Test questions on theoretical mechanics: collection of tests 9786010444096

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AL-FARABI KAZAKH NATIONAL UNIVERSITY

TEST QUESTIONS ON THEORETICAL MECHANICS Collection of tests

Almaty «Qazaq University» 2019

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UDC 531 (075.8) LBC 22.31я73 T 22 Recommended for use in educational organizations by the educational and methodical association of the Republican Educational and Methodical Council on the basis of al-Farabi KazNU as a collection of test tasks for university students enrolled in the specialty "5B060300-Mechanics" (Protocol №3 dated 26.04.2019); by the Academic Council of the Faculty of mechanics and mathematics and editorial-review board of Al-Farabi KazNU (Protocol №1 dated 13.11.2019) Reviewers: Doctor in Physical and Mathematical Sciences, acting professor K.S. Zhilisbayeva Doctor in Technical Sciences A.A. Dzhomartov Candidate in Physical and Mathematical Sciences, associate professor K.B. Tulegenova Compilers: Z.B. Rakisheva, G.M. Mayemerova, A.S. Sukhenko

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Test questions on theoretical mechanics: collection of tests / comp.: Z.B. Rakisheva, G.M. Mayemerova, A.S. Sukhenko. – Almaty: Qazaq University, 2019. – 102 p. ISBN 978-601-04-4409-6 Educational publication contains the test tasks for the discipline "Theoretical Mechanics" and covers all the topics of the Typical Curriculum of 2016 on this discipline. Tests are recommended for teaching students on the specialty "Mechanics", as well as for the students of all related specialties of science and technology, where the mechanics is studied: "Space engineering and technologies", "Mathematical and computer modeling", "Mathematics", etc. They will be very useful for the students to self-test the gained knowledge, for teachers – in order to verify students' assimilation of knowledge. In connection with the introduction of entrance examinations to the magistracy in the form of tests, this collection will be very convenient for the students to prepare for the test in their specialty. Published in authorial release.

UDC 531 (075.8) LBC 22.31я73 ISBN 978-601-04-4409-6

© Comp.: Rakisheva Z.B., Mayemerova G.M., Sukhenko A.S., 2019 © Al-Farabi KazNU, 2019

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INTRODUCTION Theoretical mechanics is the science about the general laws of mechanical interactions between material bodies, as well as general laws of motion of bodies relative to each other. Therefore, theoretical mechanics is one of the basic core disciplines for all specialties in area of natural science and area of engineering and technology. The suggested collection of test tasks covers all the topics of the classical course of theoretical mechanics: kinematics, statics, dynamics of a mass point, dynamics of a mechanical system, dynamics of a rigid bodies, and corresponds to the Typical Curriculum of 2016 on this discipline. The collection consists of the 300 test questions with five, six and seven answers according to the modern requirements to the test tasks. The number of correct answers in the test tasks: one correct answer is in 100 questions, two correct ones – in 100 questions, three correct ones – in 100 questions. The table of correct answers is adduced at the end of the book. Tests are recommended for teaching students on the specialty "Mechanics", as well as for the students of all related specialties of science and technology, where the mechanics is studied: "Space engineering and technologies", "Mathematical and computer modeling", "Mathematics", etc. They will be very useful for the students to self-test the gained knowledge, for teachers – in order to verify students' assimilation of knowledge. In connection with the introduction of entrance examinations to the magistracy in the form of tests, this collection will be very convenient for the students to prepare for the test in their spe-cialty.

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TEST QUESTIONS 1. What kind of forces are equivalent? A. system of forces with identical direction of principle vectors B. system of forces, where principle vectors are determined by the same magnitude C. in case the number of forces of the first system matches the number of forces of the second system D. in case they have the same mechanical action on the same free rigid body E. system of forces with the principle moments that have the same direction 2. State of a rigid body is not change if the force will be transferred: A. parallel to itself to any other point B. in the parallel plane C. along the line of its action to any other point D. in the plane of its action to any other point E. perpendicular to its direction 3. The following can be added without changing the state of the body: A. equilibrated system of forces B. one force C. a pair of forces D. optional plane system of forces E. equivalent system of forces 4. If any system of forces acting on the rigid body is replaced by a single force and the state of the body does not change, this force is called: A. main point of system of forces B. resultant C. main vector of system of forces D. gravity E. no correct answer 4

5. If the two forces act on a rigid body, this body will be in equilibrium when these forces are: A. equal in magnitude, oriented along the same straight line in opposite directions B. equal in magnitude, parallel and have the same direction C. equal in magnitude and oriented perpendicular to each other D. equal in magnitude, parallel to each other and oriented in opposite directions E. equal in magnitude, oriented along the same line in the same direction 6. Two forces applied to an absolutely rigid body are mutually balanced only if: A. F1  F2





B. F1  F2 and F2  F1 C. F2   F1









D. F2  F1 and both forces are in the same line of action E. F2  F1 and both forces are parallel 7. Forces are called converging: A. if its lines of action crisscross B. if its lines of action are parallel C. if its lines of action intersect at one point D. if its lines of action are perpendicular E. no correct answer 8. In which of these cases the moment of force relative to the axis is always zero? A. if the vector of force vector does not lie in one plane with this axis B. if the vector of force is parallel to this axis C. if the vector of force lies in a plane perpendicular to this axis D. if the vector of force is not perpendicular to this axis E. if the vector of force lies in one plane with this axis 9. In which case the moment of force relative to the axis is always zero? 5

A. B. C. D. E.

if the line of action of force crosses this axis if the force vector lies in a plane perpendicular to this axis if the force vector is not perpendicular to this axis if the force vector is not parallel to this axis if the force vector is parallel to this axis

10. What is called a pair of forces? A. two forces of equal modulus and oriented along a straight line in opposite directions B. two forces of equal modulus and oriented perpendicular to each other C. two forces applied at one point and of equal modulus D. two parallel forces equal in modulus and with opposite directions E. two parallel forces equal in modulus and with the same direction 11. Under the action of a pair of forces the body: A. performs plane motion B. performs linear motion C. performs rotational motion D. performs translational motion E. is in equilibrium 12. When reducing any arbitrary system of forces acting on the rigid body to some center O this system of forces can be replaced by: A. balanced system of forces B. a force C. a pair of forces D. one force and one pair of forces E. two parallel forces 13. The system is called statically indeterminated if: A. number of unknown reactions exceeds the number of equations of statics B. number of unknown reactions is less than the number of equations of statics 6

C. number of unknown reactions is equal to the number of equations of statics D. number of unknown reactions is less in half of the number of equations of statics E. number of equations of statics is greater than the number of unknown reactions 14. Terms equilibrium of parallel forces located in space: A.  M kx  0,  M ky  0,  M kz  0

B.  M kx  0,  M ky  0,  M kz  0 C.  M kx  0,  M ky  0 D.  M ky  0,  M kz  0 E.  M kx  0,  M ky  0,  M kz  0 15. Which of these systems of equations is a necessary and sufficient condition for the equilibrium of convergent forces lying in the plane Oxy?  n n n A.  Fkx  0,  Fky  0,  m O Fk  0

 

k 1 n

k 1 n

k 1 n

k 1 n

k 1

k 1

B.  Fkx  0,  Fky  0,  Fkz  0

 

 

  n C.  m A Fk  0,  m B Fk  0 k 1 n

k 1

n

D.  Fkx  0,  Fky  0 k 1 n

k 1 n

 

 

  n E.  Fkx  0,  m A Fk  0,  m B Fk  0 k 1

k 1

k 1

16. Which of these systems of equations is a necessary and sufficient condition for the equilibrium of converging spatial system of forces? n

n

n

k 1

k 1

k 1

A.  Fkx  0,  Fky  0,  Fkz  0

7

    m F   0

 n n B.  Fkx  0,  m x Fk  0 k 1 n

 

k 1

 C.  m x Fk  0, k 1 n

n

k 1

y

k

 

 n n D.  Fkx  0,  Fky  0,  m O Fk  0 k 1 n

k 1 n

k 1

k 1

 

k 1

 E.  Fkz  0,  m x Fk  0 17. Which of these systems of equations is a necessary and sufficient condition for the equilibrium of an arbitrary plane system of forces? n

n

n

k 1 n

k 1 n

k 1

k 1 n

k 1 n

k 1 n

k 1 n

k 1

k 1

n

n

k 1

k 1

A.  Fkx  0,  Fky  0,  Fkz  0 B.  Fkx  0,  Fky  0

 

 n C.  Fkx  0,  Fky  0,  m O Fk  0

 

k 1

 D.  Fkx  0,  m x Fk  0

 

E.  Fky  0,  m y Fk  0 18. Terms equilibrium of the plane system of parallel forces

  if Fk // j : A.  M ix  0,  M i  0 ;   B.  m A Fk  0,  m B Fk  0   C.  m C Fk  0,  m A Fk  0  D.  Fk  0,  m 0 Fk  0

   

 

   

E.  M iz  0,  M i  0 19. The equilibrium conditions of the spatial system of parallel forces (F1, F2…… Fn) to the axis Oz: 8

A.  Fkx  0 ,  M y Fk   0 ,  M z Fk   0

B.  Fkz  0 ,  M x Fk   0 ,  M y Fk   0 C.  Fkя  0 ,  M z Fk   0 ,  M y Fk   0

D.  Fky  0 ,  M z Fk   0 ,  M z Fk   0

E.  M x Fk   0 ,  M y Fk   0 ,  M z Fk   0

20. Which of these systems of equations is necessary and sufficient condition for the equilibrium of an arbitrary spatial system of forces?  n n n n A.  Fkx  0,  Fky  0,  Fkz  0,  m x Fk  0, k 1 n

 

k 1

k 1

k 1

 

  n  m y Fk  0,  m z Fk  0

k 1 n

k 1

n

n

k 1

k 1

 

B.  Fkx  0,  Fky  0,  Fkz  0 k 1 n

 

 

 

   n n C.  m x Fk  0,  m y Fk  0,  m z Fk  0 k 1

n

k 1

    m F   0

k 1

n

n

k 1 n

k 1 n

k 1

k 1

k 1

 

D.  Fkx  0,  m y Fk  0,  m z Fk  0 E.  Fkx  0,

x

k

21. At what point is the center of gravity of any uniform triangle? A. in one of the vertex of the triangle B. at a point on one side of the triangle C. at the point of intersection of bisectors D. at the point of intersection of the medians E. at orthocenters 22. Which of the formula determines the position vector of the center of mass of a system of particles, if they are indicated  by: rk and m k – the radius vector and the mass of k-point of the system, M – mass of the entire system? 9

A.

n  M  rk k 1

n

 mk

k 1

B. C.

n   m k rk

k 1

M M n   m k rk

k 1

n  D. M  m k rk k 1 n

E.

M  mk k 1 n 

 rk

k 1

23. Force of friction means: A. the interaction force between two bodies B. reaction force upon interaction of bodies C. resistance force of bodies to their relative sliding D. force uniformly distributed along the length of the line E. plane system of distributed forces 24. Which of the formula determines the maximum value of the force of friction, if f – coefficient of sliding friction, N – normal reaction: N A. f B. f 2 N 2

C. f 2 N D. fN f E. N 10

25. Kinematics is a section of mechanics which: A. studies the general doctrine of forces B. studies the motion of material bodies without regard to their masses and their actuating forces C. studies the movement of material bodies under the effect of forces D. studies the conditions of equilibrium of bodies under the effect of forces E. studies Newton's laws 26. How many ways of setting the motion of a mass point exist? A. 3 B. 1 C. 4 D. 5 E. 2 27. In general the vector method of setting the point's motion is defined by the formula:   A. r  xi    B. r  xi  yj   C. r  r (t )











D. r  yj  zk E. r  yj

28. Specify the formula of velocity of a mass point for the vector method of motion setting



A. r   B. r  r

 r C. r  D. r  r  E. r

11

29. Specify the formula of acceleration of a mass point for the vector method of motion setting:   A. r  r

 B. r V2 C.  D.

 E. r

x 2  y 2  z 2

30. In general the natural way of motion setting of a mass point is determined by the formulas: A.    (t ) B. S  f (t )   C. r  r (t ) 

D. W 

 E. r

 d2 r dt 2

31. Which of these expressions determines the velocity of a mass point if the motion is specified in a natural way? 2 A. V  dV B. dt C. x2  y 2  z2

D. S

 E. r

32. Which of these formulas determines the normal aceleration of mass point upon the natural way of motion specification? V2 A.  12

B.

dV dt

C.

x 2  y 2  z 2

D. S

 E. r

33. What characterizes the tangential acceleration? A. changes in the direction of the velocity vector B. changes of the full acceleration vector C. changes of the radius of curvature of the trajectory D. changes of the velocity vector modulus E. change of curvature of the trajectory 34. What characterizes the acceleration? A. changes in the direction of the velocity vector B. changes of the full acceleration vector C. changes of the radius of curvature of the trajectory D. changes of the velocity vector modulus E. change of curvature of the trajectory 35. Uniform rectilinear motion: 2

2

2

2

A. W  const , Wn  0, W  W  Wn  0 B. Wn  0, W  0, W  0 C. W  0, Wп  const , W  0 D. W  0, Wn  0, W  0 E. W  0, Wn  0, W  0 36. Uniform curvilinear motion:

A. W  const , Wn  0, W  W  Wn  0 B. Wn  const , W  0, W  0 C. W  0, Wn  0, W  0

13

D. W  0, Wn 

V2

, W  Wn

 E. W  0, W n  0, W  0

37. Uniformly accelerated curvilinear motion:

A. W  const , Wn 

V2



0

B. W  0, Wn  0 C. W  const , Wn  0 D. W  0, Wn 

V2



0

E. W  0, Wn  0, W  0 38. How can one calculate the velocity module V in a circular motion of the particle ( R – radius of the circle,  – angular velocity,  – angular acceleration)?

d dt  R 2  R   R  d 2 R 2 dt

A. V  R B. V C. V D. V E. V

39. In general, the coordinate method of motion setting of the mass point is defined by the formulas: A. x  f1 (t ), y  f 2 (t ), z  f 3 (t )

B. x  2t , y  12t 2



C. S  f (t ), V 

dS dt 14





D. r  r (t ) E. x  t , y  t 2 40. How to determine the equation of the trajectory of the mass point in the coordinate method of motion setting? A. by exclusion of point's coordinates from the equations of law of motion B. the time derivative is taken from both sides of the equations of motion C. by exclusion of time from the law of motion of the mass point D. squaring the equations of law of motion E. by exclusion of point's velocity from the equations of law of motion 41. The motion of particle is given by the equations x  b sin Rt , y  c cos Rt ( b  c ; b , c , R  const ). Establish the form of orbit of the particle: A. x  by 2 – parabola x2 у2   1 – ellipse в2 с2 C. xy  m (m0) – hyperbola D. x2  y2  c – circle

B.

E. y  bx  c – straight line 42. The motion of particle is given by the equations x  3t , y  t 2 . Define the distance of particle from the origin of coordinate system at the moment of time t  2 sec: A. 5,5 B. 10,5 C. 2,5 D. 7,21 E. 5,21 15

43. Determination of velocity of particle in coordinate method of motion setting on the plane: dS dy A. V x  , V y  , V  Vx  V y dt dt dx dy B. V x  , V y  , V  V x2  V y2 dt dt V2 d 2S C. V x  , V y  2 , V 2  V x2  V y2  dt dy dx D. V x  , Vy  , V  Vx  V y dt dt

E. V x 

V2



, Vу 

dS , V 2  V x2  V y2 dt

44. Specify an expression of acceleration module of particle for coordinate method of poin't motion setting: A. Vx2  V y2  Vz2

B.

x2  y 2  z 2

C.

x2  y 2  z2

2 D. V  d 2S E. dt 2

45. Which of the following equations are the equations of plane motion of a rigid body?

  f1 (t )   A.   f 2 (t )   f 3 (t ) x  f1 (t )   B. y  f 2 (t ) z  f 3 (t )  16

r  f1 (t )     f 2 (t ) x A  f1 (t )   D. y A  f 2 (t )    f 3 (t ) 

C.

x  f1 (t )   E.   f 2 (t )    f 3 (t ) 46. To determine the instantaneous center of velocity you only need to know: A. magnitude and direction of the velocity of any point A B. velocity directions of any two points A and B C. modules of velocities of any two points A and B D. velocity of all the points of the body E. velocity of three points of the body 47. A theorem on projections of the vector of velocities of two points of the body on the line passing through these points is: A. projection of velocities of two points of the body are equal to the velocity of point M B. projection of velocities of two points of the body are perpendicular to each other C. projection of velocities of two points of the body are parallel to each other D. projection of velocities of two points of the body on the line joining these points are equal to each other E. projection of velocities of two points of the body on the line joining these points are not equal 48. In which case the motion of a rigid body is not translational? A. trajectories of all points of the body are the same (when applied) 17

B. the rate of all the points of the body are the same C. acceleration of points of the body are different D. if the motion is considered with respect to the inertial reference system E. no correct answer 49. The law of the rotational motion of a rigid body around a fixed axis is expressed by the equation:

dw dt ds B. v  dt C. s  f (t ) D.   f (t ) A.  

E.

r  f1 (t )     f 2 (t )

50. Relative motion is called: A. the motion of moving coordinate system Oxyz with respect to the fixed coordinate system Ox1 y1 z1 B. motion of a point M with respect to the fixed coordinate system Ox1 y1 z1 C. motion of a point M with respect to the inertial coordinate system D. motion of a point M with respect to the moving coordinate system Oxyz E. movement in which two points of the body remain fixed throughout the motion 51. Which formula determines the theorem of addition of velocities at the complex motion of a particle?

   V  Vr  Ve    B. V  Vr  Ve

A.

18

   V V e C. V  r 2    D. V  Vr  Ve



 

E. V  Vr Ve







F. V  2Vr  Ve 52. Which formula determines the Coriolis theorem?    A. V B  V A  V BA







B. V  Vr  Ve







C. Wc  W A  WCA









D. W  Wr  We  Wc







E. Wc  2 e  Vr







F. Wc  e  Vr 53. Which formula determines the Coriolis acceleration vector?     A. W  W r  We  Wc







B. Wc  e  Vr



















C. Wc  2 e  Vr

D. Wc  W A  WCA E. Wc  2Vr  e







F. Wc  Wr  We



54. In what of these cases Coriolis acceleration WC  0 ?

A.





 e  Vr

B. Vr  Vе





C.  е. // Vr 19

D.  e  0 E. Vr  0 F. Vе  0

 55. When there is no WC (Coriolis acceleration)? A. if the relative motion is translational   B. if the angle between the vectors e and Vr is equal to 90° C. if the transporational motion is translational D. if the absolute velocity Vа  0 E. if the relative motion is rotational   F. if e  Vr 56. An example of the restoring force can be: A. frictional force B. lift C. Archimedes force D. elastic force of the spring E. Coriolis force F. force of gravity 57. Free oscillations of a mass point under the action of the restoring force are linear: A. non-periodic oscillations B. damped oscillations C. harmonic oscillations D. forced oscillations E. complicated fluctuations F. self-oscillation 58. The amplitude of the oscillations of the mass point is: A. the average deviation of the point from the center of oscillation B. the greatest deviation of the point from the center of oscillation C. decrement 20

D. distance between the lowermost and uppermost position of the point E. distance between two consecutive passages of the point in one direction through the provisions of point F. the smallest deviation of a point from the center of oscillations 59. Period of damped oscillations is longer than the period of:

A. B. C. D. E. F.

forced oscillations complex vibrations free oscillations isochronous vibrations harmonic oscillations non-periodic oscillations

60. Circular frequency is equal to (т – mass point, с – spring stiffness): A. k  cm

B. k 

c m

C. k 

m c

c m E. k  mc 2

D. k 

F. k 

m c

61. The period of oscillation of a mass point is determined as:

A. T  2

k B. T  2 k C. T   k 21

D. T  2 k 2 E. T 



2k F. T   k 2 62. The phase of the oscillation is determined as:

k  2 t B. kt 2   t   C. k 2 t 2   D. kt   A.

E. 2kt t F.  k 63. In the case of low-resistance n < k the mass point performs: A. free oscillations B. harmonic oscillations C. damped oscillations D. parametric oscillations E. rectilinear vibrations F. aperiodic fluctuations 64. In case of high resistance motion of the particle becomes: A. complicated B. periodic C. aperiodic D. relative E. translational F. portable 65. The differential equation of forced oscillations of mass point in the presence of the resistance has the form: A. x  k 2 x  h sin( pt   ) 22

B. x  k 2 x  h sin pt

C. x  2nx  k 2 x  h sin( pt   )

D. x  2nx  k 2 x  0 E. x  k 2 x  0 F. x  2nx  0

66. Linear momentum of a mass point is called:  A. a quantity mW C that is equal to the product of the mass point on the Coriolis acceleration 1 B. a quantity mV 2 that is equal to half the product of the 2 mass point and the square of the velocity of the point





C. a quantity r  F that is equal to the vector product of the radius vector of the point on the force



D. value m V that is equal to the product of the mass of the point by the vector of its velocity



E. quantity mW that is equal to the product of the mass point on the vector of its acceleration   F. a quantity r  F that is equal to the scalar product of the radius vector of the point on the force

  67. The value dS  Fdt is called: A. linear momentum of the point B. moment of inertia C. elementary impulse D. total impulse of force for a period of time E. kinetic energy F. angular momentum 68. Which of these quantities is an elementary impulse of force?





A. r  F   B. r  F

23





C. dS  Fdt t1







D. S  Fdt t0

  E. mW  F  dV  F. m F dt

69. The impulse-momentum theorem for a:   A. d (mV )  F dt

B. dT  d ' A     C. mW r   Fi  ( mWe )  ( mWc )





D. mWc   Fke  dG 0   E.  rF dt    F. mWr   Fi  (mWe ) 70. Linear momentum theorem for a mass point in integral form:    A. G  G0   M dt

B. T  T0  A e  Ai



C. mWc   Fke   t  D. mV  mV   Fdt 0 to













E. mWr   Fi  (mWe )  (mWc ) F. mWc   Fki 71. Linear momentum theorem for a mass point in integral form in projections on the coordinate axes:

24

t

t

t

to

to

to

A. mx  mx   F dt , my  my   F dt , mz  mz   F dt o o y o z x 2

B.

mVx mV  2 2

2 x0



F

x

 dx

M o M

d m( yz  zy )  momx F dt 2   D. d  mV x   Fx dx  2   

C.

2 E. d  mV   Fdt  2    F. m( yz  zy )  mom x F, m(zx  xz )  mom y F

72. External forces are called: A. the forces acting on a system of mass points from the outside B. the forces of interaction of points of the same system C. the two forces that are equal in magnitude and opposite in direction D. the forces acting within a single system E. invariant force F. unchangeable forces 73. Internal forces are: A. the forces of friction B. aerodynamic forces C. the forces of interaction of points of the same system D. the resistance forces E. the reaction forces F. gravity forces 74. The sum of the internal forces is: A. equal to the pair of forces B. equal to the moment of forces C. Coriolis force D. equal to zero 25

E. equal to the sum of external forces  F. mWi 75. The sum of the moments of the internal forces is: A. equal to resultant moment B. equal to a force C. equal to the pair of forces D. equal to zero E. equal to 1 F. equal to resultant force 76. The main vector of the linear momentum of the system is determined by the formula:    A. Q   ( ri  miVi ) i

 N   B. Q   (ri  Fi )  i 1  C. Q   mi ri

i   D. Q   miVi i

 N  E. Q   ri Fi

 i 1  F. Q   miWi i

77. In which case the linear momentum theorem of a system of particles is written correctly:



t





A. Q   F( e ) dr o

 (i )    dr   F( e ) dr    ( e) C.  d ( m V )   F dt B.



d ( m V )   F    



26





t







t







D. Q  Qo  F( e ) dr  F(i ) dr o

  E. d ( m V )   F( e) dt

o





t





F. Q   F(i ) dr o

78. In which case the angular momentum theorem of the system of mass points is written correctly?

 dr    A. d  ( r  m )   (r  F(e ) ) dt    dr      B. d  ( r  m  )   ( r  F( e ) )   ( r  F( i ) ) dt      d r   C.  ( r  m  )  d  ( r  F(i ) ) dt    dr    (e ) d D. ( r  m )  ( r    dt    F ) dt    dr    E.  ( r  m )  d  (r  F(e ) ) dt  

 dr    (e)   (i ) (  )  d ( r  F )  ( r r m F.          F ) dt   

79. Angular momentum theorem of the system gives a first integral if: N   A.  ri  Fi ( e )  0 i 1

B.



N

mV i 1

i

i

0

  (e) r  i  Fi  0 N

C.

i 1

27

N

D.

i 1 N

E.

  (i )

r F i

i

0

   (e) N  r  F  r  m W i i i i i 1 N

i 1

  F.  ri  Fi ( J )  0 i 1

80. The first integral of angular momentum theorem of the system is as follows:







A. G  ri  miVi  N  B. G   Fi i 1

   C. G  Fi dri









D. G  miWi E. G  miVi





F. G  G0  const 81. Angular momentum theorem in the motion relative to the center of mass is given by:

A. B. C. D. E.

dG N   (i )   ri  Fi dt  i N1   dG    ri  Fi ( e ) dt i 1  dG N  2   r  F dt  i 1N 2   d G   ri  Fi ( e ) 2 dt i 1  N   dG  r  F dt i 1

28

F.

 d 2G N   r  m V  dt 2 i 1

82. Kinetic energy of a mass point is called:





A. r  F  B. m V 2 C. D. E. F.

mV 2 2   r  mV  d (mV ) dt  mV

83. Work-energy theorem of a mass point in the differential form: A. T  T0  A e  A i





   d  r  mV  r  F dt  mV 2      F  dr C. d   2   d (mV )  F D. dt E. T  T0  A e B.

 dG 0   rF F. dt

84.

  mV 2 mV 02    F  dr – this expression determines the 2 2 M M o

theorem: A. work-energy theorem in differential form 29

B. C. D. E. F.

the theorem on the change of the inertia moment of the body work-energy theorem in the integral form angular momentum theorem linear momentum theorem energy integral

85. Work-energy theorem for the rectilinear motion of a material point:  mV 2  x   F dx A. d  x  2  



B. d m( yz  zy )  momx F

dt

  C. d (r  mV )  r  F dt

  d  ( r  mV )  F dt t t t E. mx  mx o   Fx dt , my  my o   Fy dt , mz  mz o   Fz dt to to to D.

F. m( yz  zy )  momx F , m( zx  xz)  mom y F

86. The first integral of linear momentum theorem of the system is as follows:







N

A. Q  G B. Q 



 mW i 1

 N  C. Q   Fi i 1

 N  D. Q   mV  i1 E. Q  Q0  const F.

 N   Q   ri  miWi i 1

30

87. The theorem of motion of the center of mass of the system is as follows:   A. M VC  F ( e )



    M rc  F ( i )   MWC   F(e )  (e )  Mrc  F   MWC   F(i )

B. MWC  r  F C. D. E. F.



88. The theorem of motion of the center of mass gives a first integral when: A. the sum of the external forces is zero B. Coriolis force is zero C. the sum of the internal forces is zero D. moment of the external forces is zero E. external forces are internal F. the resultant force is zero





89. The value r  m V is called: A. kinetic energy B. moment of inertia of the body C. angular momentum D. Linear momentum E. Coriolis acceleration F. torque 90. Angular momentum for a:

  t  A. mV  mV  Fdt  0 t0  d (mV )  B. F dt



2  C. m d r  F 2

dt

31



2 D. m d r2  0

dt    d  E. r  mV  r  F dt F. T  T0  A e  A i





91. Formulate the angular momentum theorem for a mass point: A. differential of angular momentum of a mass point is equal to the momentum of the elementary forces B. differential of angular momentum of the mass point is equal to the elementary work of force acting on a point C. differential of the angular momentum of the material point is equal to the elementary work D. the time derivative of the angular momentum of the mass point with respect to any center is equal to the moment of force with respect to the same center E. change of angular momentum is equal to the impulse of acting force F. change of angular momentum is equal to the work of acting force 92. The area’s law:

2  A. mom0 (V )  mV

2 2  B. mom0 (V )  d 2r  c dt  C. mom 0 (V )  0

  D. mom0 (V )  dr  c dt





E. mom0 (V )  r  V  2 d  c



 F. mom0 (V )  dS dt

dt

32

93. What ratio connects the velocities of the points A and B of a plane figure if the point P is the instantaneous center of velocity? A. V A  VB BP AP B. V A  V B  AP  BP C. V A  V B  AP  BP

VA VB  AP BP E. V A  ( AP) 2  VB  ( BP) 2 F. V A  АP  V B  ВP D.

94. What relation connects the acceleration of points A and B of a plane figure if Q is the instantaneous center of acceleration?

A.

W A WB  AQ BQ

B. W A  WB BQ AQ C. W A  WB

BQ AQ D. WA  WB ( BQ) 2 ( AQ) 2 E. WA  WB ( AQ) 2 ( BQ) 2

F. W A  АQ  W B  ВQ 95. Which formula determines the relation between the angular velocity of the body  and the number of revolutions per minute n? A.   2 n 15

B.  

30 n

33

C.   2 n D.    n 30

E.   F.  

n 60 n

180

96. Which of the systems of equations is the equation of a spherical rigid body motion? x A  f1 (t ) 

A. y  f (t )  2 A   f 3 (t )  x  f1 (t ) 

B. y  f (t ) 2

z  0    f1 (t )  C.   f 2 (t )   f 3 (t )

x  f1 (t )  D. y  f 2 (t ) z  f 3 (t ) 

E.

s  f 1 (t )   y  f 2 (t )  x  f1 (t ) 

F.   f (t )  2   f 3 (t ) 97. Solution of the fundamental problems of the dynamics can be reduced: A. to finding of first integrals f (t , x, y, z, x, y , z)  c 34

B. C. D. E. F.

to finding the moment of inertia of the body to finding the number of degrees of freedom to the determination of the equilibrium conditions to finding a moment of force to the determination of the linear momentum

98. The basic equation of the dynamics of a mass point has the form: A. x  k 2 x  0  dG 0   B. rF dt   C. m  W  F   D. m W  F







E. G  r  mV F. dT  d A

99. Differential equations of motion of the system of mass points have the form:    A. m iVi  Fi ( e )  Fi ( i )

















B. m iVi  Fi ( e ) C. m iVi  Fi ( i )    D. m i Wi  Fi( e )  Fi( i ) , i  1,..., N



E. miWi  N i  Fi ( i ) F. miWi  Ri( i ) 100. Differential equation of motion of a point in vector form: 2  A. m d r  F 2 dt

  dr B.  mV dt

 C. dG0  r  F dt

35

D. dT  d A

    dr  xi  yj  zk E. dt     d 2r  xi  yj  zk F. 2 dt 101. In which case the linear momentum theorem for a system is written correctly:      A.  d (m V )   F( i ) dr   F( e ) dr 



  B.  d ( m V )   F( e ) dt 









t







t







C. Q  Qo  F( e ) dr  F( i ) dr o

o

  D. d ( m V )   F( e ) dt 

t    (e)  E. Q  Qo   F dr o

102. Which of these values is called the angular momentum of the mass point?





A. r  mV





B. r  F



C. Fdt t1

D.



 Fdt

t0





 

E. m  r  V  sin(r , V ) 103. When does the angular momentum theorem give the first integrals?





A. when r  mV  0 36

B. when only the elastic force is acting on a mass point   C. when r  F  0





D. when r  F  const E. when only central forces is acting 104. What force applied to a mass point is called the central force? A. if the line of action of this force passes through the center when moving B. if the line of action of this force lies in the plane OXY C. if the line of action of this force is parallel to the OZ axis D. if the line of action of this force is parallel to the axis OY E. if the line of its action passes through the attracting point 105. If the motion is considered relative to the fixed coordinate system the angular momentum of the system is defined by the formula: N   A. G0   (ri  miV ) i i 1   B. G0  MVC

   i j k 

C. G 0  m  x

y z

x y z N    D. G0   (ri  Fi ) i 1

  E. G0   miVi i

106. The law of motion of the mass point is given by the equation S  0,5t 2  4t m. At what time the velocity of the mass

37

point will be equal to 10 m/sec? What will be the acceleration of the mass point at that time? A. 5 sec B. 1 m/sec2 C. 6 sec D. 4 sec E. 2 m/sec2 107. Motion of a mass point on a curved path defined by law S  5t m. Define the radius of curvature of the path at the time when the normal acceleration of a point is equal to 5 m/sec2. What will be the total acceleration of the mass point at that time? A. 5 m B. 25 m C. 1 m D. 0 m/sec2 E. 5 m/sec2 108. The velocity of the mass point is given by the equation V2  0,2t m/sec. Determine the curvilinear coordinate of a point S at time t  10 , if t 0  0 , S 0  0 : A. S(t) = 25 m B. S = 10 m C. S(t) = 100 m D. S = 15 m E. S(t) = 10 m 109. The point begins to move from the state of rest in a circle of radius r  200 m with a constant tangential acceleration 2 W  1 m/sec . Determine the total acceleration at the point when t  20 sec: A. w(t) = 2,24 m/sec2 B. w = 1,73 m/sec2 C. w(t) = 1,01 m/sec2 D. w = 1,5 m/sec2 E. w(t) = 2,24 m/sec2 38

110. The angular velocity of the flywheel is given by the law W  p(6t  t 2 ) . Determine the time of stop of the flywheel and its angular acceleration at this point. A. e = 18,8 B. e = -18,8 C. t = 9,42 D. t = 18,8 E. t = 6 111. The basic equation of dynamics of the relative motion of a mass point is:   A. mWr  Fi



    B. mWr   Fi  (  mWe )  (  mWc )    C. mWr   Fi  J e     D. mWr   Fi  J e  J c



E. m W r 



F

i

  Jc

112. Coriolis force of inertia is:   A. J  2m  e Vr sin(e,Vr )

c









B. J  2m  e Vr cos(e,Vr ) c

C. J  m We  Wr sin(We,Wr ) c

 c 

 e r  E. J  m  V c e r 

D. J  2m V

   113. It's given a plane system of forces F1  3i  2j ,    F2  5i  7j . Determine the value of the resultant of these forces R and the cosine angle between R and the positive direction Oy: A. 0,75 B. 1

39

C. 0,47 D. R = 14 E. R = 12,04 114. What factors depends the frequency of free oscillations on?

A. B. C. D. E.

the coefficient characterizing resistance force body weight gravity the coefficient characterizing the restoring force the initial conditions

115. What motion can perform a particle in the environment with resistance under the action of restoring force? A. free oscillations B. forced oscillations C. isochronous oscillations D. damped oscillations E. aperiodic motion 116. What is the ratio between the moments of inertia A, B, C correspond to the case of Kovalevskaya? Where is the center of mass of the body in this case? A. A =B = C B. A =B + C C. A =B =2C D. the center of mass lies in a plane Oxy E. the center of mass lies in a plane Оxz 117. Which of these equations are called as natural equations of motion of a particle? 2

A. m d S   F , 2 i dt B. m dS   F , i dt

m m

V

V

2

 2



 F in  F in

40

C. m dS   F

i

dt 2

D. m d S   F , 2 i dt E. mS   F , i

m

V m 2  F in



V

2



 F in

118. In which case the projection of velocity of mass point on the x-axis of Cartesian coordinate system remains constant: A. Fx  0

B. Fy  0 C. Fx  1 D. Fy  const E. in case when the vector of acting force do not have any projections on x-axis 119. The differential equation of rectilinear motion of the point: 2

A. m d x  F (t, x, x) x 2 dt

B. mV  F (t, x, x) x x C. mx  F (t, x, x) x 2

D. m  x  F (t, x, x)  x 2 t E. mx  F (t , x, x )

x

120. The tangent acceleration of a mass point is W  0,5 m/sec2. What will be the curvilinear coordinate s of a point at a time t  4 sec, if t 0  0 , s0  2 and V0  0 ? A. s = 4 m B. 2 m 41

C. s = 8 m D. 4 m E. s = 6 m 121. The tangent acceleration of a mass point is Wτ  0,2t m/sec2. At what time the velocity of a point becomes equal to 10 m/sec if at time t 0  0 , V0  2 m/sec? A. t = 10 sec B. t = 10,95 sec C. t = 8,94 sec D. t = 3,2 sec E. t = 8,94 sec 122. Which of these variables is the elementary impulse of force?   A. r  F  B. dS  Fdt



t1





C. dS  Fdt t

 0 D. mW  F E.

 Fdt

123. Linear momentum theorem for the mass point:

 d m V ( )  A. F dt B. dT  d ' A     C. mWr   Fi  (mWe )  (mWc )

  D. mWc   F ek   t  E.

mV  mV0   F dt t0

124. Work-energy theorem for linear motion of a mass point: 42

 mV 2   x   F dx x  2   

A. d 

B. d m( yz  zy )  mom F

x

dt

t

t

C. mx  mx0   Fx dt ,

my  my 0   Fy dt ,

t0

t0

D. m ( yz  zy )  mom x F ,

m( zx  xz )  mom y F

t

mz  mz 0   Fz dt t0

E.

mV

2 x1

2



mV x20 2



( M1 )

 F dx x

(M 0 )

125. Angular momentum theorem for a mass point:   t  A. mV  mV0   F dt t0 

B. d (mV )  F

dt 2 d C. m r  F dt 2

  d  (r  mV )  r  F dt    d E. K 0  mom0 F , K 0  r  mV dt

D.

126. The basic equation of dynamics of a mass point is: A. x  k 2 x  0

 dG 0   B. rF dt  C. mW  F







D. G  r  mV

43

  E. m dV  F dt 127. Differential equations of motion of system of mass points are of the form:



 (e)













A. miVi  Fi

  Fi (i )

B. miVi  Fi ( e ) C. miVi  Fi ( i ) D. miWi  Fi ( e ) , i  1,..., N

 dVi  (e)  (i ) E. mi  Fi  Fi , i  1,..., N dt

128. The differential equation of motion of a mass point in a vector form:  2 A. m d r  F dt 2

 dG0   B. rF dt C. dT  d A

    dr D.  x i  y j  zk dt





E. mr  F

129. What case of forced oscillations is called resonance? A. when the frequency of the exciting force p equal to the frequency k of natural oscillations p = k B. p ≠ k C. p = 2k D. p = 3k E. amplitude of oscillations increases with time indefinitely

44

130. The velocity of a point in the Cartesian coordinate sys    tem defined by the equation V  1,5i  1,5tj  0,5t 2 k . What will be the tangent acceleration of a point at a time t  2 sec? A. wτ = 3,5 m/sec2 B. wτ = 2,5 m/sec2 C. 1,87 m/sec2 D. wτ = 4,5 m/sec2 E. 2,5 m/sec2 131. A point moves along a curved path according to the equation s  0,5t  0,2t 2 . At time t  3 sec the radius of curvature of the trajectory is 1,5 m. Determine the total acceleration of the point at this moment: A. 2,41 m/sec2 B. w = 2,09 m/sec2 C. w = 1,55 m/sec2 D. 5,7 m/sec2 E. 1,55 m/sec2 132. The car moves along the road at a velocity of 90 km/h. Define the radius of curvature of the road when the normal acceleration of a car is W n  25 m/sec2: A. ρ = 25 m B. ρ = 324 m C. 625 m D. 324 m E. 25 m 133. The motion of the mass point is given by the equation s  5t  0,4t 2 . At what time its normal acceleration is zero? A. 0 sec B. t = 6,25 sec C. 2 sec D. t = 1,25 sec E. 6,25 sec

45

134. Let the normal acceleration is of the mass point

W n  25 m/sec2 and tangential acceleration is W  1,5 m/sec2.

Determine the total acceleration W of the mass point: A. 0 B. W = 4,01 m/sec2 C. 4,01 m/sec2 D. 2,915 m/sec2 E. W = 3,2 m/sec2 F. W = 2,915 m/sec2 135. It's given the equation of motion of a point x  3t , y  t 2 . Determine the distance S of a point from the origin at a time t  2 sec: A. 5,5 B. S = 10,5 C. S = 2,5 D. 7,21 E. 10,5 F. S = 7,21 136. It's given the equations of motion of a mass point x  cos t , x  2 sin t . Determine the distance S of a point from the origin at a time t  2 sec: A. S = 1 B. S = 2,5 C. 4,5 D. S = 5,05 E. 2,5 F. 1 137. The motion of the point is given by x  3t 2 , y  4t 2 . What is the acceleration of the point (m/sec2)? A. 5 m/sec2 B. 4 C. 3 m/sec2 D. 10 E. 5 F. 10 m/sec2 46

138. The projection of the velocity of the moving point on the axis of coordinates is V x  3t 2 , V y  4t 2 . What is the total acceleration of the point at a time t  1 sec? A. 12 B. 7 C. 10 m/sec2 D. 7 m/sec2 E. 1 m/sec2 F. 10 139. The projection of velocity of a mass point is given in the form V x  2 cos t . Determine the coordinate x of the mass point at time t  1 sec, if t 0  0 , x 0  0 : A. 2 B. х = 0 C. 10,5 D. 12,5 E. х = 2 F. 0 140. The point moves according to the equation x  cos t , y  sin t . Determine the acceleration of a point at time t  1 sec: A. w = 9,86 m/sec2 B. 3,14 m/sec2 C. 8,1 m/sec2 D. 0 m/sec2 E. w = 3,14 m/sec2 F. 9,86 m/sec2 141. The rotation of the body around a fixed axis is given by the equation   3t 2 . What is the angular velocity of the body at time t  1 sec? A. 6 B. 1,5 rad/sec C. 2 47

D. 1 rad/sec E. 1,5 F. 6 rad/sec 142. The angular velocity of the body rotating around a fixed axis is given by the equation ω  12  3t 2 rad/sec. At what time the body will stop? A. 1 B. t = 4 sec C. t = 3 sec D. 2 E. t = 1 sec F. t = 2 sec 143. The rotation of the body around a fixed axis is defined by the equation   6t 2 . What is the angular acceleration of the body? A. 12 B. 1 rad/sec2 C. 3 rad/sec2 D. 36 E. 36 rad/sec2 F. 12 rad/sec2 144. The angular velocity of the body varies according to the law ω  2  8t 2 rad/sec. Determine the time t of stop of the body: A. t = 1 sec B. t = 1,5 sec C. t = 0,5 sec D. 1 sec E. 4 sec F. 0,5 sec 145. Under the influence of this force the mass point performs the free rectilinear harmonic oscillations: A. gravity B. quasi-elastic force 48

C. D. E. F.

frictional force disturbing force resistance force elastic force

146. The work of force on any finite displacement is calculated as follows: t1





A. S   F dt t0

B. N  F  V C. A 

M1

 F ds

M0

  d ( mV )   Fk dt    E. G  r  mV F. A   ( Fx dx  Fy dy  Fz dz ) D.

M o M

147. What is equal the work of potential force acting on a point when it is moving in the closed loop to? A. arbitrary constant B.  mgh

C. 

c 2

D. zero E. M F. 0 148. When mass point moves under the influence of potential forces: A. full mechanical energy is equal to the momentum of forces B. total mechanical energy remains constant C. the total mechanical energy changes over time D. total mechanical energy depends on the distance traveled 49

E. Full mechanical energy depends on the type of motion F. the sum of kinetic and potential energies in each position of mass point remains constant 149. Energy integral: mV 2 A.   ( x, y , z )  h 2 2 2 B. mV  mV0  ( x, y, z )

2 2 2 2   mV mV 0 C.    F  dr 2 2 M M o

  D. r  mV  const  E. mV  const mV 2  U ( x, y , z )  h F. 2

150. In which case the work-energy theorem of a system of mass points is written correctly?   F( e ) dt  F( i ) dt A. dT 

 

 

t

t  (e)  B. T  T0    F dt    F( i ) dt



 t0

t0

t t   mV 2 mV02     F( e ) dt    F( i ) dt 2 2  t0  t0  (e) D. dT   F dt

C.











E. dT   F( e ) dr   F(i ) dr 

F. T1  T0  A

(e)

A

 (i )

151. Work-energy theorem for the mechanical system in differential form is: 50

A. dT 

 (i )

F i

B. dT 

 (e)

F i

C. dT 

dri

 (e)

F

i

i

dri

 dri   Fi ( i ) dri

  D. dT   Fi ( e )  Vi i

E. dT 

i

 (i )   Vi i

F i

F. dT  d ' A( e )  d ' A( i ) 152. The energy integral of a system of mass points is as follows: A. T   ( e )   ( i )  const B. T  const C. T  

1 mV 2 2  E. T  mV F. E  T   ( e )   ( i )  const

D. T 

153. The body of mass m  25 kg suspended on a rope climbs vertically with an acceleration W  0,3 m/sec. Determine the cable tension force: A. Т = 237,5 B. Т = 7,5 C. 252,5 D. Т = 0,77 E. 237,5 F. 252,5 N 154. Mass point of mass 1,8 kg is moving along a straight line according to the law x  9t 2  7t  2 . Determine the module of the resultant force applied to the point: 51

A. B. C. D. E. F.

32,4 18 19,8 18 N 7,94 N 32,4 N

155. The body with weight of 2 kg is moving in a straight line according to the law x  10sin2t under the influence of some  force F . Find the greatest value of this force: A. F = 40 N B. F = 80 N C. 40 N D. 60 N E. F = 50 N F. 80 N 156. The body moves down the inclined rough plane which forms an angle of 30 degrees with the horizon. Determine the acceleration of the body if the coefficient of friction is f  0,3 : A. 12,05 m/sec2 B. 7,02 m/sec2 C. 2,35 D. 10,3 m/sec2 E. 12,05 F. 2,35 m/sec2 157. Mass point with mass of 2 kg moves in space under the     action of force F  2i  3j  5k . Determine the module of aceleration of the point: A. 1,58 B. 20 C. 1,58 m/sec2 D. 3,08 m/sec2 E. 7,84 m/sec2 F. 3,08 52

158. The mass point of mass m  5 kg started to move from the state of rest along a horizontal line by the action of force F  10t which is directed along the same straight line. Determine the path that is passed by a point at 7 sec: A. 114,3 B. S = 74,6 m C. S = 49 m D. 245 m E. 74,6 F. S = 114,3 159. Mass point of mass 0,5 kg is moving along a straight line. Identify the module of impulse of resultant of all forces acting on the point for the first 2 sec if it moves according to the law S  4t 3 : A. 32 B. 16 C. 32 kg m/sec D. 24 E. 18 kg m/sec F. 24 kg m/sec 160. The magnitude of linear momentum of mechanical system varies as Q  4t 2 . Determine the module of principal vector of the external forces acting on the system at time t  5 sec, if   Q and F are parallel: A. 64 N B. 40 N C. 16 D. 64 E. 8 F. 40 161. A ball with the weight of 2 kg slides down from a height of 3 m along the inclined surface with length of 5 m. The angle of the surface with the horizon is 30 degrees. What is the work done by the force of gravity? 53

A. B. C. D. E. F.

84,87 J 98 58,8 J 29,36 J 98 J 58,8

162. Ball with the weight of 2 kg is rolling on a smooth horizontal surface. What is the work done by the force of gravity when the ball moved at the distance of 5 m? A. 0 B. 58,8 C. 84,87 J D. 0 J E. 58,8 J F. 24,7 163. Ball with the weight of 0,2 kg suspended on a nonstretchable thread of 2 m long starts to move from the lower position and makes a complete rotation every 5 sec. What is the work done by the force of gravity? A. 246 J B. 78,4 J C. 0 D. 25,12 E. 0 J F. 25,12 J 164. Identify the module of resultant of two converging forces F1  F2  5N forming an angle α = 135°: A. 7,07 B. 3,54 N C. 5,62 D. 7,07 N E. 9,24 F. 9,24 N 54

165. It's given the plane system of converging forces       F1  3i  4j , F2  5j , F3  3i . Find the magnitude of the resul tant force R : A. 10,8 B. 12 C. 5,6 N D. 12 N E. 0 N F. 10,8 N

 166. It's given the projections of two converging forces F1 and F2 on the Cartesian coordinate system F1x  3N, F1y  6N , F2x  5N, F2y  4N . Find the magnitude of the resultant



force R : A. 14 B. 14 N C. 12,8 D. 17,24 N E. 20 F. 12,8 N 167. Ball with the mass of 2 kg slides down from a height of 3 m along the inclined surface of 5 m. The angle of the surface with the horizon is 30 degrees. What is the work done by the force of gravity? A. 84,87 B. 49 J C. 58,8 J D. 29,36 J E. 49 F. 58,8 G. 0   168. Modules of two converging forces F1 and F2 are respectively 3N and 4N , the angle between them is 90 degrees. De termine the module of resultant force R : 55

A. B. C. D. E. F. G.

R=5N R=6N 8N R = 12 N 6 0 5

169. The resultant of plane system of converging forces      F1 , F2 , F3 and F4 is zero. The projections of the forces F2 ,   F3 and F4 on the Cartesian coordinate system are F2x  4N, F2y  7N, F3x  5N, F3y  5N, F4x  2N, F4y  0 N .



Determine the amount of force F1 : A. F1 = 3,61 B. 0 C. 4,42 D. F1 = -7 N E. 4,42 N F. F1 = 3,61 N G. 3,0



170. It's given the projection of force F on a Cartesian coordinate system F1x  F1y  210N , coordinates of the point of application of this force are x  y  0,1 m. Determine the magnitude of the moment of this force relative to the origin: A. 9 Nm B. 1 C. M0 = 0 D. 15 E. M0 = 9 Nm F. 12 G. M0 = 0 Nm    171. The projection of the resultant R of two forces F1 and F2  on the Ox axis is R x  5N , and the projection of force F1 on the sa me axis is F1x  7N . Find the projection of force F2 on the Ox axis: 56

A. B. C. D. E. F. G.

F2x F2x F2x F2x F2x F2x F2x

= 12 N = -1 N = -2 N =0 =2N = 12 = -2

   172. The system of converging forces F1 , F2 and F3 is in a   state of equilibrium. The values of forces F1 and F2 are 3N and 2N respectively. The angle between the direction of force  F1 and the positive direction of the Ox axis is 15 degrees, and  the angle between the force F2 and the axis Ox is equal to  45 degrees. Find the amount of force F3 : A. 4,84 B. 1,61 N C. 7,83 N D. 1,61 E. 2,6 N F. 4,84 N G. 0

173. The street lamp is suspended in the point B to the middle of the cable ABC fixed with its ends to the hooks A and C that are on one horizontal. Define the strains Т1 and Т2 in the parts AB and BC of cable if the weight of the lamp P = 150 N, the length of the whole cable ABC equals to 20 m and the deviation of the point of the suspension from the horizontal BD equals to 0,1 m. A. Т1 = 7,5 kN

57

B. C. D. E. F. G.

T2 = 6 kN T2 = 7,5 kN T1 = 5 kN T1 = T2 = 7 kN T1 = T2 = 6 kN T1 = T2 = 0 kN

174. Force with the moment M = 6 kNm acts on the horizontal cantilever. Vertical load P = 2 kN acting at the point C. The length of the beam AB = 3,5 m, BC = 0,5 m. Determine the reactions of the supports. A. RA = 2 kN B. RA = 1 kN C. RA = 0 kN D. RB = 4 kN E. RB = 5 kN F. RB = 1 kN G. RA = 7 kN 175. The boiler with the weight P = 40 kN that is uniformly distributed along the length and radius R = 1 m lies on the hills of stonework. The distance between walls of the stonework l = 1,6 m. Find the pressure of the boiler on the stonework in points A and B. A. NA = 33,3 kN B. NA = 11 kN C. NA = NB = 0 kN D. NB = 33,3 kN E. NB = 10 kN F. NA = NB = 30 kN G. NA = NB = 22 kN

58

176. The work of the force on any finite displacement is calculated by the formula: t1





A. A  Fdt t0

B. A  F  V C. A 

M1

 F ds

M0

 

  D. d mV   Fk dt

E. A   ( Fx dx  Fy dy Fz dz ) F.

 S  А r m V    



G. A  F  V

177. The expression Fx dx  F y dy  Fz dz is:

A. necessary and sufficient condition for the potentiality of the force field B. the value of the elementary work of the force C. power value D. kinetic energy of the point E. force work on final displacement F. the value of the elementary work of the force applied to the material point on an infinitesimal displacement G. impulse value 178. The necessary and sufficient condition for the potentiality of the force field:

A.

Fy Fx Fz Fy Fx Fz   0,   0,  0 y z z x x y

B. U   (Fx dx  Fy dy  Fz dz)  const M1

C.

 F  ds  0

M0

59

    E. F   const  F. rot ( gradF )  0   G. F   1 D. F  dr  0

179. The formula defining the gyroscopic moment: A. M   C1  2 sin 







B. M  K 0  2    C. M   2  C1 D. M   H1  2 sin  E. M  C1  2 sin  

F. M   2C1  2 G. M   K1  2 180. Euler’s angles: A. precession B. nutation C. yaw D. inclination E. roll F. angle of instantaneous rotation G. pitch 181. Find the right method for specifying motion:









A. vector method r  x(t )i  y (t ) j  z (t )k

s  s (t )   C. natural method r  r (t )

B. natural method

D. natural method x  x ( t ), y  y ( t ), z  z ( t ) E. vector method s  s( t ), y  y( t )

60







F. vector method r  x ( t ) i  y( t ) j  s( t ) G. coordinate method r  x ( t )i  y ( t ) j  z ( t ) k 182. Indicate centrifugal moments of inertia: A. J xx   m i ( y i2  z i2 ) B. J yz   m i y i z i

m z x   m (x  y  z )   m (z  x )   m (x  y )  m x ;J  m y

C. J zx 

i i i

D. J O

i

2 i

2 i

i

2 i

2 i

i

2 i

2 i

2 i i

( zx )

E. J yy F. J zz

G. J ( yz )

2 i

i

2 i

; J ( xy )   mi zi2

183. Indicate the moments of inertia relative to the planes: A. J yz   m i y i z i , J zx   m i z i x i

m x  m z  m y z ,

B. J ( yz ) 

2 i i

C. J ( xy )

2 i i

D. J yz

i

E. J xx  

J zx   mi z i xi ,

i i

m i ( y i2

z i2 ),

m i ( z i2

J yy        F.  m i ri   m i x i i  m i y i j   m i z i k

J xy   mi xi yi  x i2 ),

J zz   m i ( x i2  y i2 ) G. J O 

 m (x i

2 i

 yi2  zi2 )

184. The integrals of the Lagrange equation of the second kind: A. T2  TO    const B. T    const C. ( p 2  q 2  n 1 ) 2  const  k D. (2 pq  n 2 ) 2  const  k 61

E. p 2  q 2  n 1  2 pq  n 2  h F. Ap 2  Bq 2  n 1  2 pq  n 2  k G. T  const

185. Which of these is circular trajectory? A. ( x  5) 2  y 2  25 B. ( x  1) 2  ( y  2) 2  9 C. y  4 x 2 D. x  y 2  16 E. x  5 y  0 F. x 2  y 2  16 G. y 2  25

186. Law of motion of a point: A. S   Vx2  V y2  Vz2 dt

 S

B. S 

x 2  y 2  z 2 dt

C.

dx  dy  dz dt

D. S  dx  dy  dz E. S 



x 2  y 2  z 2 dt

F. S  dx 2  dy 2  dz 2 G. S  dx  dy  dz

187. Projections of acceleration under the natural method of specifying the motion of a point: A. Wn  B. W 

V2

 dS dt 62

dr dt dV D. W  dt V2 E. W 

C. W n 



F. W n  0 G. W  0

188. For rectilinear uniform motion: A. W  0 B.    C. W  W n  W D. W 

V2



E. V  V0  V F. W  W n G.   0

189. For a rectilinear uniformly accelerated motion: A. V  V0  Wt B. S  S 0  V0 t  C. W  W n  W D. Wb  E. W 

1 2 Wt 2

V2

 V2



F. W  W n G. W  Wn  0 63

190. The normal acceleration is determined by the formula: A. Wn   2 r B. W n 

W  2



dV C. W n  dt V2 D. Wb 

 E. Wn  V 2 

F. W  W n G. Wn  r

191. The basic equation of the dynamics of a mass point has the form:    A. G  r  mV





B. mW  F C. x  k 2 x  0 D. E. F. G.

 dG 0   rF dt   mV  F   mr  F   mr  F

192. The period of damped oscillations is greater than the period: A. T1  T  2 , n 2  c n m B. of free oscillations C. of forced oscillations D. of complex oscillations E. of harmonic oscillations 64

F. of damped oscillations 2 c G. of T1  T  , n2  n m

193. The action of forces is determined by: A. direction B. the application point C. arm D. moment E. resultant force F. angular momentum G. momentum 194. The angular momentum is determined by:    A. K 0  r  mV  i

 k

 j

B. K  x 0

y mx my    C. K 0  r  mV 2

z mz







     r  mV    2(mr  r )



D. K 0  mV  mr

E. K 0  2(mr  r )

  G. K 0

F. K 0

195. Work of gravity force: A. A( M 0 M 1 )  mgh B. A( M 0 M1 ) 



( M1 )

(M0 )

( cx )dx

с 2 2 ( x0  x1 ) 2  Ph where P is gravity force

C. A( M 0 M1 )  D. A( M 0 M 1 )

65

E. A( M 0 M1 )   F fr s F. A( M 0 M1 )  F fr s 2

2

G. A( M 0 M1 )  c( x0  x1 )

196. Work of elastic force: A. A( M 0 M1 ) 



( M1 )

(M0 )

( cx )dx

с 2 2 ( x0  x1 ) 2  mgh

B. A( M 0 M1 )  C. A( M 0 M 1 )

D. A( M 0 M1 )   F fr s

dr (M0 ) r 2   F fr s

E. A( M 0 M 1 )  km F. A( M 0 M1 )



( M1 )

G. A( M 0 M 1 )  Ph

197. Work of frictional force: A. A( M 0 M1 )   fNs B. A( M 0 M 1 )   Ph C. A( M 0 M1 ) 



( M1 )

(M0 )

( cx )dx

с 2 2 ( x0  x1 ) 2   F fr s

D. A( M 0 M1 )  E. A( M 0 M1 )

F. A( M 0 M 1 )  Ph G. A( M 0 M 1 )  mgh

198. Work of gravitational force: A. A( M 0 M 1 )  km

dr (M0 ) r 2



( M1 )

66

B. A( M 0 M1 )  Ph

1 1    r1 r0 

C. A( M 0 M1 )  mgR 2  D. A( M 0 M1 ) 



( M1 )

(M0 )

( cx )dx

с 2 2 ( x0  x1 ) 2   F fr s

E. A( M 0 M1 )  F. A( M 0 M1 )

G. A( M 0 M 1 )  mgh

199. Power is determined by:

dA dt  dr N  dF dt N  Fx x  Fy y  Fz z  dN  dF N  mgR 2  N  dF  dr N   dF dt

A. N  B. C. D. E. F. G.

200. Differential equations for the plane motion of a rigid body: A. mxC  X ie B. C. D. E.

 J    m mx   X my   Y J    m C

C

C

( Fi e )

e i

e

C

C

i

C

( Fi i Fi e ) 67

Z    ( F

F. mzC  G. J C

e i

i

i

Fi e )

201. Velocity, acceleration and the law of motion for rectilinear uniformly accelerated motion: A. W  const B. V  V0  Wt C. Wb 

V2

 1 2

D. S  S 0  V0t  Wt E. W 

2

V2



202. Angular momentum is determined by:    A. K 0  r  mV







B. K 0  ( r  mV ) / 2







C. K 0  r  mr



D. K 0



E. K 0

   i j k y z  x mx my mz    m V  mr

203. Moments of inertia relative to the planes: A. J yz   m i y i z i , J zx   m i z i x i B. J ( yz)   m i x i2 C. J ( zx )   m i y i2 68

D. J xx   m i ( y i2  z i2 ), J yy   m i (z i2  x i2 ), J zz   m i ( x i2  y i2 ) E. J ( xy )   m i z i2

204. Expression for the acceleration modulus of a point when specifying the motion by the coordinate method: A.

x 2  y 2  z 2

B.

2 V 2 V 2 V x y z

C.

W x2  W y2  Wz2

D.

x2  y2  z2

E.

V2



205. The mass point with mass 1,8 kg moves rectilinearly according to the law x  9t 2  7t  2 . Determine the modulus of the resultant forces applied to the point, the acceleration and the law of change of the point: A. 18 m/sec2 B. x  18t  7 C. 32,4 N D. 16,2 N E. 19,8 m/sec2 206. A body with mass 2 kg moves rectilinearly according to  the law x  10sin2t under the action of some force F . Determine the laws of change of velocity and acceleration. Find the greatest value of this force: A. 80 N B. x  20 cos 2t C. 50 N D. x  40 sin 2t E. x  40 sin 2t 69

207. The body moves down an inclined rough surface which forms an angle 30o with the horizon. Write differential equations of motion in coordinate form. Determine the acceleration of the body if the coefficient of sliding friction f  0,3 : A. mWx  F  G sin 30 B. 7,25 m/sec2 C. mWy  N  G cos 30 D. 2,35 m/sec2 E. 12,05 m/sec2

208. A mass point  ofmass  2 kg moves in space under the  action of force F  2i  3 j  5k . Determine the modulus of aceleration of the point and the modulus of force: A. 12,33 m/sec2 B. 3,08 m/sec2 C. 6,16 N D. 3080 mm/sec2 E. 6 N 209. The forces F1  4 kN and F2 act on the beam BC fixed in hinge A. Determine the force F2 and the reaction of the support A if AC  2 m, AB  6 m: A. F2 = 12 kN B. XA = 0 kN C. XA = 4 kN D. YA = 16 kN E. YA = 8 kN 210. On the beam АВ act forces F1  1 kN, F2  2 kN, F3  3 kN. Determine the reactions on the supports A and B if AC  CD  DE  1 m, BE  2 m: A. XA = 0 kN B. XA = 2 kN 70

C. YA = 0,8 kN D. RB = 2 kN E. RB = 1,2 kN

211. On the beam АВ act force F  5 kN and distributed load q  4 kN/m. Determine the reactions on the supports C and B if AC  3 m, BC  6 m: A. XВ = 0 kN B. XВ = 12 kN C. YВ = 2 kN D. RС = 15 kN E. RС = 17 kN 212. On the beam АВ with weight G  20 kN act distributed load q  0,5 kN/m. Determine the reactions on the supports A and B if AB  BC : A. XА = 0 kN B. YА = 10,4 kN C. YА = 21,5 kN D. YА = 1,5 kN E. RВ = 11,1 kN 213. On the beam АВ act force kN and distributed load F 9 q  3 kN/m. Determine the reactions on the supports A and B if AB  5 m, BC  2 m: A. XА = 0 kN B. XА = 1,5 kN C. YА = 4,8 kN D. YА = 15 kN E. RВ = 10,2 kN

71

214. Determine the reactions on the supports A and B if distributed load q  40 N/m, AB  4 m, BC  2 m: A. XА = 0 N B. YА = -20 N C. YА = -100 N D. RВ = 100 N E. RВ = 80 N 215. On the beam АВ act distributed load q  5 Н/m. Determine the reactions on the supports A and B if AB  9 m, AC  6 m: A. XВ = 0 kN B. YВ = 10 kN C. RА = 20 kN D. RА = 40 kN E. RА = -20 kN 216. On the cantilever beam АВ act a force F  4 N and a pair of forces with a moment M  2 Nm. Determine the reactions and moment in the seal if AB  4 m: A. XА = 0 kN B. XА = 2 kN C. YА = 4 kN D. МА = 14 kNm E. МА = 2 kNm 217. Determine the coordinates of the center of gravity of a circular homogeneous cone if the radius of the base r  0,4 m, the height of the cone h  0,8 m: A. xc = 0 B. xc = 0,4 72

C. yc = 0 D. zc = 0,5 E. zc = 0,4

218. Determine the coordinates of the center of gravity of a homogeneous body consisting of a cone and a cylinder if the height H 1  2 H  0,4 m: A. xc = 0 B. xc = 0,2 C. yc = 0 D. zc = 0,18 E. zc = 0,21 219. Determine the coordinates of the center of gravity of a homogeneous body consisting of a rectangular parallelepiped and a pyramid if the height H 1  3 H  1,2 m, the side of the base is 0,4 m: A. xc = 0,2 B. yc = 0,2 C. yc = 0 D. zc = 0,45 E. zc = 0,4 220. On the beam F  6 N and distributed Determine the reactions A and B if AC  AB / 3  1 A. XА = 0 N B. XА = -4,2 N C. YА = 4,13 N D. RB = 3,8 N E. RB = 4,07 N

AB act force load q  2 N/m. on the supports m, α = 45o:

73

221. On the beam AB act pairs of forces with moments M 1  2 kN·m and M 2  8 kN·m. Determine the reactions on the supports A and B if l  3 m. A. XА = 2 kN B. XА = 0 kN C. YА = -2 kN D. YА = 2 kN E. RB = 2 kN 222. Determine the reactions on the supports А and D if F1  84,6 N, F2  208 N, AB  1 m, BC  3 m, CD  2 m. A. XА = 59 N B. XА = -45 N C. YА = 108 N D. RD = 128 N E. RD = 108 N 223. The moment of the pair arising in the seal M A  200 Нm, length l  1 m. Determine the intensity of the distributed load q and the reaction in the seal: A. XА = 400 N B. YА = 400 N C. YА = 0 N D. q = 400 N/m E. q = -400 N/m 224. The equilibrium conditions for a plane system of forces:   A.  Fk  0,  M O ( Fk )  0

B. C.



 0,

F

 0,

F

k

kx

M

x

F

 ( Fk )  0

ky

 0,

M 74

O

 ( Fk )  0

D. E.

F M

  0,  M O ( Fk )  0   A ( Fk )  0,  M B ( Fk )  0,  Fkx  0

kx

225. Friction of rolling –…, M fr  f k N – frictional moment,

f k – coefficient of friction; has the dimension of length: A. resistance B. friction C. occurs when one body rolls over the surface of another D. appears at the deformation of the roller and the plane at the point of displacement of the normal reaction in the opposite direction of motion E. appears when deformation of the roller and the plane at the point of their contact and the displacement of the normal reaction towards possible motion 226. The angular velocity of the body varies according to the law   8t . At time t 0  0 the rotation angle of a body was 0  5 rad. Find the angle of rotation, angular acceleration, and angular velocity of the body at t  3 sec. A. φ = -41 B. φ = -31 C. ε = -8 D. ε = 8 E. n = -5 227. The rotor has 100 rotations made in the first 5 seconds. Find the angle of rotation, angular velocity and angular aceleration of the rotor. A.  = 500 B.  = 628 C.  = 125,6 D.  = 50,3 E.  = 0 228. Indicate the cases in which the Coriolis acceleration is zero: 75

A. Vr  0 B.  e  Vr C.  e // Vr D.  e  0 E.  e  Vr 229. The body is rotating by the law   t 3  2 . Determine the angular velocity, angular acceleration and the time of stop of the body. A.  = 12 B.  = -12 C.  = 12 D.  = 10 E. t = 0 230. The body rotates according to the law   2 cos t 2 . Determine the rotation angle, angular velocity and angular aceleration of the body at time t  2 sec. A.  = 1981 B.  = 25,1 C.  = 0 D.  = 78,8 E.  = 6,28 231. The body rotates under the law   t 2 . Determine the angular acceleration of the body, velocity and acceleration of the points of the body on the distance 0,5 m from the axis of rotation when   25 rad. A.  = 2 B.  = -2 C. V = 5 D. V = 10 E. W = 1 76

232. The velocity of points of the body located at a distance of 0,2 m from the axis of rotation varies in accordance with the law V  4t 2 . Determine the angular velocity, angular aceleration of the body and acceleration of the points of the body at time t  2 sec. A.  = 0 B.  = 80 C.  = 80 D. W = 16 E. W = 8 233. Normal acceleration of the point M is 6,4 m/sec2. Determine the velocity of the point M, the angular velocity and angular acceleration of the disc with the radius R  0,4 m. A.  = 0 B.  = 4 C.  = 4 D. V = 0,4 E. V = 1,6 234. The body rotates by the law   2t 3 . Determine the angular acceleration of the body, the tangent and normal aceleration of the point of the body at a distance r  0,2 m from the axis of rotation at a time t  2 sec: A.  = 6 B.  = 3 C. W = 2,4 D. W = 4,8 E. Wn = 24 F. Wn = 115 235. The body rotates by the law   1  4t . Determine the angular acceleration of the body, the tangent and normal acele-

77

ration of the points of the body at a distance r  0,2 m from the axis of rotation: A.  = 4 B.  = 0 C. W = 0 D. W = 5 E. Wn = 3,2 F. Wn = 0 236. Indicate the formula for the calculation of the work of force:

 t1  A. S   Fdt t0

 

B. N  F  V C. A 



M1

 F ds

M0





D. G  r  mV E. Fx dx  Fy dy  Fz dz F.

  F  dr

237. The load moves up by the law S  7  5 t 2 . Determine the velocity of the load,

the angular velocity and angular acceleration of the barrel of 50 cm diameter at time t  3 sec. A. V = 37 B. V = 30 C.  = 1,2 D.  = 1,5 E.  = 0,6 F.  = 0,4

78

238. Cogwheel 1 is rotated with angular acceleration 2  1  4 rad/sec . Determine the angular velocity of 1, 2 wheels and velocity of the point M at time t  2 sec. R1  0,4 m, R3  0,5 m. an

A. B. C. D. E. F.

1 = 8 2 = 8 2 = 2 VM = 3,2 VM = 6,4 2 = 0,2

239. Rigid body performs a plane-parallel motion according to the equations x A  2t 2 , y A  0,2 ,   10t 2 . Determine the velocity, the angular velocity and acceleration of the body at the time t  1 sec. A. V = 4 B. V = 10 C.  = 20 D.  = 0 E.  = 20 F.  = 0 240. The velocity of the wheel at the point A is V A  10 m/sec. Determine the velocity, the angular velocity and acceleration of the wheel with the radius r  0,2 m. A. V = 50 B. V = 20 C.  = 100 D.  = 20 E. W = 1000 F. W = 500

79

V A  10 m/sec, V B  15 m/sec, AP  60 cm. Determine the distance BP, angular velocities of points A and B. A. ВР = 2,5 B. ВР = 90 C. А = 6 D. А = 0,17 E. В = 0,17 F. В = 0

241.

242. Determine the velocity of the point B and the angular velocities of the points A and B, if V A  2 m/sec and R  1 m. A. VВ = 2 B. VВ = 1,41 C. А = 0,5 D. А = 2 E. В = 0 F. В = 2 243. The body moves according to the equation x B  2cos0,5pt , y B  0,5t ,   0,5pt . Identify the components of the vector of velocity and the angular velocity of the body. A. Vx = 0 B. Vx = 3,14 C. Vy = 0 D. Vy = 0,5 E. ω = 1,57 F. ω = 6,28 244. Addition of rotations around two parallel axes: A. rotation in one direction  = 2 + 1 B. rotation in one direction  = 1 C. rotations directed to different sides  = 2-1 D. rotations directed to different sides  = -1

80





E. a couple of rotations 2  1   F. a couple of rotations 2  1 245. The law of motion of free rigid body, velocity and acceleration of rigid body in plane-parallel motion: A. x A  f 1 ( t ), y A  f 2 ( t ), z A  f 3 ( t ),   f 4 ( t ),   f 5 ( t ),

  f 6 (t) B. x A  f 1 (t ), y A  f 2 (t ), z A  f 3 (t ),   f 4 (t )



 

C. V    r     D. V  V A    r

        a вр  a A    (  r )    r      вр  F. a  a A  a  a A    r 



 ос

E. a  a A  a

246. The complex motion of the body: A. body is involved in two movements B. body simultaneously involved in several movements C. is the absolute motion D. is not absolute motion E. motion of body is considered relative to several reference frames F. relative motion 247. The point of mass m  4 kg moves in a horizontal straight line with acceleration a  0,3t . Determine the force, velocity and path of the point at time t  3 sec. A. S = 4,05 m B. S = 8,1 m C. V = 0,3 m/sec D. V = 2,7 m/sec E. F = 3,6 N F. F = -3,6 N

81

248. The point of mass m  14 kg moves with acceleration a  lnt . Determine the force, velocity and path of the point at time t  5 sec. A. S = 40,25 m B. S = 20,1 m C. V = 0,2 m/sec D. V = 8,05 m/sec E. F = 22,5 N F. F = 159,7 N 249. Body with the weight m  50 kg rises with acceleration a  0,5 m/sec2. Determine the force, velocity and distance of the body traveled in the first 10 sec. A. S = 50 m B. S = 25 m C. V = 20 m/sec D. V = 5 m/sec E. F = 2,5 N F. F = 487,5 N 250. The mass point of mass m  12 kg moves in a straight line at a velocity V 2  e 0 ,1 t . Determine the force, acceleration and the distance traveled in time t  50 sec. A. S = 3 m B. S = 1484 m C. V = 3 m/sec D. V = 14,8 m/sec E. F = 60,4 N F. F = 178 N 251. Identify module of resultant force, velocity and aceleration at time t  6 sec. If a mass point of mass m  3 kg moves along Ox according to the equation x  0,04t 3 . A. V = 4,32 m/sec B. V = 0,72 m/sec C. a = 1,44 m/sec2 82

D. a = 0,48 m/sec2 E. F = 4,32 N F. F = 33,72 N 252. The mass point of mass m  1,4 kg is moving according to the law x  6t 2  6t  3 . Identify the module of resultant force, velocity and acceleration of the point at time t  1 sec. A. V = 0 m/sec B. V = 18 m/sec C. a = 0 m/sec2 D. a = 12 m/sec2 E. F = 16,8 N F. F = 30,52 N 253. Mass point of mass m  10 kg is moving according to the law x  5sin0,2t , y  5cos0,2t . Identify module of resultant force, velocity and acceleration of point at time t  7 sec. A. V = 1 m/sec B. V = 7,07 m/sec C. a = 0 m/sec2 D. a = 0,28 m/sec2 E. F = 2,8 N F. F = 100,8 N 254. The material point of mass m  1 kg moves on a circle of radius r  2 m with a velocity V  2t . Identify module of resultant forces, tangential and normal accelerations of the point at time t  1 sec. A. an = 1 m/sec B. an = 2 m/sec C. aτ = 0 m/sec2 D. aτ = 2 m/sec2 E. F = 2,83 N F. F = 12,63 N 255. The mass point of mass m  22 kg moves on a circle of radius R  10 m according to the equation S  0,3t 2 . Identify 83

module of resultant force, velocity and acceleration of the point at time t  5 sec. A. V = 30 m/sec B. V = 3 m/sec C. a = 0,9 m/sec2 D. a = 1,08 m/sec2 E. F = 19,8 N F. F = 23,8 N 256. The mass point of mass m  9 kg moves in space under     the action of force F  5i  6j  7k . Identify module acceleration, velocity and travelled path of the point. A. S = 0,58t2 m B. S = 1,17t2 m C. V = 1,17t m/sec D. V = 1,17 m/sec E. a = 1,17 m/sec2 F. a = 0,67 m/sec2 257. The mass point M with mass m  6 kg moves along a curve path by the force F  8 N. Determine the tangent, normal accelerations and velocity of the point M after 10 sec. A. V = 8,57 m/sec B. V = 13,3 m/sec C. an = 0,854 m/sec2 D. an = 1,02 m/sec2 E. at = 1,33 m/sec2 F. at = 0,857 m/sec2 258. The body moves along the horizontal surface and comes off from it at the point A. Determine the velocity, acceleration, and normal acceleration of the body at the moment of come off if R  6 m. A. V = 7,75 m/sec B. V = 7,67 m/sec 84

C. D. E. F.

an = 9,8 m/sec2 an = 0 m/sec2 a = 9,8 m/sec2 a = 10 m/sec2

259. Damped oscillations of a mass point are described by the equation x  2e 0,2t sin(0,5t  1) . Determine the frequency, period and amplitude of free oscillations. A. A = 2 B. A = 1 C. k = 0,539 D. k = 0,836 E. T = 11,65 F. T = 13,71 260. Damped oscillations are described by the equation x  2e 0,8t sin(4t  0,6) . Determine the cyclic frequency, period of oscillations and stiffness coefficient of spring to which is fastened the mass point with m  10 kg. A. k = 4 B. k = 16,6 C. T = 1,57 D. T = 0,38 E. c = 166 F. c = 6,4 261. Oscillations of a mass point with m  6 kg is set by the equation x  2e 0,1t sin(7t  0,6) . Determine the frequency and period of oscillations of mass point and the coefficient f if the force of resistance R   fV 2 . A. k = 0,1 B. k = 7 C. T = 10,4 D. T = 0,89 E. f = 1,2 F. f = 0,6 85

262. Disturbing force F  40sin10t acts on the body suspended from the spring. Determine the angular frequency of natural oscillations of the body, static deflection of the body and the frequency of the disturbing force if the dynamic factor f  3 and the stiffness coefficient of the spring is 1,6. A. k = 12,2 B. k = 5,3 C. Аст = 25 D. Аст = 6,25 E. р = 40 F. р = 10 263. The vibrational motion of a mass point is given in the form of 5x  320x  90sin7t . Determine the angular frequency of natural oscillations of the mass point, static deflection of the mass point and the frequency of the disturbing force. A. k = 4,24 B. k = 8 C. Аst = 0,28 D. Аst = 18 E. р = 7 F. р = 90 264. Homogeneous disk of radius R  0,5 m and mass m  20 kg rotates with an angular velocity 5 t rad/sec. Determine angular acceleration of the disk, the acceleration of the points on the rim of the disk and angular momentum of the disk in 2 sec. A. ε = 5 rad/sec2 B. ε = 0 rad/sec2 C. W = 50,06 m/sec2 D. W = 50 m/sec2 E. G = 12,5 kg·m2/sec F. G = 25 kg·m2/sec

86

265. Homogeneous rod with mass m  20 kg rotates with an angular velocity of 10 rad/sec. Determine the velocity and acceleration of the rod, the module of principle vector of the external forces if OA  1 m. A. V = 1 m/sec B. V = 10 m/sec C. a = 100 m/sec2 D. a = 0 m/sec2 E. a = -100 m/sec2 F. F = 500 N 266. The body of mass m  2 kg is moving according to the law S  2t 2  1 . Determine velocity and acceleration of the body, and the module of principle vector of external forces. A. V = 4t2 m/sec B. V = 4t m/sec C. a = 4 m/sec D. a = 2 m/sec2 E. F = 8 N F. F = 2 N 267. Determine the velocity of the slide B, kinetic energies of the crank OA rotating with angular velocity ω  2 rad/sec and rod AB of mass m  1 kg if OA  0,5 m. A. V = 1 m/sec B. V = 2 m/sec C. ТОА = 2 J D. ТОА = 0,5 J E. ТАВ = 2 J F. ТАВ = 1 J G. ТАВ = 0,5 J 87

268. The disc of radius R  1 m and mass kg rotates with the acceleration ε  2 rad/sec2. Determine the velocity, rotation angle and kinetic energy at a time t  2 sec if t 0  0 , ω0  0 ,  0  0 . A. V = 4 m/sec B. V = 2 m/sec C. φ = 4 rad D. φ = 0 rad E. Т = 240 J F. Т = 120 J G. Т = 60 J m  30

269. Homogeneous rod of mass m  3 kg and length AB  1 m revolves by the law   2t 3 . Determine the kinetic energy, the velocity of the rod and the moment of inertia of the rod about the axis of rotation at time t  1 sec. A. T = 54 J B. T = 18 J C. T = 9 J D. Iz = 1/4 kg·m2 E. Iz = 1 kg·m2 F. V = 6 m/sec G. V = 2 m/sec 270. For the given equation of rotation of the rod   5t 2  2 with an axial inertia moment I z  0,125 kg·m2 identify the principle moment of the external forces, the angular velocity and kinetic energy of the rod at time t  1 sec. A. ME = 1,25 Nm B. ME = 12,5 Nm C. ω = 10 rad/sec D. ω = 8 rad/sec E. T = 6,25 J F. T = 12,5 J G. T = 0,625 J 88

271. For a given equation of rotation of the rod   3t 2  t rod with an axial moment I z  1/6 kg·m2 identify the principle moment of the external forces, angular momentum and kinetic energy of the rod at time t  2 sec. A. ME = 0,17 Nm B. ME = 1 Nm C. К = 2,17 kg·m2/s D. К = 1,8 kg·m2/s E. T = 12 J F. T = 10,08 J G. T = -12 J 272. The shaft rotates with an angular velocity of ω  90e 20t  85(1  e 20t ) . Define the main moment of external forces, the angular momentum and kinetic energy at t  0,1 sec if the axial moment of inertia of the shaft is I z  1 kg·m2. A. ME = -13,5 Nm B. ME = 13,5 Nm C. К = 85,7 kg·m2/s D. К = -85,7 kg·m2/s E. T = 3670 J F. T = -3670 J G. T = 42,8 J 273. Determine the kinetic energy, angular momentum and angular velocity of the ring at t  1 sec if M z  3t 2 , I z  0,375 kg·m2, t0 = 0: ω0  16 rad/sec. A. T = 48 J B. T = 65,6 J C. К = 6 kg·m2/s D. К = 7 kg·m2/s E. ω = 13,3 rad/sec F. ω = 18,7 rad/sec G. ω = 2,67 rad/sec 89

274. Determine the magnitude and direction of the force vector if we know its projection Px  30 N, Py  40 N.

A. B. C. D. E. F. G.

Р = 50 N Р = 60 N cos (Р,^х) = 0,7 cos (Р,^х) = 0,6 cos (P,^у) = 0,9 cos (P,^у) = 0,8 cos (P,^у) = 0,7

275. The pipe of the mass of m  1 kg slides down by the action of force of gravity. Determine the acceleration of the pipe's center of mass and force vector projections. A. а = 2,45 m/sec2 B. а = 8,33 m/sec2 C. Fx = 0 N D. Fx = 4,9 N E. Fx = 8,33 N F. Fy = 9,8 N G. Fy = 4,9 N 276. Oscillations of mechanical system are described by equation 2q  3q  2sin5t . Determine amplitude, frequency and period of oscillations. A. А = 42,6 B. А = 2 C. А = 12 D. k = 1,22 E. k = 1,5 F. T = 6 G. T = 7 277. Inclined rod with the axial moment of I z  0,05 kg·m2 rotates in accordance with the law 2   2(t 2  1) . Define the main moment of external forces, the angular moment and kinetic energy of the rod at the time of t  2 sec. 90

A. B. C. D. E. F. G.

ME = 0,4 Nm ME = 0,2 Nm ME = 1,6 Nm К = 0,4 kg·m2/s К = 0,2 kg·m2/s T = 1,6 J T = 0,4 J

278. The law of rotation of a plate with inertia moment I z  0,125 kg·m2 is   5t 2  2 . Define the main moment of external forces, the velocity of rotation and kinetic energy of the plate at the time of t  1 sec. A. ME = 1,25 Nm B. ME = 12,5 Nm C. ME = 0,0125 Nm D. V = 10 m/sec E. V = 8 m/sec F. T = 6,25 J G. T = 0,625 J 279. Under the influence of mutually counterbalanced forces the mass point (body) is: A. in motion B. at rest C. moves rectilinearly D. moves rectilinearly and uniformly accelerated E. moves rectilinearly and uniformly F. moves curvilinearly and uniformly accelerated G. slowed motion 280. Force is the vector value which is characterized by: A. numeric value B. vector value C. projective value D. direction E. coordinates F. point of application G. projection of a point of application 91

281. Two forces applied to a rigid body will be counterbalanced in only case when they: A. are unequal by modulus B. are equal by modulus C. their values are equal D. act on parallel lines E. they act along the same straight line F. directed in opposite directions G. directed to one side 282. The resultant of two crossed forces: A. is applied in a point of their crossing B. is attached to the end of the first force C. is attached at the initial of the first force D. is represented by the diagonal of parallelogram E. is represented by the parallel side of parallelogram F. is expressed by the total vector G. is equal to R  F12  F22  2 F1 F2 cos 

 n  283. Resultant of converging forces R   Fi : i 1

A. is equal to the geometric sum of these forces B. is equal to the vector sum of these forces C. is attached at their point of intersection D. is attached in parallel to a point of their intersection E. is defined geometrically by creation of a force polygon F. is defined analytically by creation of a force polygon G. is defined geometrically by projecting the force on the coordinate axes 284. Center of parallel forces: A. is a point through which the line of action of the resultant of parallel forces passes B. a point which goes parallel to the line of action of the resultant of parallel forces C. a point through which the line of action passes that is parallel to the resultant of parallel forces D. a rotation about the points of application of forces in the same direction and at the same angle 92

E. a rotation about the points of application of forces in the same direction by a double angle

F x F  Fkx  xk 

F. xc 

kx

k

kx

G. xc

Fx

285. The bar mounted in the hinge A act the forces F1  4 kN and F2 . Determine the force F2 and the reaction of hinge A if AC  2 m, AB  6 m. A. F2 = 12 kN B. F2 = 4 kN C. F2 = 48 kN D. XA = 0 kN E. XA = 4 kN F. YA = 16 kN G. YA = 8 kN 286. On the beam AB with weight kN acts the distributed load q  0,5 kN/m. Determine the reaction in the supports A and B if AB  6 m, AC  BC . A. XА = 0 kN B. XА = 1,5 kN C. YА = 10,4 kN D. YА = 21,5 kN E. RВ = 11,1 kN F. RВ = 21,5 kN G. RВ = 1,5 kN G  20

287. On the beam AB acts the force kN and distributed load F 9 q  3 kN/m. Determine the reaction in the supports A and B if AB  5 m, BC  2 m. A. XА = 0 kN 93

B. C. D. E. F. G.

XА = 1,5 kN YА = 4,8 kN YА = 25,2 kN RВ = 10,2 kN RВ = 21,5 kN RВ = 1,5 kN

288. Determine the reaction in the supports A and B if the distributed load q  40 N/m, the sizes of the beam are AB  4 m, BC  2 m. A. XА = 0 N B. XА = 80 N C. YА = -20 N D. YА = 20 N E. YА = -100 N F. RВ = 100 N G. RВ = -100 N 289. On the beam AB acts the distributed load q  5 kN/m. Determine the reaction in the supports A and B if the lengths are AB  9 m, AC  6 m. A. XВ = 0 kN B. XВ = 30 kN C. YВ = -10 kN D. YВ = 30 kN E. YВ = 10 kN F. RА = 20 kN G. RА = -20 kN 290. On the beam AB acts the force F  6 N and distributed load q  2 N/m. Determine the reaction in the supports A and B if the lengths are AC  AB/3  1 m and an angle is 45о. A. XА = 0 N B. XА = -4,2 N

94

C. D. E. F. G.

YА = 4,2 N YА = 4,07 N YА = 4,13 N RB = 4,07 N RB = 4,13 N

291. The moment of couple of forces appeared in the end restraint is M A  200 Nm, length l  1 m. Determine the intensity of a distributed load q and reaction in the end restraint. A. XА = 400 N B. XА = 0 N C. YА = 400 N D. YА = 0 N E. YА = -400 N F. q = 400 N/m G. q = -400 N/m 292. Define the reactions of supports A and B of the beam under action of one concentrated force P1  4 kN and couple of forces P2  6 kN. A. ХA = 2 kN B. YA = -4,32 kN C. ХA = 3 kN D. YB = 7,78 kN E. YA = 4,32 kN F. YB = 8 kN G. ХA = 2,78 kN 293. Define the reactions of cantilever under action of concentrated force P1  2 kN and couple of forces P2  3 kN shown at the picture. A. X = 1,73 kN 95

B. C. D. E. F. G.

X = 1 kN Y = 2 kN Y = 1,73 kN M = 1 kN·m M = 0,47 kN·m M = 0,47 kN·m

294. The force couple (P, P) acts on the double cantilever. Distributed force with the intensity q acts on the left cantilever. Vertical load Q acts in the point D. Determine the value of distributed load Q acting on the part CA and the reactions of the supports if P  1 kN, Q  2 kN, q  2 kN/m, a  0,8 m. A. Q = 1,6 kN B. RA = 3 kN C. RA = 1,5 kN D. Q = 2 kN E. RB = 2 kN F. RB = 3 kN G. Q = 1 kN 295. Define the reactions of cantilever under action of distributed force q  1,5 kN/m, concentrated force P1  4 kN and couple of forces P2  2 kN shown at the picture. A. X = 2,8 kN B. Y = 1 kN C. Y = 1,7 kN D. M = -5,35 kN·m E. M = 1,7 kN·m 96

F. X = 2 kN G. X = 2 kN 296. Define the reactions of supports A and B of the beam under action of two concentrated forces P1  6 kN, P2  8 kN and distributed load q  3 kN/m. A. XA = 2,6 kN B. XA = 2 kN C. YA = 4,2 kN D. YA = 3 N E. XB = 15,6 kN F. XB = 10 kN G. XA = 5 kN 297. Define the reactions of cantilever and the value of distributed load q , if M A  400 Nm, AB  2 m, BC  4 m. A. q = 25 N/m B. q = 20 N/m C. XA = 0 N D. XA = 1 N E. YA = 100 N F. YA = 90 N G. q = 10 N 298. Center of gravity of a rigid body: A. is a point unrelated to this body B. is a point unrelated to this body C. is a the point through which the line of action of the resultant forces of gravity of body particles passes for any position of the body in space  p kx  x k  p  yk D. x C  , yC  P P  p kx  x k  p kx  y k  p kx  z k E. x C  , yC  , zC  P P P 97

 p ky  y k  p kx  x k  p kz  z k , yC  , zC  P P P  p kx  p kx  p kx G. x C  , yC  , zC  P P P

F. x C 

299. The center of gravity of a plane figure:  x k  Fk A. x C  F  x k  Fk B. x C  F C. geometric point D. vector E. material point 1 F. x C   xdF F ( F)

G. x C 

1  xdF x ( F)

300. Acceleration for rectilinear uniform motion: A. W  0

dV 0 dt C. W  Wn  W dr D. Wn  dt E. V  V0  V B. W 

F. W 

V2

G.   



98

KEYS TO THE TEST ↓ → 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30

0 D A B C D B D F E A B,E A,D B,E A,F E,F B,F C,G A,B A,B A,B A,C,E B,C,E A,C,E B,C,E B,D,F B,D,E A,C,E A,D,F B,E,F A,B,G

1

2

3

4

5

6

7

8

9

D C D D B A A A B D B,E B,D C,E C,E A,F C,F C,F C,G A,B B,F A,B,D A,C,D B,C,E A,C,E B,C,D A,C,E B,D,E B,D,F B,E,F A,D,F

C D B A D D D A C E A,E A,D B,E A,E D,F A,F A,D A,F B,C A,B A,C,D A,B,E B,C,D B,C,D B,D,F B,D,E A,C,E A,C,E A,D,G A,B,D

A A C D B C C C C D C,E A,E A,E B,E A,F C,F C,E A,C B,C A,B B,C,E A,C,E A,C,D A,C,E B,D,E A,D,E B,C,E B,D,F A,C,E B,D,F

B B D A C A C D C A A,E B,D A,E D,F C,F A,F E,F A,D A,B A,B A,B,C A,B,D A,C,E A,D,F A,C,E B,D,E A,C,F A,D,F A,D,F A,B,E

A D B B D C C D A D A,C D,E D,E D,F B,F B,F A,F A,D A,B A,D A,B,C A,B,C A,C,E B,C,E A,D,E B,D,F B,D,F A,C,F A,D,F A,C,D

D A A D B D D D E C B,C C,D C,E A,F C,F C,F C,F C,E A,B A,B A,B,E A,C,D B,C,E C,E,F B,C,E A,C,E B,C,E A,D,F A,C,E A,C,E

C C C A D C C C D A A,E A,E D,E D,F D,F D,F C,F B,F A,D A,E A,C,D A,C,D B,C,D B,C,E A,D,E B,D,F A,D,G B,D,F A,C,E A,C,E

B D A C C B C D A D B,E A,E A,E C,F B,F A,F A,G A,F A,B A,C B,C,D A,C,D A,C,D A,C,E B,D,E B,C,E A,C,F A,D,F A,C,F A,C,F

A B B A D C A C C D A,E A,E A,E B,F A,F D,F A,F A,B A,B A,C A,B,D A,B,D A,C,E A,C,E B,D,E A,C,E B,E,F B,C,E A,E,F A,C,F

99

BIBLIOGRAPHY 1. Bat' M.I. i dr. Teoreticheskaya mekhanika v primerakh i zadachakh. Ucheb. posob. dlya vuzov. V 2-kh t. / M.I. Bat', G.Yu. Dzhanelidze, A.S. Kel'zon.-9-ye izd., pererab. – M.: Nauka, 2007. – 670 p. 2. Butenin N.V. i dr. Kurs teoreticheskoy mekhaniki: Ucheb. posobiye dlya studentov vuzov po tekhn. spets.: v 2-kh t. / N.V. Butenin, YA.L. Lunts, D.R. Merkin. SPb.: Lan'. – 5-ye izd., ispr. 2008. – 729 p. 3. Golubev Yu.F. Osnovy teoreticheskoy mekhaniki. M.: Izd-vo MGU, 2000. – 719 p. 4. Gross D., Hauger W., Schroder J., Wall W.A., Rajapakse N. Engineering Mechanics. P.1 Springer, Verlag. – 2012. – 301 p. 5. Gross D., Hauger W., Schroder J., Wall W.A., Govindjee S. Engineering Mechanics. P.3. Springer, Verlag. – 2011. – 359 p. 6. Markeyev A.P. Teoreticheskaya mekhanika: Uchebnik dlya universitetov. 3-ye izd. – M.: Izhevsk, 2007. – 592 p. 7. Meshcherskiy I.V. Zadachi po teoreticheskoy mekhanike: Ucheb. posobiye dlya stud. vuzov, obuch. po tekhn. spets. / I.V. Meshcherskiy; pod red. V.A. Pal'mova, D.D. Merkina. – 45-ye izd., ster. – SPb. i dr.: Lan', 2009. – 447 p. 8. Polyakhov N.N., Zegzhda S.A., Yushkov M.P. Teoreticheskaya mekhanika. – M.: Vysshaya shkola, 2000. – 592 p. 9. Rakisheva Z.B., Sukhenko A.S., Textbook on Theoretical Mechanics. – 2nd ed. – Almaty: Qazaq university, 2017. – 354 p. 10. Sbornik zadaniy dlya kursovykh rabot po teoreticheskoy mekhanike: Ucheb. posobiye dlya stud. Vuzov / [A.A. Yablonskiy, S.S. Noreyko, S.A. Vol'fson i dr.]; pod obshch. red. A.A. Yablonskogo. – 11-ye izd. ster. – M.: Integral-Press, 2008. – 382 p. 11. Targ S. Theoretical mechanics a short course. Translated from Russian. – М.: High School, 2010. – 416 p. 12. Teoreticheskaya mekhanika. Terminologiya. Bukvennyye oboznacheniya velichin: Sbornik rekomenduyemykh terminov. Vyp. 102. M.: Nauka, 2007. – 48 p. 13. Yablonskiy A.A., V.M. Nikiforova Kurs teoreticheskoy mekhaniki. Ucheb. posobiye dlya vuzov: 13-ye izd., isprav. – M.: Integral-Press, 2009. – 603 p.

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CONTENT Introduction ........................................................................................................ 3 Test questions ..................................................................................................... 4 Keys to the test ................................................................................................... 99 Bibliography ....................................................................................................... 100

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Educational publication

Rakisheva Zaure Bayanovna Mayemerova Gulnara Maratovna Sukhenko Anna Sergeyevna

TEST QUESTIONS ON THEORETICAL MECHANICS Collection of tests Typesetting U. Moldasheva Cover design Ya. Gorbunov Cover design used photos from sites www.Sanfoundry

IB №13276

Signed for publishing 26.12.2019. Format 60x84 1/16. Offset paper. Digital printing. Volume 6,37 printer’s sheet. 100 copies. Order №8874. Publishing house «Qazaq University» Al-Farabi Kazakh National University KazNU, 71 Al-Farabi, 050040, Almaty Printed in the printing office of the «Qazaq University» publishing house.

    

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