the corresponding horizontal and vertical components of reaction at pin O. Neglect the weight and thickness of the brake bar. The drum has a mass of 25 kg.

300 mm

700 mm

B O

125 mm 500 mm

M P

A

SOLUTION Equations of Equilibrium: From FBD (b), a + ©MO = 0

35 - FB (0.125) = 0

FB = 280 N

From FBD (a), a + ©MA = 0;

P(1.00) + 280(0.5) - NB (0.7) = 0

Friction: When the drum is on the verge of rotating, FB = msNB 280 = 0.4NB NB = 700 N Substituting NB = 700 N into Eq. [1] yields Ans.

P = 350 N Equations of Equilibrium: From FBD (b), + c ©Fy = 0;

Oy - 245.25 - 700 = 0

+ ©F = 0; : x

280 - Ox = 0

Ans.

Oy = 945 N

Ans.

Ox = 280 N

Ans: P = 350 N Oy = 945 N Ox = 280 N 789

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8–41. Determine the minimum force P needed to push the tube E up the incline. The force acts parallel to the plane, and the coefficients of static friction at the contacting surfaces are mA = 0.2, mB = 0.3, and mC = 0.4. The 100-kg roller and 40-kg tube each have a radius of 150 mm.

E A

P

B

30

C

Solution Equations of Equilibrium. Referring to the FBD of the roller, Fig. a, + ΣFx = 0;  P - NA cos 30° - FA sin 30° - FC = 0      S

(1)

  + c ΣFy = 0;  NC + FA cos 30° - NA sin 30° - 100(9.81) = 0

(2)

a + ΣMD = 0;  FA(0.15) - FC (0.15) = 0

(3)

Also, for the FBD of the tube, Fig. b,     +QΣFx = 0;  NA - FB - 40(9.81) sin 30° = 0

(4)

+a ΣFy = 0;  NB - FA - 40(9.81) cos 30° = 0

(5)

 a + ΣME = 0;  FA(0.15) - FB(0.15) = 0

(6)

Friction. Assuming that slipping is about to occur at A. Thus

(7)

FA = mA NA = 0.2 NA

Solving Eqs. (1) to (7)

Ans.

P = 285.97 N = 286 N

NA = 245.25 N  NB = 388.88 N  NC = 1061.15 N  FA = FB = FC = 49.05 N Since FB 6 (FB)max = mB NB = 0.3(388.88) = 116.66 N and FC < (FC)max = mC NC = 0.4(1061.15) = 424.46 N, slipping indeed will not occur at B and C. Thus, the assumption was correct.

Ans: 286 N 790

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8–42. Investigate whether the equilibrium can be maintained. The uniform block has a mass of 500 kg, and the coefficient of static friction is ms = 0.3. 5

4

200 mm

3

A B

600 mm 800 mm

Solution Equations of Equilibrium. The block would move only if it slips at corner O. Referring to the FBD of the block shown in Fig. a, 4 a + ΣMO = 0;  T a b(0.6) - 500(9.81)(0.4) = 0  T = 4087.5 N 5 + ΣFx = 0;  N - 4087.5a 3 b = 0  N = 2452.5 N S 5

4 + c ΣFy = 0;  F + 4087.5a b - 500(9.81) = 0  F = 1635 N 5

Friction. Since F 7 (F)max = ms N = 0.3(2452.5) = 735.75 N, slipping occurs at O. Thus, the block fails to be in equilibrium.

Ans: The block fails to be in equilibrium. 791

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8–43. The uniform rod has a mass of 10 kg and rests on the inside of the smooth ring at B and on the ground at A. If the rod is on the verge of slipping, determine the coefficient of static friction between the rod and the ground.

C B 0.5 m

0.2 m

30⬚ A

SOLUTION a + ©mA = 0;

Ans.

NB(0.4) - 98.1(0.25 cos 30°) = 0 NB = 53.10 N

+ c ©Fy = 0;

NA - 98.1 + 53.10 cos 30° = 0 NA = 52.12 N

+ ©Fx = 0; :

m(52.12) - 53.10 sin 30° = 0 m = 0.509

Ans: m = 0.509 792

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*8–44. If the coefficient of static friction between the collars A and B and the rod is ms = 0.6, determine the maximum angle u for the system to remain in equilibrium, regardless of the weight of cylinder D. Links AC and BC have negligible weight and are connected together at C by a pin.

A

B 15

15 u

u

C

D

Solution Free-Body Diagram. Due to the symmetrical loading and system, collars A and B will slip simultaneously. Thus, it is sufficient to consider the equilibrium of either collar. Here, the equilibrium of collar B will be considered. Since collar B is required to be on the verge of sliding down the rod the friction force FB must act up the rod such that FB = msNB = 0.6NB as indicated on the free-body diagram of the colar shown in Fig. a. Equations of Equilibrium. +

S ΣFy = 0; NB - FBC sin175° - u 2 = 0 +

NB = FBC sin175° - u 2

S ΣFx = 0; 0.6[FBC sin175° - u 2] - FBC cos 175° - u 2 = 0 tan175° - u 2 = 1.6667 u = 16.0°

Ans.

Ans: u = 16.0° 793

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8–45. The beam AB has a negligible mass and thickness and is subjected to a triangular distributed loading. It is supported at one end by a pin and at the other end by a post having a mass of 50 kg and negligible thickness. Determine the two coefficients of static friction at B and at C so that when the magnitude of the applied force is increased to P = 150 N, the post slips at both B and C simultaneously.

800 N/m A

B 2m

P 5

400 mm

4

3

300 mm C

SOLUTION Member AB: a + ©MA = 0;

4 - 800a b + NB (2) = 0 3 NB = 533.3 N

Post: + c ©Fy = 0;

3 NC - 533.3 + 150 a b - 50(9.81) = 0 5 NC = 933.83 N

a + ©MC = 0;

4 - (150)(0.3) + FB (0.7) = 0 5 FB = 51.429 N

+ ©F = 0; : x

4 (150) - FC - 51.429 = 0 5 FC = 68.571 N

mC =

FC 68.571 = 0.0734 = NC 933.83

Ans.

mB =

FB 51.429 = 0.0964 = NB 533.3

Ans.

Ans: mC = 0.0734 mB = 0.0964 794

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8–46. The uniform crate has a mass of 150 kg. If the coefficient of static friction between the crate and the floor is ms = 0.2, determine whether the 85-kg man can move the crate. The coefficient of static friction between his shoes and the floor is m′s = 0.4. Assume the man only exerts a horizontal force on the crate.

2.4 m 1.6 m

1.2 m

Solution Equations of Equilibrium. Referring to the FBD of the crate shown in Fig. a, + ΣFx = 0;  P - FC = 0 S

(1)

  + c ΣFy = 0;  NC - 150(9.81) = 0  NC = 1471.5 N a + ΣMO = 0;  150(9.81)x - P(1.6) = 0

(2)

Also, from the FBD of the man, Fig. b,  + c ΣFy = 0;  Nm - 85(9.81) = 0  Nm = 833.85 N + ΣFx = 0;  Fm - P = 0 S

(3)

Friction. Assuming that the crate slips before tipping. Then

FC = ms NC = 0.2(1471.5) = 294.3 N

Solving Eqs. (1) to (3) using this result,

Fm = P = 294.3 N  x = 0.32 m

Since x < 0.6 m, the crate indeed slips before tipping as assumed. Also since  Fm 6 (Fm)max = ms′ Nm = 0.4(833.85) = 333.54 N, the man will not slip. Therefore, he is able to move the crate.

Ans: He is able to move the crate. 795

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8–47. The uniform crate has a mass of 150 kg. If the coefficient of static friction between the crate and the floor is ms = 0.2, determine the smallest mass of the man so he can move the crate. The coefficient of static friction between his shoes and the floor is m′s = 0.45. Assume the man exerts only a horizontal force on the crate.

2.4 m 1.6 m

1.2 m

Solution Equations of Equilibrium. Referring to the FBD of the crate shown in Fig. a, + ΣFx = 0;  P - FC = 0   S

(1)

  + c ΣFy = 0;  NC - 150(9.81) = 0  NC = 1471.5 N a + ΣMO = 0;  150(9.81)x - P(1.6) = 0

(2)

Also, from the FBD of the man, Fig. b,  + c ΣFy = 0;  Nm - m(9.81) = 0  Nm = 9.81 m

(3)

+ ΣFx = 0;  Fm - P = 0   S

(4)

Friction. Assuming that the crate slips before tipping. Then

FC = ms NC = 0.2(1471.5) = 294.3 N

Also, it is required that the man is on the verge of slipping. Then

Fm = ms′ Nm = 0.45 Nm

(5)

Solving Eqs. (1) to (5) using the result of FC,

Fm = P = 294.3 N  x = 0.32 m  Nm = 654 N



m = 66.667 kg = 66.7 kg

Ans.

Since x < 0.6 m, the crate indeed slips before tipping as assumed.

Ans: m = 66.7 kg 796

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*8–48. Two blocks A and B, each having a mass of 5 kg, are connected by the linkage shown. If the coefficient of static friction at the contacting surfaces is ms = 0.5, determine the largest force P that can be applied to pin C of the linkage without causing the blocks to move. Neglect the weight of the links.

P 30 B

C 30 30

A

Solution Equations of Equilibrium. Analyze the equilibrium of Joint C Fig. a, + c ΣFy = 0;  FAC sin 30° - P cos 30° = 0  FAC = 23 P

+ ΣFx = 0;  FBC - P sin 30° - ( 23P) cos 30° = 0  FBC = 2 P S

Referring to the FBD of block B, Fig. b

+ a ΣFx = 0;  2 P cos 30° - FB - 5(9.81) sin 30° = 0

(1)

+Q ΣFy = 0;  NB - 2 P sin 30° - 5(9.81) cos 30° = 0

(2)

Also, the FBD of block A, Fig. C + ΣFx = 0;   23P cos 30° - FA = 0 S

(3)

+ c ΣFy = 0;  NA - 23P sin 30° - 5(9.81) = 0

(4)

Friction. Assuming that block A slides first. Then

(5)

FA = ms NA = 0.5 NA

Solving Eqs. (1) to (5)

Ans.

P = 22.99 N = 23.0 N NA = 68.96 N  FA = 34.48 N  FB = 15.29 N  NB = 65.46 N

Since FB 6 (FB)max = ms NB = 0.5(65.46) = 32.73 N, Block B will not slide. Thus, the assumption was correct.

Ans: 23.0 N 797

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8–49. The beam AB has a negligible mass and thickness and is subjected to a triangular distributed loading. It is supported at one end by a pin and at the other end by a post having a mass of 50 kg and negligible thickness. Determine the minimum force P needed to move the post.The coefficients of static friction at B and C are mB = 0.4 and mC = 0.2, respectively.

800 N/m A

B 2m

P 5

400 mm

4

3

300 mm C

SOLUTION Member AB: a + ©MA = 0;

4 - 800 a b + NB (2) = 0 3 NB = 533.3 N

Post: Assume slipping occurs at C; FC = 0.2 NC a + ©MC = 0;

4 - P(0.3) + FB(0.7) = 0 5

+ ©F = 0; : x

4 P - FB - 0.2NC = 0 5

+ c ©Fy = 0;

3 P + NC - 533.3 - 50(9.81) = 0 5 Ans.

P = 355 N NC = 811.0 N FB = 121.6 N

(O.K.!)

(FB)max = 0.4(533.3) = 213.3 N 7 121.6 N

Ans: P = 355 N 798

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8–50. The homogenous semicylinder has a mass of 20 kg and mass center at G. If force P is applied at the edge, and r = 300 mm, determine the angle u at which the semicylinder is on the verge of slipping. The coefficient of static friction between the plane and the cylinder is ms = 0.3. Also, what is the corresponding force P for this case?

4r 3p P u r

G

Solution Equations of Equilibrium. Referring to the FBD of the semicylinder shown in Fig. a, + ΣFx = 0;  P sin u - F = 0  S

(1)

 + c ΣFy = 0;  N - P cos u - 20(9.81) = 0

(2)

a+ ΣMA = 0;  P[0.3(1 - sin u)] - 20(9.81) c

P =

4(0.3)

sin u 261.6 a b p 1 - sin u

3p

sin u d = 0 (3)

Friction. Since the semicylinder is required to be on the verge to slip at point A,

(4)

F = ms N = 0.3 N

Substitute Eq. (4) into (1),

(5)

P sin u - 0.3 N = 0

Eliminate N from Eqs. (2) and (5), we obtain

P =

58.86  sin u - 0.3 cos u

(6)

Equating Eq. (3) and (6)

261.6 sin u 58.56 a b = p 1 - sin u sin u - 0.3 cos u

sin u(sin u - 0.3 cos u + 0.225 p) - 0.225 p = 0

Solving by trial and error

Ans.

u = 39.50° = 39.5°

Substitute the result into Eq. 6

P =

58.86 sin 39.50° - 0.3 cos 39.50°



= 145.51 N



= 146 N

Ans.

Ans: 146 N 799

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8–51. The coefficient of static friction between the shoes at A and B of the tongs and the pallet is ms = 0.5, and between the pallet and floor ms = 0.4. If a horizontal towing force of P = 300 N is applied to the tongs, determine the largest mass that can be towed.

75 mm

75 mm

60°

A

P 60°

20 mm

C B

30 mm

Solution P

Chain: + c ΣFy = 0;

2T sin 60° - 300 = 0 T = 173.2 N

Tongs: a+ ΣMc = 0; F = mN;

- 173.2 cos 60° 1752 - 173.2 sin 60° 1502 + NA 1752 - FA 1202 = 0 FA = 0.5NA FA = 107.7 N

Crate: +

S ΣFx = 0; F = mN; + c ΣFy = 0;

F = 21107.72 = 215.3 N F = 0.4 N N = 538.3 N W = 538.3 N 538.3 m = = 54.9 kg 9.81

Ans.

Ans: m = 54.9 kg 800

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*8–52. Beam AB has a negligible mass and thickness, and supports the 200-kg uniform block. It is pinned at A and rests on the top of a post, having a mass of 20 kg and negligible thickness. Determine the minimum force P needed to move the post. The coefficients of static friction at B and C are mB = 0.4 and mC = 0.2, respectively.

A

P

B 1.5 m

5

1.5 m 1m 0.75 m

4

3

C

Solution Equations of Equilibrium. Referring to the FBD of member AB shown in Fig. a, a + ΣMA = 0;  NB (3) - 200(9.81)(1.5) = 0  NB = 981 N Then consider the FBD of member BC shown in Fig. b, 3 + c ΣFy = 0;  NC + P a b - 981 - 20(9.81) = 0 5

(1)

4 a + ΣMC = 0;  FB (1.75) - P a b(0.75) = 0 5

(2)

4 a + ΣMB = 0;  P a b(1) - FC (1.75) = 0 5

(3)

Friction. Assuming that slipping occurs at C. Then

(4)

FC = mC NC = 0.2 NC

Solving Eqs. (1) to (4)

Ans.

P = 407.94 N = 408 N NC = 932.44 N  FC = 186.49 N  FB = 139.87

Since FB 6 (FB)max = mB NB = 0.4(981) N = 392.4. Indeed slipping will not occur at B. Thus, the assumption is correct.

Ans: P = 408 N 801

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8–53. Beam AB has a negligible mass and thickness, and supports the 200-kg uniform block. It is pinned at A and rests on the top of a post, having a mass of 20 kg and negligible thickness. Determine the two coefficients of static friction at B and at C so that when the magnitude of the applied force is increased to P = 300 N, the post slips at both B and C simultaneously.

A

P

B 1.5 m

5

1.5 m 1m 0.75 m

4

3

C

Solution Equations of Equilibrium. Referring to the FBD of member AB shown in Fig. a, a+ ΣMA = 0;  NB(3) - 200(9.81)(1.5) = 0 NB = 981 N Then consider the FBD of member BC shown in Fig. b, 3   + c ΣFy = 0;  NC + 300 a b - 981 - 20(9.81) = 0 NC = 997.2 N 5

4 a + ΣMC = 0;  FB (1.75) - 300 a b(0.75) = 0      FB = 102.86 N 5

4 a + ΣMB = 0;  300 a b(1) - FC (1.75) = 0 5

    

 

FC = 137.14 N

Friction. It is required that slipping occurs at B and simultaneously. Then   mB = 0.1048 = 0.105

Ans.

FC = mC NC;  137.14 = mC (997.2)  mC = 0.1375 = 0.138

Ans.

FB = mB NB;  102.86 = mB (981) 

Ans: mB = 0.105 mC = 0.138 802

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8–54.

200 N

The beam AB has a negligible mass and thickness and is subjected to a force of 200 N. It is supported at one end by a pin and at the other end by a spool having a mass of 40 kg. If a cable is wrapped around the inner core of the spool, determine the minimum cable force P needed to move the spool. The coefficients of static friction at B and D are mB = 0.4 and mD = 0.2, respectively.

2m

1m

1m B

A 0.1 m

0.3 m D

P

SOLUTION Equations of Equilibrium: From FBD (a), a + ©MA = 0;

NB 132 - 200122 = 0

NB = 133.33 N

From FBD (b), + c ©Fy = 0

ND - 133.33 - 392.4 = 0

+ ©F = 0; : x

P - FB - FD = 0

(1)

a + ©MD = 0;

FB 10.42 - P10.22 = 0

(2)=

ND = 525.73 N

Friction: Assuming the spool slips at point B, then FB = ms BNB = 0.41133.332 53.33 N. Substituting this value into Eqs. (1) and (2) and solving, we have FD = 53.33 N Ans.

P = 106.67 N = 107 N

Since FD max = ms DND = 0.2 525.73 = 105.15 N 7 FD , the spool does not slip at point D. Therefore the above assumption is correct.

Ans: P = 107 N 803

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8–55. The friction hook is made from a fixed frame which is shown colored and a cylinder of negligible weight. A piece of paper is placed between the smooth wall and the cylinder. If u = 20°, determine the smallest coefficient of static friction m at all points of contact so that any weight W of paper p can be held.

u p

W

SOLUTION Paper: + c ©Fy = 0;

F = 0.5W

F = mN;

F = mN N =

0.5W m

Cylinder: a + ©MO = 0;

F = 0.5W

+ ©F = 0; : x

0.5W m = 0.176 = 0 N cos 20° + F sin 20° m

+ c ©Fy = 0;

N sin 20° - F cos 20° - 0.5 W = 0

F = mN;

m2 sin 20° + 2m cos 20° - sin 20° = 0 m = 0.176

Ans.

Ans.

Ans: m = 0.176 804

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*8–56. Determine the greatest angle X so that the ladder does not slip when it supports the 75-kg man in the position shown. The surface is rather slippery, where the coefficient of static friction at A and B is ms = 0.3.

C

0.25 m

G 2.5 m

u

2.5 m

A

B

SOLUTION Free-Body Diagram: The slipping could occur at either end A or B of the ladder. We will assume that slipping occurs at end B. Thus, FB = msNB = 0.3NB . Equations of Equilibrium: Referring to the free-body diagram shown in Fig. b, we have + ©Fx = 0; :

FBC sin u>2 - 0.3NB = 0 (1)

FBC sin u>2 = 0.3NB + c ©Fy = 0;

NB - FBC cos u>2 = 0 FBC cos u>2 = NB(2)

Dividing Eq. (1) by Eq. (2) yields tan u>2 = 0.3 Ans.

u = 33.40° = 33.4°

Using this result and referring to the free-body diagram of member AC shown in Fig. a, we have a + ©MA = 0;

FBC sin 33.40°(2.5) - 75(9.81)(0.25) = 0

FBC = 133.66 N

+ ©Fx = 0; :

FA - 133.66 sin ¢

FA = 38.40 N

+ c ©Fy = 0;

NA + 133.66 cos ¢

33.40° ≤ = 0 2 33.40° ≤ - 75(9.81) = 0 2

NA = 607.73 N

Since FA 6 (FA) max = msNA = 0.3(607.73) = 182.32 N, end A will not slip. Thus, the above assumption is correct.

Ans: u = 33.4° 805

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8–57. The disk has a weight W and lies on a plane which has a coefficient of static friction m. Determine the maximum height h to which the plane can be lifted without causing the disk to slip.

z

h a y 2a

x

SOLUTION Unit Vector: The unit vector perpendicular to the inclined plane can be determined using cross product. A = (0 - 0)i + (0 - a)j + (h - 0)k = - aj + hk B = (2a - 0)i + (0 - a)j + (0 - 0)k = 2ai - aj Then i 3 N = A * B = 0 2a n =

j -a -a

k h 3 = ahi + 2ahj + 2a2k 0

ahi + 2ahj + 2a2k N = N a25h2 + 4a2

Thus cos g =

2a

sin g =

hence

25h2 + 4a2

25h 25h2 + 4a2

Equations of Equilibrium and Friction: When the disk is on the verge of sliding down the plane, F = mN. ©Fn = 0;

N - W cos g = 0

N = W cos g

(1)

©Ft = 0;

W sin g - mN = 0

N =

W sin g m

(2)

Divide Eq. (2) by (1) yields sin g = 1 m cos g 15h 15h2 + 4a2

ma 2

2a 2

2

5h + 4a

h =

2 25

b

= 1

Ans.

am

Ans: h =

806

2 25

am

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8–58. If P = 250 N, determine the required minimum compression in the spring so that the wedge will not move to the right. Neglect the weight of A and B. The coefficient of static friction for all contacting surfaces is ms = 0.35. Neglect friction at the rollers.

k  15 kN/m B

P

A

10

SOLUTION Free-Body Diagram: The spring force acting on the cylinder is Fsp = kx = 15(103)x. Since it is required that the wedge is on the verge to slide to the right, the frictional force must act to the left on the top and bottom surfaces of the wedge and their magnitude can be determined using friction formula. (Ff)1 = mN1 = 0.35N1

(Ff)2 = 0.35N2

Equations of Equilibrium: Referring to the FBD of the cylinder, Fig. a, + c ©Fy = 0;

N1 - 15(103)x = 0

N1 = 15(103)x

Thus, (Ff)1 = 0.35315(103)x4 = 5.25(103)x Referring to the FBD of the wedge shown in Fig. b, + c ©Fy = 0;

N2 cos 10° - 0.35N2 sin 10° - 15(103)x = 0 N2 = 16.233(103)x

+ ©Fx = 0; :

250 - 5.25(103)x - 0.35316.233(103)x4cos 10° - 316.233(103)x4sin 10° = 0 Ans.

x = 0.01830 m = 18.3 mm

Ans: x = 18.3 mm 807

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8–59. Determine the minimum applied force P required to move wedge A to the right. The spring is compressed a distance of 175 mm. Neglect the weight of A and B. The coefficient of static friction for all contacting surfaces is ms = 0.35. Neglect friction at the rollers.

k = 15 kN/m B

P

A 10°

SOLUTION Equations of Equilibrium and Friction: Using the spring formula, Fsp = kx = 1510.1752 = 2.625 kN. If the wedge is on the verge of moving to the right, then slipping will have to occur at both contact surfaces. Thus, FA = msNA = 0.35NA and FB = msNB = 0.35NB. From FBD (a), + c ©Fy = 0;

NB - 2.625 = 0

NB = 2.625 kN

From FBD (b), + c ©Fy = 0;

NA cos 10° - 0.35NA sin 10° - 2.625 = 0 NA = 2.841 kN

+ ©F = 0; : x

P - 0.3512.6252 - 0.3512.8412 cos 10° - 2.841 sin 10° = 0 Ans.

P = 2.39 kN

Ans: P = 2.39 kN 808

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*8–60.

k

The coefficient of static friction between wedges B and C is ms = 0.6 and between the surfaces of contact B and A and C and D, ms ¿ = 0.4. If the spring is compressed 200 mm when in the position shown, determine the smallest force P needed to move wedge C to the left. Neglect the weight of the wedges.

A

15 B

500 N/ m

15 C 15

P

D

SOLUTION Wedge B: + ©F = 0; : x

NAB - 0.6NBC cos 15° - NBC sin 15° = 0

+ c ©Fy = 0;

NBC cos 15° - 0.6NBC sin 15° - 0.4NAB - 100 = 0 NBC = 210.4 N NAB = 176.4 N

Wedge C: + c ©Fy = 0;

NCD cos 15° - 0.4NCD sin 15° + 0.6(210.4) sin 15° - 210.4 cos 15° = 0 NCD = 197.8 N

+ ©F = 0; : x

197.8 sin 15° + 0.4(197.8) cos 15° + 210.4 sin 15° + 0.6(210.4) cos 15° - P = 0 Ans.

P = 304 N

Ans: P = 304 N 809

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8–61. If the beam AD is loaded as shown, determine the horizontal force P which must be applied to the wedge in order to remove it from under the beam.The coefficients of static friction at the wedge’s top and bottom surfaces are mCA = 0.25 and mCB = 0.35, respectively. If P = 0, is the wedge self-locking? Neglect the weight and size of the wedge and the thickness of the beam.

4 kN/m

A

D

10°

C B 3m

4m

SOLUTION Equations of Equilibrium and Friction: If the wedge is on the verge of moving to the right, then slipping will have to occur at both contact surfaces. Thus, FA = ms A NA = 0.25NA and FB = ms B NB = 0.35NB . From FBD (a), a + ©MD = 0;

NA cos 10°172 + 0.25NA sin 10°172 - 6.00122 - 16.0152 = 0 NA = 12.78 kN

From FBD (b), + c ©Fy = 0;

NB - 12.78 sin 80° - 0.25112.782 sin 10° = 0 NB = 13.14 kN

+ ©F = 0; : x

P + 12.78 cos 80° - 0.25112.782 cos 10° - 0.35113.142 = 0 Ans.

P = 5.53 kN

Since a force P 7 0 is required to pull out the wedge, the wedge will be self-locking when P = 0. Ans.

Ans: P = 5.53 kN yes 810

P

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8–62. The wedge is used to level the member. Determine the horizontal force P that must be applied to begin to push the wedge forward. The coefficient of static friction between the wedge and the two surfaces of contact is ms = 0.2. Neglect the weight of the wedge.

2m 500 N/m

P

A B 5

1m C

Solution Equations of Equilibrium and Friction. Since the wedge is required to be on the verge to slide to the right, then slipping will have to occur at both of its contact surfaces. Thus, FA = ms NA = 0.2 NA and FB = ms NB. Referring to the FBD ­diagram of member AC shown in Fig. a a+ ΣMC = 0;  500(2)(1) - NA cos 5°(2) - NA sin 5°(1)

-0.2 NA cos 5°(1) + 0.2 NA sin 5°(2) = 0



NA = 445.65 N

Using this result and the FBD of the wedge, Fig. b,  + c ΣFy = 0;  NB - 445.65 cos 5° + 0.2(445.65) sin 5° = 0

NB = 436.18 N + ΣFx = 0;    P - 0.2(445.65) cos 5° - 445.65 sin 5° - 0.2(436.18) = 0 S



Ans.

P = 214.87 N = 215 N

Ans: P = 215 N 811

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8–63. Determine the largest angle X that will cause the wedge to be self-locking regardless of the magnitude of horizontal force P applied to the blocks. The coefficient of static friction between the wedge and the blocks is ms = 0.3. Neglect the weight of the wedge.

P

u

P

SOLUTION Free-Body Diagram: For the wedge to be self-locking, the frictional force F indicated on the free-body diagram of the wedge shown in Fig. a must act downward and its magnitude must be F … msN = 0 .3N . Equations of Equilibrium: Referring to Fig. a, we have + c ©Fy = 0;

2N sin u>2 - 2F cos u>2 = 0 F = N tan u>2

Using the requirement F … 0 .3N, we obtain N tan u>2 … 0 .3N Ans.

u = 33 .4°

Ans: u = 33.4° 812

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*8–64. If the coefficient of static friction between all the surfaces of contact is ms, determine the force P that must be applied to the wedge in order to lift the block having a weight W.

B

A a

P

C

Solution Equations of Equilibrium and Friction. Since the wedge is required to be on the verge sliding to the left, then slipping will have to occur at both of its contact ­surfaces. Thus, FA = ms NA, FB = ms NB and FC = ms NC. Referring to the FBD of the wedge shown in Fig. a. + ΣFx = 0;  m s NC + m s NA cos a + NA sin a - P = 0 S

(1)

+ c ΣFy = 0;  NC + ms NA sin a - NA cos a = 0

(2)

Also, from the FBD of the block, Fig. b + ΣFx = 0;  NB - NA sin a - ms NA cos a = 0  S

(3)

+ c ΣFy = 0;  NA cos a - ms NA sin a - ms NB - W = 0

(4)

Solving Eqs. (1) to (4) W



NA =



NC = c



cos a ( 1 -

) - 2ms sin a

ms2

cos a + ms sin a

  NB = c

cos a ( 1 - ms2) - 2ms sin a P = c

sin a + ms cos a cos a ( 1 - ms2) - 2ms sin a

d W

dW

2ms cos a + sin a ( 1 - ms2) cos a ( 1 - ms2) - 2ms sin a

Ans.

d W

Ans: P = c 813

2ms cos a + sin a ( 1 - ms2) cos a ( 1 - ms2) - 2ms sin a

dW

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8–65. The column is used to support the upper floor. If a force F = 80 N is applied perpendicular to the handle to tighten the screw, determine the compressive force in the column. The square-threaded screw on the jack has a coefficient of static friction of ms = 0.4, mean diameter of 25 mm, and a lead of 3 mm.

0.5 m F

SOLUTION M = W1r2 tan1fs + up2 fs = tan-110.42 = 21.80° up = tan-1 c

3 d = 2.188° 2p112.52

8010.52 = W10.01252 tan121.80° + 2.188°2 W = 7.19 kN

Ans.

Ans: W = 7.19 kN 814

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8–66. If the force F is removed from the handle of the jack in Prob. 8–65, determine if the screw is self-locking.

0.5 m F

SOLUTION fs = tan-110.42 = 21.80° up = tan-1 c

3 d = 2.188° 2p112.52 Ans.

Since fs 7 up , the screw is self locking.

Ans: The screw is self-locking. 815

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8–67. A turnbuckle, similar to that shown in Fig. 8–17, is used to tension member AB of the truss. The coefficient of the static friction between the square threaded screws and the turnbuckle is ms = 0.5. The screws have a mean radius of 6 mm and a lead of 3 mm. If a torque of M = 10 N # m is applied to the turnbuckle, to draw the screws closer together, determine the force in each member of the truss. No external forces act on the truss.

D

B

4m M

SOLUTION

l 3 b = tan-1 c d = 4.550°, 2pr 2p162 -1 -1 M = 10 N # m and fs = tan ms = tan 10.52 = 26.565°. Since friction at two screws must be overcome, then, W = 2FAB . Applying Eq. 8–3, we have Frictional Forces on Screw: Here, u = tan-1 a

C

A 3m

M = Wr tan 1u + fS2 10 = 2FAB10.0062 tan 14.550° + 26.565°2 Ans.

FAB = 1380.62 N 1T2 = 1.38 kN 1T2

Note: Since fs 7 u, the screw is self-locking. It will not unscrew even if moment M is removed. Method of Joints: Joint B: + ©F = 0; : x

3 1380.62 a b - FBD = 0 5 Ans.

FBD = 828.37 N1C2 = 828 N 1C2 + c ©Fy = 0;

4 FBC - 1380.62 a b = 0 5 Ans.

FBC = 1104.50 N 1C2 = 1.10 kN 1C2 Joint A: + ©F = 0; : x

3 FAC - 1380.62 a b = 0 5 Ans.

FAC = 828.37 N 1C2 = 828 N 1C2 + c ©Fy = 0;

4 1380.62 a b - FAD = 0 5 Ans.

FAD = 1104.50 N 1C2 = 1.10 kN 1C2 Joint C: + ©F = 0; : x

3 FCD a b - 828.37 = 0 5 Ans.

FCD = 1380.62 N 1T2 = 1.38 kN 1T2 + c ©Fy = 0;

4 Cy + 1380.62 a b - 1104.50 = 0 5

Ans: FAB = 1.38 kN (T), FBD = 828 N (C) FBC = 1.10 kN (C), FAC = 828 N (C) FAD = 1.10 kN (C), FCD = 1.38 kN (T)

Cy = 0 (No external applied load. check!)

816

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*8–68. A turnbuckle, similar to that shown in Fig. 8–17, is used to tension member AB of the truss. The coefficient of the static friction between the square-threaded screws and the turnbuckle is ms = 0.5. The screws have a mean radius of 6 mm and a lead of 3 mm. Determine the torque M which must be applied to the turnbuckle to draw the screws closer together, so that the compressive force of 500 N is developed in member BC.

D

B

4m M

C

A 3m

SOLUTION Method of Joints: Joint B: + c ©Fy = 0;

4 500 - FAB a b = 0 5

FAB = 625 N 1C2

l 3 b = tan-1 c d = 4.550° 2pr 2p162 and fs = tan-1ms = tan-110.52 = 26.565°. Since friction at two screws must be overcome, then, W = 2FAB = 216252 = 1250 N. Applying Eq. 8–3, we have

Frictional Forces on Screws: Here, u = tan-1 a

M = Wr tan1u + f2 = 125010.0062 tan14.550° + 26.565°2 = 4.53 N # m

Ans.

Note: Since fs 7 u, the screw is self-locking. It will not unscrew even if moment M is removed.

Ans: M = 4.53 N # m 817

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8–69. If the clamping force at G is 900 N, determine the horizontal force F that must be applied perpendicular to the handle of the lever at E. The mean diameter and lead of both single square-threaded screws at C and D are 25 mm and 5 mm, respectively. The coefficient of static friction is ms = 0.3.

200 mm G

200 mm A

B

C

D E 125 mm

SOLUTION Referring to the free-body diagram of member GAC shown in Fig. a, we have FCD = 900 N ©MA = 0; FCD(0.2) - 900(0.2) = 0 L b = Since the screw is being tightened, Eq. 8–3 should be used. Here, u = tan-1 a 2pr 5 -1 d = 3.643°; tan c 2p(12.5) fs = tan - 1 ms = tan - 1(0.3) = 16.699° and M = F(0.125). Since M must overcome the friction of two screws, M = 2[Wr tan(fs + u)] F(0.125) = 2 [900(0.0125)tan(16.699° + 3.643°)] Ans.

F = 66 .7 N Note: Since fs 7 u, the screw is self-locking.

Ans: F = 66.7 N 818

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8–70. If a horizontal force of F = 50 N is applied perpendicular to the handle of the lever at E, determine the clamping force developed at G. The mean diameter and lead of the single square-threaded screw at C and D are 25 mm and 5 mm, respectively. The coefficient of static friction is ms = 0.3.

200 mm G

200 mm A

B

C

D E 125 mm

SOLUTION Since the screw is being tightened, Eq. 8–3 should be used. Here, u = tan-1 a tan-1 c

5 d = 3.643°; 2p(12.5)

L b = 2pr

fs = tan-1ms = tan-1(0.3) = 16.699°; and M = 50(0.125). Since M must overcome the friction of two screws, M = 2[Wr tan(fs + u)] 50(0.125) = 2[FCD(0.0125)tan(16.699° + 3.643°)] Ans.

FCD = 674.32 N

Using the result of FCD and referring to the free-body diagram of member GAC shown in Fig. a, we have ©MA = 0; 674.32(0.2) - FG(0.2) = 0 Ans.

FG = 674 N Note: Since fs 7 u, the screws are self-locking.

Ans: FCD = 674.32 N FG = 674 N 819

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8–71. Determine the clamping force on the board A if the screw of the “C” clamp is tightened with a twist of M = 8 N # m. The single square-threaded screw has a mean radius of 10 mm, a lead of 3 mm, and the coefficient of static friction is ms = 0.35.

M

A

SOLUTION fs = tan - 1(0.35) = 19.29° up = tan - 1 c

3 d = 2.734° 2p(10)

M = W(r) tan (fs + up) 8 = P(0.01) tan (19.29° + 2.734°) Ans.

P = 1978 N = 1.98 kN

Ans: P = 1.98 kN 820

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*8–72. If the required clamping force at the board A is to be 50 N, determine the torque M that must be applied to the handle of the “C” clamp to tighten it down. The single square-threaded screw has a mean radius of 10 mm, a lead of 3 mm, and the coefficient of static friction is ms = 0.35.

M

A

SOLUTION fs = tan - 1(0.35) = 19.29° up = tan - 1 a

3 l b = tan - 1 c d = 2.734° 2pr 2p(10)

M = W(r) tan (fs + up) = 50(0.01) tan (19.29° + 2.734°) = 0.202 N # m

Ans.

Ans:

M = 0.202 N # m 821

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8–73. If couple forces of F = 35 N are applied to the handle of the machinist’s vise, determine the compressive force developed in the block. Neglect friction at the bearing A. The guide at B is smooth. The single square-threaded screw has a mean radius of 6 mm and a lead of 8 mm, and the coefficient of static friction is ms = 0.27.

F 125 mm A

B

125 mm F

SOLUTION f = tan - 1 (0.27) = 15.11° u = tan - 1 a

8 b = 11.98° 2p(6)

M = Wr tan (u + f) 35 (0.250) = P (0.006) tan (11.98° + 15.11°) Ans.

P = 2851 N = 2.85 kN

Ans: P = 2.85 kN 822

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8–74. Prove that the lead l must be less than 2prms for the jack screw shown in Fig. 8–15 to be “self-locking.”

W

M

h

r

SOLUTION For self–locking, fs 7 uP or tan fs 7 tan up; ms 7

l ; 2p r

Q.E.D.

l 6 2prms

823

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8–75. The square-threaded bolt is used to join two plates together. If the bolt has a mean diameter of d = 20 mm and a lead of l = 3 mm, determine the smallest torque M required to loosen the bolt if the tension in the bolt is T = 40 kN. The coefficient of static friction between the threads and the bolt is ms = 0.15. M

d

SOLUTION f = tan-1 0.15 = 8.531° u = tan-1

3 = 2.734° 2p(10)

M = r W tan (f - u) = (0.01)(40 000) tan (8.531° - 2.734°) M = 40.6 N # m

Ans.

Ans: M = 40.6 N # m 824

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*8–76. The shaft has a square-threaded screw with a lead of 8 mm and a mean radius of 15 mm. If it is in contact with a plate gear having a mean radius of 30 mm, determine the resisting torque M on the plate gear which can be overcome if a torque of 7 N # m is applied to the shaft. The coefficient of static friction at the screw is mB = 0.2. Neglect friction of the bearings located at A and B.

15 mm

B

M 30 mm

A

7N·m

SOLUTION l 8 b = tan-1 c d = 4.852°, 2pr 2p1152 -1 -1 W = F, M = 7 N # m and fs = tan ms = tan 10.22 = 11.310°. Applying Eq. 8–3, we have Frictional Forces on Screw: Here, u = tan-1 a

M = Wr tan 1u + f2 7 = F10.0152 tan 14.852° + 11.310°2 F = 1610.29 N Note: Since fs 7 u, the screw is self-locking. It will not unscrew even if force F is removed. Equations of Equilibrium: a+©MO = 0;

1610.2910.032 - M = 0 M = 48.3 N # m

Ans.

Ans: M = 48.3 N # m 825

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8–77. 5 N·m

The braking mechanism consists of two pinned arms and a square-threaded screw with left and righthand threads. Thus when turned, the screw draws the two arms together. If the lead of the screw is 4 mm, the mean diameter 12 mm, and the coefficient of static friction is ms = 0.35, determine the tension in the screw when a torque of 5 N # m is applied to tighten the screw. If the coefficient of static friction between the brake pads A and B and the circular shaft is msœ = 0.5, determine the maximum torque M the brake can resist.

300 mm 200 mm M A B 300 mm C

D

SOLUTION l 4 b = tan-1 c d = 6.057°, 2pr 2p162 -1 -1 M = 5 N # m and fs = tan ms = tan 10.352 = 19.290°. Since friction at two screws must be overcome, then, W = 2P. Applying Eq. 8–3, we have Frictional Forces on Screw: Here, u = tan-1 a

M = Wr tan1u + f2 5 = 2P10.0062 tan16.057° + 19.290°2 Ans.

P = 879.61 N = 880 N

Note: Since fs 7 u, the screw is self-locking. It will not unscrew even if moment M is removed. Equations of Equilibrium and Friction: Since the shaft is on the verge to rotate about point O, then, FA = ms ¿NA = 0.5NA and FB = ms ¿NB = 0.5NB . From FBD (a), a + ©MD = 0;

879.61 10.62 - NB 10.32 = 0

NB = 1759.22 N

From FBD (b), a + ©MO = 0;

230.511759.222410.22 - M = 0

M = 352 N # m

Ans.

Ans: P = 880 N M = 352 N # m 826

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8–78. Determine the clamping force on the board A if the screw is tightened with a torque of M = 8 N # m. The squarethreaded screw has a mean radius of 10 mm and a lead of 3 mm, and the coefficient of static friction is ms = 0.35.

M

A

Solution Frictional Forces on Screw. Here u = tan - 1 a

l 3 b = tan - 1 c d = 2.7336°, 2pr 2p (10)

W = F and f s = tan - 1ms = tan - 1(0.35) = 19.2900°.



M = Wr tan (u + f s)



8 = F(0.01) tan (2.7336° + 19.2900°)



F = 1977.72 N = 1.98 kN

Ans.

Note: Since f s > u, the screw is “self-locking”. It will not unscrew even if the torque is removed.

Ans: F = 1.98 kN 827

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8–79. If the required clamping force at the board A is to be 2 kN, determine the torque M that must be applied to the screw to tighten it down. The square-threaded screw has a mean radius of 10 mm and a lead of 3 mm, and the coefficient of static friction is ms = 0.35.

M

A

Solution Frictional Forces on Screw. Here u = tan - 1 a

l 3 b = tan - 1 c d = 2.7336°, 2pr 2p (10)

W = 2000 N and f s = tan - 1ms = tan - 1(0.35) = 19.2900°.



M = Wr tan (u + f s) = 2000 (0.01) tan (2.7336° + 19.2900°) = 8.09 N # m

Ans.

Ans: 8.09 N # m 828

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*8–80. The square-threaded screw has a mean diameter of 20 mm and a lead of 4 mm. If the mass of the plate A is 2 kg, determine the smallest coefficient of static friction between the screw and the plate so that the plate does not travel down the screw when the plate is suspended as shown.

A

SOLUTION Frictional Forces on Screw: This requires a “self-locking” screw where fs Ú u. l 4 b = tan-1 c d = 3.643°. Here, u = tan-1 a 2pr 2p1102 fs = tan-1ms ms = tan fs

where fs = u = 3.643° Ans.

= 0.0637

Ans: ms = 0.0637 829

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8–81. If a horizontal force of P = 100 N is applied perpendicular to the handle of the lever at A, determine the compressive force F exerted on the material. Each single squarethreaded screw has a mean diameter of 25 mm and a lead of 7.5 mm. The coefficient of static friction at all contacting surfaces of the wedges is ms = 0.2, and the coefficient of static friction at the screw is msœ = 0.15.

A 15

C 15

250 mm

B

SOLUTION Since the screws are being tightened, Eq. 8–3 should be used. Here, L 7 .5 b = tan -1 c d = 5 .455°; u = tan -1 a 2pr 2p(12 .5)

fs = tan - 1ms = tan - 1(0.15) = 8.531°; M = 100(0.25) = 25 N # m; and W = T, where T is the tension in the screw shank. Since M must overcome the friction of two screws, M = 2[Wr tan(fs + u)4 25 = 23T(0 .0125) tan (8.531° + 5 .455°)4 Ans.

T = 4015.09 N = 4.02 kN

Referring to the free-body diagram of wedge B shown in Fig. a using the result of T, we have + ©Fx = 0; : + c ©Fy = 0;

4015 .09 - 0 .2N¿ - 0 .2N cos 15° - N sin 15° = 0

(1)

N¿ + 0 .2N sin 15° - N cos 15° = 0

(2)

Solving, N = 6324.60 N

N¿ = 5781.71 N

Using the result of N and referring to the free-body diagram of wedge C shown in Fig. b, we have + c ©Fy = 0;

2(6324.60) cos 15° - 230 .2(6324.60) sin 15°4 - F = 0 F = 11563.42 N = 11 .6 kN

Ans.

Ans: T = 4.02 kN F = 11.6 kN 830

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8–82. Determine the horizontal force P that must be applied perpendicular to the handle of the lever at A in order to develop a compressive force of 12 kN on the material. Each single square-threaded screw has a mean diameter of 25 mm and a lead of 7.5 mm. The coefficient of static friction at all contacting surfaces of the wedges is ms = 0.2, and the coefficient of static friction at the screw is msœ = 0.15.

A 15

C 15

250 mm

B

SOLUTION Referring to the free-body diagram of wedge C shown in Fig. a, we have + c ©Fy = 0;

2N cos 15° - 230 .2N sin 15°4 - 12000 = 0 N = 6563.39 N

Using the result of N and referring to the free-body diagram of wedge B shown in Fig. b, we have + c ©Fy = 0;

N¿ - 6563 .39 cos 15° + 0 .2(6563 .39) sin 15° = 0 N¿ = 6000 N

+ ©Fx = 0; :

T - 6563 .39 sin 15° - 0 .2(6563 .39) cos 15° - 0 .2(6000) = 0 T = 4166 .68 N

Since the screw is being tightened, Eq. 8–3 should be used. Here, u = tan - 1 c

L 7 .5 d = tan-1 c d = 5.455°; 2pr 2p(12 .5)

fs = tan-1ms = tan-1(0 .15) = 8.531°; M = P(0 .25); and W = T = 4166.68N. Since M must overcome the friction of two screws, M = 23Wr tan (fs + u)4 P(0.25) = 234166 .68(0.0125) tan (8 .531° + 5.455°)4 P = 104 N

Ans.

Ans: P = 104 N 831

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8–83. A force of P = 25 N is just sufficient to prevent the 20-kg cylinder from descending. Determine the required force P to begin lifting the cylinder. The rope passes over a rough peg with two and half turns.

P

SOLUTION The coefficient of static friction ms between the rope and the peg when the cylinder is on the verge of descending requires T2 = 20(9.81) N, T1 = P = 25 Nand b = 2 .5(2p) = 5p rad . Thus, T2 = T1emsb 20(9 .81) = 25ems(5p) In 7 .848 = 5pms ms = 0.1312 In the case of the cylinder ascending T2 = P and T1 = 20(9.81) N. Using the result of ms, we can write T2 = T1emsb P = 20(9.81)e0.1312(5p) = 1539.78 N Ans.

= 1.54 kN

Ans: P = 1.54 kN 832

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*8–84. The 20-kg cylinder A and 50-kg cylinder B are connected together using a rope that passes around a rough peg two and a half turns. If the cylinders are on the verge of moving, determine the coefficient of static friction between the rope and the peg.

75 mm A B

SOLUTION In this case, T1 = 50(9.81)N, T2 = 20(9.81)N and b = 2.5(2p) = 5p rad. Thus, T1 = T2emsb 50(9 .81) = 20(9 .81)ems(5p) ln 2.5 = m s(5p) Ans.

ms = 0 .0583

Ans: ms = 0 .0583 833

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8–85. A cylinder having a mass of 250 kg is to be supported by the cord which wraps over the pipe. Determine the smallest vertical force F needed to support the load if the cord passes (a) once over the pipe, b = 180°, and (b) two times over the pipe, b = 540°. Take ms = 0.2.

F

SOLUTION Frictional Force on Flat Belt: Here, T1 = F and T2 = 250(9.81) = 2452.5 N. Applying Eq. 8–6, we have a) If b = 180° = p rad T2 = T1 e mb 2452.5 = Fe 0.2p Ans.

F = 1308.38 N = 1.31 kN b) If b = 540° = 3 p rad T2 = T1 e mb 2452.5 = Fe 0.2(3p)

Ans.

F = 372.38 N = 372 N

Ans: F = 1.31 kN F = 372 N 834

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8–86. A cylinder having a mass of 250 kg is to be supported by the cord which wraps over the pipe. Determine the largest vertical force F that can be applied to the cord without moving the cylinder. The cord passes (a) once over the pipe, b = 180°, and (b) two times over the pipe, b = 540°. Take ms = 0.2.

F

SOLUTION Frictional Force on Flat Belt: Here, T1 = 250(9.81) = 2452.5 N and T2 = F. Applying Eq. 8–6, we have a)

If b = 180° = p rad T2 = T1 e mb F = 2452.5e 0.2 p Ans.

F = 4597.10 N = 4.60 kN b)

If b = 540° = 3 p rad T2 = T1e mb F = 2452.5e 0.2(3 p) Ans.

F = 16 152.32 N = 16.2 kN

Ans: F = 4.60 kN F = 16.2 kN 835

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8–87. The truck, which has a mass of 3.4 Mg, is to be lowered down the slope by a rope that is wrapped around a tree. If the wheels are free to roll and the man at A can resist a pull of 300 N, determine the minimum number of turns the rope should be wrapped around the tree to lower the truck at a constant speed. The coefficient of kinetic friction between the tree and rope is mk = 0.3.

A

20

SOLUTION Q+ ©Fx = 0;

T2 - 33 354 sin 20° = 0 T2 = 11 407.7 T2 = T1 e mb 11 407.7 = 300 e 0.3 b b = 12.1275 rad Ans.

Approx. 2 turns (695°)

Ans: Approx. 2 turns (695°) 836

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*8–88. The cord supporting the 6-kg cylinder passes around three pegs, A, B, C, where ms = 0.2. Determine the range of values for the magnitude of the horizontal force P for which the cylinder will not move up or down.

P

C

45 B 45 A

SOLUTION Given: M

6 kg

T

45 °

Ps g

0.2 9.81 m> s 2

Total angle Forces

Answer

E P min

P min

5 2

S  4T  Ps E

Mge

E

270.00° P max

15.9 N < P < P max

P sE Mge

Ans.

217.4 N

Ans:

15.9 N < P < 217.4 N 837

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8–89. A cable is attached to the 20-kg plate B, passes over a fixed peg at C, and is attached to the block at A. Using the coefficients of static friction shown, determine the smallest mass of block A so that it will prevent sliding motion of B down the plane.

mC  0.3 C

A mA  0.2 mB  0.3 B

30

SOLUTION Block A: + Q©Fx = 0;

T1 - 0.2 NA - WA sin 30° = 0

(1)

+ a©Fy = 0;

NA - WA cos 30° = 0

(2)

+Q©Fx = 0;

T2 - 20(9.81) sin 30° + 0.3 NB + 0.2 NA = 0

(3)

+ a©Fy = 0;

NB - NA - 20(9.81) cos 30° = 0

(4)

T2 = T1 e 0.3p

(5)

Plate B:

Peg C: T2 = T1 e mb;

Solving Eqs. (1)–(5) yields T1 = 14.68 N;

T2 = 37.68 N;

NA = 18.89 N;

NB = 188.8 N;

WA = 21.81 N

Thus, mA =

21.81 = 2.22 kg 9.81

Ans.

Ans: mA = 2.22 kg 838

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8–90. The smooth beam is being hoisted using a rope which is wrapped around the beam and passes through a ring at A as shown. If the end of the rope is subjected to a tension T and the coefficient of static friction between the rope and ring is ms = 0.3, determine the angle of u for equilibrium.

T

A

θ

SOLUTION Equation of Equilibrium: + c ©Fx = 0;

T - 2T¿ cos

u = 0 2

Frictional Force on Flat Belt: Here, b = T2 = T1 emb, we have

T = 2T¿cos

u 2

(1)

u , T = T and T1 = T¿. Applying Eq. 8–6 2 2

T = T¿e0.31u>22 = T¿e0.15 u

(2)

Substituting Eq. (1) into (2) yields 2T¿cos

u = T¿e0.15 u 2

e0.15 u = 2 cos

u 2

Solving by trial and error Ans.

u = 1.73104 rad = 99.2°

The other solution, which starts with T' = Te 0.3(0/2) based on cinching the ring tight, is 2.4326 rad = 139°. Any angle from 99.2° to 139° is equilibrium.

Ans: u = 92.2° 839

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8–91. Determine the smallest force P required to lift the 40-kg crate. The coefficient of static friction between the cable and each peg is ms = 0.1.

200 mm

200 mm

A

C

200 mm B

P

SOLUTION Since the crate is on the verge of ascending, T1 = 40(9.81) N and T2 = P. From the geometry shown in Figs. a and b, the total angle the rope makes when in contact with 135° 90° the peg is b = 2b 1 + b 2 = 2 a pb + a pb = 2p rad. Thus, 180° 180° T2 = T1ems b P = 40(9.81)e0.1(2p) Ans.

= 736 N

Ans: P = 736 N 840

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*8–92. Block A has a mass of 50 kg and rests on surface for which ms = 0.25. If the coefficient of static friction between the cord and the fixed peg at C is ms = 0.3, determine the greatest mass of the suspended cylinder D without causing motion.

C 5

3

4

0.25 m

0.3 m

SOLUTION

A

0.4 m

Given: mA

50 kg

P sAB P sC

D

0.25 0.3

a

0.3 m

b

0.25 m

c

0.4 m

d

3

f

4

g

9.81 m> s 2

The initial guesses:

Given

mD g

NB

Since x

T

100 N

P sCE Te

50 N

© d¹

mD

x

1 kg

Find  NB  T  mD  x

0

§ NB · ¨ ¸ ©T ¹

b 2

0

f d § § · · § b· ¨ 2 2 ¸ T a  ¨ 2 2 ¸ T¨© 2 ¸¹  NB x © f d ¹ © f d ¹

§ 413.05 · ¨ ¸N © 129.08 ¹

mD x

51.6 mm
2).

Impending motion

a b

T2

SOLUTION FBD of a section of the belt is shown. Proceeding in the general manner: ©Fx = 0;

- (T + dT) cos

du du + T cos + 2 dF = 0 2 2

©Fy = 0;

-(T + dT) sin

du a du - T sin + 2 dN sin = 0 2 2 2

Replace

sin

du du by , 2 2

cos

du by 1, 2

dF = m dN Using this and (dT)(du) : 0, the above relations become dT = 2m dN T du = 2 a dN sin

a b 2

Combine dT du = m T sin a2 Integrate from u = 0, T = T1 to u = b, T = T2 we get, mb a ab T2 = T1 sin 2

Q.E.D

845

T1

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8–97. A V-belt is used to connect the hub A of the motor to wheel B. If the belt can withstand a maximum tension of 1200 N, determine the largest mass of cylinder C that can be lifted and the corresponding torque M that must be supplied to A. The coefficient of static friction between the hub and the belt is ms = 0.3, and between the wheel and the belt is ms ¿ = 0.20. Hint: See Prob. 8–96.

200 mm 150 mm

300 mm

15⬚ A

a ⫽ 60⬚

B

M

C

SOLUTION In this case, the maximum tension in the belt is T2 = 1200 N. Referring to the freebody diagram of hub A, shown in Fig. a and the wheel B shown in Fig. b, we have a + ©MO = 0;

M + T1 (0.15) - 1200(0.15) = 0 (1)

M = 0.15(1200 - T1) a + ©MO¿ = 0;

1200(0.3) - T1(0.3) - MC (9.81)(0.2) = 0 (2)

1200 - T1 = 6.54MC If hub A is on the verge of slipping, then T2 = T1emsb1>sin(a>2) where b 1 = a

90° + 75° b p = 0.9167p rad 180°

1200 = T1e0.3(0.9167p)>sin 30° T1 = 213.19 N Substituting T1 = 213.19 N into Eq. (2), yields MC = 150.89 kg If wheel B is on the verge of slipping, then T2 = T1ems¿b1>sin(a>2) where b 2 = a

180° + 15° bp = 1.0833p rad 180°

1200 = T1e0.2(1.0833p)>sin 30° T1 = 307.57 N Substituting T1 = 307.57 N into Eq. (2), yields Ans.

MC = 136.45 kg = 136 kg (controls!) Substituting T1 = 307.57 N into Eq. (1), we obtain M = 0.15(1200 - 307.57) = 134 N # m

Ans.

Ans: M C = 136 kg (controls!) M = 134 N # m 846

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8–98. Blocks A and B have a mass of 7 kg and 10 kg, respectively. Using the coefficients of static friction indicated, determine the largest vertical force P which can be applied to the cord without causing motion.

µD = 0.1

300 mm

D

µB = 0.4

B

400 mm P

µC = 0.4 A

C

µA = 0.3

SOLUTION Frictional Forces on Flat Belts: When the cord pass over peg D, b = 180° = p rad and T2 = P. Applying Eq. 8–6, T2 = T1 e mb, we have P = T1 e 0.1 p

T1 = 0.7304P

When the cord pass over peg C, b = 90° = Applying Eq. 8–6, T2 ′ = T1 ′e mb, we have 0.7304P = T1 ′e 0.4(p>2)

p rad and T2 ′ = T1 = 0.7304P. 2

T1 ′ = 0.3897P

Equations of Equilibrium: From FDB (b), + c ΣFy = 0;

NB - 98.1 = 0

+ ΣFx = 0; S

FB - T = 0

(1)

a+ ΣMO = 0 ;

T(0.4) - 98.1(x) = 0

(2)

NB = 98.1 N

From FDB (b), + c ΣFy = 0;

NA - 98.1 - 68.67 = 0

+ ΣFx = 0; S

0.3897P - FB - FA = 0

NA = 166.77 N (3)

Friction: Assuming the block B is on the verge of tipping, then x = 0.15 m. A1 for motion to occur, block A will have slip. Hence, FA = (ms)ANA = 0.3(166.77) = 50.031 N. Substituting these values into Eqs. (1), (2) and (3) and solving yields Ans.

P = 222.81 N = 223 N FB = T = 36.79 N

Since (FB)max = (ms)B NB = 0.4(98.1) = 39.24 N 7 FB, block B does not slip but tips. Therefore, the above assumption is correct.

Ans: P = 223 N 847

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8–99. Granular material, having a density of 1.5 Mg>m3, is transported on a conveyor belt that slides over the fixed surface, having a coefficient of kinetic friction of mk = 0.3. Operation of the belt is provided by a motor that supplies a torque M to wheel A. The wheel at B is free to turn, and the coefficient of static friction between the wheel at A and the belt is mA = 0.4. If the belt is subjected to a pretension of 300 N when no load is on the belt, determine the greatest volume V of material that is permitted on the belt at any time without allowing the belt to stop. What is the torque M required to drive the belt when it is subjected to this maximum load?

mk  0.3 100 mm B

mA  0.4 M

100 mm A

SOLUTION Wheel A: a + ©MA = 0; T2 = T1 emb;

-M - 300 (0.1) + T2(0.1) = 0 T2 = 300e 0.4(p) = 1054.1 N

Thus, M = 75.4 N # m

Ans.

Belt, + ©F = 0; : x

1054.1 - 0.3 (m) (9.81) - 300 = 0 m = 256.2 kg V =

256.2 m = = 0.171 m3 p 1500

Ans.

Ans: M = 75.4 N # m V = 0.171 m3 848

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*8–100. The uniform bar AB is supported by a rope that passes over a frictionless pulley at C and a fixed peg at D. If the coefficient of static friction between the rope and the peg is mD = 0.3, determine the smallest distance x from the end of the bar at which a 20-N force may be placed and not cause the bar to move.

C

D

20 N x B

A 1m

SOLUTION a + ©MA = 0;

-20 (x) + TB (1) = 0

+ c ©Fy = 0;

TA + TB - 20 = 0

T2 = T1 e mb ;

TA = TB e 0.3( 2 ) = 1.602 TB

p

Solving, TA = 12.3 N TB = 7.69 N Ans.

x = 0.384 m

Ans: x = 0.384 m 849

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8–101. The wheel is subjected to a torque of M = 50 N # m. If the coefficient of kinetic friction between the band brake and the rim of the wheel is mk = 0.3, determine the smallest horizontal force P that must be applied to the lever to stop the wheel.

P

400 mm

M

C A 150 mm

50 mm 25 mm

B

100 mm

SOLUTION Wheel: a + ©MO = 0; T2 = T1 e mb ;

- T2 (0.150) + T1 (0.150) + 50 = 0 b T2 = T1 e 0.3a 3p 2

T1 = 107.14 N Link: a + ©MB = 0;

107.14 (0.05) - F (0.025) = 0 F = 214.28 N

Lever: a + ©MA = 0;

- P (0.4) + 214.28 (0.1) = 0 Ans.

P = 53.6 N

Ans: P = 53.6 N 850

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8–102. Blocks A and B have a mass of 20 kg and 10 kg, respectively. Using the coefficients of static friction indicated, determine the greatest mass of block D without causing motion.

B

m  0.5

mBA  0.6

20 A D

C

mAC  0.4

Solution Equations of Equilibrium. Assumming that blocks A and B slide as a unit on surface C. Then FC = mCNC = 0.4 NC. Referring to the FBD of blocks A and B, Fig. a, + c ΣFy = 0; +

S ΣFx = 0;

NC - (20 + 10)(9.81) = 0 0.4(294.3) - TB = 0

NC = 294.3 N

TB = 117.72 N

Consider the equilibrium of block B with TB = 117.72 N by referring to its FBD, Fig. b, +

S ΣF = 0; F cos 20° - N sin 20° - 117.72 = 0 x B B

(1)

+ c ΣFy = 0; FB sin 20° + NB cos 20° - 10(9.81) = 0

(2)

Solving Eqs. (1) and (2). NB = 51.92 N

FB = 144.17 N

Since FB 7 mBANB = 0.6(51.92) = 31.15 N, block B slides on block C. Therefore the assumption was wrong. Then FB = mBANB = 0.6 NB. Again, by referring to the FBD of block B, Fig. b. + c ΣFy = 0; +

S ΣF = 0; x

NB cos 20° + 0.6 NB sin 20° - 10(9.81) = 0

NB = 85.68 N

0.6(85.68)cos 20° - 85.68 sin 20° - TB = 0

TB = 19.00 N

Belt Friction. Here, the rope is required to slide over a rough peg where B = m>2 rad with T2 = MDg = MD(9.81) and T1 = TB = 19.00 N. Applying Fig. 8–6, T2 = T1 e mB;

MD(9.81) = 19.00 e 0.5(p>2) Ans.

MD = 4.249 kg = 4.25 kg

Ans: MD = 4.25 kg 851

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8–103. A conveyer belt is used to transfer granular material and the frictional resistance on the top of the belt is F = 500 N. Determine the smallest stretch of the spring attached to the moveable axle of the idle pulley B so that the belt does not slip at the drive pulley A when the torque M is applied. What minimum torque M is required to keep the belt moving? The coefficient of static friction between the belt and the wheel at A is ms = 0.2.

0.1 m

M A

F = 500 N

0.1 m B k = 4 kN/m

SOLUTION Frictional Force on Flat Belt: Here, b = 180° = p rad and T2 = 500 + T and T1 = T. Applying Eq. 8–6, we have T2 = T1 emb 500 + T = Te0.2p T = 571.78 N Equations of Equilibrium: From FBD (a), M + 571.7810.12 - 1500 + 578.1210.12 = 0

a + ©MO = 0;

M = 50.0 N # m

Ans.

From FBD (b), + ©F = 0; : x

Fsp - 21578.712 = 0

Fsp = 1143.57 N

Thus, the spring stretch is x =

Fsp k

=

1143.57 = 0.2859 m = 286 mm 4000

Ans.

Ans: M = 50.0 N # m x = 286 mm 852

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*8–104. A 10-kg cylinder D, which is attached to a small pulley B, is placed on the cord as shown. Determine the largest angle X so that the cord does not slip over the peg at C. The cylinder at E also has a mass of 10 kg, and the coefficient of static friction between the cord and the peg is ms = 0.1.

A

u

u

C

B E D

SOLUTION Since pully B is smooth, the tension in the cord between pegs A and C remains constant. Referring to the free-body diagram of the joint B shown in Fig. a, we have + c ©Fy = 0;

T =

2T sin u - 10(9.81) = 0

49.05 sin u

49.05 In the case where cylinder E is on the verge of ascending, T2 = T = and sin u p T1 = 10(9.81) N. Here, + u, Fig. b. Thus, 2 T2 = T1e msb p 49.05 = 10(9.81) e 0.1 a 2 sin u

ln

+ ub

p 0.5 = 0.1 a + u b sin u 2

Solving by trial and error, yields u = 0.4221 rad = 24.2° In the case where cylinder E is on the verge of descending, T2 = 10(9.81) N and 49.05 p . Here, + u. Thus, T1 = sin u 2 T2 = T1e m s b 10(9.81) =

49.05 0.1 a p e 2 sin u

ln (2 sin u) = 0.1 a

+ ub

p + ub 2

Solving by trial and error, yields u = 0.6764 rad = 38.8° Thus, the range of u at which the wire does not slip over peg C is 24.2° 6 u 6 38.8° Ans.

umax = 38.8°

Ans: umax = 38.8° 853

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8–105. The 20-kg motor has a center of gravity at G and is pinconnected at C to maintain a tension in the drive belt. Determine the smallest counterclockwise twist or torque M that must be supplied by the motor to turn the disk B if wheel A locks and causes the belt to slip over the disk. No slipping occurs at A. The coefficient of static friction between the belt and the disk is ms = 0.3.

M A

B

50 mm

G 50 mm

150 mm

C

100 mm

SOLUTION Equations of Equilibrium: From FBD (a), a + ©MC = 0;

T2 11002 + T1 12002 - 196.211002 = 0

(1)

M + T1 10.052 - T2 10.052 = 0

(2)

From FBD (b), a + ©MO = 0;

Frictional Force on Flat Belt: Here, b = 180° = p rad. Applying Eq. 8–6, T2 = T1 emb, we have T2 = T1 e0.3p = 2.566T1

(3)

Solving Eqs. (1), (2), and (3) yields M = 3.37 N # m T1 = 42.97 N

Ans.

T2 = 110.27 N

Ans: M = 3.37 N # m 854

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8–106. The belt on the portable dryer wraps around the drum D, idler pulley A, and motor pulley B. If the motor can develop a maximum torque of M = 0.80 N # m, determine the smallest spring tension required to hold the belt from slipping. The coefficient of static friction between the belt and the drum and motor pulley is ms = 0.3.

30°

50 mm

M = 0.8 N⋅m

A B

D

45° C

50 mm

20 mm

SOLUTION a + ©MB = 0;

-T1 10.022 + T2 10.022 - 0.8 = 0

T2 = T1 emb;

T2 = T1 e10.321p2 = 2.5663T1

T1 = 25.537 N T2 = 65.53 N a + ©MC = 0;

-Fs10.052 + 125.537 + 25.537 sin 30°210.1 cos 45°2 + 25.537 cos 30°10.1 sin 45°2 = 0 Fs = 85.4 N

Ans.

Ans: Fs = 85.4 N 855

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8–107. The collar bearing uniformly supports an axial force of P = 5 kN. If the coefficient of static friction is ms = 0.3, determine the smallest torque M required to overcome friction.

P M

150 mm 200 mm

Solution Bearing Friction. With R2 = 0.1 m, R1 = 0.075 m, P = 5 ( 103 ) N, and ms = 0.3, R23 - R13 2 b ms P a 2 3 R2 - R12



M =



=



= 132 N # m

2 0.13 - 0.0753 b (0.3) 3 5 ( 103 )4 a 2 3 0.1 - 0.0752

Ans: M = 132 N # m 856

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*8–108. The collar bearing uniformly supports an axial force of P = 8 kN. If a torque of M = 200 N # m is applied to the shaft and causes it to rotate at constant velocity, determine the coefficient of kinetic friction at the surface of contact.

P M

150 mm 200 mm

Solution

Bearing Friction. With R2 = 0.1 m, R1 = 0.075 m, M = 300 N # m, and P = 8 ( 103 ) N,

M =



200 =



R23 - R13 2 mk P a 2 b 3 R2 - R12

2 0.13 - 0.0753 b mk 3 8 ( 103 )4 a 2 3 0.1 - 0.0752

Ans.

mk = 0.284

Ans: mk = 0.284 857

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8–109. The double-collar bearing is subjected to an axial force P = 4 kN. Assuming that collar A supports 0.75P and collar B supports 0.25P, both with a uniform distribution of pressure, determine the maximum frictional moment M that may be resisted by the bearing. Take ms = 0.2 for both collars.

P M

20 mm B A

10 mm 30 mm

SOLUTION M =

R32 - R31 2 ms P ¢ 2 ≤ 3 R2 - R21

M =

(0.03)3 - (0.01)3 (0.02)3 - (0.01)3 2 (0.75) (4000) + (0.25) (4000) ≤ (0.2) ¢ 5 (0.03)2 - (0.01)2 (0.02)2 - (0.01)2

= 16.1 N # m

Ans.

Ans: M = 16.1 N # m 858

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8–110. The double-collar bearing is subjected to an axial force P = 16 kN. Assuming that collar A supports 0.75P and collar B supports 0.25P, both with a uniform distribution of pressure, determine the smallest torque M that must be applied to overcome friction. Take ms = 0.2 for both collars.

P 100 mm

M A B

50 mm

75 mm 30 mm

Solution Bearing Friction. Here (RA)2 = 0.1 m, (RA)1 = 0.05 m, PA = 0.75 316 ( 103 ) N4 = 12 ( 103 ) N, (RB)2 = 0.075 m, (RB)1 = 0.05 m and PB = 0.25 316 ( 103 ) N4 = 4 ( 103 ) N. (RA)23 - (RA)13 (RB)23 - (RB)13 2 2 ms PA c d + m P c d s B 3 3 (RA)22 - (RA)12 (RB)22 - (RB)12



M =



=



= 237.33 N # m = 237 N # m

2 2 0.13 - 0.053 0.0753 - 0.053 b + (0.2) 3 4 ( 103 )4 a b (0.2) 312 ( 103 )4 a 2 2 3 3 0.1 - 0.05 0.0752 - 0.052

Ans.

Ans: M = 237 N # m 859

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8–111. P

Assuming that the variation of pressure at the bottom of the pivot bearing is defined as p = p01R2>r2, determine the torque M needed to overcome friction if the shaft is subjected to an axial force P. The coefficient of static friction is ms. For the solution, it is necessary to determine p0 in terms of P and the bearing dimensions R1 and R2.

M

R2

R1

r

p

p0 p0 R2 r

SOLUTION R2

2p

©Fz = 0;

P =

dN =

LA

2p

=

L0

L0

R2

LR1

LR1

p0 a

pr dr du

R2 b r dr du r

= 2p p0 R2 (R2 - R1) Thus, p0 =

P

C 2pR2 (R2 - R1) D R2

2p

©Mz = 0;

M =

r dF =

LA

2p

=

L0

L0

LR1

R2

LR1

ms p0 a

= ms (2p p0) R2

ms pr 2 dr du

R2 2 b r dr du r

1 A R22 - R21 B 2

Using Eq. (1): M =

1 ms P (R2 + R1) 2

Ans.

Ans: M = 860

1 m P(R 2 + R 1) 2 s

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*8–112. The pivot bearing is subjected to a pressure distribution at its surface of contact which varies as shown. If the coefficient of static friction is m, determine the torque M required to overcome friction if the shaft supports an axial force P.

P M

R r

p0

r p = p0 cos π 2R

SOLUTION dF = m dN = m p0 cos a M =

LA

rm p0 cos a

pr b r dr du 2R 2p

R

= m p0

L0

= m p0 B = mp0 ¢

pr b dA 2R

ar2 cos a 2r

pr du b drb 2R L0 p 2 2 A 2R B r -2

pr cos a b + 2R

p 2 A 2R B

p 3 A 2R B

sin a

pr R b d 12p2 2R 0

p 2 16R3 c a b - 2d 2 ≤ 2 p

= 0.7577m p0 R3 2p

R

P =

LA

dN =

= p0 B

1 p 2 A 2R B

L0

p0 a cos a

cos a

= 4p0 R 2 a1 -

pr b + 2R

pr b rdrb du 2R L0 r p A 2R B

sin a

pr R b R 12p2 2R 0

2 b p

= 1.454p0 R2 Thus,

Ans.

M = 0.521 PmR

Ans: M = 0.521 PmR 861

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8–113. The conical bearing is subjected to a constant pressure distribution at its surface of contact. If the coefficient of static friction is ms, determine the torque M required to overcome friction if the shaft supports an axial force P.

P M

R

u

SOLUTION The differential area (shaded) dA = 2pr ¢ P =

L

p cos u dA =

L

p cos u ¢

2prdr dr ≤ = cos u cos u R

2prdr rdr ≤ = 2pp cos u L0

P pR2

P = ppR 2

p =

dN = pdA =

2P 2prdr P rdr ¢ ≤ = 2 pR2 cos u R cos u

M =

L

rdF =

L

ms rdN = =

2ms P 2

R cos u L0

R

r2 dr

2ms PR 2ms P R 3 = 2 3 3 cos u R cos u

Ans.

Ans: M = 862

2ms PR 3 cos u

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8–114. A 200-mm diameter post is driven 3 m into sand for which ms = 0.3. If the normal pressure acting completely around the post varies linearly with depth as shown, determine the frictional torque M that must be overcome to rotate the post.

M 200 mm

3m

600 Pa

SOLUTION Equations of Equilibrium and Friction: The resultant normal force on the post is 1 N = 1600 + 021321p210.22 = 180p N. Since the post is on the verge of rotating, 2 F = ms N = 0.31180p2 = 54.0p N. a + ©MO = 0;

M - 54.0p10.12 = 0 M = 17.0 N # m

Ans.

Ans: M = 17.0 N # m 863

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8–115. The plate clutch consists of a flat plate A that slides over the rotating shaft S. The shaft is fixed to the driving plate gear B. If the gear C, which is in mesh with B, is subjected to a torque of M = 0.8N # m, determine the smallest force P, that must be applied via the control arm, to stop the rotation. The coefficient of static friction between the plates A and D is ms = 0.4. Assume the bearing pressure between A and D to be uniform.

D A F

100 mm S 125 mm

P

200 mm

150 mm E

150 mm

B

30 mm M

0.8 N m

C

SOLUTION F =

0.8 = 26.667 N 0.03

M = 26.667(0.150) = 4.00 N # m M =

R32 - R31 2 b m P¿ a 2 3 R2 - R21

4.00 =

(0.125)3 - (0.1)3 2 (0.4) (P¿) a b 3 (0.125)2 - (0.1)2

P¿ = 88.525 N a + ©MF = 0;

88.525(0.2) - P(0.15) = 0 Ans.

P = 118 N

Ans: P = 118 N 864

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*8–116. The pivot bearing is subjected to a parabolic pressure distribution at its surface of contact. If the coefficient of static friction is ms, determine the torque M required to overcome friction and turn the shaft if it supports an axial force P.

P M

R r

2

SOLUTION

p0

r ) p  p0 (1 –– R2

The differential are dA = (rdu)(dr) P = P =

p dA =

p0 ¢ 1 -

L

L

pR2 p0 2

p0 =

dN = pdA = M =

R

2P pR2

r2 2P 1 - 2 ≤ (rdu)(dr) 2 ¢ pR R ms rdN =

rdF = L

2p

r2 r2 (rdu)(dr) = p0 du r ¢ 1 - 2 ≤ dr 2≤ R R L0 L0

L =

2ms P

2p

2

pR L0

du

R

L0

r2 ¢ 1 -

r2 ≤ dr R2

8 m PR 15 s

Ans.

Ans: M = 865

8 m PR 15 s

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8–117. The collar fits loosely around a fixed shaft that has a radius of 50 mm. If the coefficient of kinetic friction between the shaft and the collar is mk = 0.3, determine the force P on the horizontal segment of the belt so that the collar rotates counterclockwise with a constant angular velocity. Assume that the belt does not slip on the collar; rather, the collar slips on the shaft. Neglect the weight and thickness of the belt and collar. The radius, measured from the center of the collar to the mean thickness of the belt, is 56.25 mm.

P 56.25 mm 50 mm

100 N

Solution Given: r = 50 mm mk = 0.3 R = 56.25 mm F = 100 N f k = atan(mk) rf = r # sin (f k)

f k = 16.699 ° rf = 14.3674 mm

Equilibrium : + c ΣFy = 0; + S ΣFx = 0;

Guess Given

Ry - F = 0 Ry = F Ry = 100.00 N P - Rx = 0 Rx = P R = 2R2x + R2y = 2P2 + F 2

P = 1N - 2P2 + F 2 # rf + F # R - P # R = 0

P = Find(P)

P = 68.97 N

Ans.

Ans: P = 68.97 N 866

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8–118. The collar fits loosely around a fixed shaft that has a radius of 50 mm. If the coefficient of kinetic friction between the shaft and the collar is mk = 0.3, determine the force P on the horizontal segment of the belt so that the collar rotates clockwise with a constant angular velocity. Assume that the belt does not slip on the collar; rather, the collar slips on the shaft. Neglect the weight and thickness of the belt and collar. The radius, measured from the center of the collar to the mean thickness of the belt, is 56..25 mm.

P 56.25 mm 50 mm

100 N

Solution r = 50 mm mk = 0.3 R = 56.25 mm F = 100 N f k = atan(mk) rf = r # sin (f k)

f k = 16.699 ° rf = 14.367 mm

Equilibrium : + c ΣFy = 0; + S ΣFx = 0;

Guess Given

Ry - F = 0 Ry = F Ry = 100.00 N P - Rx = 0 Rx = P R = 2Rx2 + Ry2 = 2P2 + F 2

P = 1N 2P2 + F 2 # rf + F # R - P # R = 0

P = Find(P)

P = 145.0 N

Ans.

Ans: P = 145.0 N 867

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8–119. A disk having an outer diameter of 120 mm fits loosely over a fixed shaft having a diameter of 30 mm. If the coefficient of static friction between the disk and the shaft is ms = 0.15 and the disk has a mass of 50 kg, determine the smallest vertical force F acting on the rim which must be applied to the disk to cause it to slip over the shaft.

F

SOLUTION Frictional Force on Journal Bearing: Here, fs = tan-1ms = tan-10.15 = 8.531°. Then the radius of friction circle is rf = r sin fs = 0.015 sin 8.531° = 2.225110 -32 m Equation of Equilibrium: a + ©MP = 0;

490.512.2252110-32 - F30.06 - 12.2252110-324 = 0 Ans.

F = 18.9 N

Ans: F = 18.9 N 868

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*8–120. The 5-kg pulley has a diameter of 240 mm and the axle has a diameter of 40 mm. If the coefficient of kinetic friction between the axle and the pulley is mk = 0.15, determine the vertical force P on the rope required to lift the 80-kg block at constant velocity. 120 mm

Solution

P

Given: a

120 mm

M

5 kg

D

40 mm

Pk

0.15

MB

80 kg

Ik

atan  P k

rf

§ D · sin  I ¨ ¸ k ©2¹

6Mp = 0; MB g a  rf  M g rf  P a  rf

P

MB g a  rf  M g rf a  rf

0

P

826 N

Ans.

Ans:

P 869

826 N

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8–121. Solve Prob. 8–120 if the force P is applied horizontally to the right.

120 mm

Solution Given: a

P

120 mm

M

5 kg

D

40 mm

Pk

0.15

MB

80 kg

g

9.81 m>s2

Ik

atan  P k

rf

D sin  I k 2

Guesses P

1N

R

D

1N

1°

Given R cos  D  MB g  M g P  R sin  D

0

MB g a  P a  R rf

§P · ¨R ¸ ¨ ¸ ©D ¹

0

Find  P  R  D

0

P

814 N

Ans.

Ans:

P 870

814 N

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8–122. The uniform disk fits loosely over a fixed shaft having a diameter of 40 mm. If the coefficient of static friction between the disk and the shaft is ms = 0.15, determine the smallest vertical force P, acting on the rim, which must be applied to the disk to cause it to slip on the shaft. The disk has a mass of 20 kg.

150 mm

40 mm

P

Solution Frictional Force on Journal Bearing. Here, f k = tan - 1 ms = tan - 1 0.15 = 8.5308°. Then the radius of the friction circle is

rf = r sin f s = 0.02 sin 8.5308° = 2.9668 ( 10 - 3 ) m

Equations of Equilibrium. Referring to the FBD of the disk shown in Fig. a, a + ΣMP = 0;  20(9.81) 3 2.9668 ( 10-3 ) 4 - P 3 0.075 - 2.9668 ( 10 - 3 ) 4 = 0



P = 8.08 N

Ans.

Ans: 8.08 N 871

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8–123. A pulley having a diameter of 80 mm and mass of 1.25 kg is supported loosely on a shaft having a diameter of 20 mm. Determine the torque M that must be applied to the pulley to cause it to rotate with constant motion. The coefficient of kinetic friction between the shaft and pulley is mk = 0.4. Also calculate the angle u which the normal force at the point of contact makes with the horizontal. The shaft itself cannot rotate.

M 40 mm

SOLUTION Frictional Force on Journal Bearing: Here, fk = tan-1 mk = tan-10.4 = 21.80°. Then the radius of friction circle is rf = r sin fk = 0.01 sin 21.80° = 3.714110-32 m. The angle which the normal force makes with horizontal is Ans.

u = 90° - fk = 68.2° Equations of Equilibrium: + c ©Fy = 0;

R - 12.2625 = 0

a + ©MO = 0;

12.262513.7142110 -32 - M = 0

R = 12.2625 N

M = 0.0455 N # m

Ans.

Ans: u = 68.2, M = 0.0455 N # m 872

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*8–124. The connecting rod is attached to the piston by a 20-mm-diameter pin at B and to the crank shaft by a 50-mm-diameter bearing A. If the piston is moving downwards, and the coefficient of static friction at these points is ms = 0.2, determine the radius of the friction circle at each connection.

B

A

Solution 1rf 2 A = rAms = 1rf 2 B = rBms =

5010.22 2 20(0.2) 2

Ans.

= 5 mm

Ans.

= 2 mm

873

Ans: 1rf 2 A = 5 mm 1rf 2 B = 2 mm

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8–125. The connecting rod is attched to the piston by a 20-mm-diameter pin at B and to the crank shaft by a 50-mm-diameter bearing A. If the piston is moving upwards, and the coefficient of static friction at these points is ms = 0.3, determine the radius of the friction circle at each connection. B

A

Solution Ans.

1rf 2 A = rAms = 2510.32 = 7.50 mm

Ans.

1rf 2 B = rBms = 1010.32 = 3 mm

874

Ans: 1rf 2 A = 7.50 mm 1rf 2 B = 3 mm

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8–126. The 5-kg skateboard rolls down the 5° slope at constant speed. If the coefficient of kinetic friction between the 12.5-mm diameter axles and the wheels is mk = 0.3, determine the radius of the wheels. Neglect rolling resistance of the wheels on the surface. The center of mass for the skateboard is at G.

75 mm G 5 250 mm

300 mm

SOLUTION Referring to the free-body diagram of the skateboard shown in Fig. a, we have ©Fx¿ = 0;

Fs - 5(9.81) sin 5° = 0

Fs = 4.275 N

©Fy¿ = 0;

N - 5(9.81) cos 5° = 0

N = 48.86 N

The effect of the forces acting on the wheels can be represented as if these forces are acting on a single wheel as indicated on the free-body diagram shown in Fig. b.We have ©Fx¿ = 0;

Rx¿ - 4.275 = 0

Rx¿ = 4.275 N

©Fy¿ = 0;

48.86 - Ry¿ = 0

Ry¿ = 48.86 N

Thus, the magnitude of R is R = 2Rx¿ 2 + Ry¿ 2 = 24.2752 + 48.862 = 49.05 N fs = tan-1 ms = tan-1(0.3) = 16.699°. Thus, the moment arm of R from point O is (6.25 sin 16.699°) mm. Using these results and writing the moment equation about point O, Fig. b, we have a + ©MO = 0;

4.275(r) - 49.05(6.25 sin 16.699° = 0) Ans.

r = 20.6 mm

Ans: r = 20.6 mm 875

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8–127. Determine the force P required to overcome rolling resistance and pull the 50-kg roller up the inclined plane with constant velocity. The coefficient of rolling resistance is a = 15 mm.

P

300 mm

30⬚

30⬚

Solution From the geometry indicated on the free-body diagram of the roller shown in Fig. a, 15 u = sin - 1a b = 2.866°. 300

We have +

S ΣFx′ = 0; P cos 30° - 5019.812 sin 30° - R sin 2.866° = 0 +

S Σ y′ = 0; P sin 30° + R cos 2.866° - 5019.812 cos 30° = 0 Solving, Ans.

P = 299 N R = 275.58 N

P can also be obtained directly by writing the moment equation of equilibrium about point A. Referring to Fig. a, a+ ΣMA = 0;

5019.812 sin 130° + 2.866°) 13002 - P cos 130° - 2.866°) 13002 = 0 Ans. P = 299 N

Ans: P = 299 N 876

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*8–128. Determine the force P required to overcome rolling resistance and support the 50-kg roller if it rolls down the inclined plane with constant velocity. The coefficient of rolling resistance is a = 15 mm.

P

300 mm

30⬚

30⬚

Solution From the geometry indicated on the free-body diagram of the roller shown in Fig. a, 15 u = sin - 1a b = 2.866°. 300 +

S ΣFx′ = 0; P cos 30° + R sin 2.866° - 5019.812 sin 30° = 0 +

S Σ y′ = 0; P sin 30° + R cos 2.866° - 5019.812 cos 30° = 0 Solving, Ans.

P = 266 N R = 291.98 N

P can also be obtained directly by writing the moment equation of equilibrium about point A. Referring to Fig. a, a+ ΣMA = 0;

5019.812 sin 130° - 2.866°) 13002 - P cos 130° + 2.866°) 13002 = 0 Ans. P = 266 N

Ans: P = 266 N 877

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8–129. A large crate having a mass of 200 kg is moved along the floor using a series of 150-mm-diameter rollers for which the coefficient of rolling resistance is 3 mm at the ground and 7  mm at the bottom surface of the crate. Determine the horizontal force P needed to push the crate forward at a constant speed. Hint: Use the result of Prob. 8–130.

P

SOLUTION Rolling Resistance: Applying the result obtained in Prob. 8–130. P = with aA

W1aA + aB2

2r = 7 mm, aB = 3 mm, W = 20019.812 = 1962 N, and r = 75 mm, we have P =

1962 7 + 3 2 75

= 130.8 N = 131 N

,

Ans.

Ans: P = 131 N 878

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8–130. The cylinder is subjected to a load that has a weight W. If the coefficients of rolling resistance for the cylinder’s top and bottom surfaces are aA and aB, respectively, show that a horizontal force having a magnitude of P = [W(aA + aB)]>2r is required to move the load and thereby roll the cylinder forward. Neglect the weight of the cylinder.

W P A

r B

SOLUTION + ©F = 0; : x

(RA)x - P = 0

(RA)x = P

+ c ©Fy = 0;

(RA)y - W = 0

(RA)y = W

a + ©MB = 0;

(1)

P(r cos fA + r cos fB) - W(aA + aB) = 0

Since fA and fB are very small, cos fA - cos fB = 1. Hence, from Eq. (1) P =

W(aA + aB) 2r

(QED)

879

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8–131. The 1.4-Mg machine is to be moved over a level surface using a series of rollers for which the coefficient of rolling resistance is 0.5 mm at the ground and 0.2 mm at the bottom surface of the machine. Determine the appropriate diameter of the rollers so that the machine can be pushed forward with a horizontal force of P = 250 N. Hint: Use the result of Prob. 8–130.

P

SOLUTION P =

W(aA + aB) 2r

250 =

1400 (9.81) (0.2 + 0.5) 2r

r = 19.2 mm Ans.

d = 38.5 mm

Ans: d = 38.5 mm 880

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9–1.

y

Locate the center of mass of the homogeneous rod bent into the shape of a circular arc.

30⬚

300 mm x

SOLUTION dL = 300 d u 30⬚

' x = 300 cos u ' y = 300 sin u

x =

L

2p 3

' x dL =

L

L-2p3

300 cos u (300du) 2p 3

dL

L-2p3

300d u

2p

=

(300)2 C sin u D -3 2p3 4 300 a pb 3

= 124 mm y = 0

Ans. Ans.

(By symmetry)

Ans: x = 124 mm, y = 0 881

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9–2.

y

Locate the centroid ( x, y ) of the uniform rod. Evaluate the integrals using a numerical method.

px y ⫽ a cos –– a

a

SOLUTION Differential Element. The length of the element shown shaded in Fig. a is x

dL = 2dx2 + dy2 =

Here,

dL =

C

1 + a

a –– 2

dy 2 b dx dx

dy p p p = a a - sin x b a b = - p sin x a a a dx

C

2 p p x b dx = 1 + p2 sin2 x dx a a C

1 + a -p sin

The centroid of the element is located at xc = x and yc = y = a cos Eq. 9–5 we have a>2

x =

LL

x~ dL =

LL

dL

x

L0

a>2

B

1 + p 2 sin

a>2

y =

LL

y~ dL

LL

= dL

L0

p x dx a

p 1 + p sin x dx a B 2

L0

2

a cos

2

0.3444a2 = 0.299a 1.1524a

p 2 p x 1 + p 2 sin x dx a B a

a>2

L0

=

B

1 + p 2 sin 2

p x dx a

p x. Applying a

=

Ans.

0.6191a 2 = 0.537a Ans. 1.1524a

Ans: x = 0.299a y = 0.537a 882

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9–3. y

Locate the center of gravity x of the homogeneous rod. If the rod has a weight per unit length of 100 N>m, determine the vertical reaction at A and the x and y components of reaction at the pin B.

1m B

1m y  x2 A

x

Solution Length

And

Arm. The length of the differential element is dy dy 2 dL = 2dx2 + dy2 = c 1 + a b d dx and its centroid is ~ x = x. Here = 2x. A dx dx Perform the integration L =

Moment

LL

dL =

L0

= 2

1m

L0

= cx

21 + 4x2 dx

1m

A

A

x2 +

x2 +

= 1.4789 m LL

x~ dL =

L0

= 2

1 dx 4

1 1 1 1m + ln ax + x2 + b d 4 4 A 4 0

1m

x21 + 4x2 dx

L0

1m

x

A

x2 +

1 dx 4

2 1 3>2 1 m = c ax2 + b d 3 4 0

= 0.8484 m2 Centroid. x =

~ 0.8484 m2 1L x dL = 0.5736 m = 0.574 m = 1.4789 m 1L dL

Ans.

Equations of Equilibrium. Refering to the FBD of the rod shown in Fig. a + ΣFx = 0; S

Ans.

Bx = 0

a+ΣMB = 0;  100(1.4789) (0.4264) - Ay(1) = 0 Ans.

Ay = 63.06 N = 63.1 N a+ΣMA = 0;  By(1) - 100(1.4789) (0.5736) = 0

Ans.

By = 84.84 N = 84.8 N

Ans: x = 0.574 m Bx = 0 Ay = 63.1 N By = 84.8 N 883

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*9–4. y

Locate the center of gravity y  of the homogeneous rod.

1m B

1m y  x2 A

x

Solution Length

And

Arm. The length of the differential element is dy dy 2 dL = 2dx2 + dy2 = c 1 + a b d dx and its centroid is ~ y = y. Here = 2x. A dx dx Perform the integration, L =

Moment

LL

L0

dL =

= 2

1m

21 + 4x2 dx

L0

= cx

1m

A

A

x2 +

= 1.4789 m LL

y~ dL =

L0

= 2

1 dx 4

1 1 1 1m + ln ax + x2 + b d 4 4 A 4 0

1m

x2 21 + 4x2 dx

L0

= 2c

x2 +

1m

x

2

A

x2 +

1 dx 4

x 1 1 1 1 1m 1 3 2 x x2 + ln ax + x2 + b d ax + b 4A 32 A 4 128 A 4 0 4

= 0.6063 m2 Centroid.

0.6063 m2 1L y dL y = = 0.40998 m = 0.410 m = 1.4789 m 1L dL ~



Ans.

Ans: y = 0.410 m 884

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9–5. y

Determine the distance y  to the center of gravity of the homogeneous rod.

1m

2m

Solution Length

And

y  2x3

x

Moment

Arm. The length of the differential element is

dy 2 y = y. Here dL = 2dx2 + dy2 = a 1 + a b b dx and its centroid is at ~ A dx dy = 6x2. Evaluate the integral numerically, dx L = LL

~

LL

dL =

y dL =

L0

L0

1m

1m

21 + 36x4 dx = 2.4214 m

3

2x 21 + 36x4 dx = 2.0747 m2

Centroid. Applying Eq. 9–5, y =

~ 2.0747 m2 1L y dL = 0.8568 = 0.857 m = 2.4214 m 1L dL

Ans.

Ans: y = 0.857 m 885

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9–6. Locate the centroid y of the area.

y

y

1 1 – – x2 4

1m x 2m

SOLUTION Area and Moment Arm: The area of the differential element is y 1 1 1 ' = a 1 - x2 b . dA = ydx = a1 - x2 b dx and its centroid is y = 4 2 2 4 Centroid: Due to symmetry x = 0

Ans.

Applying Eq. 9–4 and performing the integration, we have 2m

' ydA

y =

LA

=

1 2 1 1 ¢ 1 - x ≤ ¢ 1 - x2 ≤ dx 2 4 4 L- 2m 2m

dA

LA

L- 2m

=

¢

¢1 -

1 2 x ≤ dx 4

x x3 x 5 2m + ≤` 2 12 160 - 2m x ¢x ≤` 12 - 2m 3

2m

=

2 m 5

Ans.

Ans: y = 886

2 m 5

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9–7. Locate the centroid x of the shaded area. Solve the problem by evaluating the integrals using Simpson’s rule.

y y = 0.5ex2

SOLUTION At x = 1 m 2

y = 0.5e1 = 1.359 m 1

LA

dA =

L0

1

(1.359 - y) dx =

L0

x

2

a 1.359 = 0.5 ex b dx = 0.6278 m2

1m

x = x 1

x dA =

LA

L0

2

x a 1.359 - 0.5 ex b dx

= 0.25 m3

x =

x dA LA LA

dA

=

0.25 = 0.398 m 0.6278

Ans.

Ans: x = 0.398 m 887

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*9–8. Locate the centroid y of the shaded area. Solve the problem by evaluating the integrals using Simpson’s rule.

y y = 0.5ex2

SOLUTION 1

dA =

L0

LA

y =

1

(1.359 - y) dx =

L0

2

2

a1.359 - 0.5ex b dx = 0.6278 m

1.359 + y 2

x 1m

1

y dA =

LA

L0

a

x2

1.359 + 0.5 e 2 b A 1.359 - 0.5 ex B dx 2

1

=

y =

1 2 a 1.847 - 0.25 e2x b dx = 0.6278 m3 2 L0

y dA LA dA

=

0.6278 = 1.00 m 0.6278

Ans.

LA

Ans: y = 1.00 m 888

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9–9. Locate the centroid x of the parabolic area. y

h y  ax2

SOLUTION

x b

From the figure, h = ab2, and therefore, a = h b 2 Differential Element: The area element parallel to the x axis shown shaded in Fig. a will be considered. The area of the element is

= dA xdy =

b 12

h

y1 2 dy

Centroid: The centroid of the element is located at x=

= ∫ dA A

h

b

y dy ∫0= h1 2 12

( )

x b 12 = y and y= y. 2 2h1 2

h

2b 2 = y3 2 bh 12 3 3h 0 h

h

b 1 2  b 1 2  xdA ∫0  2h1 2 y  h1 2 y dy  ∫    A = x = = 2 dA bh ∫A 3

b2  y 2    y dy 2h  2   0 3 2h= = b 2 2 8 bh bh 3 3

h b2

∫0

Ans.

Ans: x = 889

3 b 8

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9–10. Locate the centroid of the shaded area.

y

y  a cospx L a

L 2

x

L 2

Solution Area And Moment Arm. The area of the differential element shown shaded in y p a p Fig. a is dA = ydx = a cos x dx and its centroid is at y~ = = cos x. 2 2 2 L Centroid. Perform the integration

y =

~ 1A y dA

1A dA

L>2

=

p p a a cos xbaa cos x dxb L L L-L>2 2 L>2

L-L>2

L>2

=

=

L-L>2

p x dx L

a2 2p x + 1b dx acos 4 L L>2

L-L>2

a cos

p x dx L

L>2 a2 L 2p a sin x + xb ` 4 2p L -L>2

a

=

a cos

L>2 aL p sin xb ` p L -L>2

a2 L>4

2aL>p

=

p a 8

Ans.

x = 0

Ans.

Due to Symmetry,

Ans: p a 8 x = 0 y =

890

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9–11. y

Locate the centroid x of the shaded area.

4m y

1 2 x 4 x

4m

Solution Area And Moment Arm. The area of the differential element shown shaded in Fig. a 1 is dA = x dy and its centroid is at ~ x = x. Here, x = 2y1>2 2 Centroid. Perform the integration

x =

~ 1A x dA

1A dA

=

=

L0

4m

1 a2y1>2 ba2y1>2 dyb 2 L0

4m

2y1>2 dy

3 m 2

Ans.

Ans: x = 891

3 m 2

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*9–12. y

Locate the centroid y of the shaded area.

4m y

1 2 x 4 x

4m

Solution Area And Moment Arm. The area of the differential element shown shaded in Fig. a y = y. Here, x = 2y1>2. is dA = x dy and its centroid is at ∼ Centroid. Perform the integration

y =

~ 1A y dA

1A dA

=

=

=

L0

4m

L0

y a2y1>2 dyb

4m

2y1>2 dy

4m 4 a y 5>2 b ` 5 0 4m 4 a y3>2 b ` 3 0

12 m 5

Ans.

Ans: y = 892

12 m 5

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9–13. y

Locate the centroid x of the area.

h y  —2 x2 b

h

x b

SOLUTION dA = y dx ' x = x

x =

LA

b

' x dA

LA

= dA

h 3 x dx 2 L0 b b

h 2 x dx 2 L0 b

=

B

h 4 b x R 4b2 0

h B 2 x3 R 3b 0

b

=

3 b 4

Ans.

Ans: x = 893

3 b 4

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9–14. Locate the centroid y of the shaded area.

y

h y  —2 x2 b

h

SOLUTION

x b

dA = y dx y ' y = 2

y =

LA

b

' y dA

LA

= dA

h2 4 x dx 4 L0 2b b

h 2 x dx 2 b L0

=

B

h2 5 b x R 10b4 0

h B 2 x3 R 3b 0

b

=

3 h 10

Ans.

Ans: y = 894

3 h 10

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9–15. y

Locate the centroid x of the shaded area.

4m

y

4

1 2 –– x 16 8m

SOLUTION dA = 14 - y2dx = a

1 2 x b dx 16

' x = x

x =

LA

8

' xdA

LA x = 6m

= dA

L0 L0

xa 8

a

x2 b dx 16

1 2 x b dx 16 Ans.

Ans: x = 6m 895

x

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*9–16. y

Locate the centroid y of the shaded area.

4m

y

4

1 2 –– x 16 8m

SOLUTION dA = 14 - y2dx = a y =

y =

1 2 x b dx 16

4 + y 2

LA

8

' ydA

LA

= dA

x2 x2 1 b a b dx ¢8 2 L0 16 16 8

L0

a

1 2 x b dx 16

y = 2.8 m

Ans.

Ans: y = 2.8 m 896

x

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9–17. Locate the centroid y of the shaded area.

y

y

h xn –– an h

SOLUTION dA = y dx

x a

y y = 2 ' ydA

y =

LA

1 2

= dA

LA

a

L0 L0

h2 a 2n

x 2n dx =

a h an

xn dx

h2(a 2n + 1) 2a 2n(2n + 1) h(a n + 1) a n(n + 1)

=

hn + 1 2(2n + 1)

Ans.

Ans: y = 897

hn + 1 2(2n + 1)

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9–18.

y

Locate the centroid y of the area. 2 ––

y ⫽ x3 4m

x 8m

SOLUTION Area: Integrating the area of the differential element gives A =

LA

dA =

8m

8m

x

L0

2>3

3 dx = c x5>3 d 2 0 5

= 19.2 m2

Ans.

' 1 Centroid: The centroid of the element is located at y = y>2 = x2>3. Applying 2 Eq. 9–4, we have 8m '

y =

y dA LA

L0

=

dA LA

8m

1 2>3 2>3 1 4>3 x A x B dx x dx L0 2 2 = 19.2 19.2

8m

=

3 7>3 c x d2 14 0 19.2

= 1.43 m

Ans.

m

Ans: A = 19.2 m2, y = 1.43 m 898

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9–19. y

Locate the centroid x of the shaded area.

y   h2 x2h a h

x

a

Solution Area And Moment Arm. The area of the differential element shown shaded in Fig. a h is dA = y dx = a - 2 x2 + hbdx and its centroid is at ~ x = x. a Centroid. Perform the integration,

x =

~ 1A x dA

1A dA

=

L0

a

L0

x aa

a-

h 2 x + hbdx a2

h

a

2

2

x + hbdx

h 4 h 2 a x + x b` 2 4a2 0 = a h 3 a - 2 x + hxb ` 3a 0 a-

=

3 a 8

Ans.



Ans: x = 899

3 a 8

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*9–20. y

Locate the centroid y of the shaded area.

y   h2 x2h a h

x

a

Solution Area And Moment Arm. The area of the differential element shown shaded in Fig. a y 1 h2 h is dA = y dx = a - 2 x2 + hbdx and its centroid is at ~ y = = a - x2 + hb. 2 2 a a Centroid. Perform the integration,

y =

a

~ 1A y dA

1 h h a - 2 x2 + hba - 2 x2 + hbdx a a L0 2 = a h dA 2 1A a - 2 x + hbdx L0 a a 1 h2 5 2h2 3 2 a 4x x + h xb ` 2 5a 3a2 0 = a h 3 a - 2 x + hxb ` 3a 0

=

2 h 5

Ans.



Ans: y = 900

2 h 5

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9–21.

y

Locate the centroid x of the area.

xy ⫽ c2

SOLUTION Differential Element: The element parallel to the y axis shown shaded in Fig. a will be considered. The area of the element is

x a

c2 dA = y dx = dx x

b '

Centroid: The centroid of the element is located at x = x Area: Integrating, A =

LA

dA =

b 2

b c b dx = c2 ln x 2 = c2 ln a x a La

We have b

b

b

c2 c 2x 2 dx b c2 dx x b - a a LA La La = = x = = = b b b b dA c2 ln c2 ln c2 ln ln a a a a LA ' x dA

xa

Ans.

Ans: x =

901

b - a b ln a

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9–22.

y

Locate the centroid y of the area.

xy ⫽ c2

SOLUTION

x

Differential Element: The element parallel to the y axis shown shaded in Fig. a will be considered. The area of the element is

a b

c2 dA = y dx = dx x '

Centroid: The centroid of the element is located at y =

2 y c = . 2 2x

Area: Integrating, A =

LA

dA =

b 2

b c b dx = c2 ln x 2 = c2 ln a a La x

Ans.

We have b

b

y =

LA

yc dA

LA

= dA

c4 - 3 2x c2 ln

c4 c2 c2 dx b a dx b 2 x 2x La 2x = b b c2 ln c2 ln a a

b

a

=

La

a

b a

=

c2(b - a) b 2ab ln a

Ans.

Ans: b a c2(b - a) y = b 2ab ln a

A = c2ln

902

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9–23. y

Locate the centroid x of the area.

y

h

h n — nx a

h

SOLUTION

x

a

dA = y dx ' x = x

x =

LA

LA

=

h 2

B hx a

x =

=

L0

dA

B x2 -

¢ hx -

a

' x dA

a

L0

¢h -

h1xn + 22 a 1n + 22 n

h1x

n+1

2

a 1n + 12 n

h h ba 2 2 n + 2

h ba ah n + 1

=

R

R

h n+1 x ≤ dx an h n x ≤ dx an

a 0

a 0

a(1 + n) 2(2 + n)

Ans.

Ans: x = 903

a(1 + n) 2(2 + n)

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*9–24. y

Locate the centroid y of the area.

y

h

h n — x an

h

x

a

SOLUTION dA = y dx y ' y = 2

y =

LA

LA

=

y =

h2 1 h2 ¢ h2 - 2 n xn + 2n x2n ≤ dx 2 L0 a a a

' y dA =

a

dA

L0

¢h -

h n x ≤ dx an

2h21xn + 12 h21x 2n + 12 a 1 2 + 2n Bh x - n R 2 a 1n + 12 a 12n + 12 0

B hx -

h1x n + 12 a 1n + 12 n

2n2 h 21n + 1212n + 12 n n + 1

=

R

a 0

hn 2n + 1

Ans.

Ans: y = 904

hn 2n + 1

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9–25.

y

x2 ⫹ y2 ⫽ r 2

If the density at any point in the quarter circular plate is defined by r = r0xy, where r0 is a constant, determine the mass and locate the center of mass (x, y) of the plate. The plate has a thickness t. r

x

SOLUTION Differential Element: The element parallel to the y axis shown shaded in Fig. a will be considered. The mass of this element is dm = r dV = r0 xy(ty dx) = r0 txy2dx = r0 t(r 2x - x 3) dx Mass: Integrating, r

m =

dm =

Lm

L0

r0 t(r2x - x3) dx = r0 ta

r2x2 x4 2 r 1 b = r0 r4 t 2 4 0 4

Ans.

Center of Mass: The center of mass of the element is located at xc = x. Applying Eq. 9–2 we have r

x =

Lm

x~ dm =

Lm

=

L0

1 r r4 t 4 0

dm

r0 t a

r

x cr0 t(r2x - x3)dx d

r2x3 x5 2 r b 3 5 0 1 r0 r4t 4

=

=

L0

r0 t ar2x2 - x4 bdx 1 r r4t 4 0

8 r 15

Ans.

By considering the element parallel to the x axis shown shaded in Fig. b, dm = r dV = r0xy(tx dy) = r0 tx2y dy = r0 t(r 2y - y 3) dy. In this case, yc = y. Applying Eq. 9–2, we have r

y =

Lm

y~ dm

Lm

= dm

L0

r

y c r0 t(r2y - y3)dy d 1 r r4 t 4 0

=

L0

r0 t(r2y2 - y4) dy =

1 r r 4t 4 0

8 r 15

Ans.

Ans: 1 r r4t 4 0 8 x = r 15 8 y = r 15 m =

905

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9–26. Locate the centroid x of the area.

px y ⫽ a cos –– a

a

SOLUTION Differential Element: The element parallel to the y axis shown shaded in Fig. a will be considered. The area of the element is dA = y dx = a cos

x a –– 2

p x dx a

' Centroid: The centroid of the element is located at x = x. Area: Integrating, a>2

A =

LA

dA =

a cos

L0

p a2 p 2 a>2 a2 x x dx = sin = a p a 0 p

Ans.

We have a>2

x =

L0 LA

=

a>2

' x dA

L0

=

x a a cos

p x dx b a

a2 p

dA

p a3 p a>2 a2 x sin x + 2 cos x 2 p a a p 0 2

a p

= a

a>2

=

L0

ax cos

p x dx a

a2 p

p - 2 ba 2p

Ans.

Ans: A =

906

a2 p - 2 ,x = a ba p 2p

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9–27. Locate the centroid y of the area.

px y ⫽ a cos –– a

a

SOLUTION Differential Element: The element parallel to the y axis shown shaded in Fig. a will be considered. The area of the element is dA = y dx = a cos

x a –– 2

p x dx a

y a p ' Centroid: The centroid of the element is located at y = = cos x. a 2 2 Area: Integrating, a>2

A =

LA

dA =

L0

a cos

p a2 p a>2 a2 x dx = sin x 2 = a p a 0 p

Ans.

We have

y =

LA

LA

=

a>2

' y dA

L0

=

p p a a cos x b a a cos x dx b a a 2 a2 p

dA

a2 4 L0

a>2

a cos

2p x + 1 b dx a 2

a p

=

a>2 2

=

L0

a p cos2 x dx a 2 a2 p

a>2 2p a2 a a sin x + xb 2 a 4 2p 0 2

a p

=

p a 8

Ans.

Ans: A =

907

a2 p ,y = a p 8

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*9–28. Locate the centroid x of the shaded area.

y

y  a sin

x a

a x

ap

SOLUTION Area and Moment Arm: The area of the differential element is x dA = ydx = a sin dx and its centroid are x = x a

x =

LA

pa

' xdA

LA

= dA

L0

x a a sin pa

L0 c a3 sin =

a sin

x dxb a x dx a

pa x x - x aa2 cos b d ` a a 0

a - a2 cos =

x pa b` a 0

p a 2

Ans.

Ans: x = 908

p a 2

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9–29. y

Locate the centroid y of the shaded area.

y

a sin

x a

a x

ap

SOLUTION Area and Moment Arm: The area of the differential element is y a x x = sin . dA = ydx = a sin dx and its centroid are y = a a 2 2 pa 1 2x 1 2 x x a bd` c a a x - a sin sin aa sin dxb a 4 2 a a 0 pa L0 2 y = LA = = = pa pa 8 x x 2 dA a sin dx b a a cos ` a LA L0 a 0 pa

ydA

Ans.

Ans: y = 909

pa 8

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9–30. The steel plate is 0.3 m thick and has a density of 7850 kg>m3. Determine the location of its center of mass. Also compute the reactions at the pin and roller support.

y y2  2x 2m

x A

SOLUTION 2m

y1 = -x1 y 22

y  x

= 2x2

B

dA = 1y2 - y12 dx =

A 22x + x B dx

2m

' x = x y2 + y1 22x - x ' = y = 2 2

x=

LA

=

LA

y =

LA

2

' x dA dA

c =

A 22x + x B dx

2 2 22 5>2 1 x + x3 d 5 3 0

2 222 3>2 1 c x + x2 d 3 2 0

2

= dA

2

L0

' y dA

LA

L0

x A 22x + x B dx

22x - x A 22x + x B dx 2 L0 2

L0

c =

A 22x + x B dx

= 1.2571 = 1.26 m Ans.

2 x2 1 - x3 d 2 6 0

2 222 3>2 1 c x + x2 d 3 2 0

= 0.143 m

Ans.

A = 4.667 m2 W = 785019.81214.667210.32 = 107.81 kN a + ©MA = 0;

- 1.25711107.812 + NB A 2 22 B = 0 NB = 47.92 = 47.9 kN

+ ©F = 0; : x

Ans.

- A x + 47.92 sin 45° = 0 A x = 33.9 kN

+ c ©Fy = 0;

Ans.

A y + 47.92 cos 45° - 107.81 = 0 A y = 73.9 kN

Ans.

Ans: x = 1.26 m y = 0.143 m NB = 47.9 kN Ax = 33.9 kN Ay = 73.9 kN 910

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9–31. y

Locate the centroid x of the shaded area.

h y  h  —n xn a h

h yh— a x x

a

Solution

h n h x and y1 = h - x. Thus, the an a area of the differential element shown shaded in Fig. a is dA = ( y2 - y1 ) dx Area And Moment Arm. Here, y2 = h -

= ch -

h n h h h x - ah - xb d dx = a x - n xn bdx and its centroid is ~ x = x. an a a a

Centroid. Perform the integration

x =

a

~ 1A x dA

h h x a x - n xn bdx a a L0 = a h h n dA 1A a x - n x bdx a L0 a =

c

c

a h 3 h x - n xn + 2 d ` 3a a (n + 2) 0

a h 2 h x - n xn + 1 b ` 2a a (n + 1) 0

ha2 (n - 1) =

3(n + 2) ha(n - 1) 2(n + 1)

= c

2(n + 1) 3(n + 2)

da

Ans.



Ans: x = c 911

2(n + 1) 3(n + 2)

da

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*9–32. y

Locate the centroid y of the shaded area.

h y  h  —n xn a h

h yh— a x x

a

Solution

h n h x and y1 = h - x. Thus, the an a area of the differential element shown shaded in Fig. a is dA = ( y2 - y1 ) dx h h h h = c h - n xn - ah - xb d dx = a x - n xn bdx and its centroid is at a a a a Area And Moment Arm. Here, y2 = h -

∼

y = y1 + a

y2 - y1 1 1 h h 1 h h b = ( y2 + y1 ) = ah - n xn + h - xb = a2h - n xn - xb. 2 2 2 a a 2 a a

Centroid. Perform the integration

y =

a

~ 1A y dA

1 h h h h a2h - n xn - xba x - n xn bdx 2 a a a a L0 = a h n h dA 1A a x - n x bdx a a L0 =

a 1 h2 2 h2 2h2 h2 x2n + 1 d ` c x - 2 x3 - n xn + 1 + 2n 2 a a (n + 1) 3a a (2n + 1) 0

c

=

h2a c

= c

a h 2 h x - n xn + 1 b ` 2a a (n + 1) 0

(4n + 1) (n - 1)

6(n + 1)(2n + 1) hac

n - 1 d 2(n + 1)

(4n + 1)

3(2n + 1)

d

 Ans.

d h

Ans: y = c 912

(4n + 1) 3(2n + 1)

d h

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9–33. y

Locate the centroid x of the shaded area.

a

h

h y  –– ax

y(

h )(xb) ab x

b

Solution

a a - b b y + b. Thus the area y and x2 = a h h a - b a by + b - y d dy = of the differential element is dA = ( x2 - x1 ) dy = c a h h x -x b a b b 1 ( b - y ) dy and its centroid is at ~x = x1 + 2 1 = ( x2 + x1 ) = y - y + . h h 2h 2 2 2

Area And Moment arm. Here x1 =

Centroid. Perform the integration,

x =

h

~ 1A x dA

a b b b a y y + b c ab - ybdy d h 2h 2 h L0 = h b dA 1A ab - ybdy h L0 =

c

h b b b2 (a - b)y2 + (b - 2a)y3 + yd ` 2 2h 2 6h 0 b 2 h aby y b` 2h 0

bh (a + b) 6 = bh 2 1 = (a + b) 3



Ans.

Ans: x = 913

1 (a + b) 3

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9–34. y

Locate the centroid y of the shaded area.

a

h

h y  –– ax

y(

h )(xb) ab x

b

Solution

a - b a by + b. Thus the area y and x2 = a h h a a - b by + b - y d dy = of the differential element is dA = ( x2 - x1 ) dy = c a h h b ab - ybdy and its centroid is at ~ y = y. h Centroid. Perform the integration,

Area And Moment arm. Here, x1 =

y =

~ 1A y dA

1A dA

=

=

L0

h

L0

y ab -

h

ab -

b ybdy h b ybdy h

b b 3 h a y2 y b` 2 3h 0 aby -

b 2 h y b` 2h 0

1 2 bh h 6 = = 1 3 bh 2

Ans.



Ans: y = 914

h 3

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9–35. y

Locate the centroid x of the shaded area.

yx

100 mm

y  1 x2 100 x 100 mm

Solution

1 2 x . Thus the area of the 100 1 2 x ) dx differential element shown shaded in Fig. a is dA = ( y2 - y1 ) dx = ( x 100 ~ and its centroid is at x = x. Area And Moment arm. Here, y2 = x and y1 =

Centroid. Perform the integration

x =

~ 1A x dA

1A dA

=

=

L0

100 mm

L0 a a

x ax -

100 mm

ax -

1 2 x bdx 100

1 2 x bdx 100

x3 1 4 100 mm x b` 3 400 0 x2 1 3 100 mm x b` 2 300 0

= 50.0 mm

Ans.



Ans: x = 50.0 mm 915

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*9–36. Locate the centroid y of the shaded area.

y

yx

100 mm

y  1 x2 100 x 100 mm

Solution Area And Moment arm. Here, x2 = 10y 1>2 and x1 = y. Thus, the area of the differential element shown shaded in Fig. a is dA = ( x2 - x1 ) dy = ( 10y1>2 - y ) dy and its centroid is at ~ y = y. Centroid. Perform the integration,

y =

~ 1A y dA

1A dA

=

=

L0

100 mm

L0

y a10y1>2 - ybdy

100 mm

a4y 5>2 -

a

1>2

a10y

- ybdy

y3 100 mm b` 3 0

y2 100 mm 20 3>2 y - b` 3 2 0

= 40.0 mm

Ans.



Ans: y = 40.0 mm 916

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9–37. y

Locate the centroid x of the circular sector.

r a

C

x

a x

Solution Area And Moment Arm. The area of the differential element shown in Fig. a is 1 2 dA = r 2 du and its centroid is at x~ = r cos u. 2 3 Centroid. Perform the integration

x =

=

~ 1A x dA

1A dA

L-a a

=

2 1 a r cos u ba r 2 du b 3 2 1 2 r du 2 L-a a

a 1 a r 3 sin u b ` 3 -a a 1 a r2 ub ` 2 -a

2 3 r sin a 3 = r2 a =

2 r sin a a b 3 a

Ans.

Ans: x = 917

2 r sin a a b 3 a

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9–38. Determine the location r of the centroid C for the loop of the lemniscate, r2 = 2a2 cos 2u, ( -45° … u … 45°).

r2 = 2a2 cos 2θ r O

θ C _ r

SOLUTION 1 1 (r) r du = r2 du 2 2

dA =

45°

A = 2

L0

1 (2a2 cos 2u)du = a2 C sin 2u D 45° = a2 0 2 45°

2

x dA

x =

LA

L0

=

A 23 r cos u B A 12r2du B

dA

2 3

=

a2

45°

L0

r3 cos u du 2

a

LA

x dA =

LA

x =

2 3 L0

45°

3

r cos u du =

2 3 L0

45°

A 2a2 B 3/2 cos u (cos 2 u)3/2du = 0.7854 a3

0.7854 a 3 = 0.785 a a2

Ans.

Ans: x = 0.785 a 918

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9–39. Locate the center of gravity of the volume. The material is homogeneous.

z

2m

SOLUTION

y 2 = 2z

2m

Volume and Moment Arm: The volume of the thin disk differential element is ' dV = py2dz = p12z2dz = 2pzdz and its centroid z = z.

y

Centroid: Due to symmetry about z axis - = y -= 0 x

Ans.

Applying Eq. 9–3 and performing the integration, we have

Lv z = Lv

=

z12pzdz2

L0

dV

2m

L0

z3 3

2m

z2 2p 2

2m

2p =

2m

' z dV

0

=

2pzdz

4 m 3

Ans.

0

Ans: x = y = 0 4 z = m 3 919

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*9–40. Locate the centroid y of the paraboloid.

z

z2 = 4y 4m y

SOLUTION Volume and Moment Arm: The volume of the thin disk differential element is ' dV = pz2dy = p14y2dy and its centroid y = y.

4m

Centroid: Applying Eq. 9–3 and performing the integration, we have

y =

LV

4m

' ydV

LV

= dV

L0

y3p14y2dy4 4m

p14y2dy

L0 4p =

y3 3

4m

2

4m

y 4p 2

0

Ans.

= 2.67 m

0

Ans: y = 2.67 m 920

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9–41. Locate the centroid z of the frustum of the right-circular cone.

z r

h R

SOLUTION Volume y =

and

Moment

1r - R2z + Rh h

Arm:

From

the

geometry,

y - r h - z = , R - r h

y

x

. The volume of the thin disk differential element is

dV = py2dz = pc a =

1r - R2z + Rh h

2

b ddz

p c1r - R22z2 + 2Rh1r - R2z + R2h2 ddz h2

and its centroid z = z. Centroid: Applying Eq. 9.3 and performing the integration, we have

z =

LV

LV

zb

h

' z dV = dV

=

=

L0

p 31r - R22z2 + 2Rh1r - R2z + R2h24dz r h2

h

p 3r - R22z2 + 2Rh1r - R2z + R2h24dz 2 L0 h 4 2 h z3 p 2 z 2 2 z 1r R2 + 2Rh1r R2 + R h B ¢ ≤ ¢ ≤ ¢ ≤ R ` 4 3 2 h2 0 3 p 2 z 1r R2 3 h2

+ 2Rh1r - R2

z2 2

+ R2h21z2 `

R2 + 3r2 + 2rR h 4 R2 + r2 + rR

h 0

Ans.

Ans: z =

921

R2 + 3r 2 + 2rR h 4(R2 + r 2 + rR)

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9–42. Locate the centroid of the solid. z

a h 2 z2 = h– y a

Solution x = y = 0 L

dV =

L

zdV =

Ans.

(By symmetry)

L0

x

h 2

h

py2dz = p

L0

h

y

pa2 5 h pa2h 4 z dz = c z d = 4 5 5h4 0 L0 h a

h

py2zdz =

pa2 pa2h2 pa2 6 h 5 z dz = c z d = 6 h4 L0 6h4 0

pa2h2 6 5 Lv = z = = h 6 pa2h dV 5 Lv zdV

Ans.

Ans: z =

922

5 h 6

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9–43. Locate the centroid z of the volume.

z 1m

y2  0.5z 2m

Solution y

Volume And Moment arm. The volume of the thin disk differential element shown shaded in Fig. a is dV = py2 dz = p(0.5z)dz and its centroid is at ~ z = z. Centroid. Perform the integration

z =

~ 1V z dV

1V dV

=

=

=

L0

2m

L0

x

z[p(0.5z)dz]

2m

p(0.5z)dz

0.57 3 2 m z ` 3 0

0.5p 2 2 m z ` 2 0

4 m 3

Ans.

Ans: z = 923

4 m 3

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*9–44. Locate the centroid of the quarter-cone.

z

h

SOLUTION

a

' z = z r =

y x

a (h - z) h

dV =

p 2 p a2 r dz = (h - z)2 dz 4 4 h2

dV =

p a2 2 z3 h p a2 d (h2 - 2hz + z2) dz = ch z - hz2 + 2 2 3 0 4 h L0 4h

h

L

=

pa2 h p a2 h3 a b = 2 3 12 4h h

2

z3 z4 h p a 2 2 z2 pa ' 2 2 2h + d (h 2hz + z ) z dz = ch z dV = 2 2 3 4 0 4 h2 4 h L0 L =

pa2h2 p a2 h4 b = a 48 4 h2 12

p a2 h2 h 48 L z = = = 4 p a2h dV 12 L ' z dV

h

Ans.

h

pa2 pa 2 4r 4a ' (h - z)2 dz = (h3 - 3h2 z + 3hz2 - z3) dz xdV = 2 2 3p 3 4 h 4 h L0 L0 p h L =

3h4 h4 p a2 4a 4 4 + h b a h 2 4 4 h2 3ph

=

a3 h p a2 a h3 b = a 12 4 h2 3p ' xdV

a3 h 12 a L = x = y = = p p a2h dV 12 L

Ans.

Ans: h 4 a x = y = p

z =

924

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9–45. Locate the centroid of the ellipsoid of revolution.

z

y2 z2  1 b2 a 2 a x

SOLUTION

b

dV = p z2 dy b

dV = L

L0

' ydV = L

p a2 a 1 b

L0

y

y2 b

p a2ya 1 -

b dy = p a2 c y 2 y b

2

b dy = p a2 c 2

y3

b

3b

0

d = 2

2pa2b 3

2

y4 b y p a2b2 d = 2 2 4 4b 0

pa2b2 3 4 LV = b = y = 8 2pa2b dV 3 LV ' ydV

x = z = 0

Ans.

(By symmetry)

Ans.

Ans: 3 b 8 x = z = 0

y =

925

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9–46. z

The hemisphere of radius r is made from a stack of very thin plates such that the density varies with height r = kz, where k is a constant. Determine its mass and the distance to the center of mass G. G

_ z

r

SOLUTION

y

Mass and Moment Arm: The density of the material is r = kz. The mass of the thin disk differential element is dm = rdV = rpy2dz = kz3p(r2 - z2) dz4 and its ' centroid z = z. Evaluating the integrals, we have

x

r

m =

Lm

dm =

L0

kz3p(r2 - z2) dz4

= pk ¢

Lm

' z dm =

r2z2 z4 r pkr4 - ≤` = 2 4 0 4

Ans.

r

L0

z5kz3p(r2 - z2) dz46

= pk ¢

2pkr5 r2z3 z5 r - ≤` = 3 5 0 15

Centroid: Applying Eq. 9–2, we have

z =

Lm

' z dm

Lm

= dm

2pkr5>15 pkr4>4

=

8 r 15

Ans.

Ans:

pkr 4 4 8 z = r 15

m =

926

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9–47. Determine the location z of the centroid for the tetrahedron. Hint: Use a triangular “plate” element parallel to the x–y plane and of thickness dz.

z

b a

SOLUTION z = c a1 -

1 1 yb = c a 1 - x b a b c

L

dV =

c y

c

1 z abc z 1 (x)(y)dz = a a 1 - b ba 1 - b dz = c c 2 L0 6 L0 2

x

c

z a b c2 z 1 ' z a a1 - b ba 1 - bdz = z dV = c c 2 L0 24 L a b c2 c 24 L = z = = abc 4 dV 6 L ' z dV

Ans.

Ans: z = 927

c 4

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z

*9–48. The king’s chamber of the Great Pyramid of Giza is located at its centroid. Assuming the pyramid to be a solid, prove that this point is at z = 14 h, Suggestion: Use a rectangular differential plate element having a thickness dz and area (2x)(2y).

h

SOLUTION

x

dV = 12x212y2 dz = 4xy dz x = y =

a

a

a 1h - z2 h

h

L L

dV =

' z dV =

4 a2 4 a2 2 z3 h 4 a2 h 2 2 1h z2 dz = h z hz + = B R 2 3 0 3 h2 L0 h h

4 a2 4 a2 z2 a2 h2 z3 z4 h 1h - z22z dz = 2 B h2 - 2h + R = 2 2 3 4 0 3 h L0 h

' L z =

' z dV

L

= dV

a 2 h2 3 4 a2 h 3

=

h 4

(QED)

928

a

a

y

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9–49. Locate the center of gravity z of the frustum of the paraboloid. The material is homogeneous. z

z

SOLUTION

6.25 (0.25

y2)

0.3 m

Given: 1m

a

1m

b

0.5 m

c

0.3 m

y 0.5 m x

a

V

´ µ S ªb2  z b2  c2 º dz « » µ a ¬ ¼ ¶





0

a

z

1´ µ zS ªb2  z b2  c2 º dz « » Vµ a ¬ ¼ ¶





0

z

0.422 m

Ans.

Ans: z = 0.422 m 929

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9–50. Locate the centroid z of the spherical segment.

z

z2

a2

y2 1a — 2

C a

SOLUTION

z

2

2

2

dV = py dz = p(a - z ) dz y

z = z

z =

LV

a

' z dV

LV

p =

L a 2

a

p

dV

=

x

z (a2 - z2)dz

La2

(a 2 - z2)dz

p B a2 a

z4 a z2 b - a bR a 2 4 2

z3 a p B a (z) - a b R a 3 2 2

=

pB

a4 a4 a4 a4 + R 2 4 8 64

a3 a3 a3 + pBa R 3 2 24 3

z = 0.675 a

=

pB

9a 4 R 64

pB

5a 3 R 24 Ans.

Ans: z = 0.675a 930

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9–51. The steel and aluminum plate assembly is bolted together and fastened to the wall. Each plate has a constant width in the z direction of 200 mm and thickness of 20 mm. If the density of A and B is rs = 7.85 Mg>m3, and for C, ral = 2.71 Mg>m3, determine the location x of the center of mass. Neglect the size of the bolts.

y

100 mm 200 mm A x C

B 300 mm

SOLUTION ©m = 2 C 7.85(10)3(0.3)(0.2)(0.02) D + 2.71(10)3(0.3)(0.2)(0.02) = 22.092 kg ' ©xm = 150{2 C 7.85(10)3(0.3)(0.2)(0.02) D } +350 C 2.71(10)3(0.3)(0.2)(0.02) D = 3964.2 kg.mm x =

' 3964.2 ©xm = = 179 mm ©m 22.092

Ans.

Ans: x = 179 mm 931

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*9–52. Locate the center of gravity G1x, y2 of the streetlight. Neglect the thickness of each segment. The mass per unit length of each segment is as follows: rAB = 12 kg>m, rBC = 8 kg>m, rCD = 5 kg>m, and rDE = 2 kg>m.

1m

y

90 1 m 1m

1.5 m

D

E

1m C G (x, y)

3m

B

SOLUTION 2(1) p ' b a b(5) ©xm = 0(4)(12) + 0(3)(8) + 0(1)(5) + a1 p 2

4m

+ 1.5 (1) (5) + 2.75 (1.5) (2) = 18.604 kg # m

A x

p ©m = 4 (12) +3 (8)+ 1(5) + (5)+ 1(5) + 1.5 (2) = 92.854 kg 2 x =

' ©xm 18.604 = = 0.200 m ©m 92.854

Ans.

2(1) p ' b a b(5) ©ym = 2 (4) (12) + 5.5 (3)(8) + 7.5(1) (5) + a8 + p 2 + 9 (1) (5) + 9(1.5) (2) = 405.332 kg # m y =

' ©ym 405.332 = = 4.37 m ©m 92.854

Ans.

Ans: x = 0.200 m y = 4.37 m 932

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9–53. Locate the centroid (x, y) of the metal cross section. Neglect the thickness of the material and slight bends at the corners.

y

50 mm

150 mm x 50 mm 100 mm 100 mm 50 mm

SOLUTION Centroid: The length of each segment and its respective centroid are tabulated below. L (mm)

' y (mm)

' y L (mm2)

1

50p

168.17

26415.93

2

180.28

75

13520.82

3

400

0

0

4

180.28

75

13520.82

©

917.63

Segment

53457.56

- = 0 Due to symmetry about y axis, x

y =

Ans.

' ©yL 53457.56 = = 58.26 mm = 58.3 mm ©L 917.63

Ans.

Ans: x = 0 y = 58.3 mm 933

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9–54. Determine the location y of the centroidal axis x - x of the beam’s cross-sectional area. Neglect the size of the corner welds at A and B for the calculation.

150 mm 15 mm B y 15 mm

150 mm

C

x

SOLUTION

A

' ©yA = 7.5(15) (150) + 90(150) (15) + 215(p) (50)2

50 mm

= 1 907 981.05 mm2 ©A = 15(150) + 150(15) + p(50)2 = 12 353.98 mm2 y =

' ©yA 1 907 981.05 = = 154 mm ©A 12 353.98

Ans.

Ans: y = 154 mm 934

x

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9–55. The truss is made from five members, each having a length of 4 m and a mass of 7 kg>m. If the mass of the gusset plates at the joints and the thickness of the members can be neglected, determine the distance d to where the hoisting cable must be attached, so that the truss does not tip (rotate) when it is lifted.

y

d B 4m

4m

C

4m 4m 60

A

4m

x D

SOLUTION ' ©xM = 4(7)(1+ 4 + 2 + 3 + 5) = 420 kg # m ©M = 4(7)(5) = 140 kg d = x =

' 420 ©xM = = 3m ©M 140

Ans.

Ans: d = 3m 935

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*9–56. A rack is made from roll-formed sheet steel and has the cross section shown. Determine the location 1x, y2 of the centroid of the cross section. The dimensions are indicated at the center thickness of each segment.

y 30 mm

80 mm 50 mm

SOLUTION

x

©L = 15 + 50 + 15 + 30 + 30 + 80 + 15 = 235 mm ' ©xL = 7.5(15) + 0(50) + 7.5(15) + 15(30) + 30(30) + 45(80) + 37.5(15) = 5737.50 mm2

15 mm

15 mm

' ©yL = 0(15) + 25(50) + 50(15) + 65(30) + 80(30) + 40(80) + 0(15) = 9550 mm2 x =

' ©xL 5737.50 = = 24.4 mm ©L 235

Ans.

y =

' ©yL 9550 = = 40.6 mm ©L 235

Ans.

Ans: x = 24.4 mm y = 40.6 mm 936

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9–57. z

Determine the location (x, y, z) of the centroid of the homogeneous rod.

200 mm x 30 600 mm 100 mm

Solution

y

Centroid. Referring to Fig. a, the length of the segments and the locations of their respective centroids are tabulated below Segment L(mm)

x~(mm)

y~(mm)

z~(mm)

x~ L(mm2)

0

0

100

0

1

200

2

600

300 cos 30° 300 sin 30°

0

3

100

600 cos 30° 600 sin 30°

–50

a

900

y~ L(mm2) z~ L(mm2) 0

155.88 ( 103 ) 90.0 ( 103 )

20.0 ( 103 ) 0

51.96 ( 103 ) 30.0 ( 103 ) - 5.0 ( 103 ) 207.85 ( 103 ) 120.0 ( 103 ) 15.0 ( 103 )

Thus,

x =

207.85 ( 103 ) mm2 Σ~ xL = = 230.94 mm = 231 mm ΣL 900 mm

Ans.



y =

120.0 ( 103 ) mm2 Σ~ yL = = 133.33 mm = 133 mm ΣL 900 mm

Ans.



z =

15.0 ( 103 ) mm2 Σ~ zL = = 16.67 mm = 16.7 mm ΣL 900 mm2

Ans.

Ans: x = 231 mm y = 133 mm z = 16.7 mm 937

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9–58. Locate the center of gravity 1x, y, z2 of the homogeneous wire.

z

400 mm

SOLUTION 2 (300) p ' a b (300) = 165 000 mm2 ©xL = 150(500) + 0(500) + p 2

y 300 mm

p ©L = 500 + 500 + a b (300) = 1471.24 mm 2 x =

' ©xL 165 000 = = 112 mm ©L 1471.24

x

Ans.

Due to symmetry, y = 112 mm

Ans.

p ' © z L = 200(500) + 200(500) + 0a b (300) = 200 000 mm2 2 z =

' © zL 200 000 = = 136 mm ©L 1471.24

Ans.

Ans: x = 112 mm y = 112 mm z = 136 mm 938

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9–59. To determine the location of the center of gravity of the automobile it is first placed in a level position, with the two wheels on one side resting on the scale platform P. In this position the scale records a reading of W1. Then, one side is elevated to a convenient height c as shown. The new reading on the scale is W2. If the automobile has a total weight of W, determine the location of its center of gravity G(x, y).

G

– y c

P x– b

W2

SOLUTION Equation of Equilibrium: First,we will consider the case in which the automobile is in a level position. Referring to the free-body diagram in Fig. a and writing the moment equation of equilibrium about point A, a + ©MA = 0;

W1(b) - W(x) = 0

x =

W1 b W

Ans.

c 2b2 - c2 and cos u = . Using the result of x b b and referring to the free-body diagram in Fig. b, we can write the moment equation From the geometry in Fig. c, sin u = of equilibrium about point A¿ . a + ©MA¿ = 0;

W2 B b ¢

y =

c 2b2 - c2 2b2 - c2 W1 ≤ R - W¢ ≤ a bb - Wa b y = 0 b b W b

b(W2 - W1) 2b2 - c2 cW

Ans.

Ans: W1 b W b(W 2 - W 1)2b2 - c2 y = cW x =

939

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* 9–60. Locate the centroid y of the cross-sectional area of the built-up beam.

y

200 mm

200 mm

20 mm 50 mm 150 mm 10 mm

Solution

300 mm 20 mm

10 mm 20 mm

Centroid: The centroid of each composite segment is shown in Fig. a. y =

x

225(450)(40) + 23400(200)(10) 4 + 510(400)(20) ΣyA = ΣA 450(40) + 2(200)(10) + 400(20) Ans.

= 324 mm

Ans: y = 324mm 940

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9–61. Locate the centroid y of the cross-sectional area of the beam constructed from a channel and a plate. Assume all corners are square and neglect the size of the weld at A.

20 mm y 350 mm C A

10 mm 70 mm

SOLUTION

325 mm

325 mm

Centroid: The area of each segment and its respective centroid are tabulated below. Segment

A (mm2)

' y (mm)

' y A (mm3)

1

350(20)

175

1 225 000

2

630(10)

355

2 236 500

3

70(20)

385

539 000

©

14 700

4 000 500

Thus, ' ©yA 4 000 500 ' y = = = 272.14 mm = 272 mm ©A 14 700

Ans.

Ans: y = 272 mm 941

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9–62.

y

Locate the centroid (x, y) of the member’s cross-sectional area.

30 mm

40 mm

SOLUTION

50 mm

Centroid: The area of each segment and its respective centroid are tabulated below. Segment

A (mm2)

' x (mm)

' y (mm)

' x A (mm3)

' y A (mm3)

1

1 13021902 2

20

30

27 000

40 500

2

30(90)

45

45

121 500

121 500

3

100(50)

110

25

550 000

125 000

©

9 050

698 500

287 000

x 60 mm

100 mm

Thus, x =

' 698 500 ©xA = = 77.18 mm = 77.2 mm ©A 9 050

Ans.

y =

' ©yA 287 000 = = 31.71 mm = 31.7 mm ©A 9 050

Ans.

Ans: x = 77.2 mm, y = 31.7 mm 942

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9–63. Determine the location y of the centroid C of the beam having the cross-sectional area shown.

150 mm 15 mm B

150 mm

y C

x

15 mm 15 mm

A 100 mm

Solution Centroid. The locations of the centroids measuring from the x axis for segments 1, 2 and 3 are indicated in Fig. a. Thus

y =

7.5(15)(150) + 90(150)(15) + 172.5(15)(100) Σ~ yA = ΣA 15(150) + 150(15) + 15(100) Ans.

= 79.6875 mm = 79.7 mm

Ans: y = 79.7 mm 943

x

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*9–64.

y

Locate the centroid y of the beam’s cross-sectional area.

50 mm

50 mm 10 mm

10 mm 10 mm 10 mm

10 mm 75 mm x

SOLUTION

100 mm

100 mm

Centroid: The centroid of each composite segment is shown in Fig. a. We have y =

' ©y A 5(180)(10) + 2[37.5(75)(10)] + 2[70(40)(10)] = = 29.6 mm ©A 180(10) + 2(75)(10) + 2(40)(10)

Ans.

Ans: y = 29.6 mm 944

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9–65. Determine the location y of the centroid of the beam’s crosssectional area. Neglect the size of the corner welds at A and B for the calculation. 35 mm A

SOLUTION

110 mm 2

35 35 ' b = 393 112 mm3 ©yA = p(25)2(25) + 15(110)(50 + 55) + p a b a50 + 110 + 2 2 ©A = p(25)2 + 15(110) + pa

35 2 b = 4575.6 mm2 2

C 15 mm B

y 50 mm

' ©yA 393 112 = = 85.9 mm y = ©A 4575.6

Ans.

Ans: y = 85.9 mm 945

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9–66. An aluminum strut has a cross section referred to as a deep hat. Locate the centroid y of its area. Each segment has a thickness of 10 mm.

30 mm

30 mm

– y C

x'

100 mm

80 mm

Solution Centroid: The area of each segment and its respective centroid are tabulated below. Segment 1 2 3 Σ

A(mm2) 40(10) 100(20) 60(10) 3 000

y(mm) 5 50 95

yA(mm3) 2 000 100 000 57 000 159 000

Thus, y =

ΣyA 159 000 = = 53.0 mm ΣA 3 000

Ans.

Ans: y = 53.0 mm 946

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9–67. Locate the centroid y of the concrete beam having the tapered cross section shown.

300 mm 300 mm 300 mm y C

SOLUTION

80 mm

360 mm

1 ' © y A = 900(80) (40) + 100(360) (260) + 2c (100) (360) (200) d = 19.44(106) mm3 2

x

100 mm

1 ©A = 900(80) + 100(360) + 2 c (100) (360) d = 0.144(106) mm2 2 y =

' © yA 19.44(106) = 135 mm = ©A 0.144(106)

Ans.

Ans: y = 135 mm 947

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*9–68. Locate the centroid y for the beam’s cross-sectional area.

120 mm

240 mm x y

Solution

240 mm

Centroid. The locations of the centroids measuring from the x axis for segments 1  and  2  are indicated in Fig. a. Thus 300(120)(600) + 120(240)(120) Σy~A    y = = ΣA 120(600) + 240(120)

120 mm

240 mm

= 248.57 mm = 249 mm

Ans:  y = 249 mm 948

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9–69. The gravity wall is made of concrete. Determine the location (x, y) of the center of gravity G for the wall.

y 1.2 m

_ x 3m

G _ y

0.4 m

x

2.4 m

Solution

0.6 m

0.6 m

1 1 ΣxA = 1.8(3.6)(0.4) + 2.1(3)(3) - 3.4a b(3)(0.6) - 1.2a b(1.8)(3) 2 2 = 15.192 m3

1 1 ΣyA = 0.2(3.6)(0.4) + 1.9(3)(3) - 1.4a b(3)(0.6) - 2.4a b(1.8)(3) 2 2 = 9.648 m3

ΣA = 3.6(0.4) + 3(3) -

1 1 (3)(0.6) - (1.8)(3) 2 2

= 6.84 m2 x =

ΣxA 15.192 = = 2.22 m ΣA 6.84

Ans.

y =

ΣyA 9.648 = = 1.41 m ΣA 6.84

Ans.

Ans: x = 2.22 m, y = 1.41 m 949

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9–70. Locate the centroid y for the cross-sectional area of the angle. a

a

–y

C

SOLUTION

t

t

Centroid : The area and the centroid for segments 1 and 2 are A 1 = t1a - t2 t t 22 a - t ' + b cos 45° + = 1a + 2t2 y1 = a 2 2 2cos 45° 4 A 2 = at t t 22 a ' y2 = a - bcos 45° + = 1a + t2 2 2 2cos 45° 4 Listed in a tabular form, we have Segment

A

' y

' yA

1

t1a - t2

22 1a + 2t2 4

22t 2 1a + at - 2t22 4

2

at

22 1a + t2 4

22t 2 1a + at2 4

©

t12a - t2

22t 2 1a + at - t22 2

Thus, ' ©yA y = = ©A =

22t 2 1a + at - t22 2 t12a - t2 22 a2 + at - t2

Ans.

2 2a - t

Ans: y = 950

22 ( a2 + at - t 2 ) 2(2a - t)

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9–71. Uniform blocks having a length L and mass m are stacked one on top of the other, with each block overhanging the other by a distance d, as shown. Show that the maximum number of blocks which can be stacked in this manner is n 6 L>d.

y 2d d

x L

Solution n = 2: x =

n = 3: x =

L L d + d = + 2a b 2 2 2 ad +

In general: x =

L L bW + a2d + bW 2 2 L d = + 3a b 2W 2 2

d L + na b 2 2

For stable stack: x = n …

L d + na b … L 2 2 L d

Ans.

Ans: n … 951

L d

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*9–72. A toy skyrocket consists of a solid conical top, rt = 600 kg>m3, a hollow cylinder, rc = 400 kg>m3, and a stick having a circular cross section, rs = 300 kg>m3. Determine the length of the stick, x, so that the center of gravity G of the skyrocket is located along line aa.

a

20 mm

5 mm

100 mm

3 mm 10 mm a G

x

SOLUTION 20 1 x ' ©xm = a b c a b p (5)2 (20) d (600) - 50 C p A 52 - 2.52 B (100) D (400) - C (x) p (1.5)2 D (300) 4 3 2 = - 116.24 A 106 B - x2(1060.29) kg # mm4>m3 1 ©m = c p (5)2 (20) d(600) + p A 52 - 2.52 B (100)(400) + C xp (1.5)2 D (300) 3 = 2.670 A 106 B + 2120.58x kg # mm3/m3 x =

' - 116.24(106) - x2(1060.29) ©xm = - 100 = ©m 2.670(106) + 2120.58x

- 116.24 A 106 B - x2 (1060.29) = - 267.0 A 106 B - 212.058 A 103 B x 1060.29x2 - 212.058 A 103 B x - 150.80 A 106 B = 0 Solving for the positive root gives x = 490 mm

Ans.

Ans:  x = 490 mm 952

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9–73. Determine the location x of the centroid C of the shaded area which is part of a circle having a radius r.

y

r a

SOLUTION

x

2r 1 1 2 ' ©xA = r 2 aa sin a b - (r sin a) (r cos a) a r cos a b 2 3a 2 3 r3 r3 sin a sin a cos2 a 3 3

=

r3 3 sin a 3

©A = =

x

a

Using symmetry, to simplify, consider just the top half:

=

C

1 1 2 r a - (r sin a) (r cos a) 2 2 1 2 sin2a r aa b 2 2

' ©xA = x = ©A

r3 3 1 2

sin3 a

r2 A a -

sin 2 a 2

B

=

2 3

r sin3a

a -

Ans.

sin 2a 2

Ans: x =

953

2 3r

sin3 a

a -

sin 2a 2

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9–74.

225 mm

225 mm

Locate the centroid y of the bulb-tee cross section. 50 mm

400 mm

SOLUTION

75 mm C

Centroid: The area of each segment and its respective centroid are tabulated below. Segment

A 1mm22

' y 1mm2

' y A 1mm32

1

450(50)

600

13 500 000

2

475(75)

337.5

12 023 437.5

3

1 122521752 2

125

1 054 687.5

4

300(100)

50

1 500 000

©

96 562.5

_ y

75 mm 100 mm x 300 mm

28 078 125

Thus, y =

' ©yA 28 078 125 = = 290.78 mm = 291 mm ©A 96 562.5

Ans.

Ans: y = 291 mm 954

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9–75. Determine the distance x to the centroid of the solid which consists of a cylinder with a hole of length h = 50 mm bored into its base.

y 120 mm

40 mm

20 mm

h

Solution ΣV = p(40)2(120) - p(20)2(50) = 172(103)p mm3 ΣxV = 60(p)(40)2(120) - 25(p)(20)2(50) - 11.02(106)p mm4 x =

11.02(106)p ΣxV = 64.1 mm = ΣV 172(103)p

Ans.

Ans: x = 64.1 mm 955

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*9–76. Determine the distance h to which a hole must be bored into the cylinder so that the center of mass of the assembly is located at x = 64 mm. The material has a density of 8 Mg>m3.

y 120 mm

40 mm x 20 mm

h

Solution ΣV = pr 22d - pr 21h ΣxV =

d h (p) 1r 22 2d - 1p2 1r 21 2h 2 2

d2 h2 1p2 1r 22 2 1p2 1r 21 2 ΣxV 2 2 x = = ΣV pr 22d - pr 21h 2x1r 22d2 - 2x1r 21h2 = d 2r 22 - h2r 21 r2 2 h2 - 2xh + d(2x - d)a b = 0 r1 Set

x = 64 mm,

r2 = 40 mm,

r1 = 20 mm, d = 120 mm

h2 - 128h + 3840 = 0 Solving, h = 80 mm

Ans.

or

h = 48 mm

Ans.

Ans: h = 48 mm 956

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z

9–77. Locate the center of mass z of the assembly. The material has a density of r = 3 Mg/m3. There is a 30-mm diameter hole bored through the center.

40 mm 20 mm 30 mm

100 mm –z y

Solution Centroid: Since the density is the same for the whole material, the centroid of the volume coincide with centroid of the mass. The volume of each segment and its respective centroid are tabulated below. Segment 1 2 3 4 Σ

V(mm3) 1 p(402)(60) 3 p(402)(100) 1 - p(202)(30) 3 -p(152)(130) 158.75p(103)

z(mm)

zV(mm4)

115

3.68p(106)

50

8.00p(106)

137.5

- 0.550p(106)

65

- 1.90125p(106) 9.22875p(106)

x

Thus, z =

9.22875p(106) ΣzV = = 58.13 mm = 58.1 mm ΣV 158.75p(103)

Ans.

Ans: z = 58.1 mm 957

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9–78.

z

Locate the centroid z of the homogenous solid formed by boring a hemispherical hole into the cylinder that is capped with a cone.

300 mm

SOLUTION Centroid: Since the solid is made of a homogeneous material, its center of mass coincides with the centroid of its volume. The centroid of each composite segment is shown in Figs. a and b. Since segment (3) is a hole, its volume should be considered negative. We have

150 mm

x

400 mm 150 mm y

1 3 2 p(0.2)(0.152)(0.4) + p(0.4 + 0.075)(0.152)(0.3) + a (0.15) b a - p(0.153)b ' 3 8 3 ©z V = z = ©V 1 2 p(0.152)(0.4) + p(0.152)(0.3) + a - p(0.153) b 3 3 =

2.7422(10 - 3)p 9(10 - 3)p

= 0.3047 m = 305 mm

Ans.

Ans: z = 305 mm 958

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9–79. Locate the center of mass z of the solid formed by boring a hemispherical hole into a cylinder that is capped with a cone. The cone and cylinder are made of materials having densities of 7.80 Mg>m3 and 2.70 Mg>m3, respectively.

300 mm

SOLUTION

150 mm

Centroid: The center of mass of each composite segment is shown in Figs. a and b. Since segment (3) is a hole, its mass should be considered negative. We have

x

400 mm 150 mm y

1 3 2 0.2 c 2700c p(0.152)(0.4) d d + (0.4 + 0.075)7800c p(0.152)(0.3) d + a (0.15) b c -2700 a p(0.153)b d ' 3 8 3 ©z m a = z = ©m 1 2 2700 cp(0.152)(0.4) d + 7800c p(0.152)(0.3) d + c -2700a p(0.153) b d 3 3 =

12.8545p = 0.3593 m = 359 mm 35.775p

Ans.

Ans: z = 359 mm 959

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*9–80. Locate the center of mass (x, y, z) of the homogeneous block assembly.

z

250 mm 200 mm

x

SOLUTION

100 mm

150 mm

Centroid: Since the block is made of a homogeneous material, the center of mass of the block coincides with the centroid of its volume. The centroid of each composite segment is shown in Fig. a.

150 mm 150 mm

1 (75)(150)(150)(550) + (225)(150)(150)(200) + (200) a b(150)(150)(100) ' 2.165625(109) 2 ©xV = 120 mm x = = = ©V 1 18(106) (150)(150)(550) + (150)(150)(200) + (150)(150)(100) 2

Ans.

1 (275)(150)(150)(550) + (450)(150)(150)(200) + (50) a b (150)(150)(100) ' ©yV 5.484375(109) 2 = 305 mm y = = = ©V 1 18(106) (150)(150)(550) + (150)(150)(200) + (150)(150)(100) 2

Ans.

1 (75)(150)(150)(550) + (75)(150)(150)(200) + (50) a b (150)(150)(100) ' 1.321875(109) 2 © zV z = = 73.4 mm = = ©V 1 18(106) (150)(150)(550) + (150)(150)(200) + (150)(150)(100) 2

Ans.

y

Ans: x = 120 mm y = 305 mm z = 73.4 mm 960

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9–81. Determine the distance z to the centroid of the shape which consists of a cone with a hole of height h = 50 mm bored into its base.

z

500 mm

SOLUTION 1 0.5 0.05 ' b - p (0.05)2 (0.05) a b © z V = p (0.15)2 A 0.5 B a 3 4 2 -3

C

h 50 mm

4

= 1.463(10 ) m ©V =

1 p (0.15)2 (0.5) - p (0.05)2 (0.05) 3

z

y

150 mm

x

3

= 0.01139 m z =

' 1.463 (10 - 3) ©z V = = 0.12845 m = 128 mm ©V 0.01139

Ans.

Ans: z = 128 mm 961

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9–82. z

Determine the distance h to which a 100-mm diameter hole must be bored into the base of the cone so that the center of mass of the resulting shape is located at z = 115 mm. The material has a density of 8 Mg>m3.

500 mm

SOLUTION

C

h

1 0.5 2 3 p10.152 10.52 4 1 2 3 p10.152 10.52

A B - p10.052 1h2 A B 2

h 2

- p10.05221h2

50 mm

_ z 150 mm

= 0.115 x

0.4313 - 0.2875 h = 0.4688 - 1.25 h2 h2 - 0.230 h - 0.0300 = 0 Choosing the positive root, h = 323 mm

Ans.

Ans: h = 323 mm 962

y

© Pearson Education Limited 2017. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. z

9–83. Locate the center of mass z of the assembly. The assembly consists of a cylindrical center core, A, having a density of 7.90 Mg/m3, and a cylindrical outer part, B, and a cone cap, C, each having a density of 2.70 Mg> m3. C

400 mm

A

400 mm

SOLUTION

100 mm

Center of mass: The assembly is broken into four composite segments, as shown in Figs. a, b, and c. Since segment (3) is a hole to segments (1) and (2), its mass should be considred negative. We have

z =

=

g zcm = gm

600 mm B

x

300 mm

y

1 2700(0.3)c p(0.32)(0.6) d + 2700(0.6 + 0.2)c p(0.32)(0.8) d + (0.5) c -2700 cp(0.12)(1) d d + 7900(0.5)c p(0.12)(1) d 3 1 2700 c p(0.32)(0.6) d + 2700 c p(0.32)(0.8) d + c - 2700cp(0.12)(1) d d + 7900cp(0.12)(1) d 3

121.58p = 0.4630 m = 463 mm 262.6p

Ans.

Ans: z = 463 mm 963

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*9–84. Locate the center of mass z of the assembly. The cylinder and the cone are made from materials having densities of 5 Mg>m3 and 9 Mg>m3, respectively.

z

0.6 m

0.4 m

SOLUTION Center of mass: The assembly is broken into two composite segments, as shown in Figs. a and b.

z =

=

' © zm = ©m

0.2 m

1 5000(0.4) C p(0.2 2)(0.8) D + 9000(0.8 + 0.15)c p(0.4 2)(0.6) d 3

0.8 m

x

1 5000 C p(0.2 2)(0.8) D + 9000c p(0.4 2)(0.6) d 3

1060.60 = 0.754 m = 754 mm 1407.4

y

Ans.

Ans: z = 754 mm 964

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9–85.

a 2

a 2

A hole having a radius r is to be drilled in the center of the homogeneous block. Determine the depth h of the hole so that the center of gravity G is as low as possible.

a 2

a 2 h

r G

a

SOLUTION ©V = a3 - pr2h ©zV = z =

A a2 B a3 - A h2 B C pr2 h D = 12(a4 - pr2 h2)

©zV = ©V

1 4 2 (a 3

- pr2 h2)

a - pr h 2

=

a4 - pr2h2 2(a3 - pr2h)

1 (a3 - pr2h)( -2pr2h) - (a4 - pr2h2)( - pr2) dz = B R = 0 dh 2 (a3 - pr2h)2 pr2h2 - 2a3 h + a4 = 0 Solving for the smaller root, h = =

2a3 - 2(2a3)2 - 4(pr2)a4 2(pr2) a3 - a2 2a2 - pr2 pr2

Ans.

Ans: h =

965

a3 - a2 2a2 - pr2 pr2

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9–86. The composite plate is made from both steel (A) and brass (B) segments. Determine the mass and location 1x, y, z2 of its mass center G. Take rst = 7.85 Mg>m3 and rbr = 8.74 Mg>m3.

z

A

225 mm

G 150 mm B 150 mm 30 mm

SOLUTION

x

1 1 ©m = ©rV = c 8.74 a (0.15)(0.225)(0.03)b d + c 7.85 a (0.15)(0.225)(0.03) b d 2 2 + [7.85(0.15)(0.225)(0.03)] = C 4.4246 A 10 - 3 B D + C 3.9741 A 10 - 3 B D + C 7.9481 A 10 - 3 B D = 16.347 A 10 - 3 B = 16.4 kg

2 1 (0.150) b(4.4246) A 10 - 3 B + a 0.150 + (0.150) b (3.9741) A 10 - 3 B 3 3

©xm = a0.150 + +

Ans.

1 (0.150)(7.9481)(10 -3) = 2.4971(10 -3) kg # m 2

1 2 0.225 b (7.9481) A 10 - 3 B ©zm = a (0.225) b(4.4246) A 10 - 3 B + a (0.225)b (3.9471) A 10 - 3 B + a 3 3 2 = 1.8221 A 10 - 3 B kg # m x =

2.4971(10 - 3) ©xm = = 0.153 m = 153 mm ©m 16.347(10 - 3)

Ans.

Due to symmetry: y = - 15 mm z =

Ans.

1.8221(10 - 3) ©zm = 0.1115 m = 111 mm = ©m 16.347(10 - 3)

Ans.

Ans: Σm = 16.4 kg x = 153 mm y = -15 mm z = 111 mm 966

y

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9–87. The assembly is made from a steel hemisphere, rst = 7.80 Mg>m3, and an aluminum cylinder, ral = 2.70 Mg>m3. Determine the mass center of the assembly if the height of the cylinder is h = 200 mm.

z 80 mm

G _ z

h

160 mm y

SOLUTION © z m = C 0.160 - 38 (0.160) D A 23 B p(0.160)3 (7.80) + A 0.160 +

x 0.2 2

B p(0.2)(0.08)2(2.70)

= 9.51425(10 - 3) Mg # m ©m =

A 23 B p(0.160)3(7.80) + p (0.2)(0.08)2(2.70)

= 77.7706(10 - 3) Mg z =

9.51425(10 - 3) © zm = = 0.122 m = 122 mm ©m 77.7706(10 - 3)

Ans.

Ans: z = 122 mm 967

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*9–88. The assembly is made from and an rst = 7.80 Mg>m3, ral = 2.70 Mg>m3. Determine the so that the mass center of the z = 160 mm.

a steel hemisphere, aluminum cylinder, height h of the cylinder assembly is located at

z 80 mm

G _ z

h

160 mm y

SOLUTION

x

© z m = C 0.160 - 38 (0.160) D A 23 B p(0.160)3 (7.80) + A 0.160 +

h 2

B p (h)(0.08)2(2.70)

= 6.691(10 - 3) + 8.686(10 - 3) h + 27.143(10 - 3) h2 ©m =

A 23 B p(0.160)3(7.80) + p (h)(0.08)2(2.70)

= 66.91(10 - 3) + 54.29(10 - 3) h z =

6.691(10 - 3) + 8.686(10 - 3) h + 27.143(10 - 3) h2 © zm = 0.160 = ©m 66.91(10 - 3) + 54.29(10 - 3) h

Solving h = 0.385 m = 385 mm

Ans.

Ans: h = 385 mm 968

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9–89.

z

Locate the center of mass of the block. Materials 1, 2, and 3 have densities of 2.70 Mg> m3, 5.70 Mg> m3, and 7.80 Mg> m3, respectively.

20 mm 30 mm

1

40 mm

2 y

3

SOLUTION

60 mm 20 mm

Centroid: We have

60 mm

x

1 0.01c 2700 a (0.03)(0.06)(0.02)b d + 0.01 C 5700 A 0.04(0.06)(0.02) B D + 0.03 C 7800 A 0.06(0.06)(0.02) B D ' gxm 2 x = = gm 1 2700 a (0.03)(0.06)(0.02)b + 5700 A 0.04(0.06)(0.02) B + 7800 A 0.06(0.06)(0.02) B 2 =

20.07(10 - 3) = 0.02271 m = 22.7 mm 0.8838

Ans.

1 0.02 c 2700 a (0.03)(0.06)(0.02)b d + 0.03 C 5700 A 0.04(0.06)(0.02) B D + 0.03 C 7800 A 0.06(0.06)(0.02) B D ' gy m 2 y = = gm 1 2700 a (0.03)(0.06)(0.02) b + 5700 A 0.04(0.06)(0.02) B + 7800 A 0.06(0.06)(0.02) B 2 =

26.028(10 - 3) = 0.02945 m = 29.5 mm 0.8838

Ans.

1 0.07 c 2700 a (0.03)(0.06)(0.02)b d + 0.04 C 5700 A 0.04(0.06)(0.02) B D + 0.01 C 7800 A 0.06(0.06)(0.02) B D ' gz m 2 z = = gm 1 2700 a (0.03)(0.06)(0.02) b + 5700 A 0.04(0.06)(0.02) B + 7800 A 0.06(0.06)(0.02) B 2 19.962(10 - 3) = 0.02259 m = 22.6 mm = Ans. 0.8838

Ans: x = 22.7 mm y = 29.5 mm z = 22.6 mm 969

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9–90. Determine the outside surface area of the hopper.

1.5 m

4m

1.2 m

SOLUTION Surface Area: Applying Theorem of Pappus and Guldinus, Eq. 9–9, with and u = 2p rad, L1 = 4m, L2 = 21.2 2 + 1.32 = 23.13 m, N1 = 1.5 m 0.2 + 1.5 = 0.85 m as indicated in Fig. a, N2 = 2

0.2 m

A = u©NL = 2p [1.5(4) + 0.85 23.13] = 47.15 m2 = 47.1 m2

Ans.

Ans: A = 47.1 m2 970

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9–91.

z

The hopper is filled to its top with coal. Determine the volume of coal if the voids (air space) are 30 percent of the volume of the hopper.

1.5 m

4m

1.2 m 0.2 m

SOLUTION 1 V = ©u~ r A = 2p [(0.75)(1.5)(4) + (0.1)(.2)(1.2) + (0.6333)a b(1.3)(1.2)](0.70) 2 V = 22.1 m3

Ans.

Ans: V = 22.1 m3 971

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*9–92. A ring is generated by rotating the quartercircular area about the x axis. Determine its volume. a

SOLUTION

2a

Volume: Applying the theorem of Pappus and Guldinus, Eq. 9.8 , with u = 2p, 4a p 6p + 4 r = 2a + = a and A = a2, we have 3p 3p 4 V = urA = 2p

6p + 4 a 3p

p 2 a 4

=

p(6p + 4) 3 a 6

Ans.

x

Ans: V = 972

p(6p + 4) 3 a 6

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9–93. A ring is generated by rotating the quartercircular area about the x axis. Determine its surface area. a

SOLUTION

2a

Surface Area: Applying the theorem of Pappus and Guldinus, Eq. 9.9, with u = 2p , 2(p + 1) pa 5 , r1 = 2a, r 2 = L1 = L3 = a, L2 = a and r3 = a, we have p 2 2 A = u©rL = 2p 2a(a) +

2(p + 1) pa 5 a a b + a (a) p 2 2

x

= p(2p + 11)a2

Ans.

Ans: A = p(2p + 11)a2 973

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9–94. The water tank AB has a hemispherical top and is fabricated from thin steel plate. Determine the volume within the tank.

B 1.6 m

1.5 m

1.6 m A 0.2 m

Solution Volume: The perpendicular distance measured from the z axis to the centroid of each area segment is indicated in Fig. a. V = 2pΣrA = 2pc a

4(1.6) 3p

= 2p(4.064) = 25.5 m3

ba

p(1.62) 4

1 b + 0.8(1.6)(1.5) + 0.6667a b(1.4)(1.6) + 0.1(0.2)(1.6) d 2

Ans.

Ans: V = 25.5 m2 974

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9–95. The water tank AB has a hemispherical roof and is fabricated from thin steel plate. If a liter of paint can cover 3 m2 of the tank’s surface, determine how many liters are required to coat the surface of the tank from A to B.

B 1.6 m

1.5 m

1.6 m A 0.2 m

Solution Surface Area: The perpendicular distance measured from the z axis to the centroid of each line segment is indicated in Fig. a. A = 2pΣrL = 2pc a

211.62 p11.62 b + 1.611.52 + 0.9 1 21.42 + 1.62 2 d ba p 2

= 2p16.87342 = 43.18 m2

Thus, the amount of paint required is Number of liters =

43.18 = 14.4 liters 3

Ans.

Ans: Number of liters = 14.4 liters 975

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*9–96.

A

The rim of a flywheel has the cross section A–A shown. Determine the volume of material needed for its construction.

A

60 mm 300 mm

20 mm 40 mm

20 mm Section A–A

SOLUTION ' V = ©ur A = 2p(350)(60)(20) + 2p(320)(40)(20) V = 4.25(106) mm3

Ans.

Ans: V = 4.25(106) mm3 976

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9–97.

3m

3m

The process tank is used to store liquids during manufacturing. Estimate the outside surface area of the tank. The tank has a flat top and the plates from which the tank is made have negligible thickness.

6m

4m

SOLUTION Surface Area: Applying Theorem of Pappus and Guldinus, Eq. 9–9 with u = 2p, L1 = 3 m, L2 = 6 m, L3 = 232 + 4 2 = 5 m, N1 = 1.5 m, N2 = 3 m and N3 = 1.5 m as indicated in Fig. a, A = u©NL = 2p [1.5(3) + 3(6) + 1.5(5)] = 188.49 m2 = 188 m2

Ans.

Ans: A = 188 m2 977

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9–98.

3m

The process tank is used to store liquids during manufacturing. Estimate the volume of the tank. The tank has a flat top and the plates from which the tank is made have negligible thickness.

3m

6m

4m

SOLUTION ~ = 2p B 1 a 1 b (3) (4) + 1.5(3)(6)R V = ©urA 2

V = 207.3 m3 = 207 m3

Ans.

Ans: V = 207 m 978

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9–99. A ring is formed by rotating the area 360° about the x – x axes. Determine its surface area. 80 mm

50 mm 30 mm

30 mm

100 mm x

x

Solution Surface Area. Referring to Fig. a, L1 = 110 mm, L2 = 2302 + 802 = 27300 mm L3 = 50 mm, r1 = 100 mm, r2 = 140 mm and r3 = 180 mm. Applying the theorem of pappus and guldinus, with u = 2p rad,

A = uΣrL



= 2p 3 100(110) + 2 (140) ( 17300 2 + 180 (50) 4



= 276 ( 103 ) mm2

= 275.98 ( 103 ) mm2

Ans.

Ans: A = 276(103) mm2 979

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*9–100. A ring is formed by rotating the area 360° about the x – x axes. Determine its volume. 80 mm

50 mm 30 mm

30 mm

100 mm x

x

Solution

1 (60)(80) = 2400 mm2, A2 = 50(80) = 4000 mm2, 2 r1 = 126.67 mm and r2 = 140 mm. Applying the theorem of pappus and guldinus,

Volume. Referring to Fig. a, A1 = with u = 2p rad,

V = uΣrA = 2p 3 126.67(2400) + 140(4000) 4



= 5.429 ( 106 ) mm3



= 5.43 ( 106 ) mm3

Ans.

Ans: V = 5.43 ( 106 ) mm3 980

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9–101. Determine the volume of concrete needed to construct the curb. 100 mm 150 mm

30

4m

150 mm 150 mm

Solution p p 1 V = Σu A r = a b[(0.15)(0.3)(4.15)] + a b c a b(0.15)(0.1)(4.25)d 6 6 2 V = 0.114 m3



Ans.

Ans: V = 0.114 m3 981

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9–102. Determine the surface area of the curb. Do not include the area of the ends in the calculation. 100 mm 150 mm

30

4m

150 mm 150 mm

Solution A = ΣurL =

p {4(0.15) + 4.075(0.15) + (4.15 + 0.075) ( 20.152 + 0.12 ) 6 + 4.3(0.25) + 4.15(0.3)}

A = 2.25 m2

Ans.

Ans: A = 2.25 m2 982

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9–103. Determine the surface area and the volume of the ring formed by rotating the square about the vertical axis.

b

a

a

SOLUTION ~ = 2 c2p a b A = ©urL

= 4p cba -

45

a a sin 45°b(a) d + 2 c 2pa b + sin 45°b(a) d 2 2

a2 a2 sin 45° + ba + sin 45° d 2 2 Ans.

= 8pba Also A = ©urL = 2p(b)(4a) = 8pba ~ = 2p(b)(a)2 = 2pba 2 V = ©urA

Ans.

Ans: A = 8pba V = 2pba2 983

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*9–104.

y 2m

Using integration, determine both the area and the centroidal distance x of the shaded area. Then, using the second theorem of Pappus–Guldinus, determine the volume of the solid generated by revolving the area about the y axis.

C y2 = 2x

2m x

_ x

SOLUTION x ' x = 2 ' y = y dA = x dy 2

A =

L

' xdA =

L

dA =

y2 y3 2 dy = c d = 1.333 = 1.33 m2 6 0 L0 2

Ans.

2

y4 y5 2 dy = c d = 0.8 m3 40 0 L0 8

' L x =

' x dA

L

= dA

0.8 = 0.6 m 1.333

Ans.

Thus, ' V = ur A = 2p 0.6 1.333 = 5.03 m3

Ans.

Ans: V = 5.03 m3 984

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9–105 .

y

Determine the volume of an ellipsoid formed by revolving the shaded area about the x axis using the second theorem of Pappus–Guldinus. The area and centroid y of the shaded area should first be obtained by using integration.

y2 x2  –– –– 1 a2 b2 b x

SOLUTION

a

Area and Centroid: The differential element parallel to the x axis is shown shaded in Fig. a. The area of this element is given by dA = xdy =

a 2b2 - y2 dy b

Integrating, b

A =

dA =

LA

a pab 2b2 - y2 dy = b 4 L0

' The centroid y can be obtained by applying Eq. 9–4 with y = y. ' ydA

b

a y c 2b2 - y2 dy d b 4b LA L0 y = = = pab 3p dA 4 LA Volume: Applying the second theorem of Pappus–Guldinus and using the results obtained above, we have V = 2pyA = 2pa

4b pab 2 ba b = pab2 3p 4 3

Ans.

Ans: V =

985

2 pab2 3

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9–106. The heat exchanger radiates thermal energy at the rate of 2500 kJ h for each square meter of its surface area. Determine how many joules (J) are radiated within a 5-hour period.

0.5 m

0.75 m

0.75 m

1.5 m

0.75 m

SOLUTION A = ©u r L = (2p) B 2a

0.75 + 0.5 b 2(0.75)2 + (0.25)2 + (0.75)(1.5) + (0.5)(1) R 2

1m

= 16.419 m2 Q = 2500 A 103 B a

h

#

J m2

b A 16.416 m2 B (5 h) = 205 MJ

Ans.

0.5 m

Ans: Q = 205 MJ 986

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9–107. Determine the height h to which liquid should be poured into the cup so that it contacts three-fourths the surface area on the inside of the cup. Neglect the cup’s thickness for the calculation.

40 mm

160 mm h

Solution Surface Area. From the geometry shown in Fig. a,

r 40 = ; h 160

r =

1 h 4

1 217 1 2 2 h, Fig. b. Applying the theorem of pappus and h and L = a hb + h = A 4 8 4 Guldinus, with u = 2p rad,

Thus, r =

1 217 p217 2 A = u Σr L = 2p a hba hb = h 8 4 16

For the whole cup, h = 160 mm. Thus Ao = a

It is required that A =

p217 b 1 1602 2 = 1600p217 mm2 16

3 3 A = 1 1600p217 2 = 1200p217 mm2. Thus 4 o 4 1200p217 =



p217 2 h 16

Ans.

h = 138.56 mm = 139 mm

Ans: h = 139 mm 987

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*9–108. Determine the interior surface area of the brake piston. It consists of a full circular part. Its cross section is shown in the figure.

40 mm 60 mm

80 mm

SOLUTION

20 mm

A = © u r L = 2 p [20(40) + 55 2(30)2 + (80)2 + 80(20) + 90(60) + 100(20) + 110(40)]

40 mm 30 mm 20 mm

A = 119(103) mm2

Ans.

Ans: A = 119(103) mm2 988

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9–109. Determine the height h to which liquid should be poured into the cup so that it contacts half the surface area on the inside of the cup. Neglect the cup’s thickness for the calculation.

30 mm 50 mm

h

SOLUTION

10 mm

A = uz ~ rL = 2p{202(20)2 + (50)2 + 5(10)} = 2p(1127.03) mm2 x =

20h 2h = 50 5

2p b 5(10) + a 10 +

2h 2 h 1 b a b + h2 r = (2p)(1127.03) 5 B 5 2

10.77h + 0.2154h2 = 513.5 Ans.

h = 29.9 mm

Ans: h = 29.9 mm 989

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9–110. Determine the volume of material needed to make the casting.

20 mm

60 mm

60 mm

40 mm

Solution Given:

Side View

Front View

r1 = 40 mm r2 = 60 mm r3 = r2 - r1 4 # r3 4 # r2 r2 p p V = 2 # p # c 2 # a b # r 22 # a # b + 2 # r2 # 12 # r3 2 # a b - 2 # a b # r 23 # ar2 - # b d 4 3 p 2 2 3 p V = 1.4031106 2 mm3

Ans.

Ans: V = 1.403 (106) mm3 990

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9–111. The water tank has a paraboloid-shaped roof. If one liter of paint can cover 3 m2 of the tank, determine the number of liters required to coat the roof.

y

1 (144  x2) y  –– 96 x 2.5 m 12 m

SOLUTION Length and Centroid: The length of the differential element shown shaded in Fig. a is dL = 2dx2 + dy2 = where

A

1 + a

dy 2 b dx dx

dy 1 = - x. Thus, dx 48 dL =

B

2 x2 1 1 x b dx = 1 + 2 dx = 2482 + x2 dx 48 B 48 48

1 + a-

Integrating, 12 m

L =

LL

dL =

L0

1 2482 + x2 dx = 12.124 m 48

The centroid x of the line can be obtained by applying Eq. 9–5 with xc = x. 12 m

x =

LL

x~ dL =

L0

xc

1 2482 + x2 dx d 48 73.114 = = 6.031 m 12.124 12.124

dL LL Surface Area: Applying the first theorem of Pappus and Guldinus and using the results obtained above with r = x = 6.031 m, we have A = 2prL = 2p(6.031)(12.124) = 459.39 m2 Thus, the amount of paint required is Number of liters =

459.39 = 153 liters 3

Ans.

Ans: 153 liters 991

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*9–112. Half the cross section of the steel housing is shown in the figure. There are six 10-mm-diameter bolt holes around its rim. Determine its mass. The density of steel is 7.85 Mg m3. The housing is a full circular part.

30 mm

20 mm 40 mm

10 mm 30 mm 10 mm

10 mm

10 mm

SOLUTION V = 2p[ (40)(40)(10) + (55)(30)(10) + (75)(30)(10)] - 6 C p (5)2(10) D = 340.9 A 103 B mm3 m = rV = a7850 = 2.68 kg

kg m3

b (340.9) A 103 B A 10-9 B m3 Ans.

Ans: m = 2.68 kg 992

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9–113. Determine the surface area of the roof of the structure if it is formed by rotating the parabola about the y axis.

y y

16

(x2/16)

16 m

SOLUTION Centroid: The = ¢

C

1 + a

x

length

of

the

differential

element

is

dL = 2dx2 + dy 2

16 m

dy dy x b ≤ dx and its centroid is x = x. Here, = - . Evaluating the dx dx 8 2

integrals, we nave 16 m

L =

LL

L

dL =

' xdL =

L0

¢

16 m

' x¢

L0

C

C

1 +

1 +

x2 ≤ dx = 23.663 m 64

x2 ≤ dx = 217.181 m2 64

Applying Eq. 9–5, we have

x =

LL

' xdL

LL

= dL

217.181 = 9.178 m 23.663

Surface Area: Applying the theorem of Pappus and Guldinus, Eq. 9–7, with u = 2p, L = 23.663 m, r = x = 9.178 m, we have A = urL = 2p(9.178) (23.663) = 1365 m2

Ans.

Ans: A = 1365 m2 993

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9–114. A steel wheel has a diameter of 840 mm and a cross section as shown in the figure. Determine the total mass of the wheel if r = 5 Mg>m3.

100 mm A

30 mm

60 mm 420 mm 250 mm 30 mm 840 mm 80 mm

SOLUTION Volume: Applying the theorem of Pappus and Guldinus, Eq. 9.10, with u = 2p, r1 = 0.095 m, r2 = 0.235 m, r3 = 0.39 m, A 1 = 0.110.032 = 0.003 m2, 2 A 2 = 0.2510.032 = 0.0075 m and A 3 = 10.1210.062 = 0.006 m2, we have

A

Section A–A

V = u©rA = 2p30.09510.0032 + 0.23510.00752 + 0.3910.00624 = 8.775p110-32m3 The mass of the wheel is m = rV = 51103238.775110-32p4 = 138 kg

Ans.

Ans: m = 138 kg 994

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9–115 .

p 4 kPa

The loading acting on a square plate is represented by a parabolic pressure distribution. Determine the magnitude of the resultant force and the coordinates 1x, y2 of the point where the line of action of the force intersects the plate. Also, what are the reactions at the rollers B and C and the ball-and-socket joint A? Neglect the weight of the plate.

p

2y1/2

y C B

A 4m

2m

4m x

SOLUTION y = y dA = p dy x = 0

Ans.

(Due to symmetry) 4

L L

4 4 2y1/2 dy = B y 3/2 R = 10.67 kN/m 3 0 L0

dA =

' ydA =

y =

L

4

L0

2y

' y dA

L

= dA

3/2

4 5/2 4 dy = B y R = 25.6 kN 5 0

25.6 = 2.40 m 10.67

Ans.

Ans.

FR = 10.67(4) = 42.7 kN © My = 0;

By = Cy

©Mx = 0;

42.67(2.40) - 2By (4) = 0 Ans.

By = Cy = 12.8 kN + c ©F = 0;

A y - 42.67 + 12.8 + 12.8 = 0 Ans.

A y = 17.1 kN

Ans: x = 0 y = 2.40 m B y = Cy = 12.8 kN A y = 17.1 kN 995

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*9–116. The load over the plate varies linearly along the sides of the plate such that p = 23[x14 - y2] kPa. Determine the resultant force and its position 1x, y2 on the plate.

p 8 kPa

y 3m

x

4m

SOLUTION Resultant Force and its Location: The volume of the differential element is 2 ' ' dV = d FR = pdxdy = (xdx)[(4 - y)dy] and its centroid is at x = x and y = y. 3 3m

FR =

LFR =

dFR =

LFR

= ' ydFR =

LFR

=

x =

L0

(4 - y)dy

2 2 (x dx) 3 L0

Ans.

4m

(4 - y)dy

y2 4 m x3 3 m 2 b 2 R = 48.0 kN # m a4y Ba b 2 3 3 0 2 0 3m

L0

2 (xdx) 3 L0

4m

y(4 - y)dy

y3 4 m 2 x2 3 m B a b 2 a 2y2 - b 2 R = 32.0 kN # m 3 2 0 3 0 LFR

' xdFR

FR L

y =

4m

y2 4 m x2 3 m 2 a 4y b 2 R = 24.0 kN Ba b 2 3 2 0 2 0 3m

xdFR =

L0

2 (xdx) 3 L0

FR L

=

48.0 = 2.00 m 24.0

Ans.

=

32.0 = 1.33 m 24.0

Ans.

dFR

' ydFR

LFR

dFR

Ans: FR = 24.0 kN x = 2.00 m y = 1.33 m 996

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9–117. p

The load over the plate varies linearly along the sides of the plate such that p = (12 - 6x + 4y) kPa. Determine the magnitude of the resultant force and the coordinates ( x, y ) of the point where the line of action of the force intersects the plate.

12 kPa 18 kPa

x 1.5 m

6 kPa 2m

Solution Centroid. Perform the double integration. FR =

LA

r(x, y)dA = = =

L0

1.5 m

L0

1.5 m

L0

1.5 m

L0

2m

(12 - 6x + 4y)dxdy

1 12x

- 3x2 + 4xy 2 `

dy

0

(8y + 12)dy

= (4y2 + 12y) `



2m

1.5 m 0

Ans.

= 27.0 kN LA

xr(x, y)dA = = = =

L0

1.5 m

L0

1.5 m

L0

1.5 m

L0

2m

1 12x

(6x2 - 2x3 + 2x2y) `

LA

= =

2m

dy

0

(8y + 8)dy

1 4y2

+ 8y 2 `

= 21.0 kN # m yr(x, y)dA =

- 6x2 + 4xy 2 dx dy

L0

1.5 m

L0

1.5 m

L0

1.5 m

L0

2m

1.5 m 0

1 12y

1 12xy

- 6xy + 4y2 2 dx dy

- 3x2y + 4xy2 2 `

(8y2 + 12y)dy

1.5 m 8 = a y3 + 6y2 b ` 3 0

= 22.5 kN # m

997

2m

dy

0

y

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9–117. Continued

Thus,



x =

y =

LA

xp(x, y)dA

LA LA

=

21.0 kN # m 7 = m = 0.778 m  27.0 kN 9

Ans.

=

22.5 kN # m = 0.833 m  27.0 kN

Ans.

p(x, y)dA

yp(x, y)dA

LA

p(x, y)dA

Ans: FR = 27.0 kN x = 0.778 m y = 0.833 m 998

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9–118. A wind loading creates a positive pressure on one side of the chimney and a negative (suction) pressure on the other side, as shown. If this pressure loading acts uniformly along the chimney’s length, determine the magnitude of the resultant force created by the wind.

p p  p0 cos u

Solution FRx =

p 2l 2p L- 2

u

l

(p0 cos u) cos u r du =

p 2 2rlp0 p L- 2

p = 2rlp0 a b 2

cos2 u du Ans.



FRx = plrp0 FRy = 2l

p 2 p L- 2

(p0 cos u) sin u r du = 0

Thus,

Ans.

FR = plrp0

Ans: p FRx = 2rlp0 a b 2 FR = plrp0 999

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9–119. p

The rectangular plate is subjected to a distributed load over its entire surface. The load is defined by the expression p = p0 sin (px>a) sin (py>b), where p0 represents the pressure acting at the center of the plate. Determine the magnitude and location of the resultant force acting on the plate.

p0

y x

SOLUTION

a b

Resultant Force and its Location: The volume of the differential element is py px dy b . dx b a sin dV = dFR = pdxdy = p0 a sin a b a

FR =

dFR = p0

LFR

L0

a sin

= p0 B a =

b py px dx b a sin dy b a b L0

a px 2 a b px 2 b cos b a - cos b R p a p b 0 0

4ab p0 p2

Ans.

Since the loading is symmetric, the location of the resultant force is at the center of the plate. Hence, x =

a 2

y =

b 2

Ans.

Ans: FR =

4ab p0 p2

a 2 b y = 2

x =

1000

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*9–120. The concrete dam in the shape of a quarter circle. Determine the magnitude of the resultant hydrostatic force that acts on the dam per meter of length. The density of water is rw = 1 Mg>m3. 3m

SOLUTION Loading: The hydrostatic force acting on the circular surface of the dam consists of the vertical component Fv and the horizontal component Fh as shown in Fig. a. Resultant Force Component: The vertical component Fv consists of the weight of water contained in the shaded area shown in Fig. a. For a 1-m length of dam, we have Fv = rgAABCb = (1000)(9.81) B (3)(3) -

p 2 (3 ) R (1) = 18947.20 N = 18.95 kN 4

The horizontal component Fh consists of the horizontal hydrostatic pressure. Since the width of the dam is constant (1 m), this loading can be represented by a triangular distributed loading with an intensity of wC = rghCb = 1000(9.81)(3)(1) = 29.43 kN>m at point C, Fig. a. Fh =

1 (29.43)(3) = 44.145 kN 2

Thus, the magnitude of the resultant hydrostatic force acting on the dam is FR = 3F h 2 + F v 2 = 344.1452 + 18.952 = 48.0 kN

Ans.

Ans: FR = 48.0 kN 1001

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9–121. The concrete “gravity” dam is held in place by its own weight. If the density of concrete is rc = 2.5 Mg>m3, and water has a density of rw = 1.0 Mg>m3, determine the smallest dimension d that will prevent the dam from overturning about its end A.

1m

6m

A d–1

d

Solution Loadings. The computation will be based on b = 1 m width of the dam. The pressure at the base of the dam is. P = rgh = 1000(9.81)(6) = 58.86 ( 103 ) pa = 58.86 kPa Thus w = pb = 58.86(1) = 58.86 kN>m The forces that act on the dam and their respective points of application, shown in Fig. a, are W1 = 2500 3 1(6)(1) 4 (9.81) = 147.15 ( 103 ) N = 147.15 kN

1 W2 = 2500c (d - 1)(6)(1) d (9.81) = 73.575 ( d - 1 )( 103 ) = 73.575(d - 1) kN 2 1 2

( FR )v = 1000c (d - 1)(6)(1) d (9.81) = 29.43 ( d - 1 )( 103 ) = 29.43(d - 1) kN ( FR ) h =

1 (58.86)(6) = 176.58 kN 2

x1 = 0.5  x2 = 1 + y =

1 1 2 1 (d - 1) = (d + 2)  x3 = 1 + (d - 1) = (2d + 1) 3 3 3 3

1 (6) = 2 m 3

Equation of Equilibrium. Write the moment equation of equilibrium about A by referring to the FBD of the dam, Fig. a, 1 a+ΣMA = 0;  147.15(0.5) + [73.575(d - 1)]c (d + 2) d 3

1 + [29.43(d - 1)] c (2d + 1) d - 176.58(2) = 0 3

44.145d 2 + 14.715d - 338.445 = 0

Solving and chose the positive root

Ans.

d = 2.607 m = 2.61 m

Ans: d = 2.61 m 1002

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9–122. y

The factor of safety for tipping of the concrete dam is defined as the ratio of the stabilizing moment due to the dam’s weight divided by the overturning moment about O due to the water pressure. Determine this factor if the concrete has a density of rconc = 2.5 Mg>m3 and for water rw = 1 Mg>m3.

1m

6m

O

x 4m

Solution Loadings. The computation will be based on b = 1m width of the dam. The pressure at the base of the dam is   P = pwgh = 1000(9.81)(6) = 58.86(103)pa = 58.86 kPa Thus,    w = Pb = 58.86(1) = 58.86 kN>m The forces that act on the dam and their respective points of application, shown in Fig. a, are    W1 = (2500)[(1)(6)(1)](9.81) = 147.15(103) N = 147.15 kN 1   W2 = (2500) c (3)(6)(1) d (9.81) = 220.725(103) N = 220.725 kN 2   FR =

1 (58.86)(6) = 176.58 kN 2

  x1 = 3 +

1 2 1 (1) = 3.5 m  x2 = (3) = 2m  y = (6) = 2 m 2 3 3

Thus, the overturning moment about O is   MOT = 176.58(2) = 353.16 kN # m And the stabilizing moment about O is   Ms = 147.15(3.5) + 220.725(2) = 956.475 kN # m Thus, the factor of safety is   F.S. =

Ms 956.475 = = 2.7083 = 2.71 MOT 353.16

Ans.

Ans: F.S. = 2.71 1003

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9–123. The storage tank contains oil having a specific weight of go = 9 kN/m3. If the tank is 6 m wide, calculate the resultant force acting on the inclined side BC of the tank, caused by the oil, and specify its location along BC, measured from B. Also compute the total resultant force acting on the bottom of the tank.

10 m

A B

2m

8m

C 4m

D 4m

3m

Solution Given: g = 9 kN>m3 w = 6m a = 10 m b = 2m

c = 8m d = 4m e = 3m f = 4m

wB = w # g # b

wB = 108 kN>m

Fh1 = wB # c

Fh2 =

wC = w # g # 1b + c2

1# 1wC - wB 2 # c 2

wC = 540 kN>m

Fv1 = g # w # b # e

Fv2 =

1# # # # g w c e 2

The resultant force FRx = Fh1 + Fh2

FRy = Fv1 + Fv2

FR = 2F 2Rx + F 2Ry

FR = 2.77 MN

Ans.

The location h measured from point B. Guess h = 1 m. Given Fv1 #

e 2#e c 2#c c#h e#h + Fv2 # + Fh1 # + Fh2 # = FRx # + FRy # 2 2 2 3 2 3 2c + e 2c 2 + e 2

On the bottom of the tank Fbot = g # w # f # 1b + c + d2

h = Find (h)

Ans.

h = 5.22 m

Ans.

Fbot = 3.02 MN

Ans: FR = 2.77 MN

h = 5.22 m Fbot = 3.02 MN 1004

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*9–124. When the tide water A subsides, the tide gate automatically swings open to drain the marsh B. For the condition of high tide shown, determine the horizontal reactions developed at the hinge C and stop block D. The length of the gate is 6 m and its height is 4 m. rw = 1.0 Mg/m3.

C 4m A

3m

B

2m

D

SOLUTION Fluid Pressure: The fluid pressure at points D and E can be determined using Eq. 9–13, p = rgz. pD = 1.01103219.812122 = 19 620 N>m2 = 19.62 kN>m2 pE = 1.01103219.812132 = 29 430 N>m2 = 29.43 kN>m2 Thus, wD = 19.62162 = 117.72 kN>m wE = 29.43162 = 176.58 kN>m Resultant Forces: FR1 =

1 1176.582132 = 264.87 kN 2

FR2 =

1 1117.722122 = 117.72 kN 2

Equations of Equilibrium: a + ©MC = 0;

264.87132 - 117.7213.3332 - Dx 142 = 0 Ans.

Dx = 100.55 kN = 101 kN + ©F = 0; : x

264.87 - 117.72 - 100.55 - Cx = 0 Ans.

Cx = 46.6 kN

Ans: Dx = 101 kN Cx = 46.6 kN 1005

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9–125. The tank is filled with water to a depth of d = 4 m. Determine the resultant force the water exerts on side A and side B of the tank. If oil instead of water is placed in the tank, to what depth d should it reach so that it creates the same resultant forces? ro = 900 kg>m3 and rw = 1000 kg>m3.

2m

3m

A

B

d

SOLUTION For water At side A: wA = b rw g d = 2(1000)(9.81) (4) = 78 480 N/m FRA =

1 (78 480)(4) = 156 960 N = 157 kN 2

Ans.

At side B: wB = b rw g d = 3(1000)(9.81)(4) = 117 720 N>m FRB =

1 (117 720)(4) = 235 440 N = 235 kN 2

Ans.

For oil At side A: wA = b ro g d = 2(900)(9.81)d = 17 658 d FRA =

1 (17 658 d)(d) = 156 960 N 2

d = 4.22 m

Ans.

Ans: For water: FRA = 157 kN FRB = 235 kN For oil: d = 4.22 m 1006

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9–126. The 2-m-wide rectangular gate is pinned at its center A and is prevented from rotating by the block at B. Determine the reactions at these supports due to hydrostatic pressure. rw = 1.0 Mg>m3.

6m 1.5 m A B

SOLUTION

1.5 m

w1 = 1000(9.81)(3)(2) = 58 860 N/m w2 = 1000(9.81)(3)(2) = 58 860 N/m F1 =

1 (3)(58 860) = 88 290 N 2

F2 = (58 860)(3) = 176 580 N a + ©MA = 0;

88 290(0.5) - FB (1.5) = 0 FB = 29 430 N = 29.4 kN

+ ©F = 0; : x

Ans.

88 290 + 176 580 - 29 430 - FA = 0 FA = 235 440 N = 235 kN

Ans.

Ans: FB = 29.4 kN FA = 235 kN 1007

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9–127. The tank is filled with a liquid which has a density of 900 kg>m3. Determine the resultant force that it exerts on the elliptical end plate, and the location of the center of pressure, measured from the x axis.

y 1m

1m 4 y2

x2

1

0.5 m x 0.5 m

SOLUTION Fluid Pressure: The fluid pressure at an arbitrary point along y axis can be determined using Eq. 9–13, p = g(0.5 - y) = 900(9.81)(0.5 - y) = 8829(0.5 - y). Resultant Force and its Location: Here, x = 21 - 4y2. The volume of the differential element is dV = dFR = p(2xdy) = 8829(0.5 - y)[221 - 4y 2] dy. Evaluating integrals using Simpson’s rule, we have

the

0.5 m

FR =

LFR

d FR = 17658

L-0.5 m

(0.5 - y)(21 - 4y2)dy Ans.

= 6934.2 N = 6.93 kN 0.5 m

LFR

yd FR = 17658

y(0.5 - y)( 21 - 4y2)dy

= - 866.7 N # m ' ydFR

y =

L-0.5 m

LFR

= dFR

-866.7 = - 0.125 m 6934.2

Ans.

LFR

Ans: FR = 6.93 kN y = -0.125 m 1008

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*9–128. Determine the magnitude of the resultant force acting on the gate ABC due to hydrostatic pressure. The gate has a width of 1.5 m. rw = 1.0 Mg>m3.

1.5 m B

1.25 m C

SOLUTION

2m

w1 = 1000(9.81)(1.5)(1.5) = 22.072 kN>m

60° A

w2 = 1000(9.81)(2)(1.5) = 29.43 kN>m Fx =

1 (29.43)(2) + (22.0725)(2) = 73.58 kN 2

F1 = c (22.072) a1.25 + F2 =

2 b d = 53.078 kN tan 60°

1 2 (1.5)(2)a b (1000)(9.81) = 16.99 kN 2 tan 60°

Fy = F1 + F2 = 70.069 kN F = 2F2x + F2y = 2(73.58)2 + (70.069)2 = 102 kN

Ans.

Ans: F = 102 kN 1009

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9–129. The tank is filled to the top 1y = 0.5 m2 with water having a density of rw = 1.0 Mg/m3. Determine the resultant force of the water pressure acting on the flat end plate C of the tank, and its location, measured from the top of the tank.

y x2 + y2 = (0.5)2 x C 0.5 m

SOLUTION dF = p dA = 11219.81210.5 - y2 2x dy 0.5

F = 219.812 =

L-0.5

10.5 - y21210.522 - y2 dy

0.5 y 9.81 c y210.522 - y2 + 0.52 sin-1 a bd 2 0.5 -0.5

+

219.812 3

0.5 C 2510.522 - y263 D -0.5

F = 3.85 kN

Ans. 0.5

LA

y dF = 219.812

L-0.5

10.5y - y221 210.522 - y2dy = 19.62 b c -

0.5 0.5 2510.522 - y263 d + 3 -0.5

10.522 y 0.5 y cy 210.522 - y2 + 10.522 sin-1 a bd r C 2510.522 - y263 D 0.5 -0.5 4 8 0.5 -0.5 = - 0.481 kN m F1-d2 = d =

L

y dF

- 0.481 = - 0.125 m 3.85

Hence, measured from the top of the tank, d¿ = 0.5 + 0.125 = 0.625 m

Ans.

Ans: F = 3.85 kN d = 0.625 m 1010

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9–130. The underwater tunnel in the aquatic center is fabricated from a transparent polycarbonate material formed in the shape of a parabola. Determine the magnitude of the hydrostatic force that acts per meter length along the surface AB of the tunnel. The density of the water is rw = 1000 kg/m3.

y

2m y ⫽ 4 ⫺ x2

A

4m

Solution Resultant Force Component: The hydrostatic force acting on the parabolic surface AB of the tunnel consists of the vertical component Fv and the horizontal component Fh as shown in Fig. a.

2m

2m

B

x

The vertical component Fv represents the weight of water contained in the shaded area shown in Fig. a Fv = rwgAABCDb = 110002 19.812 c 2122 +

1 122 142 d 112 = 65 400 N = 65.4 kN 3

The horizontal component Fh represents the horizontal hydrostatic pressure. Since the width of the tunnel is constant (1 m), this horizontal loading can be represented by a trapezoidal distributed loading shown in Fig. a. The intensity of this distributed loading at points A and B are wA = rwghAb = 100019.812 122 112 = 19620 N>m and wB = rwghBb = 100019.812 162 112 = 58860 N>m = 58.86 kN>m. Thus, Fh =

1 119.62 + 58.862 142 = 156.96 kN 2

Resultant: The resultant hydrostatic force acting on the surface AB of the tunnel is therefore FR = 2F 2h + F 2v = 2156.962 + 65.42 = 170 kN

Ans.

Ans: FR = 170 kN 1011

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10–1. y

Determine the moment of inertia for the shaded area about the x axis.

y  x1/2

1m

x 1m

Solution Differential Element. The area of the differential element parallel to the y axis shown shaded in Fig. a is dA = ydx. The moment of inertia of this element about the x axis is dIx = dIx′ + dA ∼ y2

=



=



=

y 2 1 (dx)y3 + ydx a b 12 2 1 3 y dx 3

1 1 3 (x2) dx 3

1 3 x2 dx 3 Moment of Inertia. Perform the integration.



=



Ix =





dIx = L L0

1m

2 53 x2 = 15 =

1 3 x2dx 3 1m

0

2 4 m = 0.133 m4 15

Ans.

Ans: Ix = 0.133 m4 1012

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10–2. y

Determine the moment of inertia for the shaded area about the y axis.

y  x1/2

1m

x 1m

Solution Differential Element. The area of the 1differential element parallel to the y axis shown shaded in Fig. a is dA = ydx = x2dx. Moment of Inertia. Perform the integration,





Iy =

LA

x2dA =

L0

1m

2 7 = x2 3 7 =

1

x2 ( x2dx )

1m

0

2 4 m = 0.286 m4 7

Ans.

Ans: Iy = 0.286 m4 1013

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10–3. y

Determine the moment of inertia for the shaded area about the x axis.

y2  1  0.5x 1m

x 2m

Solution Differential Element. Here x = 2(1 - y2). The area of the differential element parallel to the x axis shown shaded in Fig. a is dA = xdy = 2(1 - y2)dy. Moment of Inertia. Perform the integration,



Ix =

LA

y2dA =

L0

= 2

1m

L0

y2[2(1 - y2)dy] 1m

(y2 - y4)dy

y5 y3 = 2a - b 3 3 5

1m

0

4 4 = m = 0.267 m4 15

Ans.



Ans: Ix = 0.267 m4 1014

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*10–4. y

Determine the moment of inertia for the shaded area about the y axis.

y2  1  0.5x 1m

x 2m

Solution Differential Element. Here x = 2(1 - y2). The moment of inertia of the differential element parallel to the x axis shown shaded in Fig. a about the y axis is

∼2 dIy = dIy' + dAx



=



=



=

1 x 2 (dy)x3 + xdya b 12 2 1 3 x dy 3

1 32(1 - y2) 4 3 dy 3

8 ( -y6 + 3y4 - 3y2 + 1)dy 3 Moment of Inertia. Perform the integration,



=

Iy =

L

dIy =



=



=

8 3 L0

1m

( -y6 + 3y4 - 3y2 + 1)dy

1m y7 8 3 a - + y5 - y3 + yb ` 3 7 5 0

128 4 m = 1.22 m4 105

Ans.



Ans: Iy = 1.22 m4 1015

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10–5. y

Determine the moment of inertia for the shaded area about the x axis.

100 mm

200 mm

y  1 x2 50

Solution 1

Differential Element. Here x = 250y2 . The area of the differential element parallel 1 to the x axis shown shaded in Fig. a is dA = 2x dy = 2250y2dy.

x

Moment of Inertia. Perform the integration,





Ix =

LA

y2dA =

L0

200 mm

= 2250

1

y2 c 2250y2dy d

L0

200 mm

2 7 = 2250 a y2 b 3 7

5

y2dy 200 mm

0



= 457.14(106) mm4



= 457(106) mm4

Ans.



Ans: Ix = 457 ( 106 ) mm4 1016

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10–6. y

Determine the moment of inertia for the shaded area about the y axis.

100 mm

200 mm

y  1 x2 50

x

Solution 1

Differential Element. Here x = 250 y2 . The moment of inertia of the differential element parallel to x axis shown in Fig. a about y axis is 1 1 2 2 100250 3 (dy)(2x)3 = x3dy = ( 250y2 ) 3dy = y2dy. 12 3 3 3 Moment of Inertia. Perform the integration,

dIy =

Iy =

L

dIy =

L0

200 mm

100150 3 y2dy 3

100250 2 5 3 a y2 b = 3 5 6

200 mm

0

4

= 53.33(10 ) mm = 53.3(106) mm4

Ans.

Ans: Iy = 53.3(106) mm4 1017

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10–7. y

Determine the moment of inertia about the x axis.

y  bn xn a

b

x a

Solution Differential Element. Here x =

a

1

y n . The area of the differential element parallel b a 1 to the x axis shown shaded in Fig. a is dA = (a - x)dy = aa - 1 y n bdy. bn 1 n

Moment of Inertia. Perform the integration,





Ix =

LA

y2dA = =

L0

b

L0

b

y2aa aay2 -

a 1 n

b

a 1 n

b

1

yn bdy 1

yn + 2 bdy

a a n 3n + 1 = c y3 - a 1 ba by d3 3 3n + 1 n n b



=



=

b

0

1 3 n ab - a bab3 3 3n + 1 ab3 3(3n + 1)

Ans.



Ans: Ix = 1018

ab3 3(3n + 1)

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*10–8. y

Determine the moment of inertia about the y axis.

y  bn xn a

b

x a

Solution Differential Element. The area of the differential element parallel to the y axis b shown shaded in Fig. a is dA = ydx = n xndx. a Moment of Inertia. Perform the integration,

Iy =

LA

x2dA =

L0

a

x2 a

b n x dxb an

a



=



=



=

b n+2 dx nx L0 a a b 1 b(xn + 3) ` na a n + 3 0

a3b n + 3

Ans.



Ans: Iy = 1019

a3b n + 3

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10–9. Determine the moment of inertia of the shaded area about the x axis.

y

y ⫽ a sin p ax a

SOLUTION

x

Differential Element: The area of the rectangular differential element in Fig. a is dA = y dx. The moment of inertia of this element about the x axis is '

dIx = dIx ¿ + dAy 2 =

a –– 2

a –– 2

y 2 p 3 p 1 1 a3 1 (dx)y3 + ydx ¢ ≤ = y3 dx = ¢ a sin x ≤ dx = sin3 ¢ x ≤ dx. a 12 2 3 3 3 a

Moment of Inertia: Performing the integration, we have Ix =

L

dIx =

a 3

a p sin3 ¢ x ≤ dx a 3 L0 0

=

a p 1 p a3 cos ¢ x ≤ d c sin2 a xb + 2 d r ` bca 3 3(p>a) a

=

4a4 9p

Ans.

Ans: Ix =

1020

4a4 9p

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10–10. Determine the moment of inertia of the shaded area about the y axis.

y

y ⫽ a sin p ax a

SOLUTION Differential Element: The area of the rectangular differential element in Fig. a is p dA = y dx = a sin ¢ x ≤ dx. a

x a –– 2

Moment of Inertia: Applying Eq. 10–1, we have a

Iy =

x2dA =

LA

L0

x2 ¢ a sin

a –– 2

p x ≤ dx a

a a p a2 p 2a3 p = a B - p ¢ x2 cos x ≤ + 2 ¢ 2x sin x ≤ + 3 cos x R ` a a a p p 0

= ¢

p2 - 4 4 ≤a p3

Ans.

Ans: Iy = a 1021

p2 - 4 4 ba p3

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10–11. Determine the moment of inertia about the x axis.

y

x2  4y2  4

1m

x 2m

Solution 1 24 - x2. The moment of inertia of the differential 2 element parallel to the y axis shown shaded in Fig. a about x axis is

Differential Element. Here, y =



∼2 dIx = dIx′ + dAy



=



=



=



=

y 2 1 (dx)y3 + ydx a b 12 2 1 3 y dx 3

3 1 1 a 24 - x2 b dx 3 2

1 2(4 - x2)3 dx 24

Moment of Inertia. Perform the integration.

Ix =



=



=

L

dIx =

L0

2m

1 2(4 - x2)3 dx 24

1 x 2m c x2(4 - x2)3 + 6x24 - x2 + 24 sin - 1 d ` 96 2 0 p 4 m 8

Ans.

Ans: Ix = 1022

p 4 m 8

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*10–12. y

Determine the moment of inertia about the y axis.

x2  4y2  4

1m

x 2m

Solution

1 24 - x2. The area of the differential element 2 1 parallel to the y axis shown shaded in Fig. a is dA = ydx = 24 - x2dx 2

Differential Element. Here, y =

Moment of Inertia. Perform the integration,

Iy =

LA

x2dA =



=



=



=

L0

2m

1 2 L0

1 x2 c 24 - x2dx d 2

2m

x2 24 - x2dx

2m 1 x 1 x c - 2(4 - x2)3 + ax24 - x2 + 4 sin - 1 b d ` 2 4 2 2 0

p 4 m 2



Ans.

Ans: Iy = 1023

p 4 m 2

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10–13. y

Determine the moment of inertia for the shaded area about the x axis.

8m y  1 x3 8 x

Solution

4m 1

Differential Element. Here, x = 2y3. The area of the1 differential element parallel to the x axis shown shaded in Fig. a is dA = xdy = 2y3 dy Moment of Inertia. Perform the integration,

Ix =

LA

2

y dA =

L0

8m

L0



= 2



= 2a



1

2

y (2y3 dy) 8m

7

y3 dy

3 10 8 m y3 b ` 10 0

= 614.4 m4 = 614 m4

Ans.

Ans: Ix = 614 m4 1024

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10–14. y

Determine the moment of inertia for the shaded area about the y axis.

8m y  1 x3 8 x

Solution

4m

Differential Element. The area of the differential element parallel to the y axis, 1 shown shaded in Fig. a, is dA = (8 - y)d x = a8 - x3 bdx 8

Moment of Inertia. Perform the integration,



Iy =

LA

x2dA = =

L0

4m

L0

4m

x2 a8 a8x2 -

1 3 x b dx 8

1 5 x b dx 8

8 1 6 3 x b = a x3 3 48

4m

0

= 85.33 m4 = 85.3 m4

Ans.

Ans: Iy = 85.3 m4 1025

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10–15. Determine the moment of inertia of the shaded area about the y axis.

y

x2 ⫹ y2 ⫽ r02

a –– 2 r0

SOLUTION Differential Element: The area of the differential element shown shaded in Fig. a is dA = (rdu)dr.

x

a –– 2

Moment of Inertia: Applying Eq. 10–1, we have a>2

x2dA =

Iy =

LA

a>2

=

a>2

L- a>2

L- a>2 L0

r2 cos2 u(rdu)dr

r0

L- a>2 L0 =

r0

¢

r3 cos2 udrdu

r4 r0 ≤ ` cos2 udu 4 0

a>2

=

r0 4 cos2 udu L- a>2 4

1 ( cos 2u + 1). Thus, 2 a>2 4 r0 Iy = ( cos 2u + 1)du L- a>2 8

However, cos2 u =

=

a>2 r0 4 1 r0 4 = ( sin a + a) B sin 2u + u R ` 8 2 8 - a>2

Ans.

Ans: Iy =

1026

r40 (sin a + a) 8

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*10–16. Determine the moment of inertia for the shaded area about the x axis.

y y2

h2 —x b h

SOLUTION d Ix = Ix =

x

b

1 3 y dx 3

L

d Ix

=

b 3

b y 1 h2 3>2 3>2 dx = a b x dx L0 3 b L0 3

=

1 h2 3>2 2 5>2 b a b a b x ]0 3 b 5

=

2 bh3 15

Ans.

Also, dA = (b - x) dy = (b Ix =

L

y2 dA h

=

L0

b 2 y ) dy h2

y2 (b -

b 2 y ) dy h2

b b 5 h = c y3 y d 3 5h2 0 =

2 bh3 15

Ans.

Ans: Ix = 1027

2 bh3 15

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10–17. y

Determine the moment of inertia for the shaded area about the x axis.

y2  1  x 1m x 1m 1m

Solution 1

Differential Element. Here y = (1 - x)2 . The moment of inertia of the differential element parallel to the y axis shown shaded in Fig. a about the x axis is 3 1 1 2 2 2 dIx = (dx)(2y)3 = y3dx = 3(1 - x)2 4 3 dx = (1 - x)2 dx. 12 3 3 3 Moment of Inertia. Perform the integration, Ix =

L

dIx =



=



=

L0

1m

3 2 (1 - x)2 dx 3

1m 5 2 2 c - (1 - x)2 d ` 3 5 0

4 4 m = 0.267 m4 15

Ans.

Ans: Ix = 0.267 m4 1028

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10–18. y

Determine the moment of inertia for the shaded area about the y axis.

y2  1  x 1m x 1m 1m

Solution Differential Element. Here x = 1 - y2. The moment of inertia of the differential element parallel to the x axis shown shaded in Fig. a about the y axis is ∼2 dIy = dIy′ + dAx

=



=



=

1 x 2 (dy)x3 + xdy a b 12 2 1 3 x dy 3

1 (1 - y2)3dy 3

1 ( - y6 + 3y4 - 3y2 + 1)dy 3 Moment of Inertia. Perform the integration,





=

Iy =

L

1m

dIy =



=



=

1 ( - y6 + 3y4 - 3y2 + 1)dy L-1 m 3 1m 1 y7 3 ( - + y5 - y3 + y) ` 3 7 5 -1 m

32 4 m = 0.305 m4 105

Ans.

Ans: Iy = 0.305 m4 1029

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10–19. y

Determine the moment of inertia for the shaded area about the x axis.

h

y  h x3 b3

x b

Solution Differential Element. The moment of inertia of the differential element parallel to the y axis shown shaded in Fig. a about the x axis is

∼2 dIx = dIx′ + dAy



=



=



=



=

y 2 1 (dx)y3 + ydx a b 12 2 1 3 y dx 3

1 h 3 3 a x b dx 3 b3

h3 9 x dx 3b9

Moment of Inertia. Perform the integration,

Ix =

L

b

dIx =



=



=

h3 9 x dx 9 L0 3b h3 x10 b a b` 3b9 10 0 1 3 bh 30

Ans.



Ans: Ix = 1030

1 bh3 30

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*10–20. y

Determine the moment of inertia for the shaded area about the y axis.

h

y  h x3 b3

x b

Solution Differential Element. The area of the differential element parallel to the y axis h shown shaded in Fig. a is dA = ydx = 3 x3dx b Moment of Inertia. Perform the integration, Iy =

LA

x2dA =

L0

b

x2 a b



=



=



=

h 3 x bdx b3

h x5dx b3 L0 h x6 6 a b` b3 6 0

b3h  6

Ans.

Ans: Iy = 1031

b3h 6

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10–21. y

Determine the moment of inertia for the shaded area about the x axis.

b2 y2  —x a b x2 y — a2

b

x

a

Solution

a 1 a y2 and x1 = 2 y2. Thus, the area of the 1 b b2 differential element parallel to the x axis shown shaded in Fig. a is dA = (x2 - x1) dy Differential Element. Here x2 =

a 1 a = a 1 y2 - 2 y2 b dy. b 2 b

Moment of Inertia. Perform the integration, Ix =

LA

b

y2dA =

a 1 a y2 a 1 y2 - 2 y2 bdy b 2 L0 b L0



=



= a



=

b

a 5 a a 1 y2 - 2 y4 bdy b 2 b

2a

1 2

7b

7

y2 -

3ab3 35

a 5 2b y b 5b2 0

Ans.

Ans: Ix = 1032

3ab3 35

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10–22. Determine the moment of inertia for the shaded area about the y axis.

y b2 y2  —x a b x2 y — a2

b

x

a

Solution

b 2 x . Thus, the area of the a2 a differential element parallel to the y axis shown shaded in Fig. a is dA = (y2 - y1)dx b 1 b = a 1 x2 - 2 x2 b dx a a2

Differential Element. Here, y2 =

b

1 2

1

x2 and y1 =

Moment of Inertia. Perform the integration, Iy =

LA

a

x2dA =

b b 1 x2 a 1 x2 - 2 x2 bdx a L0 a2 a

=

b 5 b a 1 x2 - 2 x4 bdx a 2 L0 a

= a

2b

7a

1 2

7

x2 -

b 5 2a x b 5a2 0

=

2 3 1 a b - a3b 7 5

=

3a3b  35

Ans.

Ans: Iy = 1033

3a3b 35

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10–23. y

Determine the moment of inertia for the shaded area about the x axis.

y2  2x yx

2m

2m

x

Solution

1 2 y . The area of the differential element 2 1 parallel to the x axis shown shaded in Fig. a is dA = (x2 - x1)dy = ay - y2 bdy. 2 Moment of Inertia. Perform the integration, Differential Element. Here x2 = y and x1 =

Ix =

LA

y2dA =

L0

L0



=



= a



2m

y2 ay -

2m

ay3 -

1 2 y bdy 2

1 4 y bdy 2

y5 2 m y4 b2 4 10 0

= 0.8 m4

Ans.

Ans: Ix = 0.8 m4 1034

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*10–24. Determine the moment of inertia for the shaded area about the y axis.

y y2  2x yx

2m

2m

x

Solution 1

Differential Element. Here, y2 = 22x2 and y1 = x. The area of the differential element parallel to the y axis shown shaded in Fig. a is dA = (y2 - y1) dx 1 = 1 22x2 - x 2 dx.

Moment of Inertia. Perform the integration, Iy =

LA

x2dA =

L0

2m

L0

2m



=



= a



=

1

x2 a 22x2 - xbdx

1 22x

5 2

- x3 2 dx

222 7 x4 2 2 m x2 b 7 4 0

4 4 m = 0.571 m4 7

Ans.

Ans: Iy = 0.571 m4 1035

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10–25. The polar moment of inertia for the area is JC = 642 (106) mm4, about the z′ axis passing through the centroid C. The moment of inertia about the y′ axis is 264 (106) mm4, and the moment of inertia about the x axis is 938 (106) mm4. Determine the area A.

y¿

C

x¿ 200 mm

Solution Applying the parallel-axis theorem with d y = 200 mm and Ix = 938 ( 106 ) mm4,

x

Ix = Ix′ + Ad 2y   938 ( 106 ) = Ix′ + A ( 2002 )

Ix′ = 938 ( 106 ) - 40 ( 103 ) A

with known polar moment of inertia about C, JC = Ix′ + Iy′

642 ( 10

6

) = 938 ( 106 ) - 40 ( 103 ) A + 264 ( 106 )

  A = 14.0 ( 103 ) mm2

Ans.

Ans: A = 14.0 ( 103 ) mm2 1036

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10–26. Determine the moment of inertia of the composite area about the x axis.

150 mm 150 mm 100 mm 100 mm x 300 mm

75 mm

Solution Composite Parts: The composite area can be subdivided into three segments as shown in Fig. a. Since segment (3) is a hole, it contributes a negative moment of inertia. The perpendicular distance measured from the centroid of each segment to the x axis is also indicated. Moment of Inertia: The moment of inertia of each segment about the x axis can be determined using the parallel - axis theorem. Thus, 9 Ix = I x′ + A1d y 2 2 = c

1 1 200 2 1 p a(300) 12003 2 + 13002 12002a b d + c 13002 12003 2 + 30012002 11002 2 d + c - 1754 2 + 36 2 3 12 4

= 7981106 2mm4

1 -p1752 22 11002 2 d

Ans.

Ans: Iy = 798(106) mm4 1037

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10–27. Determine the moment of inertia of the composite area about the y axis.

150 mm 150 mm 100 mm 100 mm x 300 mm

75 mm

Solution Composite Parts: The composite area can be subdivided into three segments as shown in Fig. a. Since segment (3) is a hole, it contributes a negative moment of inertia. The perpendicular distance measured from the centroid of each segment to the y axis is also indicated. Moment of Inertia: The moment of inertia of each segment about the y axis can be determined using the parallel - axis theorem. Thus, 9

Iy = Iy′ + A(d x)2 = c

1 1 1 p (200)(3003) + (200)(300)(200)2 d + c (200)(3003) + 200(300)(450)2 d + c - (754) + 36 2 12 4

= 10.3(109) mm4

Ans.

1 -p(752)2 (450)2 d

Ans: Iy = 10.3(109) mm4 1038

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*10–28. y

Determine the moment of inertia about the x axis. 150 mm

150 mm

20 mm 200 mm x

C 20 mm

Solution Moment of Inertia. Since the x axis passes through the centroids of the two segments, Fig. a, Ix =

200 mm

20 mm

1 1 (300)(4003) (280)(3603) 12 12

= 511.36(106) mm4 = 511(106) mm4

Ans.



Ans: Ix = 511(106) mm4 1039

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10–29. y

Determine the moment of inertia about the y axis. 150 mm

150 mm

20 mm 200 mm x

C 20 mm

Solution Moment of Inertia. Since the y axis passes through the centroid of the two segments, Fig. a, Iy =

200 mm

20 mm

1 1 (360)(203) + (40)(3003) 12 12

= 90.24(106) mm4 = 90.2(106) mm4

Ans.

Ans: Iy = 90.2 ( 106 ) mm4 1040

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10–30. Determine y, which locates the centroidal axis x′ for the cross-sectional area of the T-beam, and then find the moments of inertia Ix′ and Iy′.

y¿

75 mm

75 mm

y

20 mm C x¿

Solution

150 mm

Centroid. Referring to Fig. a, the areas of the segments and their respective centroids are tabulated below.

Thus, y =

y (mm)

∼A(mm3)

150(20)

10

30(103)

2

20(150)

95

285(103)

Σ

6(103)

Segment

A(mm2)

1

∼

y

20 mm

315(103)

∼2 315(103) Σy A = 52.5 mm = ΣA 6(103)

Ans.

Moment of Inertia. The moment of inertia about the x′ axis for each segment can be determined using the parallel axis theorem, Ix′ = Ix′ + Ad 2y. Referring to Fig. b, Segment

Ai(mm2)

(dy)i (mm) (Ix′)i (mm4)

1

150(20)

42.5

2

20(150)

42.5

(Ad 2y)i (mm4)

(Ix′)i (mm4)

1 6 (150)(203) 5.41875(10 ) 12 1 6 (20)(1503) 5.41875(10 ) 12

5.51875(106) 11.04375(106)

Thus     Ix′ = Σ(Ix′)i = 16.5625(106) mm4 = 16.6(106) mm4

Ans.

Since the y' axis passes through the centroids of segments 1 and 2, Iy′ =

1 1 (20)(1503) + (150)(203) 12 12

= 5.725(106) mm4

Ans.

Ans: y = 52.5 mm Ix′ = 16.6 ( 106 ) mm4 Iy′ = 5.725 ( 106 ) mm4 1041

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10–31. Determine the location y of the centroid of the channel’s cross-sectional area and then calculate the moment of inertia of the area about this axis.

50 mm

50 mm

x

250 mm –y

SOLUTION Centroid: The area of each segment and its respective centroid are tabulated below.

50 mm 350 mm

Segment

A (mm2)

' y (mm)

' y A (mm3)

1

100(250)

125

3.12511062

2

250(50)

25

0.312511062

©

37.511032

3.437511062

Thus, ' 3.437511062 ©yA ' = 91.67 mm = 91.7 mm y = = ©A 37.511032

Ans.

Moment of Inertia: The moment of inertia about the x¿ axis for each segment can be determined using the parallel-axis theorem Ix¿ = Ix¿ + Ad2y. Segment

Ai (mm2)

(dy)i (mm)

(Ix¿)i (mm4)

(Ad 2y)i (mm4)

(Ix¿)i (mm4)

1

100(250)

33.33

1 3 12 110021250 2

27.77811062

157.9911062

2

250(50)

66.67

1 3 12 12502150 2

55.55611062

58.1611062

Thus, Ix¿ = © Ix¿

i

= 216.15 106 mm4 = 216 106 mm4

Ans.

Ans: ∼ y = 91.7 mm Ix′ = 216(106) mm4 1042

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*10–32. Determine the moment of inertia of the beam’s

y

cross-sectional area about the x axis.

125 mm

125 mm 12 mm

12 mm 100 mm 25 mm 12 mm

12 mm 75 mm x 75 mm

Solution Composite Parts: The composite area can be subdivided into segments as shown in Fig. a. The perpendicular distance measured from the centroid of each segment to the x axis is also indicated. Moment of Inertia: The moment of inertia of each segment about the x axis can be determined using the parallel - axis theorem. Thus, Ix = Ix′ + A1d y 2 2 = c2a

1 1 1 (226) 1 123 2b + 2(226)(12) 1 1192 2 d + c 4 a 1 122 (1003) b + 41 122 (100) 1 752 2 d + c 2 a 1 12) 1 1503 2 b + 21 12) 1 150) 1 02 2 d 12 12 12

= 114.621106 2mm4 = 1151106 2mm4

Ans.

Ans:

1043

Ix = 1151106 2mm4

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10–33. Determine the moment of inertia of the beam’s cross-sectional area about the y axis.

y 125 mm

12 mm 100 mm 25 mm 12 mm

125 mm 12 mm 12 mm 75 mm x 75 mm

Solution Composite Parts: The composite cross - sectional area of the beam can be subdivided into segments as shown in Fig. a. The perpendicular distance measured from the centroid of each segment to the x axis is also indicated. Moment of Inertia: The moment of inertia of each segment about the y axis can be determined using the parallel - axis theorem. Thus, Iy = Ix′ + A1d x 2 2 = c2a

1 1 1 (12)1226 3 2 b + 2(226)112 2 10 2 2 d + c 4a (100)1123 2 b + 4(100)112 2 1119 2 2 d + c 2 a 1150)1123 2 b + 2 1150)112)1131 2 2 d 12 12 12

= 152.941106 2mm4 = 1531106 2mm4

Ans.

Ans: Iy = 153(106) mm4 1044

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10–34. y

Determine the moment of inertia Ix of the shaded area about the x axis.

100 mm

100 mm

150 mm

150 mm

150 mm

Solution

O

75 mm x

Moment of Inertia. The moment of inertia about the x axis for each segment can be determined using the parallel axis theorem, Ix = Ix′ + Ad 2y. Referring to Fig. a Segment

Ai(mm2)

(dy)i (mm)

1

200(300)

150

2

1 (150)(300) 2

100

3

- p(752)

150

Thus,

(Ix′)i (mm4)

(Ady)2i (mm4)

(Ix)i (mm4)

1 (200)(3003) 12 1 (150)(3003) 36 p(754) 4

1.35(109)

1.80(109)

0.225(109)

0.3375(109)

- 0.3976(109)

- 0.4225(109)

Ix = Σ(Ix)i = 1.715(109) mm4 = 1.72(109) mm4      Ans.

Ans: Ix = 1.72(109) mm4 1045

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y

10–35. Determine the moment of inertia Ix of the shaded area about the y axis.

100 mm

100 mm

150 mm

150 mm

150 mm

Solution

O

75 mm x

Moment of Inertia. The moment of inertia about the y axis for each segment can be determined using the parallel-axis theorem, Iy = Iy′ + Ad 2x. Referring to Fig. a Segment

Ai(mm2)

(dx)i (mm)

1

200(300)

100

2

1 (150)(300) 2

250

3

- p(752)

100

Iy′(mm4) 1 (300)(2003) 12 1 (300)(1503) 36 p(754) 4

(Ad 2x)i (mm4)

(Iy)i (mm4)

0.6(109)

0.800(109)

1.40625(109) 1.434375(109) - 0.1767(109) - 0.20157(109)

Thus, Iy = Σ(Iy)i = 2.033(109) mm4 = 2.03(109) mm4      Ans.

Ans: Iy = 2.03 ( 109 ) mm4 1046

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*10–36. Determine the moment of inertia of the beam’s crosssectional area about the y axis.

y

150 mm 150 mm

50 mm

SOLUTION

250 mm

x¿ C

Moment of Inertia: The dimensions and location of centroid of each segment are shown in Fig. a. Since the y axis passes through the centroid of both segments, the moment of inertia about y axis for each segment is simply (Iy)i = (Iy¿)i. Iy = g (Iy)i =

1 1 (50)(3003) + (250)(503) 12 12

x¿ _ y

25 mm

= 115.10(106) mm4 = 115(106) mm4

25 mm

x

Ans.

Ans: Iy = 115 ( 106 ) mm4 1047

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10–37. Determine y, which locates the centroidal axis x  for the cross-sectional area of the T-beam, and then find the moment of inertia about the x  axis.

y

150 mm 150 mm

50 mm

SOLUTION y =

250 mm

©yA 125(250)(50) + (275)(50)(300) = ©A 250(50) + 50(300)

x¿ C x¿ y

= 206.818 mm y = 207 mm Ix¿

Ans.

1 = c (50)(250)3 + 50(250)(206.818 - 125)2 d 12 +c

25 mm

25 mm

x

1 (300)(50)3 + 50(300)(275 - 206.818)2 d 12

I x¿ = 222(106) mm4

Ans.

Ans: y = 207 mm Ix′ = 222 ( 106 ) mm4 1048

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10–3 8 .

y 25 mm

Determine the distance y to the centroid of the beam’s cross-sectional area; then determine the moment of inertia about the x¿ axis.

25 mm

100 mm

C

x¿

_ y

25 mm x

50 mm 100 mm

75 mm

75 mm

50 mm

SOLUTION Centroid: The area of each segment and its respective centroid are tabulated below. Segment 1 2 3

A (mm2) 50(100) 325(25) 25(100)

©

15.625(103)

25 mm

yA (mm3) 375(103) 10l.5625(103) –125(103)

y (mm) 75 12.5 –50

351.5625(103)

Thus, y =

351.5625(103) ©yA = = 22.5 mm ©A 15.625(103)

Ans.

Moment of Inertia: The moment of inertia about the x¿ axis for each segment can be determined using the parallel-axis theorem Ix¿ + Ix¿ + Ad2y. Segment

Ai (mm2)

1

50(100)

A dy B i (mm) A Ix–B i (mm4) 52.5

2

325(25)

10

3

25(100)

72.5

1 12 1 12 1 12

A Ad2y B i (mm4)

A Ix¿ B i (mm4)

13.781(106)

17.948(106)

(325) (25 )

0.8125(106)

1.236(106)

(25) (1003)

13.141(106)

15.224(106)

(50) (1003) 3

Thus, Ix¿ = ©(Ix¿)i = 34.41 A 106 B mm4 = 34.4 A 106 B mm4

Ans.

Ans: y = 22.5 mm Ix = 34.4(106) mm4 1049

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10–39. Determine the moment of inertia of the beam’s crosssectional area about the y axis.

y 25 mm

25 mm

100 mm

C

x¿

_ y 50 mm 100 mm

25 mm x 75 mm

75 mm

50 mm

25 mm

SOLUTION Moment of Inertia: The moment of inertia about the y¿ axis for each segment can be determined using the parallel-axis theorem Iy¿ = Iy¿ + Ad2x. Segment

Ai (mm2)

A dx B i (mm)

1

2[100(25)]

100

2

25(325)

0

3

100(25)

0

2

A Iy–B i (mm4)

1 3 12 (100) (25 ) 1 3 12 (25) (325 ) 1 3 12 (100) (25 )

C

D

A Ad2x B i (mm4) A Iy–B i (mm4) 50.0(106)

50.260(106)

0

71.517(106) 0.130(106)

0

Thus, Iy¿ = ©(Iy¿)i = 121.91 A 106 B mm4 = 122 A 106 B mm4

Ans.

1050

Ans: Iy = 122 1 106 2 mm4

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*10–40. Determine the moment of inertia of the cross-

y

y¿

sectional area about the x axis. 10 mm

x

180 mm

x

C 100 mm 10 mm 10 mm

100 mm

Solution Composite Parts: The composite cross - sectional area of the beam can be subdivided into two segments as shown in Fig. a. Here, segment (2) is hole, and so it contributes a negative moment of inertia. Moment of Inertia: Since the x axis passes through the centroid of both rectangular segments, Ix = 1Ix 2 1 + 1Ix 2 2 =

1 1 11002 12003 2 1902 11803 2 12 12

= 22.91106 2mm4

Ans.

Ans:

1051

Ix = 22.91106 2mm4

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10–41. Locate the centroid x of the beam’s crosssectional area, and then determine the moment of inertia of the area about the centroidal y¿ axis.

y

10 mm

y¿ x

180 mm

x

C 100 mm 10 mm 10 mm

100 mm

Solution Composite Parts: The composite cross - sectional area of the beam can be subdivided into segments as shown in Fig. a. Centroid: The perpendicular distance measured from the centroid of each type of segment to the y axis is also indicated in Fig. a. Thus, ∼

x =

95(1011802) + 501211002 1102) 2711103 2 ΣxA = = = 71.32 mm ΣA 10(180) + 211002 1102 3.81103 2

Ans.

Moment of Inertia: The moment of inertia of each segment about the y¿ axis can be determined using the parallel - axis theorem. The perpendicular distance measured from the centroid of each type of segment to the y¿ axis is indicated in Fig. b. Iy′ = Iy′ + A1d x′ 2 2 = c

1 1 11802 1103 2 + 1801102 123.682 2 d + c 2a 1102 11003 2 b + 211002 1102 121.322 2 d 12 12

= 3.601106 2mm4

Ans.

Ans: x = 71.32 mm Iy = 3.60(106) mm4 1052

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10–42. Determine the moment of inertia of the beam’s crosssectional area about the x axis.

y

y¿

30 mm 30 mm 70 mm 140 mm

SOLUTION

30 mm

1 (170)(30)3 + 170(30)(15)2 Ix = 12

C y

x¿

30 mm 170 mm

1 + (30)(170)3 + 30(170)(85)2 12 +

x

x

1 (100)(30)3 + 100(30)(185)2 12

Ix = 154(106) mm4

Ans.

Ans: Ix = 154 ( 106 ) mm4 1053

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10–43. Determine the moment of inertia of the beam’s crosssectional area about the y axis.

y

y¿

30 mm 30 mm

140 mm

70 mm _ x C

30 mm

SOLUTION Iy =

1 (30)(170)3 + 30(170)(115)2 12 +

1 (170)(30)3 + 30(170)(15)2 12

+

1 (30)(100)3 + 30(100)(50)2 12

_ y

x¿

30 mm 170 mm x

Iy = 91.3(106) mm4

Ans.

Ans: Iy = 91.3 ( 106 ) mm4 1054

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*10–44. Determine the distance y to the centroid C of the beam’s cross-sectional area and then compute the moment of inertia Ix¿ about the x¿ axis.

y

y¿

30 mm 30 mm 70 mm 140 mm

SOLUTION

30 mm

170(30)(15) + 170(30)(85) + 100(30)(185) y = 170(30) + 170(30) + 100(30) = 80.68 = 80.7 mm Ix¿ = c

Ans.

x C y

x¿

30 mm 170 mm x

1 (170)(30)3 + 170(30)(80.68 - 15)2 d 12

+c +

1 (30)(170)3 + 30(170)(85 - 80.68)2 d 12

1 (100)(30)3 + 100(30)(185 - 80.68)2 12

Ix¿ = 67.6(106) mm4

Ans.

Ans: y = 80.7 mm Ix′ = 67.6(106) mm4 1055

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10–45. Determine the distance x to the centroid C of the beam’s cross-sectional area and then compute the moment of inertia Iy¿ about the y¿ axis.

y

y¿

30 mm 30 mm 70 mm 140 mm

SOLUTION

30 mm

170(30)(115) + 170(30)(15) + 100(30)(50) x = 170(30) + 170(30) + 100(30) Ans.

= 61.59 = 61.6 mm Iy¿ = c

x C y

x¿

30 mm 170 mm x

1 (30)(170)3 + 170(30)(115 - 61.59)2 d 12

+c +

1 (170)(30)3 + 30(170)(15 - 61.59)2 d 12

1 (30)(100)3 + 100(30)(50 - 61.59)2 12

Iy¿ = 41.2(106) mm4

Ans.

Ans: x = 61.6 mm Iy= = 41.2 ( 106 ) mm4 1056

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10–46. y

Determine the moment of inertia of the shaded area about the x axis.

θ

SOLUTION

x

θ

1 1 Ix = c r4 au - sin 2u b d 4 2 -2c

r 2

1 1 1 (r cos u)(r sin u)3 + (r cos u)(r sin u)a r sin ub d 36 2 3

=

1 1 4 1 1 4 r a u - sin 2ub r cos u sin3 u - r4 cos u sin3 u 4 2 18 9

=

r4 (6u - 3 sin 2u - 4 cos u sin3 u) 24

Ans.

Ans: Ix = 1057

r4 (6u - 3 sin 2u - 4 cos u sin3 u) 24

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10–47. Determine the moment of inertia of the shaded area about the y axis.

y

SOLUTION

θ

1 1 Iy = c r4 a u + sin 2u b d 4 2

θ

x

r

2 1 2 1 - c (2r sin u)(r cos u)3 + (2r sin u)(r cos u)a r cos u b d 36 2 3

=

1 1 4 1 r a u + sin 2u b - c r4 sin u cos3u + r3 sin u cos3 u d 4 2 18

=

1 r4 u + sin 2u - 2 sin u cos3 u 4 2

Ans.

Ans: Iy =

1058

r4 1 au + sin 2u - 2 sin u cos3 u b 4 2

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*10–48. y'

y

Determine the moment of inertia of the parallelogram about the x′ axis, which passes through the centroid C of the area.

C

a

x'

θ

SOLUTION

b

x

h = a sin u Ix′ =

1 1 1 3 bh3 = (b)(a sin u)3 = a b sin3 u 12 12 12

Ans.

Ans: 1 3 Ix′ = a b sin3 u 12 1059

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10–49. y'

y

Determine the moment of inertia of the parallelogram about the y′ axis, which passes through the centroid C of the area.

C

a

x'

θ

SOLUTION x = a cos u +

Iy′ = 2 J + =

b

x

1 b - a cos u = (a cos u + b) 2 2

2 a 2 1 1 b (a sin u)(a cos u)3 + (a sin u)(a cos u) a + cos u - a cos u b R 36 2 2 2 3

1 (a sin u)(b - a cos u)3 12

ab sin u 2 ( b + a2 cos2 u ) 12

Ans.

Ans: Iy′ = 1060

ab sin u 2 ( b + a2 cos2 u ) 12

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10–50. Locate the centroid y of the cross section and determine the moment of inertia of the section about the x¿ axis. 0.4 m

x' –y

0.05 m 0.2 m 0.2 m

SOLUTION

0.3 m

0.2 m 0.2 m

Centroid: The area of each segment and its respective centroid are tabulated below. Segment

A (m2)

y (m)

yA (m3)

1

0.3(0.4)

0.25

0.03

2

1 2 10.4210.42

0.1833

0.014667

3

1.1(0.05)

0.025

0.001375

©

0.255

0.046042

Thus, y =

©yA 0.046042 = = 0.1806 m = 0.181 m ©A 0.255

Ans.

Moment of Inertia: The moment of inertia about the x¿ axis for each segment can be determined using the parallel-axis theorem Ix¿ = Ix¿ + Ad2y.

Segment

Ai (m2)

(dy)i (m)

(Ix¿)i (m4)

1

0.3(0.4)

0.06944

2

1 2 10.4210.42

0.002778

3

1.1(0.05)

0.1556

1 3 12 10.3210.4 2 1 3 36 10.4210.4 2 1 3 12 11.1210.05 2

(Ad 2y)i (m4)

(Ix¿)i (m4)

0.5787110-32 2.1787110-32 0.6173110-62 0.7117110-32 1.3309110-32 1.3423110-32

Thus, Ix¿ = © Ix¿

i

= 4.233 10 -3 m4 = 4.23 10-3 m4

Ans.

Ans: y = 0.181 m Ix′ = 4.23(10-3) m4 1061

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10–51. Determine the moment of inertia for the beam’s crosssectional area about the x¿ axis passing through the centroid C of the cross section.

100 mm

100 mm

25 mm 200 mm 200 mm 45

SOLUTION Ix¿ =

45

1 1 1 141.4 2 b d (200)(332.8)3 + 4 c (141.4)(141.4)3 + a (141.4)(141.4) b a 3 12 36 2

C

45 45

x¿

25 mm

1 p 1 - 2 c (200)4 a - sin90° b d 4 4 2 = 520(106) mm4

Ans.

Ans: Ix= = 520 ( 106 ) mm4 1062

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*10–52. Determine the distance x to the centroid of the beam’s cross-sectional area: then find the moment of inertia about the y¿ axis.

y

y' –x

40 mm

SOLUTION

40 mm

Segment

A (mm2)

x (mm)

xA (mm3)

1

160(80)

80

1.024(106)

2

40(80)

20

64.0(103)

©

16.0(103)

x'

C

Centroid: The area of each segment and its respective centroid are tabulated below.

40 mm 40 mm x 120 mm 40 mm

1.088(106)

Thus, x =

1.088 (106) ©xA = 68.0 mm = ©A 16.0(103)

Ans.

Moment of Inertia: The moment of inertia about the y¿ axis for each segment can be determined using the parallel–axis theorem Iy¿ = Iy¿ + Ad2x. Segment

Ai (mm2)

(dx )i (mm)

(Iy ¿)i (mm4)

(Ad 2x )i (mm4)

(Iy¿)i (mm4)

1

80(160)

12.0

1 3 12 (80)(160 )

1.8432(106)

29.150(106)

2

80(40)

48.0

1 3 12 (80)(40 )

7.3728(106)

7.799(106)

Thus, Iy¿ = © A Iy¿ B i = 36.949 A 106 B mm4 = 36.9 A 106 B mm4

Ans.

Ans: Iy¿ = 36.9 A 106 B mm4 1063

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10–53. Determine the moment of inertia for the beam’s crosssectional area about the x¿ axis.

y

y' –x

40 mm 40 mm x'

C 40 mm 40 mm

x 120 mm

SOLUTION

40 mm

Moment of Inertia: The moment inertia for the rectangle about its centroidal axis 1 3 bh , given on the inside back cover of can be determined using the formula, Ix¿ = 12 the textbook. Ix¿ =

1 1 (160) A 1603 B (120) A 803 B = 49.5 A 106 B mm4 12 12

Ans.

Ans: Ix = 49.5(106) mm4 1064

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10–54.

y

Determine the product of inertia of the thin strip of area with respect to the x and y axes. The strip is oriented at an angle u from the x axis. Assume that t V l.

t

l

u

x

SOLUTION 1

l

lxy =

=

LA

xydA =

L0

(s cos u)(s sin u)tds = sin u cos ut

L0

2

s ds

1 3 l t sin 2u 6

Ans.

Ans: Ixy =

1065

1 3 l t sin 2u 6

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10–55. Determine the product of inertia for the shaded area with respect to the x and y axes.

y

h y

SOLUTION x ' x = 2

h x3 b3

x

b

' y = y dA = x dy d Ixy = Ixy =

x2y dy 2

L

d Ixy h

=

1 b 2 5>3 a 1>3 b y dy L0 2 h

=

h 1 b2 3 B a 2>3 b a b y8>3R 2 8 h 0

=

3 2 2 b h 16

Ans.

Ans: Ixy = 1066

3 2 2 b h 16

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*10–56. y

Determine the product of inertia of the parallelogram with respect to the x and y axes.

a θ

c

SOLUTION Product of Inertia of the Triangle: The product of inertia with respect to x and y axes can be determined by integration. The area of the differential element parallel h to y axis is dA = ydx = a h + x b dx [Fig. (a)]. The coordinates of the centroid for b y 1 h ' ' = this element are x = x, y = ¢h + x b. Then the product of inertia for 2 2 b this element is '' dIxy = dIx¿y¿ + dA x y = 0 + c ah + =

1 h h x bdx d1x2 B a h + xb R b 2 b

h2 2h2 2 1 2 x ≤ dx ¢ h x + 2 x3 + 2 b b

Performing the integration, we have 0

Ixy =

L

dIxy =

1 h2 2h2 2 b2h2 x ≤ dx = ¢ h2x + 2 x3 + 2 L-b b 24 b

The product of inertia with respect to centroidal axes, x¿ and y¿, can be determined by applying Eq. 10–8 [Fig. (b) or (c)]. Ixy = Ix¿y¿ + Adxdy -

1 b h b2h2 = Ix¿y¿ + bh ¢ - b a b 24 2 3 3 Ix¿y¿ =

b2h2 72

Here, b = a cos u and h = a sin u. Then, Ix¿y¿ =

a4 sin2 u cos2 u . 72

Product of inertia of the parallelogram [Fig. (d)] with respect to centroidal x¿ and y¿ axes, is Ix¿y¿ = 2 B =

a4 cos2 u sin2 u 1 3c - a cos u a sin u + 1a sin u21a cos u2 a ba bR 72 2 6 6

a3c sin2 u cos u 12

The product of inertia of the parallelogram [Fig. (d)] about x and y axes is Ixy = Ix¿y¿ + Adxdy =

a3c sin2 u cos u c + a cos u a sin u + 1a sin u21c2 a ba b 12 2 2

=

a2c sin2 u 4a cos u + 3c 12

Ans.

1067

x

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10–57.

y

Determine the product of inertia for the shaded area with respect to the x and y axes. y ⫽ bn xn a

b x

a

SOLUTION '' dIxy = dIxy + x y dA a

Ixy = 0 + = ¢ =

a y 1 b2 (x) ¢ ≤ (y dx) = a 2n bx2n + 1 dx 2 2 L0 a L0

a b2a2n + 2 b2 1 x2n + 2 ` = 2n ≤ ¢ 2n + 2 ≤ 2a 4(n + 1)a2n 0

a2b2 4(n + 1)

Ans.

Ans: Ixy =

1068

a2b2 4(n + 1)

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y

10–58. Determine the product of inertia for the shaded portion of the parabola with respect to the x and y axes.

100 mm

200 mm y

1 2 x 50 x

Solution 1

Differential Element: Here, x = 250y2 . The area of the differential element paral1 lel to the x axis is dA = 2xdy = 2250y2 dy. The coordinates of the centroid for this element are x = 0, y = y. Then the product of inertia for this element is dIxy = dIx′y′ + dAx y 1

= 0 + 12250y2 dy2 102 1y2 = 0

Product of Inertia: Performing the integration, we have Ixy =

L

Ans.

dIxy = 0

Note: By inspection, Ixy = 0 since the shaded area is symmetrical about the y axis.

Ans: Ixy = 0 1069

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10–59. Determine the product of inertia of the shaded area with respect to the x and y axes, and then use the parallel-axis theorem to find the product of inertia of the area with respect to the centroidal x¿ and y¿ axes.

y

y¿

y2

x

2m C

SOLUTION

x

Differential Element: The area of the differential element parallel to the y axis shown shaded in Fig. a is dA = y dx = x 1>2 dx. The coordinates of the centroid of y ' ' 1 1>2 this element are x = x and y = Thus, the product of inertia of this = x 2 2 element with respect to the x and y axes is

4m

~~ dIxy = dIx¿y¿ + dAx y 1 = 0 + A x1>2 dx B (x) a x 1>2 b 2 1 2 x dx 2 Product of Inertia: Performing the integration, we have =

4m

Ixy =

x¿

L

dIxy =

L0

4m 1 1 2 = 10.67 m4 = 10.7 m4 x dx = a x3 b 2 2 6 0

Ans.

Using the information provided on the inside back cover of this book, the location of 2 3 the centroid of the parabolic area is at x = 4 - (4) = 2.4 m and y = (2) = 0.75 m 5 8 2 and its area is given by A = (4)(2) = 5.333 m2. Thus, 3 Ixy = Ix¿y¿ + Adxdy 10.67 = Ix¿y¿ + 5.333(2.4)(0.75) Ix¿y¿ = 1.07 m4

Ans.

Ans: Ixy = 10.7 m4 Ix′y′ = 1.07 m4 1070

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*10–60. y

Determine the product of inertia for the shaded area with respect to the x and y axes. Use Simpson’s rule to evaluate the integral.

y

0.8e x

SOLUTION

x

x = x y =

2

1m

y 2

d A = ydx dIxy = Ixy =

xy2 dx 2

L

dIxy 1

=

1 2 x(0.8 ex )2 dx L0 2 1

= 0.32

2

L0

xe2x dx Ans.

= 0.511 m4

Ans: Ixy = 0.511 m4 1071

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10–61. Determine the product of inertia for the parabolic area with respect to the x and y axes.

y

y

b 1/2 x a1/2

b

x

SOLUTION

a

' x = x y ' y = 2 dA = y dx dIxy =

xy2 dx 2

Ixy =

d Ixy L a

=

1 b2 2 a bx dx L0 2 a

=

a 1 b2 B a b x3 R a 6 0

=

1 2 2 a b 6

Ans.

Ans: Ixy = 1072

1 2 2 ab 6

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10–62. Determine the product of inertia of the shaded area with respect to the x and y axes.

y

1

1

y = (a2 – x 2 )2 a

SOLUTION Differential Element: The area of the differential element parallel to the y axis is 1 1 2 dA = ydx = A a2 - x2 B dx. The coordinates of the centroid for this element are y 1 1 1 ' ' = A a2 - x2 B 2. Then the product of inertia for this element is x = x, y = 2 2

O

x a

'' dIxy = dIx¿y¿ + dAx y 2 1 1 1 1 1 = 0 + c A a2 - x2 B 2 dx d1x2c aa2 - x2 b d 2

=

3 3 1 5 1 3 A x + a2x + 6ax2 - 4a2x2 - 4a2x2 B dx 2

Product of Inertia: Performing the integration, we have a

Ixy =

L

dIxy =

3 3 1 5 1 ¢ x3 + a2x + 6ax2 - 4a2x2 - 4a2x2 b dx 2 L0

=

8 3 5 8 1 7 a a2 2 1 x4 + x + 2ax3 - a2x2 - a2x2 ` 2 4 2 5 7 0

=

a4 280

Ans.

Ans: Ixy = 1073

a4 280

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10–63. Determine the product of inertia of the cross-sectional area with respect to the x and y axes.

y

100 mm 20 mm

400 mm

SOLUTION

C

Product of Inertia: The area for each segment, its centroid and product of inertia with respect to x and y axes are tabulated below. Segment

Ai (mm2)

(dx)i (mm)

(dy)i (mm)

(Ixy)i (mm4)

1

100(20)

60

410

49.211062

2

840(20)

0

0

0

3

100(20)

- 60

-410

49.211062

x

400 mm

20 mm 100 mm

20 mm

Thus, Ixy = ©1Ixy2i = 98.411062mm4

Ans.

Ans: Ixy = 98.4 ( 106 ) mm4 1074

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*10–64. Determine the product of inertia of the beam’s

y

cross-sectional area with respect to the x and y axes.

10 mm

300 mm

10 mm x 10 mm 100 mm

Solution Composite Parts: The composite cross - sectional area of the beam can be subdivided into three segments. The perpendicular distances measured from the centroid of each element to the x and y axes are also indicated. Product of Inertia: Since the centroidal axes are the axes of all the segments are the axes of symmetry, then Ix′y′ = 0. Thus, the product of inertia of each segment with respect to the x and y axes can be determined using the parallel - axis theorem. Ixy = Ix′y′ + Ad xd y = Ad xd y = 90(10)(55)(295) + 300(10)(5)(150) + 90(10)(55)(5) = ΣIxy = 17.1(106)mm4

Ans.

Ans: Ixy = 17.1(106)mm4 1075

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10–65. Determine the location (x, y) of the centroid C of the angle’s cross-sectional area, and then compute the product of inertia with respect to the x′ and y′ axes.

y

18 mm

y¿

x

150 mm x¿

C y 150 mm

x 18 mm

SOLUTION Centroid: x =

y =

9(18)(150) + 84(18)(132) ΣxA = = 44.11 mm = 44.1 mm ΣA 18(150) + 18(132) ΣyA ΣA

=

75(18)(150) + 9(18)(132) 18(150) + 18(132)

= 44.11 mm = 44.1 mm

Ans.

Ans.

Product of inertia about x′ and y′ axes: Ix′y′ = 18(150)( -35.11)(30.89) + 18(132)(39.89)( -35.11) = -6.26(106) mm4

Ans.

Ans: x = y = 44.1 mm Ix=y= = -6.26 ( 106 ) mm4 1076

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10–66. Determine the product of inertia for the beam’s crosssectional area with respect to the u and v axes.

y

v 150 mm

150 mm

u 20

SOLUTION 20 mm

1 1 (300)(400)3 (280)(360)3 = 511.36(10)6 mm4 Ix = 12 12 Iy = 2 c

x

C

Moments of inertia Ix and Iy

200 mm

20 mm

1 1 (20)(300)3 d + (360)(20)3 = 90.24(10)6 mm4 12 12

The section is symmetric about both x and y axes; therefore Ixy = 0. Iuv =

I x - Iy

= a

2

sin 2u + Ixy cos 2u

511.36 - 90.24 sin 40° + 0 cos 40°b 106 2

= 135(10)6 mm4

Ans.

Ans: Iuv = 135(10)6 mm4 1077

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10–67.

y

Determine the moments of inertia Iu and Iv of the shaded area.

20 mm v

u

45°

SOLUTION

20 mm x 20 mm

200 mm

Moment and Product of Inertia about x and y Axes: Since the shaded area is symmetrical about the x axis, Ixy = 0. Ix =

1 1 1200214032 + 1402120032 = 27.7311062 mm4 12 12

Iy =

1 1 1402120032 + 4012002112022 + 1200214032 12 12

200 mm 40 mm

= 142.9311062 mm4 Moment of Inertia about the Inclined u and v Axes: Applying Eq. 10–9 with u = 45°, we have Iu =

Ix + Iy 2

= a

+

Ix - Iy 2

cos 2u - Ixy sin 2u

27.73 + 142.93 27.73 - 142.93 + cos 90° - 01sin 90°2b11062 2 2

= 85.311062 mm4 Iv =

Ix + Iy

= a

2

-

Ix - Iy 2

Ans. cos 2u + Ixy sin 2u

27.73 + 142.93 27.73 - 142.93 cos 90° - 01sin 90°2b11062 2 2

= 85.3 106 mm4

Ans.

Ans: Iu = 85.3(106) mm4 Iv = 85.3(106) mm4 1078

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*10–68.

y v

Determine the distance y to the centroid of the area and then calculate the moments of inertia Iu and Iv for the channel’s cross-sectional area. The u and v axes have their origin at the centroid C. For the calculation, assume all corners to be square.

10 mm

10 mm

C

50 mm

u

20 y

10 mm 150 mm

x

150 mm

SOLUTION y =

300(10)(5) + 2[(50)(10)(35)] = 12.5 mm 300(10) + 2(50)(10)

Ix = c

Ans.

1 (300)(10)3 + 300(10)(12.5 - 5)2 d 12

+ 2c

1 (10)(50)3 + 10(50)(35 - 12.5)2 d 12

= 0.9083(106) mm4 Iy =

1 1 (50)(10)3 + 50(10)(150 - 5)2 d (10)(300)3 + 2 c 12 12

= 43.53(106) mm4 Ixy = 0 Iu = =

(By symmetry)

Ix + Iy 2

+

Ix - Iy 2

cos 2u - Ixy sin 2u

0.9083(106) - 43.53(106) 0.9083(106) + 43.53(106) + cos 40° - 0 2 2 Ans.

= 5.89(106) mm4 Iv = =

Ix + Iy 2

-

Ix - Iy 2

cos 2u + Ixy sin 2u

0.9083(106) - 43.53(106) 0.9083(106) + 43.53(106) cos 40°+0 2 2 Ans.

= 38.5(106) mm4

Ans: Iu = 5.89(106) mm4 Iv = 38.5(106) mm4 1079

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10–69. y

Determine the moments of inertia Iu, Iv and the product of inertia Iuv for the rectangular area. The u and v axes pass through the centroid C.

v

u

30 x 30 mm

C

Solution

120 mm

Moment And Product of Inertia About x and y Axes. Since the rectangular area is symmetrical about the x and y axes, Ixy = 0. Ix =

1 1 (120)(303) = 0.270(106) mm4  Iy = (30)(1203) = 4.32(106) mm4 12 12

Moment And Product of Inertia About The Inclined u and v Axes. With u = 30°, Iu =

Ix + Iy 2

= c

+

Ix - Iy 2

cos 2u - Ixy sin 2u

0.270 + 4.32 0.27 - 4.32 + cos 60° - 0 sin 60° d (106) 2 2

= 1.2825(106) mm4 = 1.28(106) mm4

Iv =

Ix + Iy 2

= c

-

Ix - Iy 2

Ans.

cos 2u + Ixy sin 2u

0.27 + 4.32 0.27 - 4.32 cos 60° + 0 sin 60° d (106) 2 2

= 3.3075(106) mm4 = 3.31(106) mm4

Iuv =

Ix - Iy

= c

2

Ans.

sin 2u + Ixy cos 2u

0.270 - 4.32 sin 60° + 0 cos 60° d (106) 2

= - 1.7537(106) mm4 = - 1.75(106) mm4

Ans.

Ans: Iu = 1.28(106) mm4 Iv = 3.31(106) mm4 Iuv = - 1.75(106) mm4 1080

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10–70. y

Solve Prob. 10–69 using Mohr’s circle. Hint: To solve, find the coordinates of the point P(Iu, Iuv) on the circle, measured counterclockwise from the radial line OA. (See Fig. 10–19.) The point Q(Iv, -Iuv) is on the opposite side of the circle.

v

u

30 x 30 mm

C

Solution

120 mm

Moment And Product of Inertia About x And y Axes. Since the rectangular Area is symmetrical about the x and y axes, Ixy = 0. 1 1 (120)(303) = 0.270(106) mm4  Iy = (30)(1203) = 4.32(106) mm4 12 12 Construction of The Circle. The Coordinates of center O of the circle are Ix =

a

Ix + Iy 2

, 0b = a

0.270 + 4.32 , 0b(106) = (2.295, 0)(106) 2

And the reference point A is

(Ix, Ixy) = (0.270, 0)(106) Thus, the radius of the circle is R = OA =

1 2(2.295

- 0.27)2 + 02 2 (106) = 2.025(106) mm4

Using these Results, the circle shown in Fig. a can be constructed. Rotate radial line OA counterclockwise 2u = 60° to coincide with radial line OP where coordinate of point P is 1 Iu, Iuv 2. Then Iu = (2.295 - 2.025 cos 60°)(106) = 1.2825(106) mm4 = 1.28(106) mm4   Ans. Iuv = - 2.025(106) sin 60° = - 1.7537(106) mm4 = - 1.75(106) mm4   Ans.

Iv is represented by the coordinate of point Q. Thus, Iv = (2.295 + 2.025 cos 60°)(106) = 3.3075(106) mm4 = 3.31(106) mm4

Ans.

Ans: Iu = 1.28 ( 106 ) mm4 Iuv = - 1.75 ( 106 ) mm4 Iv = 3.31 ( 106 ) mm4 1081

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10–71. Determine the moments of inertia and the product of inertia of the beam’s cross sectional area with respect to the u and v axes.

u 300 mm

SOLUTION

30⬚

Moments and product of Inertia with Respect to the x and y Axes: The perpendicular distances measured from the centroid of the triangular segment to the y axis are indicated in Fig. a. Ix =

1 (400)(4503) = 1012.5(106) mm4 36

Iy = 2 B

x

C 150 mm

200 mm

200 mm

1 1 (450)(2003) + (450)(200)(66.672) R = 600(106) mm4 36 2

Since the cross-sectional area is symmetrical about the y axis, Ixy = 0. Moment and product of Inertia with Respect to the u and v Axes: Applying Eq. 10.8 with u = 30°, we have Iu =

Ix + Iy 2

= B

+

Ix - Iy 2

cos 2u - Ixy sin 2u

1012.5 + 600 1012.5 - 600 + ¢ ≤ cos 60° - 0 sin 60° R (106) 2 2

= 909.375(106) mm4 = 909(106) mm4 Iv =

Ix + Iy 2

= B

-

Ix - Iy 2

Ans.

cos 2u + Ixy sin 2u

1012.5 + 600 1012.5 - 600 - ¢ ≤ cos 60° + 0 sin 60 R (106) 2 2

= 703.125(106) mm4 = 703(106) mm4 Iuv =

Ix - Iy 2

= B¢

Ans.

sin 2u + Ixy cos 2u

1012.5 - 600 ≤ sin 60° + 0 cos 60° R (106) 2

= 178.62(106) mm4 = 179(106) mm4

Ans.

Ans: Iu = 909(106) mm4 Iv = 703(106) mm4 Iuv = 179(106) mm4 1082

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*10–72.

y v

Solve Prob. 10–71 using Mohr’s circle. Hint: Once the circle is established, rotate 2u = 60° counterclockwise from the reference OA, then find the coordinates of the points that define the diameter of the circle.

u 300 mm

SOLUTION

30⬚

Moments and product of Inertia with Respect to the x and y Axes: The perpendicular distances measured from the centroid of the triangular segment to the y axis are indicated in Fig. a. Ix =

1 (400)(4503) = 1012.5(106) mm4 36

Iy = 2 B

x

C 150 mm

200 mm

200 mm

1 1 (450)(2003) + (450)(200)(66.672) R = 600(106) mm4 36 2

Since the cross-sectional area is symmetrical about the y axis, Ixy = 0. Construction of Mohr’s Circle: The center of C of the circle lies along the I axis at a distance Iavg =

Ix + Iy 2

= a

1012.5 + 600 b (106)mm4 = 806.25(106) mm4 2

The coordinates of the reference point A are [1012.5, 0](106) mm4. The circle can be constructed as shown in Fig. b. The radius of the circle is R = CA = (1012.5 - 806.25)(106) = 206.25(106) mm4 Moment and Product of Inertia with Respect to the u and v Axes: By referring to the geometry of the circle, we obtain Iu = (806.25 + 206.25 cos 60°)(106) = 909(106) mm4

Ans.

Iv = (806.25 - 206.25 cos 60°)(106) = 703(106) mm4

Ans.

Iuv = 206.25 sin 60° = 179(106) mm4

Ans.

Ans: Iu = 909(106) mm4 Iv = 703(106) mm4 Iuv = 179(106) mm4 1083

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10–73. y

Determine the orientation of the principal axes having an origin at point C, and the principal moments of inertia of the cross section about these axes.

100 mm

80 mm C 80 mm

x 10 mm

Solution Moment and Product of Inertia About x and y Axes. Using the parallel-axis theorem by referring to Fig. a Ix = Σ ( Ix′ + Ad 2y ) ;  Ix =

10 mm

1 1 (140)(103) + 2 c (10)(1003) + 10(100)(452) d 12 12

= 5.7283 (106) mm4 Iy = Σ ( Iy′ + Ad 2x ) ;  Iy =

1 1 (10)(1403) + 2 c (100)(103) + 100(10)(752) d 12 12

= 13.5533 (106) mm4 Ixy = Σ(Ix′y′ + Ad x d y);  Ixy = 0 +

30

+ 10(100)( - 75)(45) 4

+ [0 + 10(100)(75)( - 45)] = - 6.75(106) mm4 Principal moments of Inertia. I max =

Ix + Iy

min

= £

2

{

C

a

Ix + Iy 2

2

b + I 2xy

5.7283 + 13.5533 5.7283 - 13.5533 2 { a b + ( -6.75)2 § (106) 2 C 2

= (9.6408 { 7.8019)(106)

Imax = 17.44(106) mm4 = 17.4(106) mm4

Ans.

Imin = 1.8389(106) mm4 = 1.84(106) mm4

Ans.

1084

100 mm

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10–73. Continued

The orientation of the principal axes can now be determined tan 2up =

- Ixy (Ix - Iy)>2

=

-( - 6.75) (5.7283 - 13.5533)>2

= -1.7252

2up = -59.90°  and  120.10° up = -29.95°  and   60.04° Substitute up = 60.05° into the equation for Iu, Iu =

Ix + Iy

= c

2

+

Ix - Iy 2

cos 2u - Ixy sin 2u

5.7283 + 13.5533 5.7283 - 13.5533 + cos 120.10° - ( -6.75) sin 120.10° d (106) 2 2

= 17.44(106) mm4 Thus,

Ans.

(up)1 = 60.0°  (up)2 = - 30.0°

Ans: Imax = 17.4 ( 106 ) mm4 Imin = 1.84 ( 106 ) mm4 (up)1 = 60.0° (up)2 = -30.0° 1085

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10–74. y

Solve Prob. 10–73 using Mohr’s circle.

100 mm

80 mm C 80 mm

x 10 mm

Solution Moment and Product of Inertia About x and y Axes. Using the parallel-axis theorem by referring to Fig. a Ix = Σ ( Ix′ + Ad 2y ) ;  Ix =

10 mm

1 1 (140)(103) + 2 c (10)(1003) + 10(100)(452) d 12 12

= 5.7283 (106) mm4 Iy = Σ ( Iy′ + Ad 2x ) ;  Iy =

1 1 (10)(1403) + 2 c (100)(103) + 100(10)(752) d 12 12

= 13.5533 (106) mm4 Ixy = Σ ( Ix′y′ + Ad x d y ) ;  Ixy = 0 +

30

+ 10(100)( - 75)(45) 4

+ [0 + 10(100)(75)( - 45)] = - 6.75(106) mm4 Construction of the circle. The coordinates of center O of the circle are a

Ix + Iy 2

, 0b = a

5.7283 + 13.5533 , 0b(106) = (9.6408, 0)(106) 2

And the reference point A is (Ix, Ixy) = (5.7283, -6.75)(106) Thus, the radius of the circle is R = OA = a 2(9.6408 - 5.7283)2 + ( - 6.75)2 b(106) = 7.8019(106)

Using these results, the circle shown in Fig. b can be constructed. Here, the coordinates of points B and C represent Imin and Imax respectively. Thus   Imax = (9.6408 + 7.8019)(106) = 17.44(106)mm4 = 17.4(106)mm4

Ans.

  Imin = (9.6408 - 7.8019)(106) = 1.8389(106)mm4 = 1.84(106)mm4

Ans.

1086

100 mm

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10–74. Continued

The orientation of the principal axes can be determined from the geometry of the shaded triangle on the circle 6.75   tan 2(up)2 = 9.6408 - 5.7283 2(up)2 = 59.90° (up)2 = 29.95° = 30.0° b

Ans.

And 2(up)1 = 180° - 2(up)1 2(up)1 = 180° - 59.90° = 120.10° Ans.

(up)1 = 60.04° = 60.0° d

Ans: Imax = 17.4 ( 106 ) mm4 Imin = 1.84 ( 106 ) mm4 (up)2 = 30.0° b (up)1 = 60.0° d 1087

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10–75. Determine the orientation of the principal axes, which have their origin at centroid C of the beam’s crosssectional area. Also, find the principal moments of inertia.

y

100 mm 20 mm 20 mm

150 mm x

C

Solution

150 mm

Moment and Product of Inertia with Respect to the x and y Axes: The perpendicular distances measured from each subdivided segment to the x and y axes are indicated in Fig. a. Applying the parallel - axis theorem, Ix = 2c Iy = 2c

1 1 (80)(203) + 80(20)(1402) d + (20)(3003) = 107.83(106)mm4 12 12 1 1 (20)(803) + 20(80)(502) d + (300)(203) = 9.907(106)mm4 12 12

Ixy = 80(20)( - 50)(140) + 80(20)(50)( - 140) = - 22.4(106)mm4 Principal Moment of Inertia: max

I min =

Ix + Iy

= c

2

{

Aa

Ix - Iy 2

2

b + Ixy2

107.83 + 9.907 107.83 - 9.907 2 + ( -22.4)2 d (106) { b 2 Aa 2

= 58.867 { 53.841

I max = 112.71(106) = 113(106)mm4

Ans.

I min = 5.026(106) = 5.03(106)mm4 Orientation of Principal Axes: tan 2up =

- Ixy (Ix - Iy)>2

=

- ( - 22.4)(106) (107.83 - 9.907)(106)>2

= 0.4575

2up = 24.58° and -155.42° up = 12.29° and -77.71° Substituting u = up = 12.29° Iu = =

Ix + Iy 2

+

Ix - Iy 2

cos 2u - Ixy sin 2u

107.83 + 9.907 107.83 - 9.907 + a b cos 24.58° - ( -22.4) sin 24.58° 2 2

= 112.71(106)mm4 = I max

1088

100 mm

20 mm

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10–75. Continued

This shows that I max corresponds to the principal axis orientated at I max = 1131106 2mm4

1up 2 1 = 12.3°

I min = 5.031106 2mm4

1up 2 2 = -77.7°

and I min corresponds to the principal axis orientated at

The orientation of the principal axes are shown in Fig. b.

Ans: Imax = 113(106) mm4 Imin = 5.03(106) mm4 (up)1 = 12.3 (up)2 = - 77.0 1089

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*10–76. The area of the cross section of an airplane wing  has the following properties about the x and y axes passing through the centroid C: Ix = 180 (10-6) m4, Iy = 720 (10-6) m4, Ixy = 60 (10-6) m4. Determine the orientation of the principal axes and the principal moments of inertia.

y

x

C

Solution Principal Moment of Inertia. Applying Eq. 10-11, Imax =

Ix + Iy 2

min

= c

{ Aa

Ix - Iy 2

b + Ixy2

180 + 720 180 - 720 2 { a b + 602 d (10 - 6) 2 A 2

= (450 { 276.59)(10 - 6) m4

Imax = 726.59(10 - 6) m4 = 727(10 - 6) m 4

Ans.

Imin = 173.41(10 - 6) m4 = 173(10 - 6) m 4

Ans.

Orientation of the Principal Axes. Applying Eq. 10-10, tan 2up =

- Ixy (Ix - Iy)>2

=

- 60 = 0.2222 (180 - 720)>2

2up = 12.53° and 192.53 up = 6.264° and 96.26° Substitute up = 96.26° into first of Eq. 10-9 (Iu), Iu =

Ix + Iy

= c

2

+

Ix - Iy 2

cos 2u - Ixy sin 2u

180 + 720 180 - 720 + cos 192.53° - 60 sin 192.53° d (10 - 6) m4 2 2

= 726.59(10 - 6) m4

Therefore (up)1 = 96.26° = 96.3° (up)2 = 6.264° = 6.26°

Ans.

Ans: Imax = 726.59(10 - 6) m4 = 727(10 - 6) m 4 Imin = 173.41(10 - 6) m4 = 173(10 - 6) m 4 (up)1 = 96.3° (up)2 = 6.26° 1090

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10–77.

Solve Prob. 10–76 using Mohr’s circle. y

Solution Construction of the Circle: The coordinates of center O of the circle is Oa

Ix + Iy 2

, 0b = a

x

C

180 + 720 , 0b(10 - 6) = (450,0)(10 - 6) 2

And the coordinates of reference point A is A(Ix, Ixy) = A(180, 60)(10 - 6) Thus, the radius of the circle is R = OA =

1 2(450

- 180)2 + (0 - 60)2 2 (10 - 6) = 276.59(10 - 6)

Using these results, the circle shown in Fig. a can be constructed. Here, the coordinates of points C and B represent Imax and Imin respectively. Thus Imax = (450 + 276.59)(10 - 6) = 726.59(10 - 6) m4 = 727(10 - 6) m4 -6

-6

4

-6

4

Imin = (450 - 276.59)(10 ) = 173.41(10 ) m = 173(10 ) m

Ans. Ans.

The orientation of the principal axes can be determined from the geometry of the shaded triangle on the circle. tan 2(up)2 =

60 ; 2(up)2 = 12.53° (up)2 = 6.264° = 6.26° d 450 - 180

Ans.

2(up)1 = 2(up)2 + 180° = 12.53° + 180° = 192.53° (up)1 = 96.26° = 96.3° d Ans.

Ans: Imax = (450 + 276.59)(10 - 6) = Imin = (450 - 276.59)(10 - 6) = 60 tan 2(up)2 = ; 2(up)2 450 - 180 (up)2 = 6.26 d 2(up)1 = 2(up)2 + 180 = 12.53 (up)1 = 96.26 = 96.3 d 1091

727(10 - 6) m4 173(10 - 6) m4 = 12.53 + 180 = 192.53

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10–78. 20 mm

Determine the principal moments of inertia for the angle’s cross-sectional area with respect to a set of principal axes that have their origin located at the centroid C. Use the equation developed in Section 10.7. For the calculation, assume all corners to be square.

32.22 mm 100 mm 32.22 mm

C

x

SOLUTION Ix = c

20 mm 100 mm

1 (20)(100)3 + 100(20)(50 - 32.22)2 d 12

+ c

1 (80)(20)3 + 80(20)(32.22 - 10)2 d 12

= 3.142(106) mm4 Iy = c

1 (100)(20)3 + 100(20)(32.22 - 10)2 d 12

+ c

1 (20)(80)3 + 80(20)(60 - 32.22)2 d 12

= 3.142(106) mm4 Ixy = ©xy A = - (32.22 - 10)(50- 32.22)(100)(20) - (60- 32.22)(32.22-10)(80)(20) = - 1.778(106) mm4 Imax/min =

Ix + Iy 2

;

C

a

Ix - Iy 2

2

b + I2xy

= 3.142(106) ; 20 + {( - 1.778)(106)}2 Imax = 4.92(106) mm4

Ans.

Imin = 1.36(106) mm4

Ans.

Ans: Imax = 4.92(106) mm4 Imin = 1.36(106) mm4 1092

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10–79.

y 20 mm

Solve Prob. 10–78 using Mohr’s circle.

32.22 mm 100 mm

SOLUTION

32.22 mm

C

Solve Prob. 10–78.

x 20 mm

Ix = 3.142(106) mm4

100 mm

Iy = 3.142(106) mm4 Ixy = –1.778(10 6) mm4 Center of circle: Ix + I y 2

= 3.142(106) mm4

R = 2(3.142 - 3.142)2 + ( -1.778)2(106) = 1.778(106) mm4 Imax = 3.142(106) + 1.778(106) = 4.92(106) mm4

Ans.

Imin = 3.142(106) - 1.778(106) = 1.36(106) mm4

Ans.

Ans: Imax = 4.92(106) mm4 Imin = 1.36(106) mm4 1093

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*10–80. Locate the centroid Y of the beam’s cross-sectional area and then determine the moments of inertia and the product of inertia of this area with respect to the U and V axes.

v

50 mm 450 mm

450 mm

50 mm

60

400 mm

x

C 50 mm

800 mm

SOLUTION Centroid: The perpendicular distances measured from the centroid of each subdivided segment to the bottom of the beam’s cross – sectional area are indicated in Fig. a. Thus, y =

ΣyC A ΣA

=

1225(1000)(50) + 2[1000(400)(50)] + 600(12000)(100) = 825 mm 1000(50) + 2(400)(50) + 1200(100)

Ans.

Moment and Product of Inertia with Respect to the x and y Axes: The perpendicular distances measured from the centroid of each segment to the x and y axes are indicated in Fig. b. Using the parallel – axis theorem, ⎡1 ⎡1 ⎡1 ⎤ ⎤ ⎤ Ix = ⎢ (1000)(50 3 ) + 1000(50)(400) 2 ⎥ + 2 ⎢ (50)(400 3 ) + 50(400)(175) 2 ⎥ + ⎢ (100)(1200 3 ) + 100(1200)(225) 2 ⎥ ⎣ 12 ⎣ 12 ⎣ 12 ⎦ ⎦ ⎦ = 302.44 (108) mm4 Iy =

⎡1 ⎤ 1 1 (50)(10003) + 2 ⎢ (400)(50 3 ) + 400(50)(75) 2 ⎥ + (1200)(1003) 12 12 12 ⎣ ⎦

= 45 (108) mm4 Since the cross – sectional area is symmetrical about the y axis, Ixy = 0. Moment and Product of Inertia with Respect to the u and v Axes: With ␪ = 60, Iu =

Ix + Iy 2

+

Ix – Iy 2

cos 2␪ – Ixy sin 2␪

⎡ 302.44 + 45 ⎤ 302.44 – 45 + cos 120° – 0 sin n 120°⎥ (108) = ⎢ 2 2 ⎣ ⎦ = 109.36 (108) mm4 = 109 (108) mm4 Iv =

Ix + Iy 2

–

Ix – Iy 2

Ans.

cos 2␪ + Ixy sin 2␪

⎡ 302.44 + 45 302.44 – 45 ⎤ – cos 120° + 0 sin n 120°⎥ (108) = ⎢ 2 2 ⎣ ⎦ = 238.08 (108) mm4 = 238 (108) mm4

1094

Ans.

y

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*10–80. Continued

Iuv =

Ix – Iy 2

sin 2␪ + Ixy cos 2␪

⎡ 302.44 – 45 ⎤ = ⎢ sin 120° + 0 cos 120°⎥ (108) 2 ⎣ ⎦ = 111.47 (108) mm4 = 111 (108) mm4

Ans.

500 mm 75 mm

500 mm

400 mm 1000 mm

1225 mm

50 mm

1000 mm

4000 mm

600 mm

1200 mm

50 mm

75 mm

175 mm 225 mm

y = 825 mm

Ans: y = 825 mm Iu = 109 (108) mm4 Iv = 238 (108) mm4 Iuv = 111 (108) mm4 1095

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10–81.

u

Solve Prob. 10–80 using Mohr’s circle. v

50 mm 450 mm

450 mm

50 mm

60

400 mm

x

C 50 mm

800 mm

y

SOLUTION Centroid: The perpendicular distances measured from the centroid of each subdivided segment to the bottom of the beam’s cross – sectional area are indicated in Fig. a. Thus, y =

ΣyA ΣA

=

1225(1000)(50) + 2[1000(400)(50)] + 600(1200)(100) 1000(50) + 2(400)(50) + 1200(100)

= 825 mm

Ans.

Moment and Product of Inertia with Respect to the x and y Axes: The perpendicular distances measured from the centroid of each segment to the x and y axes are indicated in Fig. b. Using the parallel – axis theorem, ⎡1 ⎤ ⎡1 ⎤ ⎡1 ⎤ Ix = ⎢ (1000)(50 3 ) + 1000(50)(400) 2 ⎥ + 2 ⎢ (50)(400 3 ) + 50(400)(175) 2 ⎥ + ⎢ (100)(1200 3 ) + 100(1200)(225) 2 ⎥ ⎣ 12 ⎦ ⎣ 12 ⎦ ⎣ 12 ⎦ = 302.44 (108) mm4 ⎡1 ⎤ 1 1 (50)(10003) + 2 ⎢ (400)(50 3 ) + 400(50)(75) 2 ⎥ + (1200)(1003) 12 12 12 ⎣ ⎦

Iy =

= 45 (108) mm4 Ixy = 60(5)(–14.35)(13.15) + 55(15)(15.65)(–14.35) = –11.837 (104) mm4 Since the cross – sectional area is symmetrical about the y axis, Ixy = 0.

(108 mm4) 500 mm

100 mm

1225 mm

50 mm

1000 mm

400 mm

600 mm

1200 mm

75 mm

75 mm

500 mm

400 mm

50 mm

175 mm 225 mm

y = 825 mm

1096

(108 mm4)

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10–81. Continued

Construction of Mohr’s Circle: The center C of the circle lies along the u axis at a distance Iavg =

Ix + Iy 2

⎛ 302.44 + 45 ⎞ 8 8 4 = ⎜ ⎟⎠ (10 ) = 173.72 (10 ) mm 2 ⎝

The coordinates of the reference point A are (302.44, 0) (108) mm4. The circle can be constructed as shown in Fig. c. The radius of the circle is R = CA = (302.44 – 173.72) (108) = 128.72 (108) mm4 Moment and Product of Inertia with Respect to the u and v Axes: By referring to the geometry of the circle, Iu = (173.72 – 128.72 cos 60°) (108) = 109 (108) mm4

Ans.

Iv = (173.72 + 128.72 cos 60°) (108) = 238 (108) mm4

Ans.

Iuv = (128.72 sin 60°) (108) = 111 (108) mm4

Ans.

Ans: y = 825 mm Iu = 109 (108) mm4 Iv = 238 (108) mm4 Iuv = 111 (108) mm4 1097

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y

10–82. Locate the centroid y of the beam’s cross-sectional area and then determine the moments of inertia of this area and the product of inertia with respect to the u and v axes. The axes have their origin at the centroid C.

25 mm

200 mm

25 mm

v

x

C 60

y

25 mm 75 mm 75 mm

Solution

u

Centroid: The perpendicular distances measured from the centroid of each subdivided segment to the bottom of the beam’s cross - sectional area are indicated in Fig. a. Thus, y =

23100(200)(25) 4 + 12.5(2.5)(100) ΣycA = = 82.5 mm ΣA 2(200)(25) + 25(100)

Ans.

Moment and Product of Inertia with Respect to the x and y Axes: The perpendicular distances measured from the centroid of each segment to the x and y axes are indicated in Fig. b. Using the parallel - axis theorem Ix = 2c

1 1 (25)(2003) + 25(200)(17.5)2 d + c (100)(253) + 100(25)(70)2 d 12 12

= 48.78(106) mm4 Iy = 2c

1 1 (200)(253) + 200(25)(62.5)2 d + (25)(1003) 12 12

= 41.67(106) mm4

Since the cross - sectional area is symmetrical about the y axis, Ixy = 0. Moment and Product of Inertia with Respect to the u and v Axes: With u = - 60°, Iu =

Ix + Iy 2

= c

+

Ix - Iy 2

cos 2u - Ixy sin 2u

48.78 + 41.67 48.78 - 41.67 + a b cos ( - 120°) - 0 sin ( -120°) d (106) 2 2

= 43.4(106) mm4 Iv =

Ix + Iy

= c

2

+

Ix - Iy 2

Ans.

cos 2u - Ixy sin 2u

48.78 + 41.67 48.78 - 41.67 + a b cos ( - 120°) + 0 sin ( -120°) d (106) 2 2

= 47.0(106) mm4

1098

Ans.

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10–82. Continued

Iuv =

Ix - Iy

= a

2

sin 2u + Ixy cos 2u

48.78 - 41.67 b sin 1 - 120° 2 + 0 cos 1 -120° 2 2

= -3.081106 2mm4

Ans.

Ans: y = 82.5 mm Iu = 43.4 (106) mm4 Iv = 47.0 (106) mm4 Iuv = -3.08 (106) mm4 1099

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y

10–83. Solve Prob. 10–8 2 using Mohr’s circle. 25 mm

200 mm

25 mm

v

x

C 60

25 mm 75 mm 75 mm

Solution Centroid: The perpendicular distances measured from the centroid of each subdivided segment to the bottom of the beam’s cross - sectional area are indicated in Fig. a. Thus, y =

23100(200)(25) 4 + 12.5(2.5)(100) Σ∼ yA = = 82.5 mm ΣA 2(200)(25) + 25(100)

Ans.

Moment and Product of Inertia with Respect to the x and y Axes: The perpendicular distances measured from the centroid of each segment to the x and y axes are indicated in Fig. b. Using the parallel - axis theorem Ix = 2c

1 1 (25)(2003) + 25(200)(17.5)2 d + c (100)(253) + 100(25)(70)2 d 12 12

= 48.78(106) mm4 Iy = 2c

1 1 (200)(253) + 200(25)(62.5)2 d + (25)(1003) 12 12

= 44.67(106) mm4

Since the cross - sectional area is symmetrical about the y axis, Ixy = 0.

1100

u

y

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10–83. Continued

Construction of Mohr’s Circle: The center C of the circle lies along the u axis at a distance Iavg =

Ix + Iy 2

= a

48.78 + 41.67 b(106) mm4 = 45.22(106) mm4 2

The coordinates of the reference point A are 348.78, 04(106) mm4. The circle can be constructed as shown in Fig. a. The radius of the circle is R = CA = (48.78 - 45.22)(106) = 3.56(106) mm4 Moment and Product of Inertia with Respect to the u and v Axes: By referring to the geometry of the circle, Iu = (45.22 - 3.56 cos 60°)(106) = 43.4(106) mm4

Ans.

Iv = (45.22 + 3.56 cos 60°)(106) = 47.0(106) mm4

Ans.

Iuv = - 3.56 sin 60° = - 3.08(106) mm4

Ans.

Ans: y = 82.5 mm Iu = 43.4 (106) mm4 Iv = 47.0 (106) mm4 Iuv = -3.08 (106) mm4 1101

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*10–84. Determine the moment of inertia of the thin ring about the z axis. The ring has a mass m.

y

R x

SOLUTION 2p

Iz =

r A(R du)R2 = 2p r A R3

L0 2p

m =

L0

r A R du = 2p r A R

Thus, Iz = m R 2

Ans.

Ans: Iz = m R2 1102

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10–85. Determine the moment of inertia of the homogenous triangular prism with respect to the y axis. Express the result in terms of the mass m of the prism. Hint: For integration, use thin plate elements parallel to the x-y plane having a thickness of dz.

z

z = –h a (x – a)

h

SOLUTION Differential Thin Plate Element: Here, x = a a 1 -

z b. The mass of the h

x

b

a

z differential thin plate element is dm = rdV = rbxdz = rab a1 - b dz. The mass h moment of inertia of this element about y axis is dIy = dIG + dmr2 =

x2 1 dmx2 + dm ¢ + z2 ≤ 12 4

=

1 2 x dm + z2 dm 3

= B =

a2 z 2 z a 1 - b + z2 R B raba 1 - b dz R 3 h h

rab 2 a2 3a2 3a2 3z3 z - 3 z3 + 3z2 b dz ¢ a + 2 z2 3 h h h h

Total Mass: Performing the integration, we have h

m =

Lm

dm =

L0

raba 1 -

z z2 h 1 b dz = rab ¢ z ≤ ` = rabh h 2h 0 2

Mass Moment of Inertia: Performing the integration, we have h

Iy =

L

dIy =

rab 2 3a2 3a2 a2 3z3 z - 3 z3 + 3z2 b dz ¢ a + 2 z2 h h h h L0 3

=

rab 2 a2 3a2 2 a2 4 3z4 h 3 z + z z ¢ a z + 2 z3 ≤` 3 2h 4h 0 h 4h3

=

rabh 2 1a + h22 12

The mass moment of inertia expressed in terms of the total mass is Iy =

1 rabh 6 2

a2 + h2 =

m 2 a + h2 6

Ans.

Ans: Iy = 1103

m 2 (a + h2) 6

y

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10–86. Determine the moment of inertia of the semi-ellipsoid with respect to the x axis and express the result in terms of the mass m of the semiellipsoid. The material has a constant density r.

y

x2 y2  1 a2 b2 b x

SOLUTION

a

2 2 Differential Disk Element: Here, y = b a 1 -

x2 b . The mass of the differential disk element is a2

x2 b dx. The mass moment of inertia of this element is a2 2 rp b4 x4 1 1 x x2 2x2 dIx = dmy2 = crp b2 a1 - 2 b dx d c b2 a 1 - 2 b d = a 4 - 2 + 1b dx. 2 2 2 a a a a dm = rdV = rp y2 dx = rp b2 a 1 -

Total Mass: Performing the integration, we have a

m =

dm =

Lm

L0

rp b2 a1 -

x2 x3 a 2 b dx = rp b a x b` a2 3a2 0 =

2 rpab2 3

Mass Moment of Inertia: Performing the integration, we have a

Ix =

dIx = L

rp b4 x4 2x2 a 4 - 2 + 1 b dx a a L0 2

=

a rp b4 x5 2x3 a 4 + x b ` 2 5a 3a2 0

=

4 rp ab4 15

The mass moment of inertia expressed in terms of the total mass is. Ix =

2 2 2 a rp ab2 b b2 = mb2 5 3 5

Ans.

Ans: Ix = 1104

2 mb2 5

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10–87. Determine the moment of inertia of the ellipsoid with respect to the x axis and express the result in terms of the mass m of the ellipsoid. The material has a constant density r.

y x2 a2

y2 b2

1 b x

SOLUTION a

dm= py2dx L d Ix =

y2dm 2 a

m =

LV

r dV =

L-a

2

rp b a 1 -

2

x a

2

b dx =

4 2 prab 3

a

Ix =

x2 2 8 1 prab4 rpb4 a 1 - 2 b dx = 2 15 a L-a

Thus, Ix =

2 mb2 5

Ans.

Ans: Ix = 1105

2 mb2 5

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*10–88. Determine the radius of gyration kx of the paraboloid. The density of the material is r = 5 Mg>m3.

y y2  50 x 100 mm x

SOLUTION

200 mm

Differential Disk Element: The mass dm = rdV = rpy2 dx = rp(50x) dx. The 1 1 is dIx = dmy2 = [rp(50x) dx](50x) = 2 2

of the differential disk element is mass moment of inertia of this element rp (2500x2) dx. 2

Total Mass: Performing the integration, we have m =

dm =

Lm

200 mm

mm rp(50x)dx = rp(25x2)|200 = 1(106)rp 0

L0

Mass Moment of Inertia: Performing the integration, we have Ix =

dIx = L =

200 mm

L0

rp (2500x2) dx 2

rp 2500x3 200 mm a b` 2 3 0

= 3.333(109)rp The radius of gyration is kx =

Ix 3.333(109)rp = = 57.7 mm Am A 1(106)rp

Ans.

Ans: kx = 57.7 mm 1106

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10–89. The paraboloid is formed by revolving the shaded area around the x axis. Determine the moment of inertia about the x axis and express the result in terms of the total mass m of the paraboloid. The material has a constant density r.

y 2

y2 = a–h x a x

SOLUTION

h 2

dm = r dV = r (p y dx)

d Ix =

1 1 dm y2 = r p y4 dx 2 2 h

Ix = =

1 a4 r p a 2 bx2 dx h L0 2 1 p ra4 h 6 h

m = = Ix =

1 a2 r p a b x dx h L0 2 1 r p a2 h 2 1 ma2 3

Ans.

Ans: Ix = 1107

1 ma2 3

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10–90. The right circular cone is formed by revolving the shaded area around the x axis. Determine the moment of inertia Ix and express the result in terms of the total mass m of the cone. The cone has a constant density r.

y y  –hr x r x

SOLUTION dm = r dV = r(p y2 dx)

h

h

1 1 r2 r2 m = r(p) ¢ 2 ≤ x2 dx = rp ¢ 2 ≤ a b h3 = rp r2h 3 3 h h L0 dIx =

1 2 y dm 2

=

1 2 y (rp y2 dx) 2

=

1 r4 r(p)a 4 b x4 dx 2 h h

Ix =

1 1 r4 r(p) a 4 b x4 dx = rp r4 h 2 10 h L0

Thus, Ix =

3 m r2 10

Ans.

Ans: Ix = 1108

3 mr 2 10

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10–91. Determine the radius of gyration kx of the solid formed by revolving the shaded area about x axis. The density of the material is r.

y

n

yn h x a h x

a

Solution Differential Disk Element. The mass of the differential disk element shown h 1 h 1 2 shaded in Fig. a is dm = rdv = rpy2dx. Here y = 1 x n . Thus, dm = rp a 1 xn b an an rph2 2 dx = xndx. The mass moment of Inertia of this element about the x axis is 2 an rxh4 4 1 1 rph2 n2 h 1 2 xn dxb a 2>n x dxb a 2 xn b = a dIx = (dm)y2 = 4 2 2 a an 2an Total Mass. Perform the integration,

m =

Lm

dm =

L0

= a



= a



a

rph2 a

rph2 a

2 n

2 n

2

(x ndx)

ba

n b axn n + 2

+ 2 n b

2

a 0

n brpah2 n + 2

Mass Moment of Inertia. Perform the integration,





Ix =

a rph4 4 dIx = (xn dx) 4 L L0 2an a

rph4

n n+4 bax b3 = a 4>n ba n n + 4 2a = c

0

n d rpah4 2(n + 4)

The radius of gyration is



Ix kx = = Am

c

n d rp ah4 2(n + 4)

n a b rp ah2 c n+4

=

n + 2 h A 2(n + 4)

Ans.

Ans: kx = 1109

n + 2 h A 2(n + 4)

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*10–92. Determine the moment of inertia Ix of the sphere and express the result in terms of the total mass m of the sphere. The sphere has a constant density r.

y

x2

y2

r2

r x

SOLUTION d Ix =

y2 dm 2

dm = r dV = r(py2 dx) = rp(r2 - x2)dx d Ix =

1 rp(r2 - x2)2 dx 2 r

Ix = =

1 rp(r2 - x2)2 dx L-r 2 8 prr5 15 r

m = =

L-r

rp(r2 - x2) dx

4 rpr3 3

Thus, Ix =

2 2 mr 5

Ans.

Ans: Ix = 1110

2 mr 2 5

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10–93.

z

Determine the moment of inertia Iz of the frustrum of the cone which has a conical depression. The material has a density of 2000 kg>m3.

200 mm

600 mm

SOLUTION

400 mm

z + 1 z = , 0.2 0.8 z = 0.333 m. The mass moment of inertia of each cone about z axis can be 3 determined using Iz = mr2. 10

Mass Moment of Inertia About z Axis: From similar triangles,

Iz = ©1Iz2i =

800 mm

3 p c 10.82211.3332120002 d10.822 10 3 -

3 p c 10.2 2210.3332120002 d 10.22 2 10 3

-

3 p c 10.2 2210.62120002 d 10.22 2 10 3

= 342 kg # m2

Ans.

Ans:

Iz = 342 kg # m2

1111

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10–94. Determine the mass moment of inertia Iy of the solid formed by revolving the shaded area around the y axis. The total mass of the solid is 1500 kg.

z

4m

z2

1 y3 –– 16

2m

y

O

SOLUTION

x

Differential Element: The mass of the disk element shown shaded in 2 1 1 dm = rdV = rpr2dy. Here, r = z = y3>2.Thus, dm = rpa y3>2 b dy = 4 4 The mass moment of inertia of this element about the y rp 4 rp 1 3>2 4 rp 6 1 1 r dy = a y b dy = y dy. dIy = dmr2 = A rpr2dy B r2 = 2 2 2 2 4 512

Fig. a is rp 3 y dy. 16 axis is

Mass: The mass of the solid can be determined by integrating dm. Thus, 4m

m =

L

dm =

L0

4

4m rp y rp 3 = 4 pr y dy = ¢ ≤` 16 16 4 0

The mass of the solid is m = 1500 kg. Thus, 1500 = 4pr

r =

375 kg>m3 p

Mass Moment of Inertia: Integrating dIy, 4m

Iy =

L

Substituting r =

dIy =

L0

rp 6 rp y7 4 m 32p y dy = r = ¢ ≤` 512 512 7 0 7

375 kg>m3 into Iy, p Iy =

32p 375 b = 1.71(103) kg # m2 a p 7

Ans.

Ans: Iy = 1.71 ( 103 ) kg # m2 1112

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10–95. The slender rods have a mass of 4 kg>m. Determine the moment of inertia of the assembly about an axis perpendicular to the page and passing through point A.

A

200 mm

Solution

100 mm

100 mm

Mass Moment of Inertia About An Axis Through A. The mass of each segment is mi = (4 kg>m)(0.2 m) = 0.8 kg. The mass moment inertia of each segment shown in Fig. a about an axis through their center of mass can be determined using 1 (IG)i = m l i2. 12 i



IA = Σ c (IG)i + mi d 2i d = c

1 1 (0.8) ( 0.22 ) + 0.8 ( 0.12 ) d + c (0.8) ( 0.22 ) + 0.8 ( 0.22 ) d 12 12



= 0.04533 kg # m2



= 0.0453 kg # m2



Ans.

Ans: IA = 0.0453 kg # m2 1113

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*10–96. The pendulum consists of a 8-kg circular disk A, a 2-kg circular disk B, and a 4-kg slender rod. Determine the radius of gyration of the pendulum about an axis perpendicular to the page and passing through point O.

A O

0.4 m

1m

0.5 m

B

0.2 m

Solution Mass Moment of Inertia About An Axis Through O. The mass moment of inertia of each rod segment and disk segment shown in Fig. a about an axis passes through 1 1 their center of mass can be determined using (IG)i = m l i2 and (IG)i = mi r i2. 12 i 2 IO = Σ c (IG)i + mi d 2i d   

= c

1 1 (4) ( 1.52 ) + 4 ( 0.252 ) d + c (2) ( 0.12 ) + 2 ( 0.62 ) d 12 2   

  

= 13.41 kg # m2

1 + c (8) ( 0.22 ) + 8 ( 1.22 ) d 2

The total mass is 8 kg + 2 kg + 4 kg = 14 kg The radius of gyration is kO =

13.41 kg # m2 IO = = 0.9787 m = 0.979 m Am C 14 kg

Ans.

Ans: kO = 0.979 m 1114

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10–97. Determine the moment of inertia Iz of the frustum of the cone which has a conical depression. The material has a density of 200 kg>m3.

0.2 m

z

0.8 m 0.6 m

SOLUTION Iz =

3 1 [ p (0.4)2(1.6)(200)](0.4)2 10 3 -

3 1 [ p (0.2)2(0.8)(200)](0.2)2 10 3

-

3 1 [ p (0.4)2(0.6)(200)](0.4)2 10 3

0.4 m

Iz = 1.53 kg # m2

Ans.

Ans: Iz = 1.53 kg # m2 1115

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10–98. The pendulum consists of the 3-kg slender rod and the 5-kg thin plate. Determine the location y of the center of mass G of the pendulum; then find the mass moment of inertia of the pendulum about an axis perpendicular to the page and passing through G.

O

y

SOLUTION y =

G

©ym 1(3) + 2.25(5) = = 1.781 m = 1.78 m ©m 3 + 5

0.5 m

Ans. 1m

IG = ©IG¿ + md2 =

2m

1 1 (3)(2)2 + 3(1.781 - 1)2 + (5)(0.52 + 12) + 5(2.25 - 1.781)2 12 12

= 4.45 kg # m2

Ans.

Ans: y = 1.78 m IG = 4.45 kg # m2 1116

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10–99. Determine the mass moment of inertia of the overhung crank about the x axis. The material is steel having a density of r = 7.85 Mg>m3.

20 mm 30 mm 90 mm 50 mm x

180 mm

20 mm x¿ 30 mm 20 mm

Solution me = 7.85(103) 1 (0.05)p(0.01)2 2

50 mm

30 mm

= 0.1233 kg

mp = 7.85(103)((0.03)(0.180)(0.02)) = 0.8478 kg 1 Ix = 2c (0.1233)(0.01)2 + (0.1233)(0.06)2 d 2 + c

1 (0.8478)((0.03)2 + (0.180)2) d 12

= 0.00325 kg # m2 = 3.25 g

#

m2

Ans.

Ans: Ix = 3.25 g # m2 1117

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*10–100.

Determine the mass moment of inertia of the overhung crank about the x¿ axis. The material is steel having a density of r = 7.85 Mg>m3.

20 mm 30 mm 90 mm 50 mm x

180 mm

20 mm x¿ 30 mm 20 mm

Solution me = 7.85(103) 1 (0.05)p(0.01)2 2

50 mm

30 mm

= 0.1233 kg

mp = 7.85(103)((0.03)(0.180)(0.02)) = 0.8478 kg 1 1 Ix′ = c (0.1233)(0.01)2 d + c (0.1233)(0.02)2 + (0.1233)(0.120)2 d 2 2 + c

1 (0.8478) 1 (0.03)2 + (0.180)2 2 + (0.8478)(0.06)2 d 12

= 0.00719 kg # m2 = 7.19 g # m2

Ans.

Ans: Ix′ = 7.19 g # m2 1118

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10–101. The thin plate has a mass per unit area of 10 kg>m2. Determine its mass moment of inertia about the y axis.

z 200 mm 100 mm

200 mm

200 mm

SOLUTION

100 mm

Composite Parts: The thin plate can be subdivided into segments as shown in Fig. a. Since the segments labeled (2) are both holes, the y should be considered as negative parts.

200 mm 200 mm x

200 mm

200 mm 200 mm

Mass moment of Inertia: The mass of segments (1) and (2) are m1 = 0.4(0.4)(10) = 1.6 kg and m2 = p(0.12)(10) = 0.1p kg. The perpendicular distances measured from the centroid of each segment to the y axis are indicated in Fig. a. The mass moment of inertia of each segment about the y axis can be determined using the parallel-axis theorem. Iy = © A Iy B G + md2 = 2c

1 1 (1.6)(0.42) + 1.6(0.22) d - 2 c (0.1p)(0.12) + 0.1p(0.22) d 12 4

= 0.144 kg # m2

Ans.

Ans: Iy = 0.144 kg # m2 1119

y

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10–102. The thin plate has a mass per unit area of 10 kg>m2. Determine its mass moment of inertia about the z axis.

z 200 mm 100 mm

200 mm

200 mm

SOLUTION

100 mm

Composite Parts: The thin plate can be subdivided into four segments as shown in Fig. a. Since segments (3) and (4) are both holes, the y should be considered as negative parts.

200 mm 200 mm x

200 mm

200 mm

200 mm

Mass moment of Inertia: Here, the mass for segments (1), (2), (3), and (4) are m1 = m2 = 0.4(0.4)(10) = 1.6 kg and m3 = m4 = p(0.12)(10) = 0.1p kg. The mass moment of inertia of each segment about the z axis can be determined using the parallel-axis theorem. Iz = © A Iz B G + md2 =

1 1 1 1 (1.6)(0.42) + c (1.6)(0.42 + 0.42) + 1.6(0.22) d - (0.1p)(0.12) - c (0.1p)(0.12) + 0.1p(0.22) d 12 12 4 2

= 0.113 kg # m2

Ans.

Ans: Iz = 0.113 kg # m2 1120

y

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10–103. Determine the moment of inertia Iz of the frustrum of the cone which has a conical depression. The material has a density of 200 kg>m3.

z

200 mm

600 mm

SOLUTION

400 mm

z + 1 z = , Mass Moment of Inertia About z Axis: From similar triangles, 0.2 0.8 z = 0.333 m. The mass moment of inertia of each cone about z axis can be 3 determined using Iz = mr2. 10 Iz = ©1Iz2i =

800 mm

3 p c 10.82211.333212002 d10.822 10 3 -

3 p c 10.2 2210.333212002 d10.2 22 10 3

-

3 p c 10.2 2210.6212002 d10.2 22 10 3

= 34.2 kg # m2

Ans.

Ans: Iz = 34.2 kg # m2 1121

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*10–104. z

Determine the mass moment of inertia of the assembly about the z axis. The density of the material is 7.85 Mg> m3.

100 mm

SOLUTION Composite Parts: The assembly can be subdivided into two circular cone segments (1) and (3) and a hemispherical segment (2) as shown in Fig. a. Since segment (3) is a hole, it should be considered as a negative part. From the similar triangles, we obtain z 0.1 = 0.45 + z 0.3

450 mm 300 mm

z = 0.225m

Mass: The mass of each segment is calculated as

300 mm

1 1 m1 = rV1 = r a pr2h b = 7.85(103) c p(0.32)(0.675) d = 158.9625p kg 3 3

x

y

2 2 m2 = rV2 = r a pr3 b = 7.85(103)c p(0.33) d = 141.3p kg 3 3 1 1 m3 = rV3 = r a pr2h b = 7.85(103) c p(0.12)(0.225) d = 5.8875p kg 3 3 Mass Moment of Inertia: Since the z axis is parallel to the axis of the cone and the hemisphere and passes through their center of mass, the mass moment of inertia can be 3 3 2 computed from (Iz)1 = m r12, (Iz)2 = m2r22, and m r32. Thus, 10 1 5 10 3 Iz = ©(Iz)i =

3 2 3 (158.9625p)(0.32) + (141.3p)(0.32) (5.8875p)(0.12) 10 5 10

= 29.4 kg # m2

Ans.

Ans: Iz = 29.4 kg # m2 1122

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10–105. The pendulum consists of a plate having a weight of 60 kg and a slender rod having a weight of 20 kg. Determine the radius of gyration of the pendulum about an axis perpendicular to the page and passing through point O.

O

1m 1m

3m

2m

Solution Given: Wp = 60 kg Wr = 20 kg

a b c d

= = = =

1m 1m 3m 2m

2 1 # c + d 1 # b 2 Wr(c + d)2 + Wr # a - cb + Wp 1a2 + b2 2 + Wp # ac + b 12 2 12 2 I0 k0 = k0 = 3.15 m A Wp + Wr

I0 =

Ans.

Ans: k0 = 3.15 m 1123

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10–106. The pendulum consists of a disk having a mass of 6 kg and slender rods AB and DC which have a mass of 2 kg > m . Determine the length L of DC so that the center of mass is at the bearing O . What is the moment of inertia of the assembly about an axis perpendicular to the page and passing through point O?

0.8 m

0.5 m

D

0.2 m A

O

L

B C

SOLUTION Location of Centroid: This problem requires x = 0.5 m. x = 0.5 =

©xm ©m 1.5(6) + 0.65[1.3(2)] + 0[L(2)] 6 + 1.3(2) + L(2)

L = 6.39 m

Ans.

Mass Moment of Inertia About an Axis Through Point O: The mass moment of inertia of each rod segment and disk about an axis passing through the center of mass can be 1 1 ml2 and (IG)i = mr2. Applying Eq. 10–15, we have determine using (IG)i = 12 2 IO = ©(IG)i + mid2i =

1 [1.3(2)] A 1.32 B + [1.3(2)] A 0.152 B 12

+

1 [6.39(2)] A 6.392 B + [6.39(2)] A 0.52 B 12

+

1 (6) A 0.22 B + 6 A 12 B 2

= 53.2 kg # m2

Ans.

Ans: L = 6.39 m Io = 53.2 kg # m2 1124

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10–107. Determine the mass moment of inertia of the thin plate about an axis perpendicular to the page and passing through point O. The material has a mass per unit area of 20 kg>m2.

O

50 mm

150 mm 50 mm

SOLUTION

150 mm

400 mm

Composite Parts: The plate can be subdivided into the segments shown in Fig. a. Here, the four similar holes of which the perpendicular distances measured from their centers of mass to point C are the same and can be grouped as segment (2). This segment should be considered as a negative part.

400 mm

150 mm 150 mm

Mass Moment of Inertia: The mass of segments (1) and (2) are m1 = (0.4)(0.4)(20) = 3.2 kg and m2 = p(0.052)(20) = 0.05p kg, respectively. The mass moment of inertia of the plate about an axis perpendicular to the page and passing through point C is IC =

1 1 (3.2)(0.4 2 + 0.4 2) - 4 c (0.05p)(0.052) + 0.05p(0.152) d 12 2

= 0.07041 kg # m2

The mass moment of inertia of the wheel about an axis perpendicular to the page and passing through point O can be determined using the parallel-axis theorem IO = IC + md2, where m = m1 - m2 = 3.2 - 4(0.05p) = 2.5717 kg and d = 0.4 sin 45°m. Thus, IO = 0.07041 + 2.5717(0.4 sin 45°)2 = 0.276 kg # m2

Ans.

Ans: IO = 0.276 kg # m2 1125

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*10–108. Each of the three slender rods has a mass m. Determine the moment of inertia of the assembly about an axis that is perpendicular to the page and passes through the center point O. a

a O

SOLUTION IO = 3 B

1 a sin 60° 2 1 ma2 + m ¢ ≤ R = ma2 12 3 2

Ans.

a

Ans: IO = 1126

1 ma2 2

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10–109. The pendulum consists of two slender rods AB and OC which have a mass of 3 kg>m. The thin plate has a mass of 12 kg>m2. Determine the location y of the center of mass G of the pendulum, then calculate the moment of inertia of the pendulum about an axis perpendicular to the page and passing through G.

0.4 m

0.4 m

A

B O

_ y G

C

SOLUTION y =

0.1 m

1.5(3)(0.75) + p(0.3)2(12)(1.8) - p(0.1)2(12)(1.8) 1.5(3) + p(0.3)2(12) - p(0.1)2(12) + 0.8(3)

= 0.8878 m = 0.888 m IG =

1.5 m

0.3 m

Ans.

1 (0.8)(3)(0.8)2 + 0.8(3)(0.8878)2 12 +

1 (1.5)(3)(1.5)2 + 1.5(3)(0.75 - 0.8878)2 12

1 + [p(0.3)2(12)(0.3)2 + [p(0.3)2(12)](1.8 - 0.8878)2 2 -

1 [p(0.1)2(12)(0.1)2 - [p(0.1)2(12)](1.8 - 0.8878)2 2

IG = 5.61 kg # m2

Ans.

Ans: y = 0.888 m IG = 5.61 kg # m2 1127

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11–1. Determine the force F needed to lift the block having a mass of 50 kg. Hint: Note that the coordinates sA and sB can be related to the constant vertical length l of the cord.

sA sB A

Solution

B

F

I = sA + 2sB dsA = - 2dsB dU = 0; WdsB + FdsA = 0 490.5dsB + F 1 - 2dsB 2 = 0 F = 245.25 N

Ans.

50 (9.81) N = 490.5 N

Ans: F = 245.25 N 1128

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11–2. Determine the force F acting on the cord which is required to maintain equlibrium of the horizontal 10-kg bar AB. Hint: Express the total constant vertical length l of the cord in terms of position coordinates s1 and s2. The derivative of this equation yields a relationship between d1 and d2.

s2

F

s1

A

B

Solution Free—Body Diagram: Only force F and the weight of link AB (98.1 N) do work. Virtual Displacements: Force F and the weight of link AB (98.1 N) are located from the top of the fixed link using position coordinates s2 and s1. Since the cord has a constant length, l, then 4s1 - s2 = l

[1]

4ds1 - ds2 = 0

Virtual—Work Equation: When s1 and s2 undergo positive virtual displacements ds1 and ds2, the weight of link AB (98.1 N) and force F do positive work and negative work, respectively. dU = 0;

[2]

98.11 - ds1 2 - F 1 -ds2 2 = 0

Substituting into Eq. [2] into [1] yields 1 -98.1 + 4F 2ds1 = 0

Since ds1 ≠ 0, then - 98.1 + 4F = 0

Ans.

F = 24.5 N

Ans: F = 24.5 N 1129

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11–3. The scissors jack supports a load P. Determine the axial force in the screw necessary for equilibrium when the jack is in the position u. Each of the four links has a length L and is pin-connected at its center. Points B and D can move horizontally.

P

SOLUTION x = L cos u,

dx = -L sin u du

y = 2L sin u,

dy = 2L cos u du

dU = 0;

- Pdy - Fdx = 0

C

D

A

B u

-P(2L cos u du) - F( - L sin u du) = 0 Ans.

F = 2P cot u

Ans: F = 2P cot u 1130

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*11–4. The assembly is used for exercise. It consist of four pin-connected bars, each of length L, and a spring of stiffness k and unstretched length a ( 62L). If horizontal forces P and - P are applied to the handles so that u is slowly decreased, determine the angle u at which the magnitude of P becomes a maximum.

A

L

L θ

θ

P

–P

k B

D L

Solution

L

C

Free Body Diagram: The system has only one degree of freedom defined by the independent coordinate u. When u undergoes a positive displacement du, the spring force Fsp and force P do work. Virtual Displacements: The spring force Fap and force P are located from the fixed point D and A using position coordinates y and x, respectively. y = L cos u x = L sin u

dy = -L sin udu

[1]

dx = L cos udu

[2]

Virtual – Work Equation: When points A, C, B and D undergo positive virtual displacement dy and dx, the spring force Fsp and force P do negative work. dU = 0;

[3]

- 2Fspdy - 2Pdx = 0

Substituting Eqs. [1] and [2] into [3] yields [4]

12Fsp sin u - 2P cos u 2Ldu = 0

From the geometry, the spring stretches x = 2L cos u - a. Then the spring force Fsp = kx = k12L cos u - a2 = 2kL cos u - ka. Substituting this value into Eq.[4] yields 14kL sin u cos u - 2ka sin u - 2P cos u 2Ldu = 0

Since Ldu ≠ 0, then

4kL sin u cos u - 2ka sin u - 2P cos u = 0 P = k12L sin u - a tan u 2

In order to obtain maximum P,

dP = 0. du

dP = k12L cos u - a sec 2u 2 = 0 du 1

u = cos

-1

a

a 3 b 2L

Ans.

Ans: u = cos

1131

-1

a

a 1 b 2L 3

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11–5. The machine shown is used for forming metal plates. It consists of two toggles ABC and DEF, which are operated by hydraulic cylinder BE. The toggles push the moveable bar FC forward, pressing the plate p into the cavity. If the force which the plate exerts on the head is P = 8 kN, determine the force F in the hydraulic cylinder when u = 30.

D 200 mm

θ = 30° E

F 200 mm

F

P

F 200 mm A

B = 30°

200 mm C

P

Solution Free—Body Diagram: The system has only one degree of freedom defined by the independent coordinate u. When u undergoes a positive displacement du, only the forces F and P do work. Virtual Displacements: The spring force F acting on joints E and B and force P are located from the fixed points D and A using position coordinates yE and yB, respectively. The location for force P is measured from the fixed point A using position coordinate xa. yE = 0.2 sin u

dyE = 0.2 cos udu

[1]

yB = 0.2 sin u

dyB = 0.2 cos udu

[2]

xG = 210.2 cos u 2 + l

[3]

dxG = -0.4 sin udu

Virtual – Work Equation: When points E, B and G undergo positive virtual displacements dyE, dyB and dyG, force F and P do negative work. dU = 0;

[4]

- FdyE - FdyB - PdxG = 0

Substituting Eqs. [1], [2] and [3] into [4] yields 10.4P sin u - 0.4F cos u 2du = 0

Since du ≠ 0, then

0.4P sin u - 0.4F cos u = 0

F = P tan u

At equilibrium position u = 30° set P = 8 kN, we have Ans.

F = 8 tan 30° = 4.62 kN

Ans: F = 4.62 kN 1132

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11–6. The bar is supported by the spring and smooth collar that allows the spring to be always perpendicular to the bar for any angle u. If the unstretched length of the spring is l0, determine the force P needed to hold the bar in the equilibrium position u. Neglect the weight of the bar.

a A

B u k C l

SOLUTION s = a sin u,

ds = a cos u du

y = l sin u,

dy = l cos u du P

Fs = k(a sin u - l0) dU = 0;

Pdy - Fsds = 0

Pl cos du - k(a sin u - l0) a cos u du = 0 P =

ka (a sin u - l0) l

Ans.

Ans: P = 1133

ka (a sin u - l0) l

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11–7. If the spring is unstretched when u = 30°, the mass of the cylinder is 25 kg, and the mechanism is in equilibrium when u = 45°, determine the stiffness k of the spring. Rod AB slides freely through the collar at A. Neglect the mass of the rods.

k A

B

SOLUTION Free-Body Diagram: When u undergoes a positive virtual angular displacement of du, the dash line configuration shown in Fig. a is formed. We observe that only the spring force Fsp and the weight W of the cylinder do work when the virtual displacement takes place. Virtual Displacement: The position of the point B at which WD and Fsp act is specified by the position coordinates xB and yB, measured from the fixed point C. xB = 0.45 sin u

dxB = 0.45 cos udu

(1)

yB = 0.45 cos u

dyB = - 0.45 sin udu

(2)

Virtual–Work Equation: In this case Fsp must be resolved into its horizontal and vertical component, i.e. (Fsp)x = Fsp cos f and (Fsp)y = Fsp sin f. Since (Fsp)x and W and act towards the negative sense of their corresponding virtual displacements, their work is negative. However, (Fsp)y does positive work since it acts towards the positive sense of its corresponding virtual displacement. Thus, dU = 0;

- Fsp cos fdxB + Fsp sin fdyB + (- WdyB) = 0

(3)

Substituting W = 25(9.81) = 245.25 N, Eqs. (1) and (2) into Eq. (3), we have -Fsp cos f(0.45 cos udu) + Fsp sin f( -0.45 sin udu) - 245.25(-0.45 sin udu) = 0 du A 110.3625 sin u - 0.45Fsp( cos u cos f + sin u sin f) B = 0 Using the indentity cos (u - f) = cos u cos f + sin u sin f, the above equation can be rewritten as du A 110.3625 sin u - 0.45Fsp cos (u - f) B = 0 Since du Z 0, then 110.3625 sin u - 0.45Fsp cos (u - f) = 0 Fsp =

245.25 sin u cos (u - f)

(4)

Applying the law of cosines to the geometry shown in Fig. b, we have AB = 20.452 + 0.62 - 2(0.45)(0.6) cos u = 20.5625 - 0.54 cos u Thus, the stretch of the spring is given by x = 20.5625 - 0.54 cos 45° - 20.5625 - 0.54 cos 30° = 0.1171 m

1134

0.6 m

u 0.45 m C

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11–7. (continued) The magnitude of Fsp computed using the spring force formula is therefore Fsp = kx = 0.1171 k The angle f at u = 45° can be obtained by referring to the geometry shown in Fig. c. 0.6 - 0.45 cos 45° 0.45 sin 45° f = 41.53°

tan f =

Substituting u = 45° and the results for Fsp and f into Eq. (4), we have 0.1171 k =

245.25 sin 45° cos(45° - 41.53°)

k = 1484 N>m = 1.48 kN>m

Ans.

Ans: k = 1.48 kN/m 1135

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*11–8. The punch press consists of the ram R, connecting rod AB, and a flywheel. If a torque of M = 75 N # m is applied to the flywheel, determine the force F applied at the ram to hold the rod in the position u = 60°.

B

200 mm

600 mm R

u

F M

A

Solution Free Body Diagram. The system has only one degree of freedom, defined by the independent coordinate u. When u undergoes a positive angular displacement du as shown in Fig. a, only force F and couple moment M do work. Virtual Displacement. The position of force F is measured from fixed point O by position coordinate xA. Applying the law of cosines by referring to Fig. b,

0.62 = x2A + 0.22 - 2xA(0.2) cos u

(1)

Differentiating the above expression, 0 = 2xAdxA + 0.4xA sin u du - 0.4 cos u dxA

dxA =

0.4xA sin u du 0.4 cos u - 2xA

(2)

Virtual–Work Equation. When point A undergoes a positive virtual displacement, and the flywheel undergoes positive virtual angular displacement du, both F and M do negative work.

(3)

- FdxA - Mdu = 0

dU = 0;

Substituting Eq. (2) into (3)

Since du ≠ 0, then

ca

0.4 xA sin u bF + M d du = 0 0.4 cos u - 2xA 0.4 xA sin u F + M = 0 0.4 cos u - 2xA



F = -a

0.4 cos u - 2xA bM (4) 0.4 xA sin u

At the equilibrium position u = 60°, Eq. (1) gives

0.62 = xA2 + 0.22 - 2xA(0.2) cos 60° xA = 0.6745 m

Substitute M = 75 N # m, u = 60° and this result into Eq. (4) F = -c

0.4 cos 60° - 2(0.6745) 0.4(0.6745) sin 60°

= 368.81 N = 369 N

d (75)

Ans.

Ans: F = 369 N 1136

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11–9. The flywheel is subjected to a torque of M = 75 N # m. Determine the horizontal compressive force F and plot the result of F (ordinate) versus the equilibrium position u (abscissa) for 0° … u … 180°.

B

200 mm

F

Solution Free Body Diagram. The system has only one degree of freedom, defined by the independent coordinate u. When u undergoes a positive angular displacement du as shown in Fig. a, only force F and couple moment M do work. Virtual Displacement. The position of force F is measured from fixed point O by position coordinate xA. Applying the law of cosines by referring to Fig. b, 0.62 = xA2 + 0.22 - 2xA(0.2) cos u

(1)

Differentiating the above expression, 0 = 2xAdxA + 0.4xA sin u du - 0.4 cos u dxA

dxA =

0.4xA sin u 0.4 cos u - 2xA

(2)

Virtual–Work Equation. When point A undergoes a positive virtual displacement and the flywheel undergoes positive virtual angular displacement du, both F and M do negative work.

(3)

- FdxA - Mdu = 0

dU = 0;

Substituting Eq. (2) into (3) ca

Since du ≠ 0, then

0.4 xA sin u bF + M d du = 0 0.4 cos u - 2xA

a

Here, M = 75 N # m. Then

0.4 xA sin u bF + M = 0 0.4 cos u - 2xA F = -M a

0.4 cos u - 2xA b 0.4 xA sin u

F = -75 a

0.4 cos u - 2xA b (4) 0.4 xA sin u

Using Eq. (1) and (4), the following tabulation can be computed. Subsequently, the graph of F vs u shown in Fig. c can be plotted u(deg.)

0

xA(m)

0.80

F(N)

∞

30

60

90

120

150

0.7648 0.6745 0.5657 0.4745 0.4184 580

369

375

524

1060

R

u M



600 mm

180 0.4 ∞

15

73.0

0.7909 0.6272 1095

1137

356

A

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11–10. The thin rod of weight W rest against the smooth wall and floor. Determine the magnitude of force P needed to hold it in equilibrium for a given angle u.

B

SOLUTION

l

Free-Body Diagram: The system has only one degree of freedom defined by the independent coordinate u. When u undergoes a positive displacement du, only the weight of the rod W and force P do work. Virtual Displacements: The weight of the rod W and force P are located from the fixed points A and B using position coordinates yC and xA, respectively yC =

1 sin u 2

dyC =

xA = l cos u

1 cos udu 2

P

A

θ

(1)

dxA = - l sin udu

(2)

Virtual-Work Equation: When points C and A undergo positive virtual displacements dyC and dxA, the weight of the rod W and force F do negative work. dU = 0; - WdyC - PdyA = 0

(3)

Substituting Eqs. (1) and (2) into (3) yields a Pl sin u -

Wl cos u b du = 0 2

Since du Z 0, then Pl sin u P =

Wl cos u = 0 2

W cot u 2

Ans.

Ans: P = 1138

W cot u 2

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11–11. When u = 30°, the 25-kg uniform block compresses the two horizontal springs 100 mm. Determine the magnitude of the applied couple moments M needed to maintain equilibrium. Take k = 3 kN>m and neglect the mass of the links.

50 mm

k D

300 mm 100 mm 200 mm

M

C 100 mm

Solution

A

u

k

u M

B

Free Body Diagram. The system has only one degree of freedom, defined by the independent coordinate u. When u undergoes a positive angular displacement du as shown in Fig. a, only spring force Fsp, the weight of the block W, and couple moment M do work. Virtual Displacement. The positions of Fsp and W are measured from fixed point B using position coordinates x and y respectively.          x = 0.3 cos u + 0.05  dx = - 0.3 sin u du

(1)

             y = 0.3 sin u + 0.1      dy = 0.3 cos udu

(2)

Virtual–Work Equation. When Fsp, W and M undergo their respective positive virtual displacement, all of them do negative work. Thus dU = 0;

(3)

- 2Fsp dx - Wdy - 2Mdu = 0

Substitute Eqs. (1) and (2) into (3), - 2Fsp( - 0.3 sin udu) - W(0.3 cos udu) - 2Mdu = 0 (0.6Fsp sin u - 0.3 W cos u - 2M)du = 0 Since du ≠ 0, then 0.6 Fsp sin u - 0.3 W cos u - 2M = 0 (4)

M = 0.3 Fsp sin u - 0.15 W cos u

When u = 30°, Fsp = kx = 3000(0.1) = 300 N. Also W = 25(9.81) = 245.25 N. Substitute these values into Eq. 4.   M = 0.3(300) sin 30° - 0.15(245.25) cos 30° = 13.14 N # m = 13.1 N # m Ans.

Ans: M = 13.1 N # m 1139

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*11–12. The members of the mechanism are pin connected. If a vertical force of 800 N acts at A, determine the angle u for equilibrium. The spring is unstretched when u = 0°. Neglect the mass of the links.

B

D

u

k  6 kN/m 1m

1m

1m

Solution

A

Free Body Diagram. The system has only one degree of freedom, defined by the independent coordinate u. When u undergoes a positive angular displacement du as shown in Fig. a, only spring force Fsp and force P do work.

800 N

Virtual Displacement. The positions of Fsp and P are measured from fixed point B using position coordinates yc and yA respectively.            yc = 1 sin u      dyc = cos u du

(1)

            yA = 3(1 sin u)  dyA = 3 cos u du

(2)

Virtual Work Equation. When Fsp and P undergo their respective positive virtual displacement, P does positive work whereas Fsp does negative work. (3)

-Fspdyc + PdyA = 0

            dU = 0; Substitute Eqs. (1) and (2) into (3)

- Fsp(cos udu) + P(3 cos udu) = 0 ( - Fsp cos u + 3P cos u)du = 0 Since du ≠ 0, and assuming u 6 90°, then - Fsp cos u + 3P cos u = 0 Fsp = 3P Here Fsp = kx = 6000(1 sin u) = 6000 sin u and P = 800 N, Then 6000 sin u = 3(800)  sin u = 0.4 Ans.

              u = 23.58° = 23.6°

Ans: u = 23.6° 1140

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11–13. The service window at a fast-food restaurant consists of glass doors that open and close automatically using a motor which supplies a torque M to each door. The far ends, A and B, move along the horizontal guides. If a food tray becomes stuck between the doors as shown, determine the horizontal force the doors exert on the tray at the position u.

a

a

a

u C

M

a u

A

M

B

D

SOLUTION x = 2a cos u, dU = 0;

dx = - 2a sin u du -M du - F dx = 0 -M du + F (2a sin u)du = 0 F =

M 2a sin u

Ans.

Ans: F = 1141

M 2a sin u

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11–14. If each of the three links of the mechanism have a mass of 4 kg, determine the angle u for equilibrium. The spring, which always remains vertical, is unstretched when u = 0°. A

M  30 N  m u 200 mm

k  3 kN/m D

200 mm

Solution

B

Free Body Diagram. The system has only one degree of freedom, defined by the independent coordinate u. When u undergoes a positive angular displacement du as shown in Fig. a, only the weights W1 = W2 = W3 = W, couple moment M, and spring force Fsp do work. Virtual Displacement. The positions of the weights W1, W2, W3 and spring force Fsp are measured from fixed point A using position coordinates y1, y2, y3 and y4 respectively

y1 = 0.1 sin u

dy1 = 0.1 cos u du

(1)



y2 = 0.2 sin u + 0.1

dy2 = 0.2 cos u du

(2)



y3 = 0.1 sin u + 0.2

dy3 = 0.1 cos u du

(3)



y4 = 0.5 sin u

dy4 = 0.2 cos u du

(4)

Virtual Work Equation. When all the weights undergo positive virtual displacement, all of them do positive work. However, Fsp does negative work when its undergoes positive virtual displacement. Also, M does positive work when it undergoes positive virtual angular displacement.

u 200 mm

dU = 0;

W1dy1 + W2dy2 + W3dy3 - Fspdy4 + Mdu = 0

(5)

Substitute Eqs (1), (2) and (3) into (5), using W1 = W2 = W3 = W. W(0.1 cos u du) + W(0.2 cos u du) + W(0.1 cos u du) - Fsp(0.2 cos u du) + Mdu = 0 (0.4 W cos u - 0.2 Fsp cos u + M)du = 0 Since d ≠ 0, then 0.4 W cos u - 0.2 Fsp cos u + M = 0

Here M = 30 N # m, W = 4(9.81)N = 39.24 N and Fsp = kx = 3000(0.2 sin u) = 600 sin u. Substitute these results into this equation, 0.4(39.24) cos u - 0.2(600 sin u)cos u + 30 = 0 15.696 cos u - 120 sin u cos u + 30 = 0

1142

C

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11–14. Continued

Solve numerically:

Ans.

u = 23.8354° = 23.8° or



Ans.

u = 72.2895° = 72.3°

Note: u = 23.8° is a stable equilibrium, while u = 72.3° is an unstable one.

Ans: u = 23.8° u = 72.3° 1143

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11–15. The “Nuremberg scissors” is subjected to a horizontal force of P = 600 N. Determine the angle u for equilibrium. The spring has a stiffness of k = 15 kN>m and is unstretched when u = 15°.

E u

200 mm A k

SOLUTION

D

B 200 mm C

Free - Body Diagram: When u undergoes a positive virtual angular displacement of du, the dash line configuration shown in Fig. a is a formed. We observe that only the spring force Fsp acting at points A and B and the force P do work when the virtual displacements take place. The magnitude of Fsp can be computed using the spring force formula, Fsp = kx = 15(103) C 2(0.2 sin u) - 2(0.2 sin 15°) D = 6000(sin u - 0.2588) N Virtual Displacement: The position of points A and B at which Fsp acts and point C at which force P acts are specified by the position coordinates yA, yB, and yC, measured from the fixed point E, respectively. yA = 0.2 sin u

dyA = 0.2 cos udu

(1)

yB = 3(0.2 sin u)

dyB = 0.6 cos udu

(2)

yC = 8(0.2 sin u)

dyB = 1.6 cos udu

(3)

Virtual Work Equation: Since Fsp at point A and force P acts towards the positive sense of its corresponding virtual displacement, their work is positive. The work of Fsp at point B is negative since it acts towards the negative sense of its corresponding virtual displacement. Thus, Fsp dyA + A - FspdyB B + PdyC = 0

dU = 0;

(4)

Substituting Fsp = 6000(sin u - 0.2588), P = 600 N, Eqs. (1), (2), and (3) into Eq. (4), 6000(sin u - 0.2588)(0.2 cos udu - 0.6 cos udu) + 600(1.6 cos udu) = 0 cos udu C -2400(sin u - 0.2588) + 960 D = 0 Since cos udu Z 0, then - 2400(sin u - 0.2588) + 960 = 0 u = 41.2°

Ans.

Ans: u = 41.2° 1144

P

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*11–16. The “Nuremberg scissors” is subjected to a horizontal force of P = 600 N. Determine the stiffness k of the spring for equilibrium when u = 60°. The spring is unstretched when u = 15°.

E u

200 mm A k

D

SOLUTION

B 200 mm C

P

Free - Body Diagram: When u undergoes a positive virtual angular displacement of du, the dash line configuration shown in Fig. a is formed. We observe that only the spring force Fsp acting at points A and B and the force P do work when the virtual displacements take place. The magnitude of Fsp can be computed using the spring force formula. Fsp = kx = k C 2(0.2 sin u) - 2(0.2 sin 15°) D = (0.4)k(sin u - 0.2588) N Virtual Displacement: The position of points A and B at which Fsp acts and point C at which force P acts are specified by the position coordinates yA, yB, and yC, measured from the fixed point E, respectively. yA = 0.2 sin u

dyA = 0.2 cos udu

(1)

yB = 3(0.2 sin u)

dyB = 0.6 cos udu

(2)

yC = 8(0.2 sin u)

dyB = 0.6 cos udu

(3)

Virtual Work Equation: Since Fsp at point A and force P acts towards the positive sense of its corresponding virtual displacement, their work is positive. The work of Fsp at point B is negative since it acts towards the negative sense of its corresponding virtual displacement. Thus, dU = 0;

Fsp dyA +

(4)

A -FspdyB B + PdyC = 0

Substituting Fsp = k(sin u - 0.2588), P = 600 N, Eqs. (1), (2), and (3) into Eq. (4), (0.4)k(sin u - 0.2588)(0.2 cos udu - 0.6 cos udu) + 600(1.6 cos udu) = 0 cos udu C - 0.16k(sin u - 0.2588) + 960 D = 0 Since cos udu Z 0, then - 0.16k(sin u - 0.2588) + 960 = 0 k =

6000 sin u - 0.2588

When u = 60°, k =

6000 = 9881N>m = 9.88 kN>m sin 60° - 0.2588

Ans.

Ans: k = 9.88 kN>m 1145

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11–17. A 5-kg uniform serving table is supported on each side by pairs of two identical links, AB and CD, and springs CE. If the bowl has a mass of 1 kg, determine the angle u where the table is in equilibrium. The springs each have a stiffness of k = 200 N>m and are unstretched when u = 90°. Neglect the mass of the links.

250 mm

150 mm

E

A

k

C

250 mm u

u

SOLUTION Free -Body Diagram: When u undergoes a positive virtual angular displacement of du, the dash line configuration shown in Fig. a is formed. We observe that only the spring force Fsp, the weight Wt of the table, and the weight Wb of the bowl do work when the virtual displacement takes place. The magnitude of Fsp can be computed using the spring force formula, Fsp = kx = 200 A 0.25 cos u B = 50 cos u N.

D B 150 mm

Virtual Displacement: The position of points of application of Wb, Wt, and Fsp are specified by the position coordinates yGb, yGt, and xC, respectively. Here, yGb and yGt are measured from the fixed point B while xC is measured from the fixed point D. yGb = 0.25 sin u + b

dyGb = 0.25 cos udu

(1)

yGt = 0.25 sin u + a

dyGt = 0.25 cos udu

(2)

xC = 0.25 cos u

dxC = -0.25 sin udu

(3)

Virtual Work Equation: Since Wb, Wt, and Fsp act towards the negative sense of their corresponding virtual displacement, their work is negative. Thus, dU = 0;

-WbdyGb +

A - WtdyGt B + A -FspdxC B = 0

(4)

1 5 Substituting Wb = a b(9.81) = 4.905 N, Wt = a b(9.81) = 24.525 N, 2 2 Fsp = 50 cos u N, Eqs. (1), (2), and (3) into Eq. (4), we have -4.905(0.25 cos udu) - 24.525(0.25 cos udu) - 50 cos u( -0.25 sin udu) = 0 du A -7.3575 cos u + 12.5 sin u cos u B = 0 Since du Z 0, then - 7.3575 cos u + 12.5 sin u cos u = 0 cos u( - 7.3575 + 12.5 sin u) = 0 Solving the above equation, cos u = 0

u = 90°

Ans.

- 7.3575 + 12.5 sin u = 0 u = 36.1°

Ans.

Ans: u = 90° u = 36.1° 1146

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11–18. A 5-kg uniform serving table is supported on each side by two pairs of identical links, AB and CD, and springs CE. If the bowl has a mass of 1 kg and is in equilibrium when u = 45°, determine the stiffness k of each spring. The springs are unstretched when u = 90°. Neglect the mass of the links.

250 mm

150 mm

E

A

k

C

250 mm u

u

SOLUTION Free -Body Diagram: When u undergoes a positive virtual angular displacement of du, the dash line configuration shown in Fig. a is formed. We observe that only the spring force Fsp, the weight Wt of the table, and the weight Wb of the bowl do work when the virtual displacement takes place. The magnitude of Fsp can be computed using the spring force formula, Fsp = kx = k A 0.25 cos u B = 0.25 k cos u.

D B 150 mm

Virtual Displacement: The position of points of application of Wb, Wt, and Fsp are specified by the position coordinates yGb, yGt, and xC, respectively. Here, yGb and yGt are measured from the fixed point B while xC is measured from the fixed point D. yGb = 0.25 sin u + b

dyGb = 0.25 cos udu

(1)

yGt = 0.25 sin u + a

dyGt = 0.25 cos udu

(2)

xC = 0.25 cos u

dxC = -0.25 sin udu

(3)

Virtual Work Equation: Since Wb, Wt, and Fsp act towards the negative sense of their corresponding virtual displacement, their work is negative. Thus, dU = 0;

-WbdyGb +

A - WtdyGt B + A -FspdxC B = 0

(4)

1 5 Substituting Wb = a b(9.81) = 4.905 N, Wt = a b(9.81) = 24.525 N, 2 2 Fsp = 0.25k cos u N, Eqs. (1), (2), and (3) into Eq. (4), we have -4.905(0.25 cos udu) - 24.525(0.25 cos udu) - 0.25k cos u(- 0.25 sin udu) = 0 du A -7.3575 cos u + 0.0625k sin u cos u B = 0 Since du Z 0, then -7.3575 cos u + 0.0625k sin u cos u = 0 117.72 k = sin u When u = 45°, then 117.72 k = = 166 N>m sin 45°

Ans.

Ans: k = 166 N>m 1147

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11–19. The disk is subjected to a couple moment M. Determine the disk’s rotation u required for equilibrium. The end of the spring wraps around the periphery of the disk as the disk turns. The spring is originally unstretched.

k  4 kN/ m

0.5 m

M  300 N  m

Solution Free Body Diagram. The system has only one degree of freedom, defined by the independent coordinate u. When u undergoes a positive angular displacement du as shown in Fig. a, only the spring force Fsp and couple moment M do work. Virtual Work Equation. When the disk undergoes a positive angular displacement du, correspondingly point A undergoes a positive displacement of dxA. As a result couple moment M does positive work whereas spring force Fsp does negative work.

dU = 0;

(1)

Mdu + ( - FspdxA) = 0

Here, Fsp = kxA = 4000(0.50) = 2000u and dxA = 0.5du. Substitute these results into Eq. (1) 300du - 2000u(0.5du) = 0 (300 - 1000u)du = 0 Since du ≠ 0, 300 - 1000u = 0 u = 0.3 rad = 17.19° = 17.2°          Ans.

Ans: u = 17.2° 1148

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*11–20.

k ⫽ 300 N⭈m/rad

If the spring has a torsional stiffness of k = 300 N # m>rad and it is unstretched when u = 90°, determine the angle u when the frame is in equilibrium.

B 0.5 m 0.5 m M ⫽ 600 N⭈m u

A

C

SOLUTION Free-Body Diagram: When u undergoes a positive virtual angular displacement of du, the dash line configuration shown in Fig. a is formed. We observe that only couple moment M and the torque M sp developed in the torsional spring do work when the virtual displacement takes place. The magnitude of M sp can be computed using the spring force formula, Msp = k(2a) = 300 c 2 a Virtual Displacement: Since a =

p - u b d = 300(p - 2u) 2

p - u, then 2 da = - du

(1)

Virtual–Work Equation: Since M and M sp act towards the negative sense of their corresponding angular virtual displacements, their work is negative. Thus, dU = 0;

- Mdu + 2( -Mspda) = 0

(2)

Substituting M = 600 N # m, Msp = 300(p - 2u), and Eq. (1) into Eq. (2), we have -600du - 2[300(p - 2u)]( - du) = 0 du[-600 + 600(p - 2u)] = 0 Since du Z 0, then -600 + 600(p - 2u) = 0 u = 1.071 rad = 61.4°

Ans.

Ans: u = 61.4° 1149

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11–21. The crankshaft is subjected to a torque of M = 50 N # m. Determine the horizontal compressive force F applied to the piston for equilibrium when u = 60°.

400 mm

100 mm u

F M

SOLUTION (0.4)2 = (0.1)2 + x2 - 2(0.1)(x)(cos u) 0 = 0 + 2x dx + 0.2x sin u du - 0.2 cos u dx dU = 0; For u = 60°,

- 50du - Fdx = 0

x = 0.4405 m dx = - 0.09769 du

Ans.

(-50 + 0.09769F) du = 0 F = 512 N

Ans.

Ans: dx = -0.09769 du F = 512 N 1150

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11–22. The crankshaft is subjected to a torque of M = 50 N # m. Determine the horizontal compressive force F and plot the result of F (ordinate) versus u (abscissa) for 0° … u … 90°.

400 mm

100 mm u

F M

SOLUTION (0.4)2 = (0.1)2 + x 2 - 2(0.1)(x)(cos u)

(1)

0 = 0 + 2x dx + 0.2x sin u du - 0.2 cos u dx dx = a dU = 0; -50du - Fa F =

0.2x sin u b du 0.2 cos u - 2x -50du - Fdx = 0

0.2x sin u b du = 0, 0.2 cos u - 2x

du Z 0

50(2x - 0.2 cos u) 0.2x sin u

From Eq. (1) x2 - 0.2x cos u - 0.15 = 0 x =

0.2 cos u ; 20.04 cos2 u + 0.6 , 2

x =

0.2 cos u + 20.04 cos2 u + 0.6 2

F =

since20.04 cos2 u + 0.6 7 0.2 cos u

500 20.04 cos2 u + 0.6

Ans.

(0.2 cos u + 20.04 cos2 u + 0.6) sin u

Ans: F =

1151

50020.04 cos2 u + 0.6 (0.2 cos u + 20.04 cos2 u + 0.6) sin u

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11–23.

C

The spring has an unstretched length of 0.3 m. Determine the angle u for equilibrium if the uniform links each have a mass of 5 kg. 0.6 m

SOLUTION Free Body Diagram: The system has only one degree of freedom defined by the independent coordinate u. When u undergoes a positive displacement du, only the spring force Fsp and the weights of the links (49.05 N) do work.

θ

θ

B 0.1 m

D A

k = 400 N/m

E

Virtual Displacements: The position of points B, D and G are measured from the fixed point A using position coordinates xB , xD and yG , respectively. xB = 0.1 sin u

dxB = 0.1 cos udu

xD = 210.7 sin u2 - 0.1 sin u = 1.3 sin u yG = 0.35 cos u

(1) dxD = 1.3 cos udu

dyG = - 0.35 sin udu

(2) (3)

Virtual–Work Equation: When points B, D and G undergo positive virtual displacements dxB , dxD and dyG , the spring force Fsp that acts at point B does positive work while the spring force Fsp that acts at point D and the weight of link AC and CE (49.05 N) do negative work. dU = 0;

21- 49.05dyG2 + Fsp1dxB - dxD2 = 0

(4)

Substituting Eqs. (1), (2) and (3) into (4) yields 134.335 sin u - 1.2Fsp cos u2 du = 0

(5)

However, from the spring formula, Fsp = kx = 4003210.6 sin u2 - 0.34 = 480 sin u - 120. Substituting this value into Eq. (5) yields 134.335 sin u - 576 sin u cos u + 144 cos u2 du = 0 Since du Z 0, then 34.335 sin u - 576 sin u cos u + 144 cos u = 0 u = 15.5°

Ans.

and u = 85.4°

Ans.

Ans: u = 15.5 u = 85.4 1152

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*11–24. The dumpster has a weight W and a center of gravity at G. Determine the force in the hydraulic cylinder needed to hold it in the general position u.

b a

G d

SOLUTION θ

s = 2a2 + c2 - 2a c cos (u + 90°) = 2a2 + c2 + 2a c sin u c

1

ds = (a2 + c2 + 2a c sin u)- 2 ac cos u du y = (a + b) sin u + d cos u dy = (a + b) cos u du - d sin u du dU = 0;

Fds - Wdy = 0 1

F(a2 + c2 + 2a c sin u)- 2 ac cos u du - W(a + b) cos u du + Wd sin u du = 0 F = a

W(a + b - d tan u) b 2a2 + c2 + 2a c sin u ac

Ans.

Ans: F = a

1153

W(a + b - d tan u) b 2a2 + c2 + 2a c sin u ac

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11–25. If the potential energy for a conservative onedegree-of-freedom system is expressed by the relation V = (3y3 + 2y2 - 4y + 50) J, where y is given in meters, determine the equilibrium positions and investigate the stability at each position.

Solution Potential Function: V = 3y3 + 2y2 - 4y + 50 Equilibrium Configuration: Taking the first derivative of V, dV = 9y2 + 4y - 4 = 0 du Thus, Ans.

y = 0.481 m, y = - 925 m Stability: The second derivative of V is d 2V = 18y + 4 du 2 At y = 0.481 m,

d 2V = 12.7 7 0 Stable du 2

Ans.

At y = -925 m,

d 2V = -12.7 6 0 Unstable du 2

Ans.

Ans: y = 0.481 m Stable y = - 925 m Unstable 1154

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11–26. If the potential function for a conservative one-degree-offreedom system is V = 110 cos 2u + 25 sin u2 J, where 0° 6 u 6 180°, determine the positions for equilibrium and investigate the stability at each of these positions.

SOLUTION V = 10 cos 2u + 25 sin u For equilibrium: dV = - 20 sin 2u + 25 cos u = 0 du 1 -40 sin u + 252 cos u = 0 u = sin-1 a

25 b = 38.7° and 141° 40

Ans.

and u = cos-1 0 = 90°

Ans.

Stability: d2V = - 40 cos 2u - 25 sin u du2 u = 38.7°,

d2V = - 24.4 6 0, du2

Unstable

Ans.

u = 141°,

d2V = - 24.4 6 0, du2

Unstable

Ans.

Stable

Ans.

u = 90°,

d 2V = 15 7 0, du2

Ans: u = 38.7° unstable u = 90° stable u = 141° unstable 1155

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11–27. If the potential function for a conservative one-degree-offreedom system is V = 18x3 - 2x 2 - 102 J, where x is given in meters, determine the positions for equilibrium and investigate the stability at each of these positions.

SOLUTION V = 8x3 - 2x2 - 10 dV = 24x2 - 4x = 0 dx 124x - 42x = 0 x = 0

Ans.

and x = 0.167 m

d2V = 48x - 4 dx2 x = 0,

d 2V = -4 6 0 dx2

x = 0.167 m,

Ans.

Unstable

d2V = 4 7 0 dx2

Ans.

Stable

Ans: x = 0.167 m d 2V = - 4 6 0 Unstable dx2 d 2V = 4 7 0 Stable dx2 1156

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*11–28. If the potential energy for a conservative onedegree-of-freedom system is expressed by the relation V = (24 sin u + 10 cos 2u) J, 0 … u … 90, determine the equilibrium positions and investigate the stability at each position.

Solution V = 24 sin u + 10 cos 2u Equilibrium Position: dV = du

24 cos u - 20 sin 2u = 0 24 cos u - 40 sin u cos u = 0 cos u 124 - 40 sin u 2 = 0 cos u = 0 u = 90° 24 - 40 sin u = 0

Ans. Ans.

u = 36.9°

Stability: d 2V = - 40 cos 2u - 24 sin u du 2 At u = 90° At u = 36.9°

d 2V = - 40 cos 180° - 24 sin 90° = 16 7 0 du 2

Ans.

stable

d 2V = -40 cos 73.7° - 24 sin 36.9° = -25.6 6 0 du 2

unstable

Ans.

Ans: stable at u = 90° Unstable at u = 36.9° 1157

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11–29. If the potential function for a conservative one-degree-offreedom system is V = (12 sin 2u + 15 cos u) J,where 0° 6 u 6 180°, determine the positions for equilibrium and investigate the stability at each of these positions.

SOLUTION V = 12 sin 2u + 15 cos u dV = 0; du

24 cos 2u - 15 sin u = 0

2411 - 2 sin2 u2 - 15 sin u = 0 48 sin2 u + 15 sin u - 24 = 0 Choosing the angle 0° 6 u 6 180° u = 34.6°

Ans.

u = 145°

Ans.

and

d2V = - 48 sin 2u - 15 cos u du2 u = 34.6°,

d 2V = - 57.2 6 0 du2

Unstable

Ans.

u = 145°,

d2V = 57.2 7 0 du2

Stable

Ans.

Ans: Unstable at u = 34.6° stable at u = 145° 1158

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11–30. The spring of the scale has an unstretched length of a. Determine the angle u for equilibrium when a weight W is supported on the platform. Neglect the weight of the members. What value W would be required to keep the scale in neutral equilibrium when u = 0°?

W

L

L k

SOLUTION Potential Function: The datum is established at point A. Since the weight W is above the datum, its potential energy is positive. From the geometry, the spring stretches x = 2L sin u and y = 2L cos u. V = Ve + Vg

L

L u

u

a

=

1 2 kx + Wy 2

=

1 (k)(2 L sin u)2 + W(2L cos u) 2

= 2kL2 sin 2 u + 2WL cos u Equilibrium Position: The system is in equilibrium if

dV = 0. du

dV = 4kL2 sin u cos u - 2WL sin u = 0 du dV = 2kL2 sin 2u- 2WLsin u = 0 du Solving, u = 0°

or

u = cos-1 a

Stability: To have neutral equilibrium at u = 0°,

W b 2kL

Ans.

d2V 2 = 0. du2 u - 0°

d 2V = 4kL2 cos 2 u - 2WL cos u du2 d 2V 2 = 4kL2 cos 0° - 2WLcos 0° = 0 du2 u - 0° Ans.

W = 2kL

Ans: u = cos-1a W = 2kL

1159

W b 2kL

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11–31. The uniform bar has a mass of 80 kg. Determine the angle u for equilibrium and investigate the stability of the bar when it is in this position. The spring has an unstretched length when u = 90°. B 4m

k  400 N/m

Solution Potential Function. The Datum is established through point A, Fig. a. Since the center of gravity of the bar is above the datum, its potential energy is positive. Here, y = 2 sin u and the spring stretches x = 4(1 - sin u) m. Thus,

A

u

V = Ve + Vg 1 2 kx + Wy 2 1 = (400) 3 4(1 - sin u) 4 2 + 80(9.81)(2 sin u) 2 =

= 3200 sin2 u - 4830.4 sin u + 3200

Equilibrium Position. The bar is in equilibrium of

dV = 0 du

dV = 6400 sin u cos u - 4830.4 cos u = 0 du cos u(6400 sin u - 4830.4) = 0 Solving,

Ans.

u = 90° or u = 49.00° = 49.0°

Using the trigonometry identify sin 2u = 2 sin u cos u, dV = 3200 sin 2u - 4830.4 cos u du d 2V = 6400 cos 2u + 4830.4 sin u du 2 d 2V d 2V Stability. The equilibrium configuration is stable if 7 0, unstable if 6 0 du 2 du 2 d 2V and neutral if = 0. du 2 At u = 90°, d 2V 2 = 6400 cos [2(90°)] + 4830.4 sin 90° = -1569.6 6 0 du 2 u = 90° Thus, the bar is in unstable equilibrium at u = 90° At u = 49.00°, d 2V = 6400 cos [2(49.00°)] + 4830.4 sin 49.00° = 2754.26 7 0 du2 Thus, the bar is in stable equilibrium at u = 49.0°.

Ans: Unstable equilibrium at u = 90° Stable equilibrium at u = 49.0° 1160

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*11–32. The uniform link AB has a mass of 3 kg and is pin connected at both of its ends. The rod BD, having negligible weight, passes through a swivel block at C. If the spring has a stiffness of k = 100 N>m and is unstretched when u = 0°, determine the angle u for equilibrium and investigate the stability at the equilibrium position. Neglect the size of the swivel block.

400 mm

D k  100 N/m

A

C u

400 mm B

SOLUTION s = 2(0.4)2 + (0.4)2 - 2(0.4)2 cos u = (0.4)22(1 - cos u) V = Vg + Ve = -(0.2)(sin u)3(9.81) +

1 (100) C (0.4)2(2)(1 - cos u) D 2

dV = -(5.886) cos u + 16(sin u) = 0 du

(1) Ans.

u = 20.2° d2V = 5.886 sin u + (16) cos u = 17.0 7 0 du2

stable

Ans.

Ans: u = 20.2° stable 1161

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11–33. If each of the three links of the mechanism has a weight W, determine the angle u for equilibrium. The spring, which always remains vertical, is unstretched when u = 0°.

k u a a

SOLUTION y1 = a sin u

dy1 = a cos u du

y2 = 2a + a sin u y3 = 2a + 2a sin u

dy2 = a cos u du

2a

dy3 = 2a cos u du 2a

Fs = ka sin u dU = 0; (W - Fs)dy1 + Wdy2 + Wdy3 = 0 (W - ka sin u)a cos u du + Wa cos u du + W(2a)cos u du = 0 Assume u 6 90°, so cos u Z 0. 4W - ka sin u = 0 u = sin-1 a

4W b ka

Ans.

or u = 90°

Ans.

Ans: u = sin-1 a u = 90° 1162

4W b ka

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11–34. A spring with a torsional stiffness k is attached to the pin at B. It is unstretched when the rod assembly is in the vertical position. Determine the weight W of the block that results in neutral equilibrium. Hint: Establish the potential energy function for a small angle u. i.e., approximate sin u L 0, and cos u L 1 - u2>2.

L 2 A L 2

SOLUTION Potential Function: With reference to the datum, Fig. a, the gravitational potential

k B

energy of the block is positive since its center of gravity is located above the datum. Here, the rods are tilted with a small angle u. Thus, y =

3 L cos u + L cos u = L cos u. 2 2

u2 .Thus, 2 u2 3WL u2 3 b = a1 - b Vg = Wy = W a L b a1 2 2 2 2

However, for a small angle u, cos u

L 2 C

1 -

The elastic potential energy of the torsional spring can be computed using 1 Ve = kb 2, where b = 2u. Thus, 2 Vg =

1 k(2u)2 = 2ku2 2

The total potential energy of the system is V = Vg + Ve =

u2 3WL a1 - b + 2ku2 2 2

Equilibrium Configuration: Taking the first derivative of V, we have 3WL 3WL dV = u + 4ku = u a + 4k b du 2 2 dV = 0. Thus, du 3WL + 4kb = 0 ua 2

Equilibrium requires

u = 0° Stability: The second derivative of V is 3WL d2V + 4k = 2 du2

To have neutral equilibrium at u = 0°, 3WL + 4k = 0 2 8k W = 3L

d2V 2 = 0. Thus, du2 u = 0°

-

Ans.

Note: The equilibrium configuration of the system at u = 0° is stable if 8k d2V 8k d 2V and is unstable if W > W< > 0 < 0 . 3L du2 3L du2 1163

Ans: W =

8k 3L

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11–35. Determine the angle u for equilibrium and investigate the stability at this position. The bars each have a mass of 3 kg and the suspended block D has a mass of 7 kg. Cord DC has a total length of 1 m.

500 mm

500 mm

u

A

u

C

500 mm

SOLUTION l = 500 mm y1 =

l sin u 2

D

y2 = l + 2l(1 - cos u) = l(3 - 2 cos u) V = 2Wy1 - WDy 2 = Wl sin u - WD l(3 - 2 cos u) dV = l(W cos u - 2WD sin u) = 0 du tan u =

3(9.81) W = 0.2143 = 2WD 14(9.81)

u = 12.1°

Ans.

d2V = l( -W sin u - 2WD cos u) du2 u = 12.1°,

d 2V = 0.5[-3(9.81) sin 12.1° - 14(9.81) cos 12.1°] du2 = - 70.2 6 0

Unstable

Ans.

Ans: u = 12.1° unstable 1164

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*11–36. The uniform bar AD has a mass of 20 kg. If the attached spring is unstretched when u = 90°, determine the angle u for equilibrium. Note that the spring always remains in the vertical position due to the roller guide. Investigate the stability of the bar when it is in the equilibrium position.

C

k  2 kN/m

A

u

B

0.5 m

Solution Potential Function. The Datum is established through point A, Fig. a. Since the center of gravity of the bar is below the datum, its potential energy is negative. Here, y = 0.75 cos u and the spring stretches x = 0.5 cos u. Thus,

1m D

V = Ve + Vg 1 2 kx + ΣWy 2 1 = (2000)(0.5 cos u)2 + 2 =

3 - 20(9.81)(0.75 cos u) 4

= 250 cos2 u - 147.15 cos u

Equilibrium Position. The bar is in equilibrium if

dV = 0. du

dV = -500 sin u cos u + 147.15 sin u = 0 du sin u(147.15 - 500 cos u) = 0 Solving,

u = 0°

Ans.

   u = 72.88° = 72.9°

Using the trigonometry identity sin 2u = 2 sin u cos u, dV = -250 sin 2u + 147.15 sin u du d 2V = -500 cos 2u + 147.15 cos u du 2 d 2V d 2V Stability. The equilibrium configuration is stable if 7 0, unstable if 6 0 2 du du 2 d 2V and neutral if = 0. du 2 d 2V 2 = - 500 cos 0° + 147.15 cos 0° = -352.85 6 0 du 2 u = 0° Thus, the bar is in unstable equilibrium at u = 0°. At u = 72.88°, d 2V 2 = - 500 cos [2(72.88°)] + 147.15 cos 72.88° = 456.69 7 0 du2 u = 72.88° Thus, the bar is in stable equilibrium at u = 72.9°. Ans: Unstable equilibrium at u = 0° Stable equilibrium at u = 72.9° 1165

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11–37. The two bars each have a mass of 8 kg. Determine the required stiffness k of the spring so that the two bars are in equilibrium when u = 60°. The spring has an unstretched length of 1 m. Investigate the stability of the system at the equilibrium position.

A u

1.5 m

B

k

Solution

1.5 m

Potential Function. The Datum is established through point A, Fig. a. Since the centers of gravity of the bars are below the datum, their potential energies are negative. Here, y1 = 0.75 cos u, y2 = 1.5 cos u + 0.75 cos u = 2.25 cos u and the spring stretches x = 2(1.5 cos u) - 1 = (3 cos u - 1) m. Thus, V = Ve + Vg =    =

1 2 kx + ΣWy 2 1 k(3 cos u - 1)2 + [ -8(9.81)(0.75 cos u)] + [ - 8(9.81)(2.25 cos u)] 2

= 4.5 k cos2 u - 3 k cos u + 0.5 k - 235.44 cos u = 4.5 k cos2 u - (3 k + 235.44) cos u + 0.5 k Equilibrium Position. The system is in equilibrium if

dV = 0 du

dV = - 9 k sin u cos u + (3 k + 235.44) sin u = 0 du sin u( - 9 k cos u + 3 k + 235.44) = 0 Since sin u ≠ 0, then

-9 k cos u + 3 k + 235.44 = 0 k =

When u = 60°,

k =

235.44 9 cos u - 3

235.44 = 156.96 N>m = 157 N>m 9 cos 60° - 3

Using this result, dV = - 9(156.96) sin u cos u + [3(156.96) + 235.44] sin u du   = - 1412.64 sin u cos u + 706.32 sin u Using the trigonometry identity sin 2u = 2 sin u cos u, dV = 706.32 sin u - 706.32 sin 2u du d 2V = 706.32 cos u - 1412.64 cos 2u du 2

1166

Ans.

C

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11–37

Continued

d 2V d 2V Stability. The equilibrium configuration is stable if 7 0, unstable if 6 0, 2 du du 2 d 2V and neutral if = 0. du 2 At u = 60°, d 2V 2 = 706.32 cos 60° - 1412.64 cos [2(60°)] = 1059.16 7 0 du 2 u = 60° Thus, the system is in stable equilibrium at u = 60°.

Ans: k = 157 N>m Stable equilibrium at u = 60° 1167

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11–38. The uniform rod has a mass of 100 kg. If the spring is unstretched when u = 60°, determine the angle u for equilibrium and investigate the stability at the equilibrium position. The spring is always in the horizontal position due to the roller guide at B

A

u

2m k  500 N/m B

2m

Solution Potential Function. The Datum is established through point A, Fig. a. Since the center of gravity of the bar is below the datum, its potential energy is negative. Here y = 2 cos u, and the spring stretches x = 2 sin 60° - 2 sin u = 2(sin 60° - sin u). Thus V = Ve + Vg =

1 2 kx + Wy 2

=

1 (500)[2(sin 60° - sin u)]2 + [ - 100(9.81)(2 cos u)] 2

= 1000 sin2 u - 100013 sin u - 1962 cos u + 750 dV = 0. Equilibrium Position. The bar is in equilibrium if du dV = 2000 sin u cos u - 100013 cos u + 1962 sin u du Using the trigonometry identity sin 2u = 2 sin u cos u, dV = 1000 sin 2u - 100013 cos u + 1962 sin u = 0 du Solved numerically, Ans.

u = 24.62° = 24.6° d 2V = 2000 cos 2u + 100013 sin u + 1962 cos u du 2

d 2V d 2V Stability. The equilibrium configuration is stable if 7 0, unstable if 6 0 2 du du 2 d 2V and neutral if = 0. du 2 At u = 24.62°, d 2V = 2000 cos[2(24.62°)] + 100013 sin 24.62° + 1962 cos 24.62° = 3811.12 7 0 du 2 Thus, the bar is in stable equilibrium at u = 24.6°.

Ans: Stable equilibrium at u = 24.6° 1168

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11–39. Determine the angle u for equilibrium and investigate the stability at this position. The bars each have a mass of 10 kg and the spring has an unstretched length of 100 mm.

500 mm

k 1.5 kN/m u

u

A

500 mm

Solution Potential Function. The Datum the centers of gravity of the energies are positive. Here x = 3 2(0.5 cos u) - 0.5 4 - 0.1 =

is established through point A, Fig. a. Since bars are above the datum, their potential y = 0.25 sin u and the spring stretches (cos u - 0.6) m. Thus

V = Ve + Vg

1 2 kx + ΣWy 2 1 = (1500)(cos u - 0.6)2 + 2[10(9.81)](0.25 sin u) 2 =

= 750 cos2 u - 900 cos u + 49.05 sin u + 270 Equilibrium Position. The system is in equilibrium if

dV = 0. du

dV = -1500 sin u cos u + 900 sin u + 49.05 cos u = 0 du Solving numerically,

u = 4.713° = 4.71° or u = 51.22° = 51.2°

Ans.

Using the trigonometry identity sin 2u = 2 sin u cos u, dV = - 750 sin 2u + 900 sin u + 49.05 cos u du d 2V = 900 cos u - 1500 cos 2u - 49.05 sin u du 2 d 2V d 2V Stability. The equilibrium configurate is stable if 7 0, unstable if 6 0 and du 2 du 2 d 2V neutral if = 0. du 2

1169

500 mm

C

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11–39. Continued

At u = 51.22°, d 2V 2 = 900 cos 51.22° - 49.05 sin 51.22° - 1500 cos [2(51.22°)] = 848.77 7 0 du 2 u = 51.22° Thus, the system is in stable equilibrium at u = 51.2°. At u = 4.713°, d 2V 2 = 900 cos 4.713° - 49.05 sin 4.713° - 1500 cos [2(4.713°)] = -586.82 6 0 du2 u = 4.713° Thus, the system is in unstable equilibrium at u = 4.71°.

Ans: Stable equilibrium at u = 51.2° Unstable equilibrium at u = 4.71° 1170

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*11–40. A conical hole is drilled into the bottom of the cylinder, which is supported on the fulcrum at A. Determine the minimum distance d in order for it to remain in stable equilibrium.

A

h

d

SOLUTION Potential Function: First, we must determine the center of gravity of the cylinder. By referring to Fig. a, d 1 h (rpr2h) - a rpr2d b ©yCm 2 4 3 6h2 - d 2 = y = = 1 ©m 4(3h - d) rpr 2h - rpr2d 3

r

(1)

With reference to the datum, Fig. a, the gravitational potential energy of the cylinder is positive since its center of gravity is located above the datum. Here, y = (y - d)cos u = B

6h2 - d2 6h2 - 12hd + 3d2 - d R cos u = B R cos u 4(3h - d) 4(3h - d)

Thus, V = Vg = Wy = W B

6h2 - 12hd + 3d2 R cos u 4(3h - d)

Equilibrium Configuration: Taking the first derivative of V, dV 6h2 - 12hd + 3d2 = -W B R sin u du 4(3h - d) Equilibrium requires

dV = 0. Thus, du

-WB

6h2 - 12hd + 3d2 R sin u = 0 4(3h - d)

sin u = 0

u = 0°

Stability: The second derivative of V is 6h2 - 12hd + 3d2 d 2V = -W B R cos u 2 4(3h - d) du To have neutral equilibrium at u = 0°, -WB

d 2V 2 = 0. Thus, d2u u = 0°

6h2 - 12hd + 3d2 R cos 0° = 0 4(3h - d)

6h2 - 12hd + 3d2 = 0 d =

12h ; 2( - 12h)2 - 4(3)(6h2) = 0.5858h = 0.586h 2(3)

Ans.

Note: If we substitute d = 0.5858h into Eq. (1), we notice that the fulcrum must be at the center of gravity for neutral equilibrium. 1171

Ans: d = 0.586h

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11–41. If the spring has a torsional stiffness k = 300 N # m>rad and is unwound when u = 90°, determine the angle for equilibrium if the sphere has a mass of 20 kg. Investigate the stability at this position. Collar C can slide freely along the vertical guide. Neglect the weight of the rods and collar C.

300 mm D

C

300 mm

k

u B

SOLUTION Potential Function: With reference to the datum, Fig. a, the gravitational potential energy of the sphere is positive its center of gravity is located above the datum. Here, y = (0.3 sin (u>2) + 0.6 sin (u>2))m = (0.9 sin (u/2)). Thus,

A

300 mm

Vg = mgy = 20(9.81)(0.9 sin u>2) = 176.58 sin (u>2) 1 The elastic potential energy of the torsional spring can be computed using Ve = kb 2, 2 p where b = - u. Thus, 2 2 1 p Ve = 300 ¢ - u ≤ = 150u2 - 150pu + 37.5p2 2 2 The total potential energy of the system is V = Vg + Ve = 176.58 sin (u/2) + 150u2 - 150pu + 37.5p2 Equilibrium Configuration: Taking the first derivative of V, we have dV = 88.29 cos (u/2) + 300u - 150p du Equilibrium requires

dV = 0, Thus, du 88.29 cos (u/2) + 300u - 150p = 0

Solving by trial and error, we obtain u = 1.340 rad = 76.78° = 76.8°

Ans.

Stability: The second derivative of V is d2V = - 44.145 sin (u/2) + 300 du2 At the equilibrium configuration of u = 76.78°, d2V 2 = - 44.14 sin 38.39° + 300 = 272.58 7 0 Stable du2 u = 76.78°

Ans.

Ans: u = 76.8 Stable 1172

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11–42. If the uniform rod OA has a mass of 12 kg, determine the mass m that will hold the rod in equilibrium when u = 30°. Point C is coincident with B when OA is horizontal. Neglect the size of the pulley at B.

B

C m 3m

SOLUTION Geometry: Using the law of cosines, 2

A

1m

2

lA¿B = 21 + 3 - 2112132 cos190° - u2 = 210 - 6 sin u

e

O

lAB = 212 + 32 = 210 m l = lAB - lA¿B = 210 - 210 - 6 sin u Potential Function: The datum is established at point O. Since the center of gravity of the rod and the block are above the datum, their potential energy is positive. Here, y1 = 3 - l = 33 - 1 210 - 210 - 6 sin u24 m and y2 = 0.5 sin u m. V = Vg = W1y1 + W2y2 = 9.81m33 - 1210 - 210 - 6 sin u24 + 117.7210.5 sin u2 = 29.43m - 9.81m1210 - 210 - 6 sin u2 + 58.86 sin u Equilibrium Position: The system is in equilibrium if dV = 0. ` du u = 30° 1 1 dV = - 9.81m c - 110 - 6 sin u2- 21 -6 cos u2 d + 58.86 cos u du 2

= -

29.43m cos u 210 - 6 sin u

+ 58.86 cos u

At u = 30°, dV 29.43m cos 30° = + 58.86 cos 30° = 0 ` du u = 30° 210 - 6 sin 30° Ans.

m = 5.29 kg

Ans: m = 5.29 kg 1173

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11–43. Each bar has a mass per length of m0. Determine the angles u and f at which they are suspended in equilibrium. The contact at A is smooth, and both are pin connected at B.

B

l

u

3l 2

Solution

A l 2

Require G for system to be at its lowest point. 0 x =

f

l l 3 a b - (0.75 l sin 26.565°)a lb 4 2 2 = -0.20937 l l 3 l + + l 2 2

lx 1 3 - a b(l) - l a b - (0.75 l cos 26.565°)a lb 2 2 2 y = = -0.66874 l l 3 l + + l 2 2 f = tan-1a

0.20937 l b = 17.38° = 17.4° 0.66874 l

Ans. Ans.

u = 26.565° - 17.38° = 9.18°

Ans: f = 17.4° u = 9.18° 1174

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*11–44. The cylinder is made of two materials such that it has a mass of m and a center of gravity at point G. Show that when G lies above the centroid C of the cylinder, the equilibrium is unstable.

G C r

SOLUTION Potential Function: The datum is established at point A. Since the center of gravity of the cylinder is above the datum, its potential energy is positive. Here, y = r + d cos u. V = Vg = Wy = mg(r + d cos u) Equilibrium Position: The system is in equilibrium if

dV = 0. du

dV = - mgd sin u = 0 du sin u = 0

u = 0°

Stability: d 2V = - mgd cos u du2 d 2V 2 = - mgd cos 0° = - mgd 6 0 du2 u = 0° Thus, the cylinder is in unstable equilibrium at u = 0°

(Q.E.D.)

1175

a

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11–45. The small postal scale consists of a counterweight W1, connected to the members having negligible weight. Determine the weight W2 that is on the pan in terms of the angles u and f and the dimensions shown. All members are pin connected.

W2

a

f

a

f

u

SOLUTION

b

y1 = b cos u

W1

y2 = a sin f = a sin (90° - u - g) where g is a constant and f = (90° - u - g) V = -W1y1 + W2y2 = -W1 b cos u + W2 a sin (90° - u - g) dV = W1 b sin u - W2 a cos (90° - u - g) du b sin u W2 = W1 a b a cos f

Ans.

Ans: b sin u W2 = W1a b a cos f 1176

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11–46. The truck has a mass of 20 Mg and a mass center at G. Determine the steepest grade u along which it can park without overturning and investigate the stability in this position.

G 3.5 m

SOLUTION Potential Function: The datum is established at point A. Since the center of gravity for the truck is above the datum, its potential energy is positive. Here, y = (1.5 sin u + 3.5 cos u) m.

1.5 m u

1.5 m

V = Vg = Wy = W(1.5 sin u + 3.5 cos u) Equilibrium Position: The system is in equilibrium if

dV = 0 du

dV = W(1.5 cos u - 3.5 sin u) = 0 du Since W Z 0, 1.5 cos u - 3.5 sin u = 0 Ans.

u = 23.20° = 23.2° Stability: d2V = W( -1.5 sin u - 3.5 cos u) du2 d2V du2

u = 23.20°

= W(- 1.5 sin 23.20° - 3.5 cos 23.20°) = -3.81W 6 0

Thus, the truck is in unstable equilibrium at u = 23.2°

Ans.

Ans: Unstable equilibrium at u = 23.2° 1177

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11–47. The triangular block of weight W rests on the smooth corners which are a distance a apart. If the block has three equal sides of length d, determine the angle u for equilibrium.

d

60

G 60 u

SOLUTION

a

AF = AD sin f = AD sin (60° - u) a AD = sin a sin 60° AD =

a (sin (60° + u)) sin 60°

AF =

a (sin (60° + u)) sin (60° - u) sin 60°

= y=

a (0.75 cos2 u - 0.25 sin2 u) sin 60° d 23

cos u - AF

V = Wy dV a ( -1.5 sin u cos u - 0.5 sin u cos u)d = 0 = Wc( -0.5774 d) sin u du sin 60° Require, sin u = 0 or

- 0.5774 d -

u = 0°

Ans.

a (- 2 cos u) = 0 sin 60° u = cos-1 a

d b 4a

Ans.

Ans: u = 0° u = cos-1a 1178

d b 4a

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*11–48.

b

A homogeneous block rests on top of the cylindrical surface. Derive the relationship between the radius of the cylinder, r, and the dimension of the block, b, for stable equilibrium. Hint: Establish the potential energy function for a small angle u, i.e., approximate sin u L 0, and cos u L 1 - u2>2.

b

r

SOLUTION Potential Function: The datum is established at point O. Since the center of gravity for the block is above the datum, its potential energy is positive. Here, b y = a r + b cos u + ru sin u. 2 b b cos u + ru sin u d 2

V = Wy = W c a r +

For small angle u, sin u = u and u = 1 V = WB ar + = Wa

[1]

u2 . Then Eq. [1] becomes 2

b u2 b ¢ 1 - ≤ + ru2 R 2 2

ru2 bu2 b + r + b 2 4 2

Equilibrium Position: The system is in equilibrium if b dV = W ar - b u = 0 du 2 Stability: To have stable equilibrium,

dV = 0. du

u = 0°

d2V 2 7 0. du2 u = 0°

d2V b 2 = W ar - b 7 0 2 2 du u = 0° ar -

b b 7 0 2 Ans.

b 6 2r

Ans: b 6 2r 1179