Engineering Mechanics, Statics & Dynamics [6 ed.] 9780138049430

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Engineering Mechanics

STATICS & DYNAMICS

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Engineering Mechanics

STATICS & DYNAMICS SIXTH EDITION

Anthony Bedford • Wallace Fowler University of Texas at Austin

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CONTENTS STATICS Preface

xvii

About the Authors

xxi

Pearson’s Commitment to Diversity, Equity, and Inclusion xxii

1 Introduction 3 1.1 Engineering and Mechanics 4 Problem Solving 4 Numbers 5 Space and Time 6 Newton’s Laws 6 International System of Units 8 US Customary Units 8 Angular Units 9 Conversion of Units 9 Results 10

1.2 Newtonian Gravitation 15 Results 16

2 Vectors 21 2.1 Scalars and Vectors 22 Vector Addition 22 Product of a Scalar and a Vector 24 Vector Subtraction 25 Unit Vectors 25 Results 25

2.2 Components in Two Dimensions 31 Manipulating Vectors in Terms of Components 31 Position Vectors in Terms of Components 32 Results 32

2.3 Components in Three Dimensions 45 Magnitude of a Vector in Terms of Components 46 Direction Cosines 47 Position Vectors in Terms of Components 48 Components of a Vector Parallel to a Given Line 48 Results 49

2.4 Dot Products 62 Definition 63 Dot Products in Terms of Components 63

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v

vi Contents

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Vector Components Parallel and Normal to a Line 64 Results 65

2.5 Cross Products 71 Definition 72 Cross Products in Terms of Components 72 Evaluating a 3 × 3 Determinant 73 Mixed Triple Products 74 Results 75

3 Forces 87 3.1 Forces, Equilibrium, and Free-Body Diagrams 88 Terminology 88 Gravitational Forces 89 Contact Forces 89 Equilibrium 93 Free-Body Diagrams 94 Results 96

3.2 Two-Dimensional Force Systems 98 3.3 Three-Dimensional Force Systems 116

4 Systems of Forces and Moments 131 4.1 Two-Dimensional Description of the Moment 132 Results 134

4.2 The Moment Vector 146 Magnitude of the Moment 146 Direction of the Moment 146 Relation to the Two-Dimensional Description 148 Varignon’s Theorem 149 Results 150

4.3 Moment of a Force about a Line 159 Definition 160 Applications 161 Results 163

4.4 Couples 174 Results 177

4.5 Equivalent Systems 184 Conditions for Equivalence 185 Representing Systems by Equivalent Systems 186 Representing a System by a Wrench 187 Results 189 Get complete eBook Order by email at [email protected]

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5 Objects in Equilibrium 211 5.1 Two-Dimensional Applications 212 The Scalar Equilibrium Equations 212 Supports 213 Free-Body Diagrams 215 Results 217

5.2 Statically Indeterminate Objects 234 Redundant Supports 234 Improper Supports 236 Results 236

5.3 Three-Dimensional Applications 240 The Scalar Equilibrium Equations 241 Supports 241 Results 247

5.4 Two-Force and Three-Force Members 260 Two-Force Members 260 Three-Force Members 262 Results 262

6 Structures in Equilibrium 273 6.1 Trusses 274 Results 276

6.2 The Method of Joints 277 Results 279

6.3 The Method of Sections 288 Results 289

6.4 Space Trusses 295 Results 297

6.5 Frames and Machines 301 Analyzing the Entire Structure 303 Analyzing the Members 303 Results 308

7 Centroids and Centers of Mass 331 7.1 Centroids of Areas 332 Results 334

7.2 Composite Areas 340 Results 342

7.3 Distributed Loads 348 Describing aGet Distributed Load 348 Order by email at [email protected] complete eBook

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viii Contents

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Determining Force and Moment 349 The Area Analogy 349 Results 350

7.4 Centroids of Volumes and Lines 355 Results 357

7.5 Composite Volumes and Lines 363 7.6 The Pappus–Guldinus Theorems 370 First Theorem 370 Second Theorem 371 Results 372

7.7 Centers of Mass of Objects 375 Results 377

7.8 Centers of Mass of Composite Objects 382

8 Moments of Inertia 395 Areas 396 8.1 Definitions 396 8.2 Parallel-Axis Theorems 402 Results 405

8.3 Rotated and Principal Axes 416 Rotated Axes 417 Moment of Inertia about the x ′-Axis 418 Moment of Inertia about the y ′-Axis 418 Principal Axes 418 Results 420

8.4 Mohr’s Circle 426 Determining I x ′ , I y ′ , and I x ′y ′ 426 Determining Principal Axes and Principal Moments of Inertia 427 Results 428

Masses 430 8.5 Simple Objects 430 Slender Bars 430 Thin Plates 431 Results 432

8.6 Parallel-Axis Theorem 436 Results

437

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9 Friction 451 9.1 Theory of Dry Friction 452 Coefficients of Friction 454 Angles of Friction 455 Results 456

9.2 Wedges 470 9.3 Threads 474 Motion Opposite to the Direction of the Axial Force 475 Motion in the Direction of the Axial Force 475 Results 476

9.4 Journal Bearings 481 Results

482

9.5 Thrust Bearings and Clutches 486 Results

488

9.6 Belt Friction 492 Results

494

9.7 Current Research on Friction 500

10 Internal Forces and Moments 507 Beams 508 10.1 Axial Force, Shear Force, and Bending Moment 508 Results 509

10.2 Shear Force and Bending Moment Diagrams 515 Results 516

10.3 Relations between Distributed Load, Shear Force, and Bending Moment 520 Construction of the Shear Force Diagram 522 Construction of the Bending Moment Diagram 523 Results 526

Cables 533 10.4 Loads Distributed Uniformly Along Straight Lines 534 Shape of the Cable 534 Tension of the Cable 535

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Length of the Cable 535 Results 536

10.5 Loads Distributed Uniformly Along Cables 540 Shape of the Cable 541 Tension of the Cable 542 Length of the Cable 542 Results 542

10.6 Discrete Loads 545 Determining the Configuration and Tensions 545 Comments on Continuous and Discrete Models 546 Results 547

Liquids and Gases 551 10.7 Pressure and the Center of Pressure 551 Center of Pressure 552 Pressure in a Stationary Liquid 553 Results 555

11 Virtual Work and Potential Energy

567

11.1 Virtual Work 568 Work 568 Principle of Virtual Work 569 Application to Structures 570 Results 571

11.2 Potential Energy 580 Examples of Conservative Forces 580 Principle of Virtual Work for Conservative Forces 581 Stability of Equilibrium 582 Results 583

APPENDICES

A Results from Mathematics 595 A.1 Algebra 595 Quadratic Equations 595 Natural Logarithms 595

A.2 Trigonometry 596 A.3 Derivatives 596 A.4 Integrals 597 A.5 Taylor Series 598 Get complete eBook Order by email at [email protected]

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A.6 Vector Analysis 598 Cartesian Coordinates 598 Cylindrical Coordinates 598

A.7 Matrices 599

B Properties of Areas and Lines 601 B.1 Areas 601 B.2 Lines 604

C Properties of Volumes and Homogeneous Objects 605 Solutions to Practice Problems 609 Answers to Even-Numbered Problems 639 Index 649

DYNAMICS 12 Introduction 3 12.1 Engineering and Mechanics Problem Solving 4 Numbers 5 Space and Time 6 Newton’s Laws 6 International System of Units US Customary Units 8 Angular Units 9 Conversion of Units 9 Results 10

4

8

12.2 Newtonian Gravitation 15 Results 16

13 Motion of a Point 21 13.1 Position, Velocity, and Acceleration

22

Results 24

13.2 Straight-Line Motion 25 Description of the Motion 25 Analysis of the Motion 26 Get complete eBook Order by email at [email protected] Results 29

Contents

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xii Contents

13.3 Straight-Line Motion When the Acceleration Depends on Velocity or Position 42 Results

44

13.4 Curvilinear Motion—Cartesian Coordinates 50 Results

52

13.5 Angular Motion 62 Angular Motion of a Line Rotating Unit Vector 62 Results 64

62

13.6 Curvilinear Motion—Normal and Tangential Components 67 Planar Motion 68 Circular Motion 71 Three-Dimensional Motion Results 73

72

13.7 Curvilinear Motion—Polar and Cylindrical Coordinates 85 Results

89

13.8 Relative Motion 100 Results

101

14 Force, Mass, and Acceleration 109 14.1 Newton’s Second Law

110

Equation of Motion for the Center of Mass Inertial Reference Frames 112 Results 114

110

14.2 Applications—Cartesian Coordinates and Straight-Line Motion 114 14.3 Applications—Normal and Tangential Components 135 14.4 Applications—Polar and Cylindrical Coordinates 148 14.5 Orbital Mechanics

155

Determination of the Orbit Types of Orbits 158 Results 159

155

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15 Energy Methods 167 15.1 Work and Kinetic Energy Principle of Work and Energy Evaluating the Work 169 Power 170 Results 171

168

168

15.2 Work Done by Particular Forces

182

Weight 183 Springs 184 Results 185

15.3 Potential Energy and Conservative Forces 198 Potential Energy 198 Conservative Forces 199 Results 202

15.4 Relationships between Force and Potential Energy 215 Results 216

16 Momentum Methods 225 16.1 Principle of Impulse and Momentum

226

Results 228

16.2 Conservation of Linear Momentum and Impacts 241 Conservation of Linear Momentum 241 Impacts 242 Results 246

16.3 Angular Momentum 258 Principle of Angular Impulse and Momentum Central-Force Motion 259 Results 260

16.4 Mass Flows

258

267

Results 268

17 Planar Kinematics of Rigid Bodies 283 17.1 Rigid Bodies and Types of Motion

284

Translation 285 Rotation about a Fixed Axis 285 Planar Motion 286 Results 286 Get complete eBook Order by email at [email protected]

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17.2 Rotation about a Fixed Axis Results

287

289

17.3 General Motions: Velocities

296

Relative Velocities 296 The Angular Velocity Vector 298 Results 300

17.4 Instantaneous Centers Results

314

315

17.5 General Motions: Accelerations Results

324

17.6 Sliding Contacts Results

321

335

338

17.7 Moving Reference Frames

349

Motion of a Point Relative to a Moving Reference Frame 349 Inertial Reference Frames 350 Results 354

18 Planar Dynamics of Rigid Bodies

371

18.1 Momentum Principles for a System of Particles 372 Force–Linear Momentum Principle 372 Moment–Angular Momentum Principles 373 Results 374

18.2 The Planar Equations of Motion 376 Rotation about a Fixed Axis 376 General Planar Motion 377 Results 378

Appendix: Moments of Inertia Simple Objects 402 Parallel-Axis Theorem

402

405

19 Energy and Momentum in Rigid-Body Dynamics 19.1 Work and Energy

419

420

Kinetic Energy 421 Work and Potential Energy Power 425 Results 426

423

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19.2 Impulse and Momentum

445

Linear Momentum 445 Angular Momentum 446 Results 449

19.3 Impacts

459

Conservation of Momentum 460 Coefficient of Restitution 460 Results 463

20 Three-Dimensional Kinematics and Dynamics of Rigid Bodies 20.1 Kinematics

483

484

Velocities and Accelerations 485 Moving Reference Frames 485 Results 487

20.2 Euler’s Equations

499

Rotation about a Fixed Point 499 General Three-Dimensional Motion Equations of Planar Motion 504 Results 505

20.3 The Euler Angles

502

524

Objects with an Axis of Symmetry Arbitrary Objects 528 Results 531

524

Appendix: Moments and Products of Inertia Simple Objects 540 Thin Plates 541 Parallel-Axis Theorems 543 Moment of Inertia about an Arbitrary Axis Principal Axes 545

540

545

21 Vibrations 561 21.1 Conservative Systems

562

Examples 562 Solutions 564 Results 567

21.2 Damped Vibrations

578

Subcritical Damping 579 Critical and Supercritical Damping Results 581

21.3 Forced Vibrations

580

589

Oscillatory Forcing Function 590 Polynomial Forcing Function 591 Get complete eBook Order by email at [email protected] Results 592

Contents

xv

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APPENDICES

A Results from Mathematics 607 A.1 Algebra 607 Quadratic Equations 607 Natural Logarithms 607

A.2 Trigonometry 608 A.3 Derivatives 608 A.4 Integrals 609 A.5 Taylor Series 610 A.6 Vector Analysis 610 Cartesian Coordinates 610 Cylindrical Coordinates 610

A.7 Matrices 611

B Properties of Areas and Lines 613 B.1 Areas 613 B.2 Lines 616

C Properties of Volumes and Homogeneous Objects 617

D Spherical Coordinates 620 E D’Alembert’s Principle 621 Solutions to Practice Problems 623 Answers to Even-Numbered Problems 651 Index 659

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Engineering Mechanics

STATICS

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How do engineers design and construct the devices we use, from simple objects such as chairs and pencil sharpeners to complicated ones such as dams, cars, airplanes, and spacecraft? They must have a deep understanding of the physics underlying the design of such devices and must be able to use mathematical models to predict their behavior. Students of engineering begin to learn how to analyze and predict the behaviors of physical systems by studying mechanics.

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CHAPTER

Introduction

1

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1.1 Engineering and Mechanics QUESTIONS TO CONSIDER • What is the science of mechanics, and how is it related to engineering? • When you tell someone how far apart two cities are, you tell them in miles or kilometers, which are units of length. What other kinds of units do we use in mechanics?

BACKGROUND How can engineers design complex systems and predict their characteristics before they are constructed? Engineers have always relied on their knowledge of previous designs, experiments, ingenuity, and creativity to develop new designs. Modern engineers add a powerful technique: They develop mathematical equations based on the physical characteristics of the devices they design. With these mathematical models, engineers predict the behavior of their designs, modify them, and test them prior to their actual construction. Aerospace engineers use mathematical models to predict the paths airplanes and spacecraft will follow in flight. Civil engineers use mathematical models to analyze the effects of loads on buildings and foundations. At its most basic level, mechanics is the study of forces and their effects. Elementary mechanics is divided into statics, the study of objects in equilibrium, and dynamics, the study of objects in motion. The results obtained in elementary mechanics apply directly to many fields of engineering. Mechanical and civil engineers designing structures use the equilibrium equations derived in statics. Civil engineers analyzing the responses of buildings to earthquakes and aerospace engineers determining the trajectories of satellites use the equations of motion derived in dynamics. Mechanics was the first analytical science. As a result, fundamental concepts, analytical methods, and analogies from mechanics are found in virtually every field of engineering. Students of chemical and electrical engineering gain a deeper appreciation for basic concepts in their fields, such as equilibrium, energy, and stability, by learning them in their original mechanical contexts. By studying mechanics, they retrace the historical development of these ideas. Mechanics consists of broad principles that govern the behavior of objects. In this book we describe these principles and provide examples that demonstrate some of their applications. Although it is essential that you practice working problems similar to these examples, and we include many problems of this kind, our objective is to help you understand the principles well enough to apply them to situations that are new to you. Each generation of engineers confronts new problems.

PROBLEM SOLVING In the study of mechanics you learn problem-solving procedures that you will use in succeeding courses and throughout your career. Although different types of Get complete eBook Order by different email atapproaches, [email protected] problems require the following steps apply to many of them:

Get complete eBook Order by email at [email protected] 1.1 Engineering and Mechanics 5 • Identify the information that is given and the information, or answer, you must determine. It’s often helpful to restate the problem in your own words. When appropriate, make sure you understand the physical system or model involved. • Develop a strategy for the problem. This means identifying the principles and equations that apply and deciding how you will use them to solve the problem. Whenever possible, draw diagrams to help visualize and solve the problem. • Whenever you can, try to predict the answer. This will develop your intuition and will often help you recognize an incorrect answer. • Solve the equations and, whenever possible, interpret your results and compare them with your prediction. This last step is a reality check. Is your answer reasonable? Students sometimes ask to be provided with a solution manual, or worked examples similar to all the different types of problems they are asked to solve. We explain that devising solutions themselves is an essential part of their educational experience. Although that may remind you of Ralphie’s comment in the film A Christmas Story, “Adults loved to say things like that, but kids knew better,” think ahead to when you are a practicing engineer. You will inevitably encounter truly new problems. Where would you find solutions to similar examples? Some of our problems involve straightforward substitution of numbers into equations that we have just discussed, but others are challenging, requiring you to find sequences of steps leading to results and conclusions in situations that may be unfamiliar to you. That after all is how engineers work, and every problem you solve yourself in this course and others is a step toward becoming one. And you will find that success builds on success. Your initial progress may be halting, but you will become increasingly proficient.

NUMBERS Engineering measurements, calculations, and results are expressed in numbers. You need to know how we express numbers in the examples and problems and how to express the results of your own calculations. Significant Digits The number of significant digits in a given number refers to how many accurate digits it contains, counting from left to right beginning with the first nonzero digit. For example, the numbers 2695, 2.695, and 0.002695 each have four significant digits if the digits 2, 6, 9, and 5 are known to be accurate. Numbers are often “rounded off” to a smaller number of significant digits for convenience in writing them or because greater accuracy is not needed or warranted. For example, the number 34,662 can be rounded off to three significant digits by writing it as 34,700. The two zeros act as “place keepers” to indicate the magnitude of the number. The third digit of this rounded-off number is 7 because 34,700 is nearer to 34,662 than 34,600. The number 34,662 can be expressed to three significant digits by writing it in scientific notation as 3.47 × 10 4. In this book we use notation now commonly used in technical software, writing such numbers as 3.47E4. How would we express the number 34,650 to three significant digits? Either 34,600 or 34,700 would be equally valid. The most common practice, which we follow, is to round off such numbers to the higher digit, so we write it as 34,700 or 3.47E4.

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Get complete eBook Order by email at [email protected] 6 Chapter 1 Introduction Use of Numbers in This Book We generally express intermediate results and answers in the examples and the answers to the problems to at least three significant digits. If you use a calculator, your results should be that accurate. Be sure to avoid round-off errors that occur if you round off intermediate results when making a series of calculations. Instead, carry through your calculations with as much accuracy as you can by retaining values in your calculator.

SPACE AND TIME Space simply refers to the three-dimensional universe in which we live. Our daily experiences give us an intuitive notion of space and the locations, or positions, of points in space. The distance between two points in space is the length of the straight line joining them. Measuring the distance between points in space requires a unit of length. We use both the International System of units, or SI units, and US customary units. In SI units, the unit of length is the meter (m). In US customary units, the unit of length is the foot (ft). Time is, of course, familiar—our lives are measured by it. The daily cycles of light and darkness and the hours, minutes, and seconds measured by our clocks and watches give us an intuitive notion of time. Time is measured by the intervals between repeatable events, such as the swings of a clock pendulum or the vibrations of a quartz crystal in a watch. In both SI units and US customary units, the unit of time is the second (s). The minute (min), hour (h), and day are also frequently used. If the position of a point in space relative to some reference point changes with time, the rate of change of its position is called its velocity, and the rate of change of its velocity is called its acceleration. In SI units, the velocity is expressed in meters per second (m/s) and the acceleration is expressed in meters per second per second, or meters per second squared (m/s 2 ). In US customary units, the velocity is expressed in feet per second (ft/s) and the acceleration is expressed in feet per second squared (ft/s 2 ).

NEWTON’S LAWS Newtonian mechanics, also referred to as classical mechanics, was established with the publication in 1687 of Philosophiae Naturalis Principia Mathematica, by Isaac Newton. Although highly original, it built on fundamental concepts developed by many others during a long and difficult struggle toward understanding (Fig. 1.1). Newton stated three “laws” of motion, which we express in modern terms: 1. When the sum of the forces acting on a particle is zero, its velocity is constant. In particular, if the particle is initially stationary, it will remain stationary. 2. When the sum of the forces acting on a particle is not zero, the sum of the forces is equal to the rate of change of the linear momentum of the particle. If the mass is constant, the sum of the forces is equal to the product of the mass of the particle and its acceleration. 3. The forces exerted by two particles on each other are equal in magnitude and opposite in direction.

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Get complete eBook Order by email at [email protected] 1.1 Engineering and Mechanics 7 Peloponnesian War Roman invasion of Britain

Coronation of Charlemagne

400 BCE Aristotle: Statics of levers, speculations on dynamics Archimedes: Statics of levers, centers of mass, buoyancy 0 Hero of Alexandria: Statics of levers and pulleys Pappus: Precise definition of center of mass CE 400 John Philoponus: Concept of inertia 800

Norman conquest of Britain Signing of Magna Carta

1200 Jordanus of Nemore: Stability of equilibrium

Bubonic plague in Europe 1400 Printing of Gutenberg Bible Voyage of Columbus Founding of Jamestown Colony

1600

Thirty Years’ War Pilgrims’ arrival in Massachusetts Founding of Harvard University Settlement of Carolina

1650

Albert of Saxony: Angular velocity Nicole d’Oresme: Graphical kinematics, coordinates William Heytesbury: Concept of acceleration Nicolaus Copernicus: Concept of the solar system Dominic de Soto: Kinematics of falling objects Tycho Brahe: Observations of planetary motions Simon Stevin: Principle of virtual work Johannes Kepler: Geometry and kinematics of planetary motions Galileo Galilei: Experiments and analyses in statics and dynamics, motion of a projectile René Descartes: Cartesian coordinates Evangelista Torricelli: Experiments on hydrodynamics Blaise Pascal: Analyses in hydrostatics John Wallis, Christopher Wren, Christiaan Huyghens: Impacts between objects

Pennsylvania grant to William Penn Salem witchcraft trials 1700

Isaac Newton: Concept of mass, laws of motion, postulate of universal gravitation, analyses of planetary motions

FIGURE 1.1 Chronology of developments in mechanics up to the publication of Newton’s Principia in relation to other events in history.

Notice that we did not define force and mass before stating Newton’s laws. The modern view is that these terms are defined by the second law. To demonstrate, suppose that we choose an arbitrary object and define it to have unit mass. Then we define a unit of force to be the force that gives our unit mass an acceleration of unit magnitude. In principle, we can then determine the mass of any object: We apply a unit force to it, measure the resulting acceleration, and use the second law to determine the mass. We can also determine the magnitude of any force: We apply it to our unit mass, measure the resulting acceleration, and use the second law to determine the force. Thus, Newton’s second law gives precise meanings to the terms mass and force. In SI units, the unit of mass is the kilogram (kg). The unit of force is the newton (N), which is the force required to give a mass of one kilogram an acceleration of one meter per second squared. In US customary units, the unit of force is the pound (lb). The unit of mass is the slug, which is the amount of mass accelerated at one foot per second squared by a force of one pound.

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Get complete eBook Order by email at [email protected] 8 Chapter 1 Introduction Although the results we discuss in this book are applicable to many of the problems met in engineering practice, there are limits to the validity of Newton’s laws. For example, they don’t give accurate results if a problem involves velocities that are not small compared to the velocity of light (3 × 10 8  m/s). Einstein’s special theory of relativity applies to such problems. Elementary mechanics is also inadequate in problems involving dimensions that are not large compared to atomic dimensions. Quantum mechanics must be used to describe phenomena on the atomic scale.

INTERNATIONAL SYSTEM OF UNITS In SI units, length is measured in meters (m) and mass in kilograms (kg). Time is measured in seconds (s), although other familiar measures such as minutes (min), hours (h), and days are also used when convenient. Meters, kilograms, and seconds are called the base units of the SI system. Force is measured in newtons (N). Recall that these units are related by Newton’s second law: One newton is the force required to give an object of one kilogram mass an acceleration of one meter per second squared: 1 N = (1 kg)(1 m/s 2 ) = 1 kg-m/s 2. Table 1.1 The common prefixes used in SI units and the multiples they represent. Prefix nano-

Abbreviation n

Multiple 10 −9

micro-

µ

10 −6

milli-

m

10 −3

kilo-

k

10 3

mega-

M

10 6

giga-

G

10 9

Because the newton is defined in terms of the base units, it is called a derived unit. Although in principle the quantities we deal with in mechanics can be expressed in terms of base units, it is often not convenient or intuitive to do so. We say an object is subjected to a force of 5 newtons, not that it is subjected to a force of 5 kilogram-meters per second squared. To express quantities by numbers of convenient size, multiples of units are indicated by prefixes. The most common prefixes, their abbreviations, and the multiples they represent are shown in Table 1.1. For example, 1 km is 1 kilometer, which is 1000 m, and 1 Mg is 1 megagram, which is 10 6  g, or 1000 kg. We frequently use kilonewtons (kN).

US CUSTOMARY UNITS In US customary units, length is measured in feet (ft) and force is measured in pounds (lb). Time is measured in seconds (s). These are the base units of the US customary system. In this system of units, mass is a derived unit. The unit of mass is the slug, which is the mass of material accelerated at one foot per second squared by a force of one pound. (US customary units are equivalent to the imperial system of units sometimes used in the UK.) Newton’s second law states that 1 lb = (1 slug)(1 ft/s 2 ). From this expression we obtain 1 slug = 1 lb-s 2 /ft.

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Get complete eBook Order by email at [email protected] 1.1 Engineering and Mechanics 9 We use other US customary units such as the mile (1 mi = 5280 ft) and the inch (1 ft = 12 in). We also use the kilopound (kip), which is 1000 lb.

ANGULAR UNITS The angle θ shown in Fig. 1.2 specifies the position of the slanted radial line relative to the horizontal line. Angles are traditionally measured in degrees. If the radial line rotates completely around to its initial position, that is defined to be an angle of three hundred and sixty degrees, denoted by θ = 360 °. If the line rotates one-fourth of the way around the circle that is θ = 90 °, halfway around is θ = 180 °, and so on. To express fractions of a degree, degrees are subdivided into sixty “minutes” of arc and minutes into sixty “seconds” of arc. The notation used, for example for sixty degrees, twenty minutes, and 10 seconds, is 60 °20 ′10 ″. However, it is often more convenient to express angles such as this as (to five significant digits) 60.336 °. Although degrees are much more familiar, in analytical work it is advantageous to express angles in terms of radians (rad). The equation for the angle θ in terms of radians is shown in Fig. 1.2. Its value is the length of the circumference of the circle subtended by θ divided by the radius. Because there are 360 ° in a complete circle, and the entire circumference of the circle is 2πR, 360 ° equals 2π radians. Equations containing angles are nearly always derived under the assumption that angles are expressed in radians. When you want to substitute the value of an angle in degrees into an equation, you usually need to convert it to radians first. An exception to this rule is when you use a calculator to evaluate functions such as sin θ. Most calculators expect angles to be input in degrees.

s u R

s u5 R

FIGURE 1.2 Definition of an angle in radians.

CONVERSION OF UNITS Many situations arise in engineering practice that require values expressed in one kind of unit to be converted into values in other units. For example, if some of the data to be used in an equation are given in SI units and some are given in US customary units, they must all be expressed in terms of one system of units before they are substituted into the equation. Converting units is straightforward, although it must be done with care. Suppose that we want to express 1 mile per hour (mi/h) in terms of feet per second (ft/s). Because 1 mile equals 5280 feet and 1 hour equals 3600 seconds, we can treat the expressions

( 5280 ft ) 1 mi

and

1 h ( 3600 s )

as ratios whose values are 1. In this way, we obtain

(

1 mi/h = ( 1 mi/h )

5280 ft 1 mi

)(

)

1 h = 1.47 ft/s. 3600 s

Some useful unit conversions are given in Table 1.2.

Table 1.2 Unit conversions. Time

1 minute 1 hour 1 day

Length 1 foot 1 mile 1 inch 1 foot

= 60 seconds = 60 minutes = 24 hours = = = =

12 inches 5280 feet 25.4 millimeters 0.3048 meters

Angle

2π  radians = 360 degrees

Mass

1 slug

= 14.59 kilograms

Force

1 pound

= 4.448 newtons

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Get complete eBook Order by email at [email protected] 10 Chapter 1 Introduction

RESULTS Identify the given information and the answer that must be determined. Develop a strategy; identify principles and equations that apply and how they will be used. Try to predict the answer whenever possible. Obtain the answer and, whenever possible, interpret it and compare it with the prediction. The number of significant digits in a given number refers to how many accurate digits it contains, counting from left to right beginning with the first nonzero digit. Numbers are often “rounded off” to a smaller number of significant digits. For example, the number 34,662 can be rounded off to three significant digits by writing it as 34,700 or 3.47E4. SI Units—The base units are time in seconds (s), length in meters (m), and mass in kilograms (kg). The unit of force is the newton (N), which is the force required to accelerate a mass of one kilogram at one meter per second squared.

Problem Solving: These steps apply to many types of problems.

Significant digits.

Systems of units.

US Customary Units—The base units are time in seconds (s), length in feet (ft), and force in pounds (lb). The unit of mass is the slug, which is the mass accelerated at one foot per second squared by a force of one pound.

s u R

s u5 R

Equivalent quantities, such as 1 hour 5 60 minutes, can be written as ratios whose values are 1: 1h 5 1, 60 min and used to convert units. For example, 15 min 5 15 min 1 h 5 0.25 h. 60 min

Definition of an angle in radians.

Conversion of units.

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Get complete eBook Order by email at [email protected] 1.1 Engineering and Mechanics 11 Practice Example 1.1

Converting Units

A man is riding a bicycle at a speed of 6 meters per second (m/s). How fast is he going in kilometers per hour (km/h)? Strategy One kilometer is 1000 meters and one hour is 60 minutes × 60 seconds = 3600 seconds. We can use these unit conversions to determine his speed in km/h. Solution Convert meters to kilometers. Convert seconds to hours. 6 m/s = 6 m/s 

Source: Courtesy of Stefan Schurr/ Shutterstock.

1 km ( 1000 m )( 3600 s ) 1 h

= 21.6 km/h. Practice Problem A man is riding a bicycle at a speed of 10 feet per second (ft/s). How fast is he going in miles per hour (mi/h)? Answer: 6.82 mi/h.

Practice Example 1.2

Converting Units of Pressure

The pressure exerted at a point of the hull of the deep submersible vehicle is 3.00 × 10 6  Pa (pascals). A pascal is 1 newton per square meter. Determine the pressure in pounds per square foot. Strategy From Table 1.2, 1 pound = 4.448 newtons and 1 foot = 0.3048 meters. With these unit conversions we can calculate the pressure in pounds per square foot. Solution The pressure (to three significant digits) is Convert

m2

to

ft 2

Source: Courtesy of WaterFrame_tfr/Alamy Stock Photo.

Convert newtons to pounds  1 lb  0.3048 m  2  3.00 × 10 6 N/m 2 = ( 3.00 × 10 6 N/m 2 )   4.448 N  1 ft  = 62,700 lb/ft 2. Practice Problem What is the pressure on the submersible in pounds per square inch? Answer: 435 lb/in 2 .

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Get complete eBook Order by email at [email protected] 12 Chapter 1 Introduction Practice Example 1.3

Determining Units from an Equation Consider the equation 1 2 mv , 2 where F is force in newtons, m is mass in kilograms, and v is velocity in m/s. (a) What are the units of d in SI base units? 1 (b) If mv 2 = 40 in SI units, what is its value in US customary units? 2 Strategy For part (a), we need to express the newton, a derived unit in the SI system, in terms of base units. Then by writing the given equation with the units of the terms shown explicitly, we can deduce the units of d . For part (b), we need to convert the units of the mass and velocity from SI units to US customary units. Fd =

Solution (a) Remember that the newton is defined using Newton’s second law: 1 N = ( 1 kg )( 1 m/s 2 ). The newton in terms of base units is 1 N = 1 kg-m/s 2. Writing the given equation with the units shown explicitly, Units of this expression kg-m are multiplied by s2 the units of d .

Units of this expression kg-m 2 . are s2

1( m  kg )( v  m/s ) 2. 2 We see that the units of d must be meters. 1 (b) The SI units of the expression mv 2 are kg-m 2 /s 2. Using unit conver2 sion factors, we obtain ( F  kg-m/s 2 ) d =

Convert m 2 to ft 2. Convert kilograms to slugs.  0.0685 slug  3.281 ft 40 kg-m 2 /s 2 = 40 kg-m 2 /s 2    1 kg 1 m

(

)

2

= 29.5 slug-ft 2 /s 2 . Practice Problem Consider the equation My σ = . I In US customary units, σ is expressed in lb/ft 2 , M in ft-lb, and y in ft. (a) What are the US customary units of I? (b) If σ = 60 lb/ft 2, what is its value in N/m 2? Answer: (a) The units of I are ft 4. (b) σ = 2870 N/m 2 .

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Problems

13

PROBLEMS (An asterisk signifies a problem whose solution is lengthier or more difficult.) 1.1 In 1967, the International Committee of Weights and Measures defined one second to be the time required for 9,192,631,770 cycles of the transition between two quantum states of the cesium-133 atom. Express the number of cycles in two seconds to four significant digits. 1.2 The base of natural logarithms is e = 2.71828183... . (a) Express e to three significant digits. (b) Determine the value of e 2 to three significant digits. (c) Use the value of e you obtained in part (a) to determine the value of e 2 to three significant digits. [Comparing the answers of parts (b) and (c) demonstrates the hazard of using rounded-off values in calculations.] 1.3 The base of natural logarithms (see Problem 1.2) is given by the infinite series 1 1 1 e = 2+ + + +  . 2! 3! 4! Its value can be approximated by summing the first few terms of the series. How many terms are needed for the approximate value rounded off to five digits to be equal to the exact value rounded off to five digits?

1.7 The length of this Boeing 737 is 110 ft  4 in and its wingspan is 117 ft  5 in. Its maximum takeoff weight is 154,500 lb. Its maximum range is 3365 nautical miles. Express each of these quantities in SI units to three significant digits.

Problem 1.7 1.8 The maglev (magnetic levitation) train from Shanghai to the airport at Pudong reaches a speed of 430 km/h. Determine its speed (a) in mi/h; (b) in ft/s.

1.4 The opening in the soccer goal is 24 ft wide and 8 ft high, so its area is 24 ft × 8 ft = 192 ft 2 . What is its area in m 2 to three significant digits?

Source: Courtesy of Qilai Shen/EPA/Shutterstock. Problem 1.8 Problem 1.4 1.5 In 2020, teams from China and Nepal, based on their independent measurements using GPS satellites, determined that the height of Mount Everest is 8848.86 meters. Determine the height of the mountain to three significant digits (a) in kilometers; (b) in miles. 1.6 The distance D = 6 in (inches). The magnitude of the moment of the force F = 12 lb (pounds) about point P is defined to be the product M P = FD. What is the value of M P in N-m (newton-meters)?

1.9 In the 2006 Winter Olympics, the men’s 15-km cross-country skiing race was won by Andrus Veerpalu of Estonia in a time of 38 minutes,  1.3 seconds. Determine his average speed (the distance traveled divided by the time required) to three significant digits (a) in km/h; (b) in mi/h. 1.10 The Porsche’s engine exerts 229 ft-lb (foot-pounds) of torque at 4600 rpm. Determine the value of the torque in N-m (newton-meters).

F

P D

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Problem 1.10

Get complete eBook Order by email at [email protected] 14 Chapter 1 Introduction 1.11 The kinetic energy of the man in Practice Example 1.1 is defined by 12 mv 2, where m is his mass and υ is his velocity. The man’s mass is 68 kg and he is moving at 6 m/s, so his kinetic energy is 12 (68 kg)(6 m/s) 2 = 1224 kg-m 2 /s 2. What is his kinetic energy in US customary units?

1.18 A typical speed of the head of a driver on the PGA Tour is 100 miles per hour. Determine the speed in ft/s, m/s,  and km/h.

1.12 The acceleration due to gravity at sea level in SI units is g = 9.81 m/s 2 . By converting units, use this value to determine the acceleration due to gravity at sea level in US customary units. 1.13 The value of the universal gravitational constant in SI units is G = 6.67E−11 m 3 /kg-s 2 . Use this value and convert units to determine the value of G in US customary units. 1.14 The density (mass per unit volume) of aluminum is 2700 kg/m 3 . Determine its density in slug/ft 3 . 1.15 The cross-sectional area of the C12×30 American Standard Channel steel beam is A = 8.81 in 2 . What is its cross-sectional area in mm 2 ?

Problem 1.18 1.19 In 2000, a carbon nanotube was shown to support a tensile stress of 63 GPa (gigapascals). A pascal is 1 N/m 2 . Determine the value of this tensile stress in lb/in 2 . (This is the amount of force in pounds that could theoretically be supported in tension by a bar of this material with a cross-sectional area of one square inch.)

y A

x

Problem 1.15 1.16 A pressure transducer measures a value of 300 lb/in 2. Determine the value of the pressure in pascals. A pascal (Pa) is one newton per square meter. 1.17 A horsepower is 550 ft-lb/s. A watt is 1 N-m/s. Determine how many watts are generated by the engines of the passenger jet if they are producing 7000 horsepower.

Problem 1.17

Source: Courtesy of Takanakai/123RF.

1.20 In dynamics, the moments of inertia of an object are expressed in units of ( mass ) × ( length ) 2. If one of the moments of inertia of a particular object is 600 slug-ft 2, what is its value in kg-m 2? 1.21 The airplane’s drag coefficient C D is defined by D CD = 1 2 , S 2 ρv where D is the drag force exerted on the airplane by the air, S is the wing area, ρ is the density (mass per unit volume) of the air, and v is the magnitude of the airplane’s velocity. At the instant shown, the drag force D = 5300 N, the wing area is S = 100 m 2 , the air density is ρ = 1.226 kg/m 3 , and the velocity is v = 60 m/s. (a) What is the value of the drag coefficient? (b) Determine the values of D, S, ρ, and v in terms of US customary units and use them to calculate the drag coefficient.

Source: Courtesy of Fasttailwind/Shutterstock. Problem 1.21

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Get complete eBook Order by email at [email protected] 1.2 Newtonian Gravitation 15

1.2 Newtonian Gravitation QUESTIONS TO CONSIDER • What is the distinction between an object’s weight and its mass? If you know how much something weighs, how do you determine its mass? • When a person is in orbit around the Earth, they are said to be “weightless.” Is that true?

BACKGROUND

m1

Newton postulated that the gravitational forces exerted on each other by two F particles of masses m1 and m 2 that are separated by a distance r (Fig. 1.3) are opposite in direction and of equal magnitude, Gm1m 2 , F = (1.1) F r r2 where G is called the universal gravitational constant. The value of G is 6.67E−11 m 3 /kg-s 2 ,  or 3.44E−8 ft 3 /slug-s 2 . Based on this postulate, he calculated the gravitational force between a particle of mass m1 and a homogeneous sphere of mass m 2 and found that it is also given by Eq. (1.1), with r denoting the distance from the particle to the center of the sphere. Although the Earth is not a homogeneous sphere, we can use this result to approximate the weight FIGURE 1.3 of an object of mass m due to the gravitational attraction of the Earth. We have The gravitational forces between two Gmm E , W = (1.2) particles are equal in magnitude and directed along the line between them. r2 where m E is the mass of the Earth, and r is the distance from the center of the Earth to the object. Notice that the weight of an object depends on its location relative to the center of the Earth, whereas the mass of the object is a measure of the amount of matter it contains and doesn’t depend on its position. When an object’s weight is the only force acting on it, the resulting acceleration is called the acceleration due to gravity. In this case, Newton’s second law states that W = ma, and from Eq. (1.2) we see that the acceleration due to gravity is Gm E a = . (1.3) r2 The acceleration due to gravity at sea level is denoted by g. Denoting the radius of the Earth by R E , we see from Eq. (1.3) that Gm E = gR E 2 . Substituting this result into Eq. (1.3), we obtain an expression for the acceleration due to gravity at a distance r from the center of the Earth in terms of the acceleration due to gravity at sea level: R 2 a   = g E2 . (1.4) r Because the weight of the object W = ma, the weight of an object at a distance r from the center of the Earth is RE 2 . W Get = mg (1.5) complete eBook Order by email at [email protected] r2

m2

Get complete eBook Order by email at [email protected] 16 Chapter 1 Introduction At sea level (r = R E ), the weight of an object is given in terms of its mass by the simple relation W = mg.

(1.6)

The value of g varies from location to location on the surface of the Earth. The values we use in examples and problems are g = 9.81 m/s 2 in SI units and g = 32.2 ft/s 2 in US customary units. In classical, or Newtonian, mechanics an object’s mass is a constant that does not depend on its location. In contrast, an object’s weight is the force exerted on it by gravity and, as we have seen, does depend on location.

RESULTS

Gm1m 2 F = . r2

a =

Gm E . r2

R 2 a = g E2 . r

(1.1)

The gravitational force between two particles of mass m1 and m 2 that are a distance r apart. The value of the universal gravitational constant G is 6.67E−11 m 3 /kg-s 2 , or 3.44E−8 ft 3 /slug-s 2 .

(1.3)

The acceleration due to gravity of the Earth (modeled as a homogeneous sphere) at a distance r from its center, where m E is the Earth’s mass.

(1.4)

The acceleration due to gravity of the Earth at a distance r from its center, where R E is the Earth’s radius and g is the acceleration due to gravity at sea level. This is an alternative form of Eq. (1.3).

R 2 W = mg E2 . r

(1.5)

Weight of an object of mass m at a distance r from the center of the Earth, where R E is the Earth’s radius and g is the acceleration due to gravity at sea level.

W = mg.

(1.6)

Weight of an object m at sea level on Earth, where g is the acceleration due to gravity at sea level.

In terms of the two systems of units we are using, Eq. (1.6) is Acceleration due to × gravity at sea level 32.2 ft/s 2

Weight at sea level in pounds

=

Weight at sea level in newtons

Acceleration due to = Mass in kilograms × gravity at sea level 9.81 m/s 2

Mass in slugs

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Get complete eBook Order by email at [email protected] 1.2 Newtonian Gravitation 17 Practice Example 1.4

Weight and Mass

The C-clamp weighs 14 oz at sea level. [16 oz (ounces) = 1 lb.] The acceleration due to gravity at sea level is g = 32.2 ft/s 2 . What is the mass of the C-clamp in slugs? Strategy We must first determine the weight of the C-clamp in pounds. Then we can use Eq. (1.6) to determine the mass in slugs. Solution  1 lb  14 oz = 14 oz  = 0.875 lb.  16 oz 

m =

0.875 lb W = = 0.0272 slug. g 32.2 ft/s 2

Convert the weight from ounces to pounds.

Use Eq. (1.6) to calculate the mass in slugs.

Practice Problem The mass of the C-clamp is 0.397 kg. The acceleration due to gravity at sea level is g = 9.81 m/s 2. What is the weight of the C-clamp at sea level in newtons? Answer: 3.89 N.

Practice Example 1.5

Weight of an Object Near the Ground

When an object is near the surface of the Earth, the dependence of its weight on distance from the center of the Earth can often be neglected, and its weight can be assumed to be its weight W = mg at sea level. To illustrate this, determine the height above sea level at which the weight of an object decreases to 0.999mg. Strategy An object’s weight at a distance r from the center of the Earth is given by Eq. (1.5). We can use it to solve for the height h above the ground at which the object’s weight is 0.999mg. Source: Courtesy of Valentin Valkov/123RF.

Solution mg

RE 2 = 0.999mg. ( RE + h )2

Equate Eq. (1.5) with r = R E + h to 0.999mg.

We solve this equation for h, obtaining h = 0.000500375 R E . The radius of the Earth, R E = 6370 km, so h = 3.19 km (10,500 ft).

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(continued)

Get complete eBook Order by email at [email protected] 18 Chapter 1 Introduction This example has an important implication. For most situations we deal with in applications of statics and dynamics, an object’s weight can be treated as a constant, its weight at sea level. Practice Problem The radius of the planet Neptune is 24,600 km and its mass is 1.02E26 kg. If you model it as a homogeneous sphere, what is the acceleration due to gravity at its surface? Answer: 11.2 m/s 2 .

Practice Example 1.6

Determining an Object’s Weight The mass of the Mars rover Curiosity is 899 kg. The acceleration due to gravity at the surface of Mars is 3.72 m/s 2 , and the radius of Mars is 3390 km. (a) What was the rover’s weight when it was at sea level on Earth? (b) What is its weight on the surface of Mars? Strategy For part (a), we know that the acceleration due to gravity at sea level on Earth is g = 9.81 m/s 2 . We can use Eq. (1.6) to determine the weight. We can also use Eq. (1.6) for part (b) by using the given acceleration due to gravity for the surface of Mars g M = 3.72 m/s 2 .

Source: Courtesy of NASA.

Solution

(a) Eq. (1.6) gives an object’s weight at sea level.

(b) Adapt Eq. (1.6) to Mars to obtain the weight at the surface.

W = mg = ( 899 kg )( 9.81 m/s 2 ) = 8820 N (1980 lb).

W M = mg M = ( 899 kg )( 3.72 m/s 2 ) = 3340 N (752 lb).

Practice Problem When the spacecraft carrying Curiosity approached Mars, the entry phase began when it was 125 km above the surface. What was the rover’s weight at that point? Answer: 3110 N (699 lb).

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Problems

19

PROBLEMS 1.22 A person has a mass of 72 kg. (a) What is their weight at sea level in newtons? (b) What is their mass in slugs? (c) What is their weight at sea level in pounds? 1.23 The 1 ft × 1 ft × 1 ft cube of iron weighs 490 lb at sea level. Determine the weight in newtons of a 1 m × 1 m × 1 m cube of the same material at sea level.

1 ft

1 ft

1 ft

Problem 1.23 1.24 The area of the Pacific Ocean is 64,186,000 square miles and its average depth is 12,925 ft. Assume that the weight per unit volume of ocean water is 64 lb/ft 3. Determine the mass of the Pacific Ocean (a) in slugs; (b) in kilograms. 1.25 Two oranges weighing 8 oz (ounces) each at sea level are 3 ft apart. (a) What is the magnitude of the gravitational force F they exert on each other? (b) If a stationary object of mass m is subjected to a constant force F and no other forces, the time required for it to reach a speed v is mv . t = F If you subjected one of the oranges to a constant force equal to the force you determined in part (a) and no other forces, how long would it take to reach a speed of 1 ft/s?

1.26 A person weighs 180 lb at sea level. The radius of the Earth is 3960 mi. What force is exerted on the person by the gravitational attraction of the Earth if he is in a space station in orbit 200 mi above the surface of the Earth? 1.27 The acceleration due to gravity on the surface of the Moon is 1.62 m/s 2 . The Moon’s radius is R M = 1738 km. (See Practice Example 1.6.) (a) What is the weight in newtons on the surface of the Moon of an object that has a mass of 10 kg? (b) Using the approach described in Practice Example 1.6, determine the force exerted on the object by the gravity of the Moon if the object is located 1738 km above the Moon’s surface. 1.28 If an object is near the surface of the Earth, the variation of its weight with distance from the center of the Earth can often be neglected. The acceleration due to gravity at sea level is g = 9.81 m/s 2. The radius of the Earth is 6370 km. The weight of an object at sea level is mg, where m is its mass. At what height above the surface of the Earth does the weight of the object decrease to 0.99 mg ? 1.29 A person has a mass of 72 kg. The radius of the Earth is 6370 km. (a) What is his weight at sea level in newtons and in pounds? (b) If he is in a space station 420 km above the surface of the Earth, what force is exerted on him by the Earth’s gravitational attraction in newtons and in pounds? 1.30 If you model the Earth as a homogeneous sphere, at what height above sea level does your weight decrease to one-half of its value at sea level? Determine the answer (a) in kilometers; (b) in miles. (The radius of the Earth is 6370 km.)

3 ft

Problem 1.25

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Source: Courtesy of Sherab/Alamy Stock Photo.

m Vectors are used to specify positions, directions, and distances, such as those of the points of the apartment building. Vectors are used to describe any quantities that have magnitude and direction, including forces, moments, velocities, and accelerations.

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Vectors

If an object is subjected to several forces that have different magnitudes and act in different directions, how can the magnitude and direction of the resulting total force on the object be determined? Forces are vectors and must be added according to the definition of vector addition. In engineering we deal with many quantities that have both magnitude and direction and can be expressed and analyzed as vectors. In this chapter we review vector operations, express vectors in terms of components, and present examples of engineering applications of vectors.

y z

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x

CHAPTER

2

22 Chapter 2 VectorsGet complete eBook Order by email at [email protected]

2.1 Scalars and Vectors QUESTIONS TO CONSIDER • What kinds of quantities must be represented by vectors? • What is meant by adding or subtracting vectors?

BACKGROUND

B A (a)

B A A

rAB

(b)

FIGURE 2.1 (a) Two points A and B of a mechanism. (b) The vector r AB from A to B.

A

F

B

FIGURE 2.2 Representing the force cable AB exerts on the tower by a vector F.

A physical quantity that is completely described by a real number is called a scalar. Time is a scalar quantity. Mass is also a scalar quantity. For example, you completely describe the mass of a car by saying that its value is 1200 kg. In contrast, you have to specify both a nonnegative real number, or magnitude, and a direction to describe a vector quantity. Two vector quantities are defined to be equal only if both their magnitudes and their directions are equal. The position of a point in space relative to another point is a vector quantity. To describe the location of a city relative to your home, it is not enough to say that it is 100 miles away. You must say that it is 100 miles west of your home. Force is also a vector quantity. When you push a piece of furniture across the floor, you apply a force of magnitude sufficient to move the furniture and you apply it in the direction you want the furniture to move. We will represent vectors by boldfaced letters, U, V, W, . . ., and will denote the magnitude of a vector U by U . A vector is represented graphically by an arrow. The direction of the arrow indicates the direction of the vector, and the length of the arrow is defined to be proportional to the magnitude. For example, consider the points A and B of the mechanism in Fig. 2.1a. We can specify the position of point B relative to point A by the vector r AB in Fig. 2.1b. The direction of r AB indicates the direction from point A to point B. If the distance between the two points is 200 mm, the magnitude r AB = 200 mm. The cable AB in Fig. 2.2 helps support the television transmission tower. We can represent the force the cable exerts on the tower by a vector F as shown. If the cable exerts an 800-N force on the tower, F = 800 N. (A cable suspended in this way will exhibit some sag, or curvature, and the tension will vary along its length. For now, we assume that the curvature in suspended cables and ropes and the variations in their tensions can be neglected. This assumption is approximately valid if the weight of the rope or cable is small in comparison to the tension. We discuss and analyze suspended cables and ropes in more detail in Chapter 10.) Vectors are a convenient means for representing physical quantities that have magnitude and direction, but that is only the beginning of their usefulness. Just as real numbers are manipulated with the familiar rules for addition, subtraction, multiplication, and so forth, there are rules for manipulating vectors. These rules provide powerful tools for engineering analysis.

VECTOR ADDITION When an object moves from one location in space to another, we say it undergoes a displacement. If we move a book (or, speaking more precisely, some point of a book) from one location on a table to another, as shown in Fig. 2.3a,

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Get complete eBook Order by email at [email protected] 2.1 Scalars and Vectors 23

V

U

U

(a)

(b)

FIGURE 2.3 U

V

U

V

W

(a) A displacement represented by the vector U. (b) The displacement U followed by the displacement V. (c) The displacements U and V are equivalent to the displacement W. (d) The final position of the book doesn’t depend on the order of the displacements.

V W

U

(d)

(c)

we can represent the displacement by the vector U. The direction of U indicates the direction of the displacement, and U is the distance the book moves. Suppose that we give the book a second displacement V, as shown in Fig.  2.3b. The two displacements U and V are equivalent to a single displacement of the book from its initial position to its final position, which we represent by the vector W in Fig.  2.3c. Notice that the final position of the book is the same whether we first give it the displacement U and then the displacement V or we first give it the displacement V and then the displacement U (Fig.  2.3d). The displacement W is defined to be the sum of the displacements U and V: U + V = W. The definition of vector addition is motivated by the addition of displacements. Consider the two vectors U and V shown in Fig. 2.4a. If we place them head to tail (Fig. 2.4b), their sum is defined to be the vector from the tail of U to the head of V (Fig. 2.4c). This is called the triangle rule for vector addition. Fig.  2.4d demonstrates that the sum is independent of the order in which the vectors are placed head to tail. From this figure we obtain the parallelogram rule for vector addition (Fig. 2.4e). V

V U

U

U V (a)

U1V

FIGURE 2.4

U

U

U1V V

V (d)

(c)

(b) V

U

U1V

(e)

(a) (b) (c) (d)

Two vectors U and V. The head of U placed at the tail of V. The triangle rule for obtaining the sum of U and V. The sum is independent of the order in which the vectors are added. (e) The parallelogram rule for obtaining the sum of U and V.

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24 Chapter 2 VectorsGet complete eBook Order by email at [email protected] The definition of vector addition implies that

W

U+V = V+U

V U1V1W

Vector addition is commutative.

and (U + V ) + W = U + ( V + W)

U

FIGURE 2.5 Sum of the three vectors U,  V, and W. V U

W

FIGURE 2.6 Three vectors U,  V, and W whose sum is zero.

(2.1)

Vector addition is associative.

(2.2)

for any vectors U, V, and W. These results mean that when two or more vectors are added, the order in which they are added doesn’t matter. The sum can be obtained by placing the vectors head to tail in any order, and the vector from the tail of the first vector to the head of the last one is the sum (Fig. 2.5). If the sum of two or more vectors is zero, they form a closed polygon when they are placed head to tail (Fig. 2.6). A physical quantity is called a vector if it has magnitude and direction and obeys the definition of vector addition. We have seen that displacement is a vector. The position of a point in space relative to another point is also a vector quantity. In Fig. 2.7, the vector r AC from A to C is the sum of r AB and rBC . A force has direction and magnitude, but do forces obey the definition of vector addition? For now we will assume that they do. When we discuss dynamics, we will show that Newton’s second law implies that force is a vector.

PRODUCT OF A SCALAR AND A VECTOR The product of a scalar (real number) a and a vector U is a vector written as aU. Its magnitude is a U , where a is the absolute value of the scalar a. The direction of aU is the same as the direction of U when a is positive and is opposite to the direction of U when a is negative. The product (−1)U is written as −U and is called “the negative of the vector U.” It has the same magnitude as U but the opposite direction. The division of a vector U by a scalar a is defined to be the product

C rBC B

rAC

rAB A

( )

U 1 = U. a a FIGURE 2.7 Arrows denoting the relative positions of points are vectors.

Fig. 2.8 shows a vector U and the products of U with the scalars 2, −1, and 1 2. The definitions of vector addition and the product of a scalar and a vector imply that a(bU) = (ab)U,

The product is associative with  respect to scalar multiplication.

(a + b)U = aU + bU,

The products are distributive  with respect to scalar addition.

(2.3)

(2.4)

and U

2U

2U 5 (21)U

U 1 5 U 2 2

FIGURE 2.8 A vector U and some of its scalar multiples.

a ( U + V ) = a U + a V,

The products are distributive  with respect to vector addition.

(2.5)

for any scalars a and b and vectors U and V. We will need these results when we discuss components of vectors.

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Get complete eBook Order by email at [email protected] 2.1 Scalars and Vectors 25 FIGURE 2.9

(21)V

V

U

V

U2V

U

(a) Two vectors U and V. (b) The vectors V and (−1) V. (c) The sum of U and (−1) V is the vector difference U − V.

(21)V (b)

(a)

(c)

VECTOR SUBTRACTION The difference of two vectors U and V is obtained by adding U to the vector (−1) V: U − V = U + (−1) V.

(2.6)

Consider the two vectors U and V shown in Fig. 2.9a. The vector (−1) V has the same magnitude as the vector V but is in the opposite direction (Fig. 2.9b). In Fig. 2.9c, we add the vector U to the vector (−1) V to obtain U − V.

UNIT VECTORS A unit vector is simply a vector whose magnitude is 1. A unit vector specifies a direction and also provides a convenient way to express a vector that has a particular direction. If a unit vector e and a vector U have the same direction, we can write U as the product of its magnitude U and the unit vector e (Fig. 2.10), U = U e. Any vector U can be regarded as the product of its magnitude and a unit vector that has the same direction as U. Dividing both sides of this equation by U yields

U

uUu

e

1 u U ue 5 U

FIGURE 2.10 Because U and e have the same direction, the vector U equals the product of its magnitude with e.

U = e, U so dividing any vector by its magnitude yields a unit vector that has the same direction.

RESULTS A physical quantity that is completely described by a real number is called a scalar. A vector has both magnitude and direction and satisfies a defined rule of addition. A vector is represented graphically by an arrow whose length is defined to be proportional to the magnitude.

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26 Chapter 2 VectorsGet complete eBook Order by email at [email protected] V

U

U1V

Vector Addition The sum of two vectors U and V is defined by the triangle rule or the equivalent parallelogram rule.

Triangle rule

U U1V V Parallelogram rule

Product of a Scalar and a Vector The product of a scalar a and a vector U is def ined to be a vector aU with magnitude ƒ a ƒƒ U ƒ. Its direction is the same as that of U when a is positive and is opposite to that of U when a is negative. The division of U by a is def ined to be the product (1>a) U.

U

2U

U

2U 5 (21)U

V

Vector Subtraction The difference of two vectors U and V is def ined by U 2 V 5 U 1 (21)V.

(21)V

U2V

Unit Vectors A unit vector is a vector whose magnitude is 1. Any vector U can be expressed as ƒU ƒe, where e is a unit vector with the same direction as U. Dividing a vector U by its magnitude yields a unit vector with the same direction as U.

U 1 5 U 2 2

U

ƒU ƒ

U

e

1

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ƒU ƒe = U

Get complete eBook Order by email at [email protected] 2.1 Scalars and Vectors 27 Practice Example 2.1

Vector Operations

The magnitudes of the vectors shown are U = 8 and  V = 3. The vector V is vertical. Graphically determine the magnitude of the vector U + 2 V. Strategy By drawing the vectors to scale and applying the triangle rule for addition, we can measure the magnitude of the vector U + 2 V.

U

V

458

Solution Drawing the vectors U and 2V to scale, place them head to tail.

6

2V

8 U 458

The measured value of ƒ U 1 2V ƒ is 13.0.

2V 13.0

U 458

Practice Problem The magnitudes of the vectors shown are U = 8 and V = 3. The vector V is vertical. Graphically determine the magnitude of the vector U − 2 V. Answer: U − 2 V = 5.7.

Example 2.2

U

V

458

Adding Vectors

Part of the roof of a sports stadium is to be supported by the cables AB and AC. The forces the cables exert on the pylon to which they are attached are represented by the vectors F AB and F AC . The magnitudes of the forces are F AB = 100 kN and F AC = 60 kN. Determine the magnitude and direction of the sum of the forces exerted on the pylon by the cables.

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(continued )

28 Chapter 2 VectorsGet complete eBook Order by email at [email protected]

B FAB

A

FAB

FAB 1 FAC

100 kN 198 60 kN

FAC

(a) Graphical solution.

308

308

FAC

C

Strategy By drawing the parallelogram rule for adding the two forces with the vectors drawn to scale, we can measure the magnitude and direction of their sum. Solution We graphically construct the parallelogram rule for obtaining the sum of the two forces with the lengths of F AB and F AC proportional to their magnitudes (Fig. a). By measuring the figure, we estimate the magnitude of the vector F AB + F AC to be 155 kN and its direction to be 19 ° above the horizontal. Critical Thinking In engineering applications, vector operations are nearly always done analytically. So why is it worthwhile to gain experience with graphical methods? Doing so enhances your intuition about vectors and helps you understand vector operations. Also, sketching out a graphical solution can often help you formulate an analytical solution.

PROBLEMS 2.1 Consider vectors U and V oriented as shown. Their magnitudes are U = 8 and V = 3. Graphically determine the magnitude of the vector 2U − 3V. 208 458

2.3 Two unit vectors e 1 and e 2 are oriented as shown. Graphically determine the magnitude of the vector U = 2e 1 + 3e 2 . 2.4 Two unit vectors e 1 and e 2 are oriented as shown. Graphically determine the magnitude of the vector U = 4 e 1 − 3e 2 .

U V

Problem 2.1 2.2 Suppose that the pylon in Example 2.2 is moved closer to the stadium so that the angle between the forces F AB and F AC is 45 ° . Draw a sketch of the structure with the cables in their new orientation. The magnitudes of the forces are F AB = 100 kN and F AC = 60 kN. Graphically determine the magnitude and direction of the sum of the forces exerted on the pylon at A by the two cables.

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e2

608

e1

Problems 2.3/2.4 2.5 Three forces act on the ring. Their sum is zero: F A + FB + FC = 0. The magnitudes FB = 100 N and FC = 160 N. Graphically determine the magnitude of F A and the angle email at α. [email protected]

Get complete eBook Order by email at [email protected] 2.6 Three forces act on the ring. Their sum is zero: F A + FB + FC = 0. The magnitude FC = 200 lb. Graphically determine the value of the angle α for which the magnitude of F A is a minimum. If α has that value, what are the magnitudes of F A and FB ? FB FA

408

Problems

29

2.9 The 2700-kg tractor is at rest on the inclined surface. The angle α = 30 ° relative to the horizontal. The forces exerted on the tractor’s tires by the inclined surface are represented by the normal force N, which is perpendicular to the surface, and the friction force f, which is tangential to the surface. The sum of the normal force, the friction force, and the tractor’s weight is zero: N + f + W = 0. Graphically determine the magnitudes of N and f in newtons. 2.10 The 2700-kg tractor is at rest on the inclined surface. The forces exerted on the tractor’s tires by the inclined surface are represented by the normal force N, which is perpendicular to the surface, and the friction force f, which is tangential to the surface. The sum of the normal force, the friction force, and the tractor’s weight is zero: N + f + W = 0. Graphically determine the value of the angle α relative to the horizontal for which the magnitudes of N and f are equal. What is their magnitude?

a

808

FC

Problems 2.5/2.6 2.7 A T-shaped structural member is suspended from cables. The member is subjected to three forces: the forces F A and FB exerted by the cables and its weight W. The weight of the member is W = 1000 lb. The sum of the forces is zero. Graphically determine the magnitudes of F A and FB . 2.8 A T-shaped structural member is suspended from cables. The member is subjected to three forces: the forces F A and FB exerted by the cables and its weight W. The sum of the forces is zero. Suppose that the magnitude of the force exerted by either cable must not exceed 1200 lb. Graphically determine the largest magnitude of the weight W that can be suspended in this way.

f W

a

N

Source: Courtesy of Tka4ko/Shutterstock. Problems 2.9/2.10 2.11 A spherical storage tank is suspended from cables The tank is subjected to three forces: the forces F A and FB exerted by the cables and its weight W. The weight of the tank is W = 800 lb. The vector sum of the forces acting on the tank equals zero. Graphically determine the magnitudes of F A and FB .

FA 408 408

308

408 F A

FB

FB

208

208

308

W

W

Problem 2.11

Problems 2.7/2.8

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30 Chapter 2 VectorsGet complete eBook Order by email at [email protected] 2.12 The rope ABC exerts forces FBA and FBC of equal magnitude on the block at B. The magnitude of the total force exerted on the block by the two forces is 200 lb. Graphically determine FBA .

2.14 A surveyor determines that the horizontal distance from A to B is 400 m and the horizontal distance from A to C is 600 m. Graphically determine the magnitude of the vector rBC and the angle α. North

FBC

C

B 208

a rBC

B

B

C 608 208

FBA

East

A

A

Problem 2.14 2.15 The figure shows three forces acting on a joint of a structure. The magnitude of FC is 60 kN, and F A + FB + FC = 0. Graphically determine the magnitudes of F A and FB .

Problem 2.12 2.13 Two snowcats tow an emergency shelter to a new location near McMurdo Station, Antarctica. (The top view is shown. The cables are horizontal.) The total force F A + FB exerted on the shelter is in the direction of the line L and has a magnitude of 300 lb. Graphically determine the magnitudes of F A and FB .

y FC FB 158

L

FA

508

308

x

408 FB

FA

Problem 2.15 2.16 By drawing sketches of the vectors, explain why

Top View

U + ( V + W) = (U + V) + W.

Problem 2.13

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Get complete eBook Order by email at [email protected] 2.2 Components in Two Dimensions 31

2.2 Components in Two Dimensions QUESTIONS TO CONSIDER • What is meant by the “components” of a vector, and how can you determine them? • What is a unit vector? What does it tell you?

BACKGROUND Vectors are much easier to work with when they are expressed in terms of mutually perpendicular vector components. Here we explain how to express vectors in cartesian components and give examples of vector manipulations using components. Consider the vector U in Fig. 2.11a. By placing a cartesian coordinate system so that the vector U is parallel to the x–y plane, we can write it as the sum of perpendicular vector components U x and U y that are parallel to the x and y axes (Fig. 2.11b): U = Ux + Uy.

U

(a) y U

Then by introducing a unit vector i defined to point in the direction of the positive x-axis and a unit vector j defined to point in the direction of the positive y-axis (Fig. 2.11c), we can express the vector U in the form U = U x i + U y j.

U =

Ux 2 + U y 2 .

Ux (b)

(2.7)

The scalars U x and U y are called scalar components of U. When we refer simply to the components of a vector, we will mean its scalar components. We will refer to U x and U y as the x and y components of U. The components of a vector specify both its direction relative to the cartesian coordinate system and its magnitude. From the right triangle formed by the vector U and its vector components (Fig. 2.11c), we see that the magnitude of U is given in terms of its components by the Pythagorean theorem: (2.8)

With this equation the magnitude of a vector can be determined when its components are known.

U

Uy 5 Uy j

Ux 5 Ux i

j i

(c)

x

FIGURE 2.11 (a) A vector U. (b) The vector components U x and U y . (c) The vector components can be expressed in terms of i and j.

The sum of two vectors U and V in terms of their components is U + V = (U x i + U y j) + (V x i + V y j) (2.9)

The components of U + V are the sums of the components of the vectors U and V. Notice that in obtaining this result we used Eqs. (2.2), (2.4), and (2.5).

x

y

MANIPULATING VECTORS IN TERMS OF COMPONENTS

= (U x + V x ) i + (U y + V y ) j.

Uy

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32 Chapter 2 VectorsGet complete eBook Order by email at [email protected] y

y

V

U1V

Vy j

U1V

U1V

(Uy 1 Vy)j

Uy j Vx i

U

Ux i (a)

x

(Ux 1 Vx)i

(b)

x

(c)

FIGURE 2.12 (a) The sum of U and V. (b) The vector components of U and V. (c) The sum of the components in each coordinate direction equals the component of U + V in that direction.

It is instructive to derive Eq. (2.9) graphically. The summation of U and V is shown in Fig. 2.12a. In Fig. 2.12b we introduce a coordinate system and show the components U and V. In Fig. 2.12c we add the x and y components, obtaining Eq. (2.9). The product of a number a and a vector U in terms of the components of U is

y

rAB A

B (xB, yB)

a U = a(U x i + U y j) = aU x i + aU y j.

(xA, yA) x (a)

POSITION VECTORS IN TERMS OF COMPONENTS

y B

yB rAB yA

(yB 2 yA)j

A (xB 2 xA)i xA

(b)

The component of aU in each coordinate direction equals the product of a and the component of U in that direction. We used Eqs. (2.3) and (2.5) to obtain this result.

xB

x

We can express the position vector of a point relative to another point in terms of the cartesian coordinates of the points. Consider point A with coordinates ( x A ,   y A ) and point B with coordinates ( x B ,   y B ). Let r AB be the vector that specifies the position of B relative to A (Fig. 2.13a). That is, we denote the vector from a point A to a point B by r AB . We see from Fig. 2.13b that r AB is given in terms of the coordinates of points A and B by r AB = ( x B − x A ) i + ( y B − y A ) j.

FIGURE 2.13 (a) Two points A and B and the position vector r AB from A to B. (b) The components of r AB can be determined from the coordinates of points A and B.

(2.10)

Notice that the x component of the position vector from a point A to a point B is obtained by subtracting the x coordinate of A from the x coordinate of B, and the y component is obtained by subtracting the y coordinate of A from the y coordinate of B.

RESULTS A vector U that is parallel to the x–y plane can be expressed as U 5 Ux i 1 Uy j, (2.7) y

where i is a unit vector that points in the positive x-axis direction and j is a unit vector that points in the positive y-axis direction. U

x

The magnitude of U is given by ƒUƒ 5

2

2

(2.8)

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MANIPULATING VECTORS IN TERMS OF COMPONENTS U + V = (U x i + U y j) + (V x i + V y j) = (U x + V x ) i + (U y + V y ) j,

(2.9)

a U = a(U x i + U y j) = aU x i + aU y j.

Vector addition (or subtraction) and multiplication of a vector by a number can be carried out in terms of components.

POSITION VECTORS IN TERMS OF COMPONENTS y

B (xB, yB)

rAB A

The position vector from A to B is given by rAB 5 (xB 2 xA)i 1 (yB 2 yA)j. (2.10)

(xA, yA) x

Practice Example 2.3

Determining Components A

The cable from point A to point B exerts a 900-N force on the top of the television transmission tower that is represented by the vector F . Express F in terms of components using the coordinate system shown. 80 m

Strategy We will determine the components of the vector F in two ways. In the first method, we will determine the angle between F and the y-axis and use trigonometry to determine the components. In the second method, we will use the given slope of the cable AB and apply similar triangles to determine the components of F. Solution

40 m

y

First Method

y A

A

Determine the angle between F and the y-axis: 40 5 26.68. a 5 arctan 80

1 2

Force exerted on the tower by cable AB

a 80 m

B

F

80 m

B

F

B

x

40 m

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x

40 m

(continued )

34 Chapter 2 VectorsGet complete eBook Order by email at [email protected] y A

Use trigonometry to determine F in terms of its components:

a

F 5 u F usin ai 2 u F ucos aj 5 900 sin 26.68 i 2 900 cos 26.68 j (N)

F

5 402i 2 805j (N). B

x

y

Second Method A

Using the given dimensions, calculate the distance from A to B:

80 m

(40 m)2 1 (80 m)2 5 89.4 m. B

x

40 m

y

Use similar triangles to determine the components of F: u Fx u u Fyu 80 m 40 m and , 5 5 u F u 89.4 m u F u 89.4 m so 40 80 (900 N)i 2 (900 N)j F5 89.4 89.4 5 402i 2 805j (N).

uFu

u Fy u

89.4 m

80 m

u Fx u x 40 m

Practice Problem The cable from point A to point B exerts a 900-N force on the top of the television transmission tower that is represented by the vector F. Suppose that you change the placement of point B so that the magnitude of the y component of F is three times the magnitude of the x component of F. Express F in terms of its components. How far along the x-axis from the origin of the coordinate system should B be placed? Answer: F = 285i − 854 j (N). Place point B at 26.7 m from the origin.

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Get complete eBook Order by email at [email protected] 2.2 Components in Two Dimensions 35 Example 2.4

Determining Components in Terms of an Angle

Hydraulic cylinders are used to exert forces in many mechanical devices. The force is exerted by pressurized liquid (hydraulic fluid) pushing against a piston within the cylinder. The hydraulic cylinder AB exerts a 4000-lb force F on the bed of the dump truck at B. Express F in terms of components using the coordinate system shown. y

B

B 308 A

F 308

A x

Strategy When the direction of a vector is specified by an angle, as in this example, we can determine the values of the components from the right triangle formed by the vector and its components. Solution We draw the vector F and its vector components in Fig. a. From the resulting right triangle, we see that the magnitude of Fx is

y

Fx = F cos30 ° = (4000 lb) cos30 ° = 3460 lb. Fx points in the negative x direction, so Fx = −3460 i  (lb). The magnitude of Fy is Fy = F sin 30 ° = (4000 lb)sin 30 ° = 2000 lb.

Fx Fy

308

x

(a) The force F and its components form a right triangle.

The vector component Fy points in the positive y direction, so Fy = 2000 j (lb). The vector F, in terms of its components, is F = Fx + Fy = −3460 i + 2000 j (lb). The x component of F is −3460 lb, and the y component is 2000 lb. Critical Thinking When you have determined the components of a given vector, you should make sure they appear reasonable. In this example you can see from the vector’s direction that the x component should be negative and the y component positive. You can also make sure that the components yield the correct magnitude. In this example, F =

F

(−3460 lb) 2 + (2000 lb) 2 = 4000 lb.

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Example 2.5

Determining an Unknown Vector Magnitude The cables A and B exert forces F A and FB on the hook. The magnitude of F A is 100 lb. The tension in cable B has been adjusted so that the total force F A + FB is perpendicular to the wall to which the hook is attached. (a) What is the magnitude of FB? (b) What is the magnitude of the total force exerted on the hook by the two cables?

A 408

Strategy The vector sum of the two forces is perpendicular to the wall, so the sum of the components parallel to the wall equals zero. From this condition we can obtain an equation for the magnitude of FB . Solution (a) In terms of the coordinate system shown in Fig. a, the components of F A and FB are

208 B

F A = F A sin 40 °i FA cos 40 ° j, FB = FB sin 20 °i − FB cos 20 ° j. The total force is

FA

408

F A + FB = ( F A sin 40 ° + FB sin 20 °) i + ( F A cos 40 ° − FB cos 20 °) j. Now we set the component of the total force parallel to the wall (the y component) equal to zero: 208

F A cos 40 ° − FB cos 20 ° = 0,

FB

We thus obtain an equation for the magnitude of FB : FB =

F A cos 40 ° (100 lb) cos 40 ° = = 81.5 lb. cos 20 ° cos 20 °

(b) Because we now know the magnitude of FB , we can determine the total force acting on the hook: F A + FB = ( F A sin 40 ° + FB sin 20 °) i

y

= [(100 lb)sin 40 ° + (81.5 lb)sin 20 °]i = 92.2 i  (lb). The magnitude of the total force is 92.2 lb.

FA 408 x

208

Critical Thinking We can obtain the solution to (a) in a less formal way. If the component of the total force parallel to the wall is zero, we see in Fig. a that the magnitude of the vertical component of F A must equal the magnitude of the vertical component of FB : F A cos 40 ° = FB cos 20 °.

FB

(a) Resolving F A and FB into components parallel and perpendicular to the wall.

Therefore, the magnitude of FB is FB =

F A cos 40 ° (100 lb) cos 40 ° = = 81.5 lb. cos 20 ° cos 20 °

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Problems

37

PROBLEMS 2.17 Two forces are given in terms of their components by F A = 60 i − 20 j (lb), FB = 30 i + 40 j (lb). (a) What are the magnitudes of the forces? (b) What is the magnitude of their sum F A + FB ? Strategy: The magnitude of a vector in terms of its components is given by Eq. (2.8). 2.18 Consider three vectors A = 12 i + 8 j, B = B x i + B y j, and C = C x i + C y j. Their sum is zero, A + B + C = 0, and the components of B and C satisfy the relations B x = −2 B y and C x = 3C y . What are the magnitudes of B and C?

2.21 The forces acting on the sailplane are its weight W = −500 j (lb), the lift L, and the drag D. The angle between the lift vector L and the vertical y-axis is 20°, and the drag vector is perpendicular to the lift vector. At the instant shown, the sum of the forces on the sailplane is zero: L + D + W = 0. Determine the magnitudes of L and D. y

2.19 The vector r = xi + yj extends to a point on the curve described by the equation shown. Determine the values of x and y for which the magnitude of the vector is a minimum. What is the minimum magnitude?

L

D

y W x (x, y)

r

Problem 2.21 x

y 5 4 2 1 x2 4

2.22 Two unit vectors e 1 and e 2 are oriented as shown. Determine the magnitude of the vector U = 2e 1 + 3e 2 . Strategy: Introduce a cartesian coordinate system and express e 1 and e 2 in terms of their components.

Problem 2.19 2.20 The forces F A = 90 i (kN) and FB = 60 j (kN). The force FC = c ( −2 i − j ) , where c is a parameter. Determine the value of c so that the magnitude of the total force exerted on the beam by the three forces is a minimum. What is the minimum magnitude?

e2

608

e1

y FB

Problem 2.22 FC FA

x

Problem 2.20

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38 Chapter 2 VectorsGet complete eBook Order by email at [email protected] 2.23 Two forces act on the support. Their magnitudes are F A = 300 lb, FB = 200 lb. Determine the magnitude F A + FB of the total force exerted on the support by the two forces.

2.26 The magnitude of the force F A is 8 kN. The magnitude of the vertical force FB is 2 kN. For what value of the angle α in the range 0 ≤ α ≤ 90 ° is the magnitude of the sum of the two forces equal to 9 kN ?

y

FB

FB FA 5

FA

5

7

11 aa x

Problem 2.23 Problem 2.26 2.24 The person exerts a 30-lb force F to push the crate onto a truck. (a) Express F in terms of components using the coordinate system shown. (b) The weight of the crate is 120 lb. What is the magnitude of the sum of the two forces exerted by the person and the crate’s weight?

y

F

2.27 The x-axis is parallel to the bar AB. The x ′-axis is parallel to the bar BC. The magnitude of the vector F is 200 lb. (a) Express F in terms of its components in terms of the x – y coordinate system. Use these components to determine the magnitude of F. (b) Express F in terms of its components in terms of the x ′ – y ′ coordinate system. Use these components to determine the magnitude of F. 2.28 Suppose that the magnitude of the component of the vector F parallel to the x ′-axis is 150 lb. (a) What is the magnitude of F ? (b) Express F in terms of its components in terms of the x – y coordinate system.

208

y9

x

y F

Problem 2.24

408

2.25 The missile’s engine exerts a 130-kN thrust F. (a) Express F in terms of components using the coordinate system shown. (b) The mass of the missile is 1800 kg. What is the magnitude of the sum of the thrust F and the missile’s weight?

5 ft

x

A

C

x9

B

10 ft

Problems 2.27/2.28 y

F 3 4

x

Problem 2.25

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Get complete eBook Order by email at [email protected] 2.29 The magnitudes A = 12 and B = 10. (a) Express A in terms of its components in terms of the x ′ – y ′ coordinate system. (b) Express B in terms of its components in terms of the x – y coordinate system. 2.30 The magnitudes A = 12 and B = 10. Determine A+B.

Problems

39

2.33 In Example 2.4, the coordinates of the fixed point A are (17, 1) ft. The driver lowers the bed of the truck into a new position in which the coordinates of point B are (9, 3) ft. The magnitude of the force F exerted on the bed by the hydraulic cylinder when the bed is in the new position is 4800 lb. Draw a sketch of the new situation. Express F in terms of components. 2.34 Using laser instruments, a surveyor determines that the coordinates of point B are x = 120 m, y = 122 m. He knows that rOA = 125 m and r AB = 62 m. Determine the vector rOA .

y9

y N

B 158

408

B

A

rAB

x9

rOA

y

Proposed roadway

A 208

x

x

O

Problems 2.29/2.30 2.31 In Practice Example 2.3, the cable AB exerts a 900-N force on the top of the tower. Suppose that the attachment point B is moved in the horizontal direction farther from the tower, and assume that the magnitude of the force F the cable exerts on the top of the tower is proportional to the length of the cable. (a) What is the distance from the tower to point B if the magnitude of the force is 1000 N ? (b) Express the 1000-N force F in terms of components using the coordinate system shown. 2.32 Determine the position vector r AB in terms of its components if (a) θ = 30 °; (b) θ = 225 ° .

Problem 2.34 2.35 The magnitude of the position vector rBA from point B to point A is 6 m and the magnitude of the position vector rCA from point C to point A is 4 m. What are the components of rBA? 2.36 In Problem 2.35, determine the components of a unit vector e CA that points from point C toward point A. Strategy: Determine the components of rCA and then divide the vector rCA by its magnitude. y

y 3m B 150 mm

60 mm

rAB B

x

rBC

u A

C

C

Problem 2.32

x

A

Problems 2.35/2.36

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40 Chapter 2 VectorsGet complete eBook Order by email at [email protected] 2.39 The link AC exerts a force F AC on the hydraulic cylinder at C. The force points from C toward A, and its magnitude is F AC = 650 N. Determine the components of F AC .

2.37 The hydraulic cylinder BC exerts a force F on the boom of the crane at C. The force points from B toward C, and its magnitude is F = 300 kN. Determine the components of F.

2.40 The link CD exerts a force FCD on the hydraulic cylinder at C. The force points from C toward D, and its magnitude is FCD = 600 N. Determine the components of FCD . y y

1m D

C 2.4 m 1m

C

Hydraulic cylinder

A

1m B

0.6 m

B

A 1.8 m

1.2 m

Scoop

0.6 m 0.15 m

7m x

Problems 2.39/2.40

Problem 2.37 2.38 The length of the bar AB is 0.6 m. Determine the components of a unit vector e AB that points from point A toward point B.

x

2.41 A surveyor finds that the length of the line OA is 1500 m and the length of the line OB is 2000 m. (a) Determine the components of the position vector from point A to point B. (b) Determine the components of a unit vector that points from point A toward point B.

y

N

A

Proposed bridge B

608 308

Problem 2.38

River x

O

Problem 2.41

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Problems

41

2.42 The magnitudes of the forces exerted by the cables are T1 = 2800 lb, T2 = 3200 lb, T3 = 4000 lb, and T4 = 5000 lb. What is the magnitude of the total force exerted by the four cables? 2.43 The tensions in the four cables are T1 = T , T2 = 1.5T , T3 = 2.5T , and T4 = 3 T . Determine the value of T so that the four cables exert a total force of 24 kN magnitude on the support. y T4

T3

518

408

T2 298

T1

98

x

Problems 2.42/2.43 2.44 The rope ABC exerts forces FBA and FBC on the block at B. Their magnitudes are equal: FBA = FBC . The magnitude of the total force exerted on the block at B by the rope is FBA + FBC = 920 N. Determine FBA by expressing the forces FBA and FBC in terms of components. FBC C

208 B

B

FBA A

Problem 2.44

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42 Chapter 2 VectorsGet complete eBook Order by email at [email protected] 2.45 The magnitude of the horizontal force F1 is 5 kN and F1 + F2 + F3 = 0. What are the magnitudes of F2 and F3? y

F3

2.48 The bracket must support the two forces shown, where F1 = F2 = 2 kN. An engineer determines that the bracket will safely support a total force of magnitude 3.5 kN in any direction. Assume that 0 ≤ α ≤ 90 ° . What is the safe range of the angle α ?

308 F2

F1 a

F1

458 F2

x

Problem 2.45 Problem 2.48 2.46 Four groups engage in a tug-of-war. The magnitudes of the forces exerted by groups B, C, and D are FB = 800 lb,  FC = 1000 lb, and FD = 900 lb. If the vector sum of the four forces equals zero, what is the magnitude of F A and the angle α ?

2.49 The figure shows three forces acting on a joint of a structure. The magnitude of FC is 80 kN, and F A + FB + FC = 0. What are the magnitudes of F A and FB ? y

y

FC

FB

FB

FC

158 708

308

FA

208

a

x

408

Problem 2.49

FD FA x

Problem 2.46 2.47 In Example 2.5, suppose that the attachment point of cable A is moved so that the angle between the cable and the wall increases from 40° to 55 °. Draw a sketch showing the forces exerted on the hook by the two cables. If you want the total force F A + FB to have a magnitude of 200 lb and be in the direction perpendicular to the wall, what are the necessary magnitudes of F A and FB ?

2.50 Four coplanar forces act on a beam. The forces FB and FC are vertical. The vector sum of the forces is zero. The magnitudes FB = 10 kN and FC = 5 kN. Determine the magnitudes of F A and FD .

FD

308 FA FB

Problem 2.50

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FC

Get complete eBook Order by email at [email protected] 2.51 Six forces act on a beam that forms part of a building’s frame. The vector sum of the forces is zero. The magnitudes FB = FE = 20 kN,  FC = 16 kN, and FD = 9 kN. Determine the magnitudes of F A and FG .

Problems

43

2.53 The three forces acting on the car are shown. The force T is parallel to the x -axis and the magnitude of the force W is 14 kN. If T + W + N = 0, what are the magnitudes of the forces T and N ?

208

y FA 708

FC

FD 408

FG 508

408

FB

T

FE

x

Problem 2.51 2.52 The total weight of the man and parasail is W = 230 lb. The drag force D is perpendicular to the lift force L. If the vector sum of the three forces is zero, what are the magnitudes of L and D?

W

208 N

Problem 2.53 2.54 The sum of the three forces acting on the beam is zero: F A + FB + FC = 0. The magnitude F A = 800 lb. Determine the magnitudes of FC and FB .

y L 5 2

FA

D

FC 508

458 808

FB

Problem 2.54 x

W

Problem 2.52

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44 Chapter 2 VectorsGet complete eBook Order by email at [email protected] 2.55 The total force exerted on the top of the mast B by the sailboat’s forestay AB and backstay BC is 180 i − 820 j (N). What are the magnitudes of the forces exerted at B by the cables AB and BC?

2.58 Determine the x and y coordinates of the collar C as functions of the distance s. y

A (14, 45) in

y B (4, 13) m

s

C

B (75, 12) in x

Problems 2.57/2.58 2.59 The vector U can be expressed in terms of its components in terms of either the x–y coordinate system or the x′–y′ coordinate system: C (9, 1) m

A (0, 1.2) m

U = Uxi + Uy j x

Problem 2.55 2.56 The structure shown forms part of a truss designed by an architectural engineer to support the roof of an orchestra shell. The members AB, AC, and AD exert forces F AB ,   F AC , and F AD on the joint A. The magnitude F AB = 4 kN. If the vector sum of the three forces equals zero, what are the magnitudes of F AC and F AD ?

= U x ′ i ′ + U y ′ j′. It can be shown that the component U x′ is given in terms of the components U x and U y and the angle θ by U x ′ = U x cos θ + U y sin θ. Determine the component U y′ in terms of the components U x and U y and the angle θ. Strategy: Draw a diagram showing the components U x , U y , U x′ , and U y′ and apply trigonometry to determine U y′ . y9

y

y U x9 B

(24, 1) m

FAB

FAC

C

FAD

A

u (4, 2) m x

D (22, 23) m

x

Problem 2.59 2.60 Let r be the position vector from point C to the point that is a distance s meters from point A along the straight line between A and B. Express r in terms of components. (Your answer will be in terms of s.) y B

Problem 2.56 2.57 The distance s = 45 in. (a) Determine the unit vector e BA that points from B toward A. (b) Use the unit vector you obtained in (a) to determine the coordinates of the collar C.

(10, 9) m

s r A (3, 4) m C (9, 3) m x

Problem 2.60

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Get complete eBook Order by email at [email protected] 2.3 Components in Three Dimensions 45

2.3 Components in Three Dimensions QUESTIONS TO CONSIDER • Can you guess how many components a vector has in three dimensions? • How can you specify the direction of a vector in three dimensions?

BACKGROUND Many engineering applications require vectors to be expressed in terms of components in a three-dimensional coordinate system. In this section we explain this technique and demonstrate vector operations in three dimensions. We first review how to draw objects in three dimensions. Consider a three-dimensional object such as a cube. If we draw the cube as it appears when the point of view is perpendicular to one of its faces, we obtain Fig. 2.14a. In this view, the cube appears two dimensional. The dimension perpendicular to the page cannot be seen. To remedy this, we move the point of view upward and to the right, obtaining Fig. 2.14b. In this oblique view, the third dimension is visible. The hidden edges of the cube are shown as dashed lines. We can use this approach to draw three-dimensional coordinate systems. In Fig. 2.14c we align the x-, y-, and z-axes of a three-dimensional cartesian coordinate system with the edges of the cube. The threedimensional representation of the coordinate system alone is shown in Fig.  2.14d. The coordinate system shown is said to be right handed.

y

y

x

z (a)

(b)

x

z (c)

(d)

FIGURE 2.14 (a) (b) (c) (d)

A cube viewed with the line of sight perpendicular to a face. An oblique view of the cube. A cartesian coordinate system aligned with the edges of the cube. Three-dimensional representation of the coordinate system.

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46 Chapter 2 VectorsGet complete eBook Order by email at [email protected] If the fingers of the right hand are pointed in the direction of the positive x-axis and then bent (as in preparing to make a fist) toward the positive y-axis, the thumb points in the direction of the positive z-axis (Fig. 2.15). Otherwise, the coordinate system is left handed. Because some equations used in mathematics and engineering do not yield correct results when they are applied using a left-handed coordinate system, we use only righthanded coordinate systems. We can express a vector U in terms of vector components U x ,  U y , and U z parallel to the x-, y-, and z-axes, respectively (Fig. 2.16), as

y

z

x

FIGURE 2.15

U = Ux + Uy + Uz.

Recognizing a right-handed coordinate system.

(We have drawn a box around the vector to help in visualizing the directions of the vector components.) By introducing unit vectors i, j, and k that point in the positive x, y, and z directions, we can express U in terms of scalar components as

y Uz

(2.12)

We will refer to the scalars U x ,   U y , and U z as the x, y, and z components of U.

Ux z

U = U x i + U y j + U z k.

Uy

U

j

(2.11)

x

i

MAGNITUDE OF A VECTOR IN TERMS OF COMPONENTS

k

FIGURE 2.16 A vector U and its vector components.

Consider a vector U and its vector components (Fig. 2.17a). From the right triangle formed by the vectors U y ,   U z , and their sum U y + U z (Fig. 2.17b), we can see that Uy + Uz

2

= Uy

2

2

+ Uz

.

(2.13)

The vector U is the sum of the vectors U x and U y + U z . These three vectors form a right triangle (Fig. 2.17c), from which we obtain U

2

= Ux

2

+ Uy + Uz

2

.

Substituting Eq. (2.13) into this result yields the equation U

y

2

= Ux

2

+ Uy

2

+ Uz

y

Uy Ux

z

y

u Uy 1 Uz u

x

uUu

u Uy u

u Ux u

x z

(a)

= U x 2 + U y 2 + U z 2.

u Uz u

Uz U

2

z (b)

(c)

FIGURE 2.17 (a) A vector U and its vector components. (b) The right triangle formed by the vectors U y ,   U z , and U y + U z . (c) The right triangle formed by the vectors U, U x , and U y + U z .

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uUy 1 Uz u

x

Get complete eBook Order by email at [email protected] 2.3 Components in Three Dimensions 47 y

y

y

uz z

uUu

U

uy

y

Uy j ux

ux

x

Ux i z

(a)

uy uz

x

x

(b)

uUu

uUu

z

Uz k z

(c)

FIGURE 2.18 (a) A vector U and the angles θ x ,   θ y , and θ z . (b) –(d) The angles θ x ,   θ y , and θ z and the vector components of U.

Thus, the magnitude of a vector U is given in terms of its components in three dimensions by U =

Ux2 + Uy2 + Uz2.

(2.14)

DIRECTION COSINES We described the direction of a vector relative to a two-dimensional cartesian coordinate system by specifying the angle between the vector and one of the coordinate axes. One of the ways we can describe the direction of a vector in three dimensions is by specifying the angles θ x ,   θ y , and θ z between the vector and the positive coordinate axes (Fig. 2.18a). In Figs. 2.18b–d, we demonstrate that the components of the vector U are respectively given in terms of the angles θ x ,   θ y , and θ z , by U x = U cos θ x , U y = U cos θ y , U z = U cos θ z .

(2.15)

The quantities cos θ x ,   cos θ y , and cos θ z are called the direction cosines of U. The direction cosines of a vector are not independent. If we substitute Eqs. (2.15) into Eq. (2.14), we find that the direction cosines satisfy the relation cos 2 θ x + cos 2 θ y + cos 2 θ z = 1.

(2.16)

Suppose that e is a unit vector with the same direction as U, so that U = U e. In terms of components, this equation is U x i + U y j + U z k = U (e x i + e y j + e z k). Thus, the relations between the components of U and e are U x = U ex , U y = U ey , U z = U ez . By comparing these equations to Eqs. (2.15), we see that cos θ x = e x ,   cos θ y = e y ,   cos θ z = e z . The direction cosines of a vector U are the components of a unit vector with the same direction as U.

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(d)

x

48 Chapter 2 VectorsGet complete eBook Order by email at [email protected] y

y

rAB

FIGURE 2.19

(zB 2 zA)k

B (xB, yB, zB)

A

(a) The position vector from point A to point B. (b) The components of r AB can be determined from the coordinates of points A and B.

B

rAB

(yB 2 yA)j

A

(xA, yA, zA)

(xB 2 xA)i x

z

x

z (b)

(a)

POSITION VECTORS IN TERMS OF COMPONENTS Generalizing the two-dimensional case, we consider a point A with coordinates ( x A ,   y A ,   z A ) and a point B with coordinates ( x B ,   y B ,   z B ). The position vector r AB from A to B, shown in Fig. 2.19a, is given in terms of the coordinates of A and B by r AB = ( x B − x A ) i + ( y B − y A ) j + ( z B − z A )k.

(2.17)

The components are obtained by subtracting the coordinates of point A from the coordinates of point B (Fig. 2.19b).

COMPONENTS OF A VECTOR PARALLEL TO A GIVEN LINE In three-dimensional applications, the direction of a vector is often defined by specifying the coordinates of two points on a line that is parallel to the vector. This information can be used to determine the components of the vector. Suppose that we know the coordinates of two points A and B on a line parallel to a vector U (Fig.  2.20a). We can use Eq. (2.17) to determine the y B

A

y

B

(xB, yB, zB) rAB

U

U

A

(xA, yA, zA)

x

x z

z (a)

(b) y

B

A

FIGURE 2.20

eAB 5

rAB u rAB u

(a) Two points A and B on a line parallel to U. z (b) The position vector from A to B. (c) (c) The unit vector e that points from A toward B. Get complete eBook AB Order by email at [email protected]

U 5 u U ueAB

x

Get complete eBook Order by email at [email protected] 2.3 Components in Three Dimensions 49 position vector r AB from A to B (Fig. 2.20b). We can divide r AB by its magnitude to obtain a unit vector e AB that points from A toward B (Fig. 2.20c). Because e AB has the same direction as U, we can determine U in terms of its scalar components by expressing it as the product of its magnitude and e AB . More generally, suppose that we know the magnitude of a vector U and the components of any vector V that has the same direction as U. Then V/ V is a unit vector with the same direction as U, and we can determine the components of U by expressing it as U = U ( V/ V ).

RESULTS Any vector U can be expressed as (2.12)

U 5 Ux i 1 Uy j 1 Uz k,

where i is a unit vector that points in the positive x-axis direction, j is a unit vector that points in the positive y-axis direction, and k is a unit vector that points in the positive z-axis direction.

y U

The magnitude of U is given by

x

uUu 5

Ux2 1 Uy2 1 Uz2 .

(2.14)

z

DIRECTION COSINES

The direction of a vector U relative to a given coordinate system can be specified by the angles ux, uy, and uz between the vector and the positive coordinate axes.

y U

uy

The components of U are given by uz

Ux 5 u U ucos ux,

ux

Uy 5 u U ucos uy, Uz 5 u U ucos uz.

(2.15)

z

The terms cos ux, cos uy, and cos uz are called the direction cosines of U. The direction cosines are the components of a unit vector with the same direction as U.

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x

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POSITION VECTORS IN TERMS OF COMPONENTS y

B (xB, yB, zB)

rAB A

The position vector from A to B is given by rAB 5 (xB 2 xA)i 1 (yB 2 yA)j 1 (zB 2 zA)k.

(2.17)

(xA, yA, zA) x

z

COMPONENTS OF A VECTOR PARALLEL TO A GIVEN LINE y B

A

The vector U is parallel to the line through points A and B. Obtain the position vector rAB from A to B in terms of its components. Divide rAB by its magnitude to obtain a unit vector eAB that is parallel to the line. Then the vector U in terms of its components is given by

(xB, yB, zB)

U (xA, yA, zA)

x

U 5 u U ueAB.

z

Practice Example 2.6

Direction Cosines The coordinates of point C of the truss are x C = 4 m,  y C = 0,   z C = 0, and the coordinates of point D are x D = 2 m, y D = 3 m,  z D = 1 m. What are the direction cosines of the position vector rCD from point C to point D ? y

D

A

C z

B

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x

Get complete eBook Order by email at [email protected] 2.3 Components in Three Dimensions 51 Strategy Knowing the coordinates of points C and D, we can determine rCD in terms of its components. Then we can calculate the magnitude of rCD (the distance from C to D ) and use Eqs. (2.15) to obtain the direction cosines. Solution y

D (2, 3, 1) m

rCD

(4, 0, 0) m C z

rCD 5 (xD 2 xC)i 1 ( yD 2 yC)j 1 (zD 2 zC)k. 5 (2 2 4)i 1 (3 2 0)j 1 (1 2 0)k (m) 5 22i 1 3j 1 k (m).

Determine the position vector rCD in terms of its components.

 rCD  5 r 2CDx 1 r 2CDy 1 r 2CDz Calculate the magnitude of rCD.

5 (22 m)2 1 (3 m)2 1 (1 m)2 5 3.74 m.

−2 m rCDx = = −0.535, rCD 3.74 m rCDy 3m cos θ y = = = 0.802, rCD 3.74 m 1m r cos θ z = CDz = = 0.267, rCD 3.74 m

cos θ x =

         

Determine the direction cosines.

Practice Problem The coordinates of point B of the truss are x B = 2.4 m, y B = 0, z B = 3 m. Determine the components of a unit vector e BD that points from point B toward point D. Answer: e BD = −0.110 i + 0.827 j − 0.551k.

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x

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Example 2.7

Determining Components in Three Dimensions The crane exerts a 600-lb force F on the caisson. The angle between F and the x -axis is 54 °, and the angle between F and the y -axis is 40 °. The z component of F is positive. Express F in terms of components.

y

408

F 548

x z

Strategy Only two of the angles between the vector and the positive coordinate axes are given, but we can use Eq. (2.16) to determine the third angle. Then we can determine the components of F by using Eqs. (2.15). Solution The angles between F and the positive coordinate axes are related by cos 2 θ x + cos 2 θ y + cos 2 θ z = (cos 54 °) 2 + (cos 40 °) 2 + cos 2 θ z = 1. Solving this equation for cos θ z , we obtain the two solutions cos θ z = 0.260 and cos θ z = −0.260, which tells us that θ z = 74.9 ° or θ z = 105.1 °. The z component of the vector F is positive, so the angle between F and the positive z -axis is less than 90 °. Therefore, θ z = 74.9 °. The components of F are Fx = F cos θ x = 600 cos 54 ° = 353 lb,   Fy = F cos θ y = 600 cos 40 ° = 460 lb, Fz = F cos θ z = 600 cos 74.9 ° = 156 lb. Critical Thinking You are aware that knowing the square of a number does not tell you the value of the number uniquely. If a 2 = 4, the number a can be either 2 or −2. In this example, knowledge of the angles θ x and θ y allowed us to solve Eq. (2.16) for the value of cos 2   θ z , which resulted in two possible values of the angle θ z . There is a simple geometrical explanation for why this happened. The two angles θ x and θ y are sufficient to define a line parallel to the vector F but not the direction of F along that line. The two values of θ z we obtained correspond to the two possible directions of F along the line. Additional information is needed to indicate the direction. In this example, the additional information was supplied by stating that the z component of F is positive.

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Get complete eBook Order by email at [email protected] 2.3 Components in Three Dimensions 53 Example 2.8

Determining Components in Three Dimensions

The tether of the balloon exerts an 800-N force F on the hook at O. The vertical line AB intersects the x–z plane at point A. The angle between the z-axis and the line OA is 60 °, and the angle between the line OA and F is 45 °. Express F in terms of components. Strategy We can determine the components of F from the given geometric information in two steps. First, we express F as the sum of two vector components parallel to the lines OA and AB. The component parallel to AB is the vector component Fy . Then we can use the component parallel to OA to determine the vector components Fx and Fz .

y

B F O

O

x A

z

Solution In Fig. a, we express F as the sum of its y component Fy and the component Fh parallel to OA. The magnitude of Fy is

y

Fy = F sin 45 ° = (800 N)sin 45 ° = 566 N,

B F

Fy

and the magnitude of Fh is

458 x

O

Fh = F cos 45 ° = (800 N) cos 45 ° = 566 N. In Fig. b, we express Fh in terms of the vector components Fx and Fz . The magnitude of Fx is

Fh

A

z

(a) Resolving F into vector components parallel to OA and OB.

Fx = Fh sin 60 ° = (566 N)sin 60 ° = 490 N, y

and the magnitude of Fz is Fz = Fh cos60 ° = (566 N) cos60 ° = 283 N.

B

The vector components Fx ,   Fy , and Fz all point in the positive axis directions, so the scalar components of F are positive: F = 490 i + 566 j + 283k  (N).

F

Fy

Fx

O

Fz

608

x

Fh A

z

Critical Thinking As this example demonstrates, two angles are required to specify a vector’s direction relative to a three-dimensional coordinate system. The two angles used may not be defined in the same way as in the example, but however they are defined, you can determine the components of the vector in terms of the magnitude and the two specified angles by a procedure similar to the one we used here.

(b) Resolving Fh into vector components parallel to the x- and z-axes.

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Example 2.9

Determining Components in Three Dimensions The rope extends from point B through a metal loop attached to the wall at A to point C. The rope exerts forces F AB and F AC on the loop at A with magnitudes F AB = F AC = 200 lb. What is the magnitude of the total force F = F AB + F AC exerted on the loop by the rope? 6 ft

y 6 ft

A

A

7 ft 4 ft

2 ft

7 ft B

6 ft

4 ft

C 2 ft

10 ft

FAB

FAC x

B

C

6 ft

10 ft

z

Strategy The force F AB is parallel to the line from A to B, and the force F AC is parallel to the line from A to C. Because we can determine the coordinates of points A, B, and C from the given dimensions, we can determine the components of unit vectors that have the same directions as the two forces and use them to express the forces in terms of scalar components. Solution Let r AB be the position vector from point A to point B and let r AC be the position vector from point A to point C (Fig. a). From the given dimensions, the coordinates of points A, B, and C are A: (6,  7,  0) ft,

B: (2,  0,  4) ft,

C: (12,  0,  6) ft.

y A rAB rAC B

x

C

z

(a) The position vectors r AB and r AC .

Therefore, the components of r AB and r AC , with the coordinates in ft, are given by r AB = ( x B − x A ) i + ( y B − y A ) j + ( z B − z A )k = (2 − 6) i + (0 − 7) j + (4 − 0)k = −4 i − 7 j + 4 k  (ft)

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Get complete eBook Order by email at [email protected] 2.3 Components in Three Dimensions 55 and r AC = ( x C − x A ) i + ( y C − y A ) j + ( z C − z A )k = (12 − 6) i + (0 − 7) j + (6 − 0)k = 6 i − 7 j + 6k  (ft). Their magnitudes are r AB = 9 ft and r AC = 11 ft. By dividing r AB and r AC by their magnitudes, we obtain unit vectors e AB and e AC that point in the directions of F AB and F AC (Fig. b ): r AB = −0.444 i − 0.778 j + 0.444 k, r AB r = AC = −0.545i − 0.636 j + 0.545k. r AC

e AB = e AC

y

eAB

A

eAC x

B

C

z

(b) The unit vectors e AB and e AC .

The forces F AB and F AC are F AB = (200 lb)e AB = −88.9 i − 155.6 j + 88.9k  (lb), F AC = (200 lb)e AC = 109.1i − 127.3 j + 109.1k  (lb). The total force exerted on the loop by the rope is F = F AB + F AC = 20.2 i − 282.8 j + 198.0 k  (lb), and its magnitude is F =

(20.2) 2 + (−282.8) 2 + (198.0) 2 = 346 lb.

Critical Thinking How do you know that the magnitude and direction of the total force exerted on the metal loop at A by the rope is given by the magnitude and direction of the vector F = F AB + F AC ? At this point in our development of mechanics, we assume that force is a vector, but have provided no proof. In the study of dynamics it is shown that Newton’s second law implies that force is a vector.

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Example 2.10

Determining Components of a Force The cable AB exerts a 50-N force T on the collar at A. Express T in terms of components. y 0.15 m 0.4 m

B

C T 0.2 m

A

0.5 m O

0.3 m x

D

0.25 m

0.2 m

z

Strategy Let r AB be the position vector from A to B. We will divide r AB by its magnitude to obtain a unit vector e AB having the same direction as the force T. Then we can obtain T in terms of scalar components by expressing it as the product of its magnitude and e AB . To begin this procedure, we must first determine the coordinates of the collar A. We will do so by obtaining a unit vector e CD pointing from C toward D and multiplying it by 0.2 m to determine the position of the collar A relative to C. Solution Determining the Coordinates of Point A D, with the coordinates in meters, is

The position vector from C to

rCD = (0.2 − 0.4) i + (0 − 0.3) j + (0.25 − 0)k 0.15 m

y

= −0.2 i − 0.3 j + 0.25k  (m). Dividing this vector by its magnitude, we obtain the unit vector e CD (Fig. a):

0.4 m B

0.3 m

A

0.5 m O

Using this vector, we obtain the position vector from C to A: x

D z

e CD = eCD

eAB

0.25 m

0.2 m

(a) The unit vectors e AB and e CD .

rCD = rCD

−0.2 i − 0.3 j + 0.25k (−0.2) 2 + (−0.3) 2 + (0.25) 2 = −0.456 i − 0.684 j + 0.570 k.

C

rCA = (0.2 m)e CD = −0.091i − 0.137 j + 0.114 k  (m).

The position vector from the origin of the coordinate system to C is rOC = 0.4 i + 0.3 j (m), so the position vector from the origin to A is rOA = rOC + rCA = (0.4 i + 0.3 j) + (−0.091i − 0.137 j + 0.114 k) = 0.309 i + 0.163 j + 0.114 k  (m). The coordinates of A are (0.309,  0.163,  0.114) m.

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Problems

57

Determining the Components of T Using the coordinates of point A, we find that the position vector from A to B is r AB = (0 − 0.309) i + (0.5 − 0.163) j + (0.15 − 0.114)k = −0.309 i + 0.337 j + 0.036k  (m). Dividing this vector by its magnitude, we obtain the unit vector e AB (Fig. a). r AB = r AB

−0.309 i + 0.337 j + 0.036k  (m) (−0.309 m) 2 + (0.337 m) 2 + (0.036 m) 2 = −0.674 i + 0.735 j + 0.079k.

e AB =

The force T is T = T e AB = (50 N)(−0.674 i + 0.735 j + 0.079k) = −33.7 i + 36.7 j + 3.9k  (N). Critical Thinking Look at the two ways unit vectors were used in this example. The unit vector e CD was used to obtain the components of the position vector rCA , which made it possible to determine the coordinates of point A. The coordinates of point A were then used to determine the unit vector e AB , which was used to express the force T in terms of its components.

PROBLEMS 2.61 A vector U = 3i − 4 j − 12k. What is its magnitude? Strategy: The magnitude of a vector is given in terms of its components by Eq. (2.14).

2.66 The position vector r AB extends from point A to point B. (a) Express r AB in terms of components. (b) Determine its magnitude r AB .

2.62 Two forces are given in terms of their components by F A = 60 i − 20 j + 30 k (lb), FB = 30 i + 40 j − 10 k (lb). (a) What are the magnitudes of the forces? (b) What is the magnitude of their sum F A + FB ? Strategy: The magnitude of a vector in terms of its components is given by Eq. (2.14).

2.67 The position vector r AB extends from point A to point B. (a) Determine the direction cosines of r AB . (b) Determine the components of a unit vector that has the same direction as r AB . y (22, 6, 3) ft

2.63 Consider three vectors A = 12 i + 8 j − 16k, B = B x i + B y j + B z k, and C = C x i + C y j + C z k. Their sum is zero, A + B + C = 0, and the components of B and C satisfy the relations B x = −2 B y , B y = −B z , and C x = 3C y . What are the magnitudes of B and C? 2.64 A vector U = U x i + U y j + U z k. Its magnitude U = 30. Its components are related by the equations U y = −2U x and U z = 4U y . Determine the components.

A rAB x (12, 22, 8) ft B

z

Problems 2.66/2.67

2.65 An object is acted on by two forces F1 = 20 i + 30 j − 24 k (kN) and F2 = −60 i + 20 j + 40 k  (kN). What is the magnitude of the total force acting on the object?

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58 Chapter 2 VectorsGet complete eBook Order by email at [email protected] 2.68 A force vector is given in terms of its components by F = 10 i − 20 j − 20 k  (N). (a) What are the direction cosines of F ? (b) Determine the components of a unit vector e that has the same direction as F.

2.74 Determine the components of the unit vector e CD that points from point C toward point D.

2.69 Express the position vector rBD from point B to point D in terms of components. What are its direction cosines? 2.70 The cable from B to C exerts a force on the bar AB at B. The force points from B toward C and its magnitude is 80 N. Express the force in terms of components.

2.76 In Example 2.7, suppose that the caisson shifts on the ground to a new position. The magnitude of the force F remains 600 lb. In the new position, the angle between the force F and the x-axis is 60° and the angle between F and the z-axis is 70 °. Express F in terms of components.

2.71* The suspended weight E exerts an 800-N downward force on bar AB at B. The cable from B to C exerts a force on the bar at B that points from B toward C, but its magnitude is unknown. The cable from B to D exerts a force on the bar that points from B toward D, and its magnitude is also unknown. If the weight of bar AB is negligible, the sum of the three forces exerted on it at B is a vector that is parallel to the bar. Use this condition to determine the magnitudes of the forces exerted by cables BC and BD.

2.77 Astronauts on the space shuttle use radar to determine the magnitudes and direction cosines of the position vectors of two satellites A and B. The vector r A from the shuttle to satellite A has magnitude 2 km and direction cosines cos θ x = 0.768, cos θ y = 0.384, cos θ z = 0.512. The vector rB from the shuttle to satellite B has magnitude 4 km and direction cosines cos θ x = 0.743, cos θ y = 0.557, cos θ z = −0.371. What is the distance between the satellites?

2.75 What are the direction cosines of the unit vector e CD that points from point C toward point D ?

y B

(0, 4, 23) m C

rB

B (4, 3, 1) m

D (0, 5, 5) m

x

x A E

y

rA

A

z

z

Problem 2.77

Problems 2.69–2.71 Refer to the following diagram when solving Problems 2.72 through 2.75. y

2.78 Archaeologists measure a pre-Columbian ceremonial structure and obtain the dimensions shown. Determine (a) the magnitude and (b) the direction cosines of the position vector from point A to point B.

D (4, 3, 1) m y 4m

A

10 m

4m 10 m

A

C (6, 0, 0) m 8m

x z

B

B (5, 0, 3) m

Problems 2.72–2.75

8m

z

2.72 Determine the components of the position vector rBD from point B to point D. Use your result to determine the distance from B to D. 2.73 What are the direction cosines of the position vector rBD from point B to point D?

b

C

Problem 2.78

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x

Get complete eBook Order by email at [email protected] 2.79 Consider the structure described in Problem 2.78. After returning to the United States, an archaeologist discovers that a graduate student has erased the only data file containing the dimension b. But from recorded GPS data he is able to calculate that the distance from point B to point C is 16.61 m. (a) What is the distance b ? (b) Determine the direction cosines of the position vector from B to C. 2.80 Observers at A and B use theodolites to measure the direction from their positions to a rocket in flight. If the coordinates of the rocket’s position at a given instant are (4, 4, 2) km, determine the direction cosines of the vectors r AR and rBR that the observers would measure at that instant.

Problems

59

2.82 The height of Mount Everest was originally measured by a surveyor in the following way. He first measured the altitudes of two points and the horizontal distance between them. For example, suppose that the points A and B are 3000 m above sea level and are 10,000 m apart. He then used a theodolite to measure the direction cosines of the vector r AP from point A to the top of the mountain P and the vector rBP from point B to P. Suppose that the direction cosines of r AP are cos θ x = 0.5179, cos θ y = 0.6906, and cos θ z = 0.5048, and the direction cosines of rBP are cos θ x = −0.3743, cos θ y = 0.7486, and cos θ z = 0.5472. Using this data, determine the height of Mount Everest above sea level.

y

z

P

y

x

B

rAR

A

rBR

Problem 2.82

A x B (5, 0, 2) km

z

Problem 2.80 2.81 Point p lies in the x–z plane. The line from point p to the tip of the vector r is parallel to the y-axis. If r = 12 i + 6 j + 10 k (m), what are the angles φ and ψ in degrees?

2.83 The distance from point O to point A is 20 ft. The straight line AB is parallel to the y -axis, and point B is in the x–z plane. Express the vector rOA in terms of components. Strategy: You can express rOA as the sum of a vector from O to B and a vector from B to A. You can then express the vector from O to B as the sum of vector components parallel to the x- and z-axes. See Example 2.8.

y

y

A

r

rOA

c x O

f

608

p

z

Problem 2.81

x

308

B

z

Problem 2.83 2.84 The magnitudes of the two force vectors are F A = 140 lb and FB = 100 lb. Determine the magnitude of the sum of the forces F A + FB .

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60 Chapter 2 VectorsGet complete eBook Order by email at [email protected] 2.85 Determine the direction cosines of the vectors F A and FB . y

2.88 The cable BC exerts an 8-kN force F on the bar AB at B. (a) Determine the components of a unit vector that points from point B toward point C. (b) Express F in terms of components.

FB

y

FA 608 308

408

B (5, 6, 1) m

x

F

508 A

z

Problems 2.84/2.85

x

2.86 The positions of two airplanes A and B relative to a control tower are determined by radar. Their distances from the control tower are r A = 16 km and rB = 12 km. Determine the components of the position vector r AB from plane A to plane B (that is, the vector that specifies the position of plane B relative to plane A ). What is the straight-line distance between the two planes? y

B

C (3, 0, 4) m z

Problem 2.88 2.89 A cable extends from point C to point E. It exerts a 50-lb force T on the plate at C that is directed along the line from C to E. Express T in terms of components. y

rB 608

rA 208

408

6 ft

A

308

z

A

E x

Problem 2.86

208

B

x

4 ft

T

2 ft z

2.87 An engineer calculates that the magnitude of the axial force in one of the beams of a geodesic dome is P = 7.65 kN. The cartesian coordinates of the endpoints A and B of the straight beam are (−12.4, 22.0, −18.4) m and (−9.2, 24.4, −15.6) m, respectively. Express the force P in terms of components.

D

C 4 ft

Problem 2.89 2.90 A cable extends from point B to point D. It exerts a 30-lb force on the bar at B that is directed along the line from B to D. Express the force in terms of components. y

B P A

A

x B 8 in

14 in

C

z D

Problem 2.87

Problem 2.90

(18, 28, 7) in

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Get complete eBook Order by email at [email protected] 2.91 The cable AB exerts a 200-lb force F AB at point A that is directed along the line from A to B. Express F AB in terms of components.

Problems

61

2.93 The 70-m -tall tower is supported by three cables that exert forces F AB ,   F AC , and F AD on it. The magnitude of each force is 2 kN. Express the total force exerted on the tower by the three cables in terms of components.

y 8 ft

A

C

y

FAD

A

8 ft FAB

FAC

6 ft

D

B x

60 m

60 m

B 40 m

FAC

FAB

A (6, 0, 10) ft

z

C 40 m

40 m z

Problem 2.93

Problem 2.91 2.92 The cable CD exerts a 6-kN force on the bar at C that is directed from C toward D. Express the force in terms of components.

2.94 The magnitudes of the two force vectors are F A = 4.6 kN and FB = 5.2 kN. Determine the magnitude of the sum of the forces F A + FB . (Notice that the angle between the vector FB and the positive x-axis is 180 ° − 57 ° = 123 °. )

y

y 1m

FB

C

408

B

578 708

448

FA

758 508

2m

z x A

z

Problem 2.94 1.5 m

x

D x

Problem 2.92

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62 Chapter 2 VectorsGet complete eBook Order by email at [email protected] 2.95 The rope BD is attached to the bar ABC at B. The distance from A to B is 3 m. Determine the components of the position vector rBD from point B to point D.

2.97 The circular bar has a 4-m radius and lies in the x–y plane. Express the position vector from point B to the collar at A in terms of components.

2.96 The rope BD is attached to the bar ABC at B. The distance from A to B is 3 m. The rope exerts a 2-kN force on the bar at B that is directed from B toward D. Determine the components of the force.

2.98 The cable AB exerts a 60-N force T on the collar at A that is directed along the line from A toward B. Express T in terms of components. y

y 3m

C 4m

6m

B

B A 3m z

4m

D

A

4m

2m x

208

Problems 2.95/2.96 4m

x

z

Problems 2.97/2.98

2.4 Dot Products QUESTIONS TO CONSIDER • What is the “dot” product of two vectors? What can you do with it? • If you know two vectors in terms of their components, what is their dot product?

BACKGROUND Two kinds of products of vectors, the dot and cross products, have been found to have applications in science and engineering, especially in mechanics and electromagnetic field theory. We use both of these products in Chapter 4 to evaluate moments of forces about points and lines. The dot product of two vectors has many uses, including determining the components of a vector parallel and perpendicular to a given line and determining the angle between two lines in space.

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Get complete eBook Order by email at [email protected] 2.4 Dot Products 63

DEFINITION Consider two vectors U and V (Fig.  2.21a). The dot product of U and V, denoted by U ⋅ V (hence the name “dot product”), is defined to be the product of the magnitude of U, the magnitude of V, and the cosine of the angle θ between U and V when they are placed tail to tail (Fig. 2.21b): U ⋅ V = U V cos θ.

(2.18)

(a)

Because the result of the dot product is a scalar, the dot product is sometimes called the scalar product. The units of the dot product are the product of the units of the two vectors. Notice that the dot product of two nonzero vectors is equal to zero if and only if the vectors are perpendicular. The dot product has the properties U ⋅ V = V ⋅ U,

U

V

The dot product is commutative.

V u U

(2.19)

(b)

a(U ⋅ V) = (aU) ⋅ V = U ⋅ (aV),

The dot product is associative  with respect to scalar  multiplication.

(2.20)

(a) The vectors U and V. (b) The angle θ between U and V when the two vectors are placed tail to tail.

and U ⋅ ( V + W) = U ⋅ V + U ⋅ W,

The dot product is associative  with respect to vector addition.

FIGURE 2.21

(2.21)

for any scalar a and vectors U, V, and W.

DOT PRODUCTS IN TERMS OF COMPONENTS In this section we derive an equation that allows you to determine the dot product of two vectors if you know their scalar components. The derivation also results in an equation for the angle between the vectors. The first step is to determine the dot products formed from the unit vectors i, j, and k. Let us evaluate the dot product i ⋅ i. The magnitude i = 1, and the angle between two identical vectors placed tail to tail is zero, so we obtain i ⋅ i = i i cos(0) = (1)(1)(1) = 1. The dot product of i and j is i ⋅ j = i j ⋅ cos(90 °) = (1)(1)(0) = 0. Continuing in this way, we obtain i ⋅ i = 1, i ⋅ j = 0, i ⋅ k = 0, j ⋅ i = 0, j ⋅ j = 1, j ⋅ k = 0, k ⋅ i = 0, k ⋅ j = 0, k ⋅ k = 1.

(2.22)

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64 Chapter 2 VectorsGet complete eBook Order by email at [email protected] The dot product of two vectors U and V, expressed in terms of their components, is U ⋅ V = (U x i + U y j + U z k) ⋅ (V x i + V y j + V z k) = U xV x ( i ⋅ i) + U xV y ( i ⋅ j) + U xV z ( i ⋅ k) + U yV x ( j ⋅ i) + U yV y ( j ⋅ j) + U yVz ( j ⋅ k) + U zV x (k ⋅ i) + U zV y (k ⋅ j) + U zV z (k ⋅ k). In obtaining this result, we used Eqs. (2.20) and (2.21). Substituting Eqs. (2.22) into this expression, we obtain an equation for the dot product in terms of the scalar components of the two vectors:

L

U ⋅ V = U xV x + U yV y + U z V z .

(2.23)

To obtain an equation for the angle θ in terms of the components of the vectors, we equate the expression for the dot product given by Eq. (2.23) to the definition of the dot product, Eq. (2.18), and solve for cos θ: cos θ =

U (a) L

U xV x + U yV y + U z V z U⋅V = . U V U V

(2.24)

VECTOR COMPONENTS PARALLEL AND NORMAL TO A LINE

Up Un u U (b)

FIGURE 2.22 (a) A vector U and line L. (b) Resolving U into components parallel and normal to L.

L

In some engineering applications a vector must be expressed in terms of vector components that are parallel and normal (perpendicular) to a given line. The component of a vector parallel to a line is called the projection of the vector onto the line. For example, when the vector represents a force, the projection of the force onto a line is the component of the force in the direction of the line. We can determine the components of a vector parallel and normal to a line by using the dot product. Consider a vector U and a straight line L (Fig. 2.22a). We can express U as the sum of vector components U p and U n that are parallel and normal to L (Fig. 2.22b). The Parallel Component In terms of the angle θ between U and the vector component U p , the magnitude of U p is U p = U cos θ.

(2.25)

Let e be a unit vector parallel to L (Fig. 2.23). The dot product of e and U is e ⋅ U = e U cos θ = U cos θ.

e

Comparing this result with Eq. (2.25), we see that the magnitude of U p is

u

U p = e ⋅ U.

U

Therefore the parallel vector component, or projection of U onto L, is

FIGURE 2.23 The unit vector e is parallel to L.

U p = (e ⋅ U)e.

(2.26)

(This equation holds even if e doesn’t point in the direction of U p. In that case, the angle θ > 90 ° and e ⋅ U is negative.) When the components of a vector and the components of a unit vector e parallel to a line L are known, we can use Eq. (2.26) to determine the component of the vector parallel to L.

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Get complete eBook Order by email at [email protected] 2.4 Dot Products 65 The Normal Component Once the parallel vector component has been determined, we can obtain the normal vector component from the relation U = Up + Un: Un = U − Up.

(2.27)

RESULTS Dot Product The dot product of two vectors U and V is def ined by U?V 5 u U uu V u cos u,

V

(2.18)

u

where u is the angle between the vectors when they are placed tail to tail. Notice that U?U 5 u U u2. If u U u fi 0 and u V u fi 0, U?V 5 0 if and only if U and V are perpendicular.

U

Dot Product in Terms of Components The dot product of U and V is given in terms of the components of the vectors by U? V 5 UxVx 1 UyVy 1 UzVz.

(2.23)

Vector Components Parallel and Normal to a Line A vector U can be resolved into a vector component Up that is parallel to a given line L and a vector component Un that is normal to L. If e is a unit vector that is parallel to L, the parallel component of U is given by (2.26) Up 5 (e? U)e. The normal component can be obtained from the relation Un 5 U 2 Up.

Practice Example 2.11

L

Up Un u U

(2.27)

Dot Products

The components of two vectors U and V are U = 6 i − 5 j − 3k  and  V = 4 i + 2 j + 2k. (a) What is the value of U ⋅ V? (b) What is the angle between U and V when they are placed tail to tail? Strategy Knowing the components of U and V, we can use Eq. (2.23) to determine the value of U ⋅ V. Then we can use the definition of the dot product, Eq. (2.18), to calculate the angle between the vectors.

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(continued )

66 Chapter 2 VectorsGet complete eBook Order by email at [email protected] Solution Use the components of the vectors to determine the value of U ⋅ V.

 U ⋅ V = U xV x + U yV y + U zVz  = (6)(4) + (−5)(2) + (−3)(2)   = 8. 

Use the definition of U ⋅ V to determine θ.

 U ⋅ V = U V cos θ,  so   U⋅V  cos θ = U V  8  =  2 2 (6) + (−5) + (−3) 2   = 0.195.   Therefore, θ = 78.7 °.

(4) 2 + (2) 2 + (2) 2

Practice Problem The components of two vectors U and V are U = 6 i − 5 j − 3k and V = V x i + 2 j + 2k. Determine the value of the component V x so that the vectors U and V are perpendicular. Answer: V x = 2.67.

Example 2.12 y

Using the Dot Product to Determine an Angle What is the angle θ between the lines AB and AC?

C (8, 8, 4) m u

A (4, 3, 2) m

Strategy We know the coordinates of the points A, B, and C, so we can determine the components of the vector r AB from A to B and the vector r AC from A to C (Fig. a). Then we can use Eq. (2.24) to determine θ.

B (6, 1, 22) m x

Solution The vectors r AB and r AC , with the coordinates in meters, are

z y

C (8, 8, 4) m

rAC A (4, 3, 2) m

u rAB

r AB = (6 − 4) i + (1 − 3) j + (−2 − 2)k = 2 i − 2 j − 4 k  (m), r AC = (8 − 4) i + (8 − 3) j + (4 − 2)k = 4 i + 5 j + 2k  (m).

B (6, 1, 22) m x

Their magnitudes are

z

(a) The position vectors r AB and r AC .

r AB =

(2 m) 2 + (−2 m) 2 + (−4 m) 2 = 4.90 m,

r AC =

(4 m) 2 + (5 m) 2 + (2 m) 2 = 6.71 m.

The dot product of r AB and r AC is r AB ⋅ r AC = (2 m)(4 m) + (−2 m)(5 m) + (−4 m)(2 m) = −10 m 2 .

cos u 1

Therefore,

0

The angle θ = arccos(−0.304) = 107.7 °.

cos θ =

21 0

(b) Graph of cos θ.

908 u

1808

r AB ⋅ r AC −10 m 2 = = −0.304. r AB r AC (4.90 m)(6.71 m)

Critical Thinking What does it mean if the dot product of two vectors is negative? From Eq. (2.18) and the graph of the cosine (Fig. b), you can see that the dot product is negative, as it is in this example, only if the enclosed angle between the two vectors is greater than 90 °.

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Get complete eBook Order by email at [email protected] 2.4 Dot Products 67 Example 2.13

Vector Components Parallel and Normal to a Line y

Suppose that you pull on the cable OA, exerting a 50-N force F at O. What are the vector components of F parallel and normal to the cable OB? Strategy Expressing F as the sum of vector components parallel and normal to OB (Fig. a), we can determine the vector components by using Eqs. (2.26) and (2.27). But to apply them, we must first express F in terms of scalar components and determine the scalar components of a unit vector parallel to OB. We can obtain the components of F by determining the components of the unit vector pointing from O toward A and multiplying them by F .

A

(6, 6, 23) m

F x (10, 22, 3) m

O z

B

Solution The position vectors from O to A and from O to B are (Fig. b)

y

rOA = 6 i + 6 j − 3k  (m),

A

rOB = 10 i − 2 j + 3k  (m). Their magnitudes are rOA = 9 m and rOB = 10.6 m. Dividing these vectors by their magnitudes, we obtain unit vectors that point from the origin toward A and B (Fig. c): e OA e OB

r 6 i + 6 j − 3k  (m) = OA = = 0.667 i + 0.667 j − 0.333k, 9 m rOA r 10 i − 2 j + 3k  (m) = OB = = 0.941i − 0.188 j + 0.282k. rOB 10.6 m

F Fn

O z

B

(a) The components of F parallel and normal to OB.

The force F in terms of scalar components is

y

F = F e OA = (50 N)(0.667 i + 0.66 j − 0.333 k)

A (6, 6, 23) m

= 33.3i + 33.3 j − 16.7k  (N).

rOA

Taking the dot product of e OB and F, we obtain e OB ⋅ F = (0.941)(33.3 N) + ( −0.188 ) (33.3 N) + (0.282)(−16.7 N) = 20.4 N.

x

Fp

O

x rOB

z

(10, 22, 3) m B

The parallel vector component of F is Fp = (e OB ⋅ F)e OB = (20.4 N)(0.941i − 0.188 j + 0.282k)

(b) The position vectors rOA and rOB .

= 19.2 i − 3.83 j + 5.75k  (N), and the normal vector component is Fn = F − Fp = 14.2 i + 37.2 j − 22.4 k  (N). Critical Thinking How can you confirm that two vectors are perpendicular? It is clear from Eq. (2.18) that the dot product of two nonzero vectors is zero if and only if the enclosed angle between them is 90 °. We can use this diagnostic test to confirm that the components of F determined in this example are perpendicular. Evaluating the dot product of Fp and Fn in terms of their components in newtons, we obtain Fp ⋅ Fn = (19.2)(14.2) + (−3.83)(37.2) + (5.75)(−22.4) = 0.

y

A eOA O z

x eOB B

(c) The unit vectors e OA and e OB .

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68 Chapter 2 VectorsGet complete eBook Order by email at [email protected]

PROBLEMS 2.99 In Practice Example 2.11, suppose that the vector V is changed to V = 4 i − 6 j − 10 k. (a) What is the value of U ⋅ V? (b) What is the angle between U and V when they are placed tail to tail?

2.105 The magnitudes U = 10 and V = 20. (a) Use Eq. (2.18) to determine U ⋅ V. (b) Use Eq. (2.23) to determine U ⋅ V. y

2.100 In Example 2.12, suppose that the coordinates of point B are changed to (6,  4,  4) m. What is the angle θ between the lines AB and AC ?

V U

2.101 Consider the vectors U = 12 i + 4 j − 8k, V = −6 i + 8 j + 14 k, W = 8i − 8 j + 8k.

458

Determine the values of the dot products U ⋅ V, U ⋅ W, and V ⋅ W. What can you deduce about the three vectors from the results? 2.102 Consider the position vectors P and Q. (a) Determine the dot product P ⋅ Q by using the definition, Eq. (2.18). (b) Determine the dot product P ⋅ Q by using Eq. (2.23).

308 x

Problem 2.105 2.106 What is the angle between the ropes OA and OB?

B

y

y A 608 408

(8, 0, 0) m

Q

O

x

358 158 x

z

P

Problem 2.106

z (12, 0, 10) m

Problem 2.102 2.103 Two perpendicular vectors are given in terms of their components by U = U x i − 4 j + 6k and V = 3i + 2 j − 3k. Use the dot product to determine the component U x .

2.107 Use the dot product to determine the angle between the forestay (cable AB) and the backstay (cable BC) of the sailboat. y B (4, 13) m

2.104 The vectors P, Q, and F = Fx i + Fy j lie in the x–y plane. The magnitudes of P and Q are 4 m and 5 m, respectively. The dot products P ⋅ F = 50 N-m (newton-meters) and Q ⋅ F = 60 N-m. Determine the components Fx and Fy . y Q

F 308 P 158

Problem 2.104

C (9, 1) m

A (0, 1.2) m

Problem 2.107 x

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x

Get complete eBook Order by email at [email protected] 2.108 Determine the angle θ between the lines AB and AC (a) by using the law of cosines (see Appendix A); (b) by using the dot product.

Problems

2.111 Segment AB of the bar is parallel to the x-axis. Determine the angle θ between the straight segments AB and BC. y

y

A

5m

A

B (4, 3, 21) m

(0, 3, 1) m

B

u

x

x

u C

(5, 21, 3) m C

z

z

(2, 0, 4) m

Problem 2.108

Problem 2.111

2.109 The ship O measures the positions of the ship A and the airplane B and obtains the coordinates shown. What is the angle θ between the lines of sight OA and OB ? y

B

2.112 The person exerts a force F = 60 i − 40 j (N) on the handle of the exercise machine. Use Eq. (2.26) to determine the vector component of F that is parallel to the line from the origin O to where the person grips the handle.

(4, 4, 24) km

150 mm

y

u x

O A

F

O

(6, 0, 3) km

z

Problem 2.109

200 mm

z

2.110 Astronauts on the space shuttle use radar to determine the magnitudes and direction cosines of the position vectors of two satellites A and B. The vector r A from the shuttle to satellite A has magnitude 2 km and direction cosines cos θ x = 0.768, cos θ y = 0.384, cos θ z = 0.512. The vector rB from the shuttle to satellite B has magnitude 4 km and direction cosines cos θ x = 0.743, cos θ y = 0.557,   cos θ z = −0.371. What is the angle θ between the vectors r A and rB ?

250 mm

x

Problem 2.112

B

rB x u

A

69

y

rA z

Problem 2.110

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70 Chapter 2 VectorsGet complete eBook Order by email at [email protected] 2.113 At the instant shown, the Harrier’s thrust vector is T = 4.22 i + 21.6 j − 3.30 k (kip) and its velocity vector is v = 24.4 i + 8.86 j + 2.44 k (ft/s). (a) Determine the components of the thrust vector parallel and normal (perpendicular) to the velocity vector. (b) The term T ⋅ v is the power being transferred to the plane by its engine. What is the power in horsepower?

2.117 The rope AB exerts a 50-N force T on collar A. Determine the vector component of T parallel to the bar CD. 2.118 In Problem 2.117, determine the vector component of T normal to the bar CD. y 0.15 m

y

0.4 m

B

C

v

T 0.2 m 0.3 m

A 0.5 m O D

T x

0.2 m

z

Problems 2.117/2.118

Problem 2.113 2.114 For the three-dimensional truss shown, determine the angle between members AB and AD. 2.115 For the three-dimensional truss shown, determine the angle between members AC and CD. 2.116 The three-dimensional truss is subjected to a force F = 25i − 100 j − 30 k (lb) at A. Determine the components of F parallel and normal (perpendicular) to member AD. y

x 0.25 m

2.119 The magnitudes of the two force vectors are F A = 200 lb and FB = 160 lb. What is the angle between the two vectors? 2.120 The magnitudes of the two force vectors are F A = 200 lb and FB = 160 lb. Determine the components of F A parallel and normal to FB .

F

y A (4, 3, 4) ft FB

B

z

FA

D (6, 0, 0) ft x 608 C (5, 0, 6) ft

Problems 2.114−2.116

308

408 508

z

Problems 2.119/2.120

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x

Get complete eBook Order by email at [email protected] Cross Products 71 2.121 An astronaut in a maneuvering unit approaches the ISS. At the present instant, the station informs him that his position relative to the origin of the station’s coordinate system is rG = 50 i + 80 j + 180 k  (m) and his velocity is v = −2.2 j − 3.6k  (m/s). The position of an airlock is r A = −12 i + 20 k  (m). Determine the angle between his velocity vector and the line from his position to the airlock’s position. 2.122 In Problem 2.121, determine the vector component of the astronaut’s velocity parallel to the line from his position to the airlock’s position.

2.123 Point P is at longitude 30 ° W and latitude 45 ° N on the Atlantic Ocean between Nova Scotia and France. Point Q is at longitude 60 ° E and latitude 20 ° N in the Arabian Sea. Use the dot product to determine the shortest distance along the surface of the earth from P to Q in terms of the radius of the earth R E . Strategy: Use the dot product to determine the angle between the lines OP and OQ; then use the definition of an angle in radians to determine the distance along the surface of the earth from P to Q. y N

P Q 458

z

308

208

O 608

G

Source: Courtesy of Marc Ward/Shutterstock. Problems 2.121/2.122

Equator x

Problem 2.123

2.5 Cross Products QUESTIONS TO CONSIDER • The cross product of two vectors is a vector. What do you know about its direction? • If you know that two vectors have a cross product equal to zero, what do you know about them?

BACKGROUND Like the dot product, the cross product of two vectors has many applications, including determining the rate of rotation of a fluid particle and calculating the force exerted on a charged particle by a magnetic field. Because of its usefulness for determining moments of forces, the cross product is an indispensable tool in mechanics. In this section we show you how to evaluate cross products and give examples of simple applications.

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72 Chapter 2 VectorsGet complete eBook Order by email at [email protected] V

U

DEFINITION Consider two vectors U and V (Fig. 2.24a). The cross product of U and V, denoted U × V, is defined by U × V = U V sin θ   e.

(a) V U

u

(b)

V U

e

(2.28) The angle θ is the angle between U and V when they are placed tail to tail (Fig. 2.24b). The vector e is a unit vector defined to be perpendicular to both U and V. Because this leaves two possibilities for the direction of e, the vectors U, V, and e are defined to be a right-handed system. The right-hand rule for determining the direction of e is shown in Fig. 2.24c. If the fingers of the right hand are pointed in the direction of the vector U (the first vector in the cross product) and then bent toward the vector V (the second vector in the cross product), the thumb points in the direction of e. Because the result of the cross product is a vector, it is sometimes called the vector product. The units of the cross product are the product of the units of the two vectors. Notice that the cross product of two nonzero vectors is equal to zero if and only if the two vectors are parallel. An interesting property of the cross product is that it is not commutative. Eq. (2.28) implies that the magnitude of the vector U × V is equal to the magnitude of the vector V × U, but the right-hand rule indicates that they are opposite in direction (Fig. 2.25). That is, U × V = −V × U.

(c)

FIGURE 2.24

The cross product is not  commutative.

(2.29)

The cross product also satisfies the relations

(a) The vectors U and V. (b) The angle θ between the vectors when they are placed tail to tail. (c) Determining the direction of e by the right-hand rule.

a( U × V ) = ( aU ) × V = U × ( aV )

The cross product is  associative with  respect to scalar  multiplication.

(2.30)

and U × ( V + W) = (U × V ) + (U × W) U3V V U

The cross product is  distributive with  respect to vector  addition.

(2.31)

for any scalar a and vectors U, V, and W.

CROSS PRODUCTS IN TERMS OF COMPONENTS V U

To obtain an equation for the cross product of two vectors in terms of their components, we must determine the cross products formed from the unit vectors i, j, and k. Because the angle between two identical vectors placed tail to tail is zero, it follows that i × i = i i sin(0)e = 0 .

V3U

FIGURE 2.25

The cross product i × j is

Directions of U × V and V × U.

i × j = i j sin 90 °e = e,

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where e is a unit vector perpendicular to i and j. Either e = k or e = −k. Applying the right-hand rule, we find that e = k (Fig. 2.26). Therefore, i × j = k.

j

Continuing in this way, we obtain i × i = 0, i × j = k, i × k = − j, j × i = −k, j × j = 0, j × k = i, k × i = j, k × j = −i, k × k = 0 .

z

(2.32)

These results can be remembered easily by arranging the unit vectors in a circle, as shown in Fig.  2.27a. The cross product of adjacent vectors is equal to the third vector with a positive sign if the order of the vectors in the cross product is the order indicated by the arrows and a negative sign otherwise. For example, in Fig. 2.27b we see that i × j = k, but i × k = − j. The cross product of two vectors U and V, expressed in terms of their components, is

k

i

x

FIGURE 2.26 The right-hand rule indicates that i × j = k.

i

U × V = (U x i + U y j + U z k) × (V x i + V y j + V z k) = U xV x ( i × i) + U xV y ( i × j) + U xVz ( i × k) +  U yV x ( j × i) + U yV y ( j × j) + U yV z ( j × k) +  U zV x (k × i) + U zV y (k × j) + U zV z (k × k).

j

By substituting Eqs. (2.32) into this expression, we obtain the equation

(a)

U × V = (U yVz − U zV y ) i − (U xVz − U zV x ) j

i

+ (U xV y − U yV x )k.

(2.33)

i3j5k

This result can be compactly written as the determinant

U×V =

i j k Ux Uy Uz Vx

Vy

i 3 k 5 2j

j

FIGURE 2.27

Vz

(a) Arrange the unit vectors in a circle with arrows to indicate their order. (b) You can use the circle to determine their cross products.

EVALUATING A 3×3 DETERMINANT A 3 × 3 determinant can be evaluated by expressing it as

Vx

Vy

Vz

= i

Uy Uz Vy

Vz

k (b)

(2.34)

.

This equation is based on Eqs. (2.32), which we obtained using a right-handed coordinate system. It gives the correct result for the cross product only if a righthanded coordinate system is used to determine the components of U and V.

i j k Ux Uy Uz

k

− j

Ux Uz Vx

Vz

+k

Ux Uy Vx

Vy

.

The terms on the right are obtained by multiplying each element of the first row of the 3 × 3 determinant by the 2 × 2 determinant obtained by crossing

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74 Chapter 2 VectorsGet complete eBook Order by email at [email protected] out that element’s row and column. For example, the first element of the first row, i, is multiplied by the 2 × 2 determinant i j k Ux Uy Uz Vx

Vy

.

Vz

Be sure to remember that the second term is subtracted. Expanding the 2 × 2 determinants, we obtain the value of the determinant: i j k Ux Uy Uz Vx

Vy

= (U yVz − U zV y ) i − (U xVz − U zV x ) j + (U xV y − U yV x )k.

Vz

MIXED TRIPLE PRODUCTS In Chapter 4, when we discuss the moment of a force about a line, we will use an operation called the mixed triple product, defined by U ⋅ ( V × W).

(2.35)

In terms of the scalar components of the vectors,

U ⋅ ( V × W) = (U x i + U y j + U z k) ⋅

i Vx

j Vy

k Vz

Wx

Wy

Wz

= (U x i + U y j + U z k) ⋅ [(V y W z − Vz W y ) i − (V x W z − V z W x ) j + (V x W y − V y W x )k] = U x (V y W z − V z W y ) − U y (V x W z − V z W x ) + U z (V x W y − V y W x ). This result can be expressed as the determinant

U ⋅ ( V × W) =

V

U W

Ux

Uy

Uz

Vx

Vy

Vz

Wx

Wy

Wz

.

(2.36)

Interchanging any two of the vectors in the mixed triple product changes the sign but not the absolute value of the result. For example, U ⋅ ( V × W) = −W ⋅ ( V × U).

FIGURE 2.28 Parallelepiped defined by the vectors U, V, and W.

If the vectors U, V, and W in Fig. 2.28 form a right-handed system, it can be shown that the volume of the parallelepiped equals U ⋅ ( V × W).

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Get complete eBook Order by email at [email protected] Cross Products 75

RESULTS V

Cross Product The cross product of two vectors U and V is def ined by U V u U uu V u sin u e . (2.28)

u

As in the dot product, u is the angle between the vectors when they are placed tail to tail. The unit vector e is defined to be perpendicular to U, perpendicular to V, and directed so that U, V, e form a right-handed system. If u U u Þ 0 and u V u Þ 0, U 3 V 5 0 if and only if U and V are parallel.

V U

e

Cross Product in Terms of Components The cross product of U and V is given in terms of the components of the vectors by U 3 V 5 (UyVz 2 UzVy)i 2 (UxVz 2 UzVx)j (2.33) (UxVy UyVx) k i

j

k

5 Ux

Uy

Uz

Vx

Vy

Vz

U

(2.34)

Mixed Triple Product The operation U? (V 3 W) is called the mixed triple product of the vectors U, V, and W. It can be expressed in terms of the components of the vectors by the determinant Ux

Uy

Uz

U?(V 3 W) 5 Vx

Vy

Vz .

(2.36)

Wx Wy Wz

V

When U, V, W form a right-handed system, the volume of the parallelepiped shown equals U ?(V 3 W).

U W

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Practice Example 2.14

Cross Products The components of two vectors U and V are U = 6 i − 5 j − k and V = 4 i + 2 j + 2k. (a) Determine the cross product U × V. (b) Use the dot product to prove that U × V is perpendicular to U. Strategy (a) Knowing the components of U and V, we can use Eq. (2.33) to determine U × V. (b) Once we have determined the components of the vector U × V, we can prove that it is perpendicular to U by showing that (U × V) ⋅ U = 0. Solution

(a) Use the components of the vectors to determine U × V.

 U × V = (U yVz − U zV y ) i − (U xVz − U zV x ) j   + (U xV y − U yV x )k  = [(−5)(2) − (−1)(2)]i − [(6)(2) − (−1)(4)] j   + [(6)(2) − (−5)(4)]k   = −8i − 16 j + 32k. 

(b) Show that (U × V) ⋅ U = 0.

 (U × V) ⋅ U = (U × V) x U x + (U × V) y U y + (U × V) z U z  = (−8)(6) + (−16)(−5) + (32)(−1)   = 0.  Practice Problem The components of two vectors U and V are U = 3i + 2 j − k and V = 5i − 3 j − 4 k. Determine the components of a unit vector that is perpendicular to U and perpendicular to V. Answer: e = −0.477 i + 0.304 j − 0.825k  or  e = 0.477 i − 0.304 j + 0.825k.

Example 2.15

Minimum Distance from a Point to a Line Consider the straight lines OA and OB. (a) Determine the components of a unit vector that is perpendicular to both OA and OB. (b) What is the minimum distance from point A to the line OB?

y B (6, 6, 23) m

O

z

x A (10, 22, 3) m

Strategy (a) Let rOA and rOB be the position vectors from O to A and from O to B (Fig. a). Because the cross product rOA × rOB is perpendicular to rOA and rOB , we will determine it and divide it by its magnitude to obtain a unit vector perpendicular to the lines OA and OB. (b) The minimum distance from A to the line OB is the length d of the straight line from A to OB that is perpendicular to OB (Fig. b). We can see that d = rOA sin θ, where θ is the angle between rOA and rOB . From the definition of the cross product, the magnitude of rOA × rOB is rOA rOB sin θ, so we can determine d by dividing the magnitude of rOA × rOB by the magnitude of rOB .

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Get complete eBook Order by email at [email protected] Cross Products 77 Solution (a) The components of rOA and rOB are

y B

rOA = 10 i − 2 j + 3k  (m),

rOB

rOB = 6 i + 6 j − 3k  (m). By using Eq. (2.34), we obtain rOA × rOB : rOA × rOB =

i j k 10 −2 3 6 6 −3

O

= −12 i + 48 j + 72k  (m 2 ).

This vector is perpendicular to rOA and rOB . Dividing it by its magnitude, we obtain a unit vector e that is perpendicular to the lines OA and OB:

A

z

(a) The vectors rOA and rOB .

rOA × rOB = rOA × rOB

−12 i + 48 j + 72k  (m 2 ) (−12 m 2 ) 2 + (48 m 2 ) 2 + (72 m 2 ) 2 = −0.137 i + 0.549 j + 0.824 k.

e =

x rOA

y B

(b) From Fig. b, the minimum distance d is

rOB

d = rOA sin θ. The magnitude of rOA × rOB is rOA × rOB = rOA rOB sin θ. Solving this equation for sin θ, we find that the distance d is  r × rOB d = rOA  OA  rOA rOB =

d

O

 rOA × rOB  =  rOB

u z

x

rOA A

(b) The minimum distance d from A to the line OB.

(−12 m 2 ) 2 + (48 m 2 ) 2 + (72 m 2 ) 2 = 9.71 m. (6 m) 2 + (6 m) 2 + (−3 m) 2

Critical Thinking This example is an illustration of the power of vector methods. Determining the minimum distance from point A to the line OB can be formulated as a minimization problem in differential calculus, but the vector solution we present is far simpler.

Example 2.16

Component of a Vector Perpendicular to a Plane

The rope CE exerts a 500-N force T on the door ABCD. What is the magnitude of the component of T perpendicular to the door? Strategy We are given the coordinates of the corners A, B, and C of the door. By taking the cross product of the position vector rCB from C to B and the position vector rCA from C to A, we will obtain a vector that is perpendicular to the door. We can divide the resulting vector by its magnitude to obtain a unit vector perpendicular to the door and then apply Eq. (2.26) to determine the component of T perpendicular to the door.

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(continued )

78 Chapter 2 VectorsGet complete eBook Order by email at [email protected]

E

y

Solution The components of rCB and rCA are (0.2, 0.4, 20.1) m

rCB = 0.35i − 0.2 j + 0.2k  (m), rCA = 0.5i − 0.2 j (m).

D

T

Their cross product is

C (0, 0.2, 0) m

(0.5, 0, 0) m A x B (0.35, 0, 0.2) m

rCB × rCA =

= 0.04 i + 0.1 j + 0.03k  (m 2 ).

Dividing this vector by its magnitude, we obtain a unit vector e that is perpendicular to the door (Fig. a):

z

e =

rCB × rCA = rCB × rCA

0.04 i + 0.1 j + 0.03k (m 2 )

(0.04 m 2 ) 2 + (0.1 m 2 ) 2 + (0.03 m 2 ) 2 = 0.358i + 0.894 j + 0.268k.

y e

To use Eq. (2.26), we must express T in terms of its scalar components. The position vector from C to E is

D C

i j k 0.35 −0.2 0.2 0.5 −0.2 0

rCA rCB

A

x

rCE = 0.2 i + 0.2 j − 0.1k  (m), so we can express the force T as

B z

(a) Determining a unit vector perpendicular to the door.

T = T

rCE 0.2 i + 0.2 j − 0.1k  (m) = (500 N)  rCE (0.2 m) 2 + (0.2 m) 2 + (−0.1 m) 2 = 333i + 333 j − 167k  (N).

The component of T parallel to the unit vector e, which is the component of T perpendicular to the door, is (e ⋅ T)e = [(0.358)(333 N) + (0.894)(333 N) + (0.268)(−167 N)]e = 373e  (N). The magnitude of the component of T perpendicular to the door is 373 N. Critical Thinking Why is it useful to determine the component of the force T perpendicular to the door? If the y-axis is vertical and the rope CE is the only thing preventing the hinged door from falling, you can see intuitively that it is the component of the force perpendicular to the door that holds it in place. We analyze problems of this kind in Chapter 5.

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Problems

79

PROBLEMS 2.124 Consider the vectors U = 12 i + 4 j − 8k, V = −6 i + 8 j + 14 k. (a) Determine the components of the vector W = U × V. (b) Evaluate the dot products U ⋅ W and V ⋅ W. What is the explanation of these two results?

2.130 (a) Determine the components of a unit vector e p that is parallel to the line AB and points from A toward B . (b) Determine the components of a unit vector e n that is normal (perpendicular) to the line AB. (Make sure your result is a unit vector.) Confirm that it is perpendicular to the line by showing that e p ⋅ e n = 0. y

2.125 Two vectors U = 3i + 2 j and V = 2 i + 4 j. (a) What is the cross product U × V ? (b) What is the cross product V × U ? 2.126 Consider the position vectors P and Q. (a) Determine the cross product P × Q by using the definition, Eq. (2.28). (b) Determine the dot product P ⋅ Q by using Eq. (2.34).

A (22, 12, 2) ft

2.127 Consider the position vectors P and Q. (a) Determine the cross product Q × P by using the definition, Eq. (2.28). (b) Determine the dot product Q ⋅ P by using Eq. (2.34).

x B (2, 26, 24) ft

z

Problem 2.130

y

2.131 The force F = 10 i − 4 j (N). Determine the cross product r AB × F. Q

(8, 0, 0) m

y

x

(6, 3, 0) m

P

A

z (12, 0, 10) m

rA B

Problems 2.126/2.127 2.128 Suppose that the cross product of two vectors U and V is U × V = 0. If U ≠ 0, what do you know about the vector V ? 2.129 The cross product of two vectors U and V is U × V = −30 i + 40 k. The vector V = 4 i − 2 j + 3k. The vector U = 4 i + U y j + U z k. Determine U y and U z .

z

x (6, 0, 4) m B F

Problem 2.131

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80 Chapter 2 VectorsGet complete eBook Order by email at [email protected] 2.132 By evaluating the cross product U × V, prove the identity sin(θ1 − θ 2 ) = sin θ1 cos θ 2 − cos θ1 sin θ 2 .

2.136 The cable BC exerts a 1000-lb force F on the hook at B. Determine r AB × F.

y

y

U

F

V

6 ft

B rAB

u1

x u2

8 ft

x

C rAC

4 ft

Problem 2.132

4 ft z

2.133 In Example 2.15, what is the minimum distance from point B to the line OA? 2.134 (a) What is the cross product rOA × rOB ? (b) Determine a unit vector e that is perpendicular to rOA and rOB . 2.135 Use the cross product to determine the length of the shortest straight line from point B to the straight line that passes through points O and A.

A

12 ft

Problem 2.136 2.137* The force F = 200 i + 180 j + 100 k ( lb ) . Determine the vector component of F that is normal (perpendicular) to the flat plate ACD. y

y B (4, 4, 24) m

C

(0, 10, 0) ft F

rOB A

O

x rOA

z

A (6, 22, 3) m

Problems 2.134/2.135

(6, 0, 0) ft z

D

(4, 0, 12) ft

Problem 2.137

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x

Get complete eBook Order by email at [email protected] 2.138 The rope AB exerts a 50-N force T on the collar at A. Let rCA be the position vector from point C to point A. Determine the cross product rCA × T.

Problems

81

2.141* Determine the minimum distance from point P to the plane defined by the three points A, B, and C. y

y

B (0, 5, 0) m

0.15 m

P (9, 6, 5) m 0.4 m

B

C

A (3, 0, 0) m

T 0.2 m 0.3 m

A

C

0.5 m O

z

z

x

(0, 0, 4) m

Problem 2.141

0.25 m

D 0.2 m

2.142 Two cartesian coordinate systems are shown. The unit vectors i ′ and j′ of the x ′y ′z ′ coordinate system are given in terms of their components in the xyz coordinate system by

Problem 2.138 2.139 In Example 2.16, suppose that the attachment point E is moved to the location (0.3,  0.3,  0) m and the magnitude of T increases to 600 N. What is the magnitude of the component of T perpendicular to the door? 2.140 The slender bar AB is 14 ft in length and is perpendicular to the flat plate ACD. (a) Determine the components of a unit vector e that is perpendicular to the plate ACD and points in the direction from A toward B. (b) Determine the coordinates ( x B , y B , z B ) of point B.

i ′ = 0.743i + 0.557 j + 0.371k, j′ = −0.408i + 0.816 j − 0.408k. (a) Determine the unit vector k′ of the x ′y ′z ′ coordinate system in terms of its components in the xyz coordinate system. (b) The force F = 40 i + 80 j + 20 k (N). Use the dot product to determine the components of F in terms of the x ′y ′z ′ coordinate system.

y9

y F x9

y

C

(0, 10, 0) ft

B (xB, yB, zB)

x z9

A x (6, 0, 0) ft z

D

x

z

Problem 2.142

(4, 0, 12) ft

Problem 2.140

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82 Chapter 2 VectorsGet complete eBook Order by email at [email protected] 2.143 For the vectors U = 6 i + 2 j − 4 k,   V = 2 i + 7 j, and W = 3i + 2k, evaluate the following mixed triple products:

2.145 By using Eqs. (2.23) and (2.34), show that Ux

(a) U ⋅ ( V × W); (b) W ⋅ ( V × U); (c) V ⋅ (W × U).

U ⋅ (V × W) =

2.144 Use the mixed triple product to calculate the volume of the parallelepiped.

Uy

Uz

Vx

Vy

Vz

Wx

Wy

Wz

.

2.146 The magnitudes of the two force vectors are F A = 200 lb and FB = 160 lb. What is the magnitude of their cross product F A × FB ?

y y FB

(140, 90, 30) mm

FA 608

(200, 0, 0) mm x

308

(160, 0, 100) mm

z

Problem 2.144

408

x

508

z

Problem 2.146

REVIEW PROBLEMS 2.147 The magnitude of F is 8 kN. Express F in terms of scalar components. y

2.148 The magnitude of the vertical force W is 600 lb, and the magnitude of the force B is 1500 lb. Given that A + B + W = 0, determine the magnitude of the force A and the angle α.

(3, 7) m

F W B a

(7, 2) m x

508

A

Problem 2.148

Problem 2.147

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Get complete eBook Order by email at [email protected] Review Problems 83 2.149 The magnitude of the vertical force vector A is 200 lb. If A + B + C = 0, what are the magnitudes of the force vectors B and C? 2.150 The magnitude of the horizontal force vector D is 280 lb. If D + E + F = 0, what are the magnitudes of the force vectors E and F ?

50 in

2.154 Determine the vector component of F that is parallel to the line AB.

2.156 Determine the vector rBA × F, where rBA is the position vector from B to A. 2.157 (a) Write the position vector r AB from point A to point B in terms of components. (b) A vector R has magnitude R = 200 lb and is parallel to the line from A to B. Write R in terms of components.

E

C D

B

2.153 What is the angle θ between the line AB and the force F ?

2.155 Determine the vector component of F that is normal to the line AB.

100 in

70 in

2.152 Determine the components of a unit vector parallel to line AB that points from A toward B.

F

A

Problems 2.149/2.150

2.158 The rope exerts a force of magnitude F = 200 lb on the top of the pole at B. (a) Determine the vector r AB × F, where r AB is the position vector from A to B. (b) Determine the vector r AC × F, where r AC is the position vector from A to C.

Refer to the following diagram when solving Problems 2.151 through 2.157.

y

B (5, 6, 1) ft

y A (4, 4, 2) ft

F 5 20i 1 10j 2 10k (lb) F

u A

B (8, 1, 22) ft x

x C (3, 0, 4) ft

z

Problems 2.151–2.157

z

Problem 2.158

2.151 What are the direction cosines of F ?

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84 Chapter 2 VectorsGet complete eBook Order by email at [email protected] 2.159 The pole supporting the sign is parallel to the x-axis and is 6 ft long. Point A is contained in the y–z plane. (a) Express the vector r in terms of components. (b) What are the direction cosines of r? y

2.161 The magnitude of the force vector FB is 2 kN. Express it in terms of components. 2.162 The magnitude of the vertical force vector F is 6 kN. Determine the vector components of F parallel and normal to the line from B to D. 2.163 The magnitude of the vertical force vector F is 6 kN. Given that F + F A + FB + FC = 0, what are the magnitudes of F A ,   FB , and FC ?

A

Bedford Falls

y

F D

r 608

(4, 3, 1) m

458

FC

FA x

O

FB

A

C

z (5, 0, 3) m

B

x (6, 0, 0) m

Problems 2.161–2.163 z

Problem 2.159 2.160 The z component of the force F is 80 lb. (a) Express F in terms of components. (b) What are the angles θ x , θ y , and θ z between F and the positive coordinate axes? y

2.164 The magnitude of the vertical force W is 160 N. The direction cosines of the position vector from A to B are cos θ x = 0.500, cos θ y = 0.866, and cos θ z = 0, and the direction cosines of the position vector from B to C are cos θ x = 0.707, cos θ y = 0.619, and cos θ z = −0.342. Point G is the midpoint of the line from B to C. Determine the vector r AG × W, where r AG is the position vector from A to G.

y

F x

208

O

m

0m

60

C G

B

608

W

600 mm

A

z

Problem 2.160

A z

x

Problem 2.164

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Get complete eBook Order by email at [email protected] Review Problems 85 2.165 The rope CE exerts a 500-N force T on the hinged door. (a) Express T in terms of components. (b) Determine the vector component of T parallel to the line from point A to point B. 2.166 In Problem 2.165, let rBC be the position vector from point B to point C. Determine the cross product rBC × T. E (0.2, 0.4, 20.1) m

y

C (0, 0.2, 0) m

T

D

A (0.5, 0, 0) m

z

x

B (0.35, 0, 0.2) m

Problems 2.165/2.166

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Source: Courtesy of Yanlev/123RF.

m The climber’s weight is supported by the forces exerted by his rope and the forces exerted on his shoes by the mountain. In this chapter we use free-body diagrams to analyze the forces acting on objects in equilibrium.

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