A 1500 kg car skids to a halt on a wet road where μk = 0.53.

Physics11thEditionbyJohnD.Cutnell.pdf
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Cutnell & Johnson Physics

Eleventh Edition

DAVID YOUNG SHANE STADLER

Louisiana State University

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iii

About the Authors

DAVID YOUNG received his Ph.D. in experimental condensed matter physics from Florida State University in 1998. He then held a

post-doc position in the Department of Chemistry and the Princeton

Materials Institute at Princeton University before joining the fac-

ulty in the Department of Physics and Astronomy at Louisiana State

University in 2000. His research focuses on the synthesis and char-

acterization of high-quality single crystals of novel electronic and

magnetic materials. The goal of his research group is to understand

the physics of electrons in materials under extreme conditions, i.e.,

at temperatures close to absolute zero, in high magnetic fi elds, and

under high pressure. He is the coauthor of over 200 research publica-

tions that have appeared in peer-reviewed journals, such as Physical Review B, Physical Review Letters, and Nature. Professor Young has taught introductory physics with the Cutnell & Johnson text since he

was a senior undergraduate over 20 years ago. He routinely lectures

to large sections, often in excess of 300 students. To engage such a

large number of students, he uses WileyPLUS, electronic response systems, tutorial-style recitation sessions, and in-class demonstra-

tions. Professor Young has received multiple awards for outstanding

teaching of undergraduates. David enjoys spending his free time with

his family, playing basketball, and working on his house.

I would like to thank my family for their continuous love and support.

—David Young

SH ANE STADLER Shane Stadler earned a Ph.D. in experi- mental condensed matter physics from Tulane University in 1998.

Afterwards, he accepted a National Research Council Postdoctoral

Fellowship with the Naval Research Laboratory in Washington, DC,

where he conducted research on artifi cially structured magnetic ma-

terials. Three years later, he joined the faculty in the Department of

Physics at Southern Illinois University (the home institution of John

Cutnell and Ken Johnson, the original authors of this textbook), be-

fore joining the Department of Physics and Astronomy at Louisiana

State University in 2008. His research group studies novel magnetic

materials for applications in the areas of spintronics and magnetic

cooling.

Over the past fi fteen years, Professor Stadler has taught the full

spectrum of physics courses, from physics for students outside the

sciences, to graduate-level physics courses, such as classical electro-

dynamics. He teaches classes that range from fewer than ten students

to those with enrollments of over 300. His educational interests are

focused on developing teaching tools and methods that apply to both

small and large classes, and which are applicable to emerging teach-

ing strategies, such as “fl ipping the classroom.”

In his spare time, Shane writes science fi ction/thriller novels.

I would like to thank my parents, George and Elissa, for their constant

support and encouragement. —Shane Stadler

C o u rt

es y D

av id

Y o u n g

C o u rt

es y S

h an

e S

ta d le

r

Dear Students and Inst ructors:

Welcome to college ph ysics! To the students:

We know there is a ne gative stigma associate

d with physics, and yo u yourself may har-

bor some trepidation a s you begin this course

. But fear not! We’re h ere to help. Whether y

ou’re worried about yo ur math profi ciency,

understanding the con cepts, or developing yo

ur problem-solving ski lls, the resources avail

able to you are designe d to address all of

these areas and more. Research has shown th

at learning styles vary greatly among student

s. Maybe some of you have a more visual

preference, or auditory preference, or some o

ther preferred learning modality. In any case,

the resources availabl e to you in this course

will satisfy all of these preferences and impro

ve your chance of succ ess. Take a moment to

explore below what th e textbook and

online course have to o ff er. We suspect that, a

s you continue to impr ove throughout the cou

rse, some of that initia l trepidation will be

replaced with exciteme nt.

To start, we have creat ed a new learning med

ium specifi c to this boo k in the form of a comp

rehensive set of LECTURE VIDE OS – one

for every section (259 in all). These animated

lectures (created and n arrated by the authors)

are 2–10 minutes in le ngth, and explain the

basic concepts and lear ning objectives of each

section. They are assig nable within WileyPLU

S and can be paired wi th follow-up ques-

tions that are gradable. In addition to supplem

enting traditional lectu ring, the videos can be

used in a variety of wa ys, including fl ipping

the classroom, a comp lete set of lectures for o

nline courses, and revi ewing for exams. Next

, we have enhanced “T he Physics of …”

examples by increasing the bio-inspired exam

ples by 40%. Although they are of general ins

tructional value, they a re also similar to what

premed students will e ncounter in the Chemical and Ph

ysical Foundations of Biological Systems Passages section of the

MCAT. Finally,

we have introduced new “team problems” in th

e end-of-chapter proble ms that are designed fo

r group problem-solvin g exercises. These

are context-rich proble ms of medium diffi cult

y designed for group c ooperation, but may al

so be tackled by the in dividual student.

One of the great streng ths of this text is the sy

nergistic relationship i t develops between pro

blem solving and conc eptual understand-

ing. For instance, avail able in WileyPLUS are animated

Chalkboard Videos, which cons ist of short (2–3 min) v

ideos demonstrat-

ing step-by-step practi cal solutions to typical

homework problems. Also available are num

erous Guided Online (GO) Tuto rials that

implement a step-by-s tep pedagogical approa

ch, which provides stu dents a low-stakes env

ironment for refi ning t heir problem solving

skills. One of the most important techniques

developed in the text f or solving problems in

volving multiple force s is the free-body

diagram (FBD). Many problems in the force-intensive

chapters, such as chap ters 4 and 18, take adv

antage of the new FBD capabilities

now available online in WileyPLUS, where students can

construct the FBD’s f or a select number of p

roblems and be graded on them.

Finally, ORION, an online adapt ive learning environme

nt, is seamlessly integr ated into WileyPLUS for Cutnell

& Johnson.

The content and functi onality of WileyPLUS, and the a

daptive learning enviro nment of ORION (see below), w

ill provide students wi th

all the resources they n eed to be successful in

the course.

• The Lecture Videos created b y the authors for each

section include questio ns with intelligent feed

back when a student en ters the

wrong answer.

• The multi-step GO Tutorial p roblems created in WileyPLUS a

re designed to provide targeted, intelligent fe

edback.

• The Free-body Diagram vecto r drawing tools provid

e students an easy way to enter answers requi

ring vector drawing, an d also

provide enhanced feed back.

• Chalkboard Video Solutions t ake the students step-b

y-step through the solu tion and the thought pr

ocess of the authors. P roblem-

solving strategies are d iscussed, and common

misconceptions and p otential pitfalls are add

ressed. The students ca n then apply these

techniques to solve sim ilar, but diff erent probl

ems.

All of these features ar e designed to encourag

e students to remain w ithin the WileyPLUS environme

nt, as opposed to pursu ing the

“pay-for solutions” we bsites that short circui

t the learning process. To the students – We s

trongly recommend th at you take this honest

approach to the course . Take full advantage o

f the many features an d learning resources th

at accompany the text and the online con-

tent. Be engaged with the material and push

yourself to work throu gh the exercises. Physi

cs may not be the easie st subject to under-

stand, but with the Wi ley resources at your d

isposal and your hard w ork, you CAN be succ

essful.

We are immensely gra teful to all of you who

have provided feedbac k as we’ve worked on

this new edition, and t o our students who

have taught us how to teach. Thank you for y

our guidance, and keep the feedback coming.

Best wishes for succes s in this course and

wherever your major m ay take you!

Sincerely,

David Young and Shan e Stadler, Louisiana St

ate University

email: dyoun14@gma il.com or stadler.ls

u.edu@gmail.com

iv

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v

Brief Contents

1 Introduction and Mathematical Concepts 1

2 Kinematics in One Dimension 27

3 Kinematics in Two Dimensions 55

4 Forces and Newton’s Laws of Motion 80

5 Dynamics of Uniform Circular Motion 121

6 Work and Energy 144

7 Impulse and Momentum 175

8 Rotational Kinematics 200

9 Rotational Dynamics 223

10 Simple Harmonic Motion and Elasticity 257

11 Fluids 289

12 Temperature and Heat 326

13 The Transfer of Heat 360

14 The Ideal Gas Law and Kinetic Theory 380

15 Thermodynamics 401

16 Waves and Sound 433

17 The Principle of Linear Superposition and Interference Phenomena 465

18 Electric Forces and Electric Fields 489

19 Electric Potential Energy and the Electric Potential 523

20 Electric Circuits 551

21 Magnetic Forces and Magnetic Fields 590

22 Electromagnetic Induction 625

23 Alternating Current Circuits 661

24 Electromagnetic Waves 684

25 The Reflection of Light: Mirrors 711

26 The Refraction of Light: Lenses and Optical Instruments 733

27 Interference and the Wave Nature of Light 777

28 Special Relativity 808

29 Particles and Waves 832

30 The Nature of the Atom 853

31 Nuclear Physics and Radioactivity 885

32 Ionizing Radiation, Nuclear Energy, and Elementary Particles 911

Contents

1 Introduction and Mathematical Concepts 1

1.1 The Nature of Physics 1 1.2 Units 2 1.3 The Role of Units in Problem Solving 3 1.4 Trigonometry 6 1.5 Scalars and Vectors 8 1.6 Vector Addition and Subtraction 10 1.7 The Components of a Vector 12 1.8 Addition of Vectors by Means of Components 15 Concept Summary 19 Focus on Concepts 19 Problems 21 Additional Problems 24 Concepts and Calculations Problems 25 Team Problems 26

2 Kinematics in One Dimension 27 2.1 Displacement 27 2.2 Speed and Velocity 28 2.3 Acceleration 31 2.4 Equations of Kinematics for Constant

Acceleration 34 2.5 Applications of the Equations of Kinematics 37 2.6 Freely Falling Bodies 41 2.7 Graphical Analysis of Velocity and

Acceleration 45 Concept Summary 47 Focus on Concepts 48 Problems 49 Additional Problems 53 Concepts and Calculations Problems 54 Team Problems 54

3 Kinematics in Two Dimensions 55 3.1 Displacement, Velocity, and Acceleration 55 3.2 Equations of Kinematics in Two Dimensions 56 3.3 Projectile Motion 60 3.4 Relative Velocity 68 Concept Summary 72 Focus on Concepts 73 Problems 74 Additional Problems 77 Concepts and Calculations Problems 78 Team Problems 79

4 Forces and Newton’s Laws of Motion 80 4.1 The Concepts of Force and Mass 80 4.2 Newton’s First Law of Motion 81 4.3 Newton’s Second Law of Motion 83 4.4 The Vector Nature of Newton’s Second Law of Motion 85 4.5 Newton’s Third Law of Motion 86 4.6 Types of Forces: An Overview 88 4.7 The Gravitational Force 88 4.8 The Normal Force 92 4.9 Static and Kinetic Frictional Forces 95 4.10 The Tension Force 101 4.11 Equilibrium Applications of Newton’s Laws

of Motion 102 4.12 Nonequilibrium Applications of Newton’s Laws

of Motion 106 Concept Summary 111 Focus on Concepts 112 Problems 114 Additional Problems 118 Concepts and Calculations Problems 119 Team Problems 120

5 Dynamics of Uniform Circular Motion 121

5.1 Uniform Circular Motion 121 5.2 Centripetal Acceleration 122 5.3 Centripetal Force 125 5.4 Banked Curves 129 5.5 Satellites in Circular Orbits 130 5.6 Apparent Weightlessness and Artificial Gravity 133 5.7 *Vertical Circular Motion 136 Concept Summary 137 Focus on Concepts 138 Problems 139 Additional Problems 141 Concepts and Calculations Problems 142 Team Problems 143

6 Work and Energy 144 6.1 Work Done by a Constant Force 144 6.2 The Work–Energy Theorem and Kinetic Energy 147 6.3 Gravitational Potential Energy 153 6.4 Conservative Versus Nonconservative Forces 155 6.5 The Conservation of Mechanical Energy 157 6.6 Nonconservative Forces and the Work–Energy

Theorem 161

Contents vii

6.7 Power 162 6.8 Other Forms of Energy and the Conservation

of Energy 164 6.9 Work Done by a Variable Force 164 Concept Summary 166 Focus on Concepts 167 Problems 168 Additional Problems 172 Concepts and Calculations Problems 173 Team Problems 174

7 Impulse and Momentum 175 7.1 The Impulse–Momentum Theorem 175 7.2 The Principle of Conservation of Linear

Momentum 179 7.3 Collisions in One Dimension 184 7.4 Collisions in Two Dimensions 189 7.5 Center of Mass 189 Concept Summary 192 Focus on Concepts 193 Problems 194 Additional Problems 197 Concepts and Calculations Problems 198 Team Problems 199

8 Rotational Kinematics 200 8.1 Rotational Motion and Angular Displacement 200 8.2 Angular Velocity and Angular Acceleration 203 8.3 The Equations of Rotational Kinematics 205 8.4 Angular Variables and Tangential Variables 208 8.5 Centripetal Acceleration and Tangential

Acceleration 210 8.6 Rolling Motion 213 8.7 *The Vector Nature of Angular Variables 214 Concept Summary 215 Focus on Concepts 216 Problems 216 Additional Problems 220 Concepts and Calculations Problems 221 Team Problems 222

9 Rotational Dynamics 223 9.1 The Action of Forces and Torques on Rigid

Objects 223 9.2 Rigid Objects in Equilibrium 226 9.3 Center of Gravity 231 9.4 Newton’s Second Law for Rotational Motion About a

Fixed Axis 236 9.5 Rotational Work and Energy 241 9.6 Angular Momentum 244

Concept Summary 246 Focus on Concepts 247 Problems 248 Additional Problems 254 Concepts and Calculations Problems 255 Team Problems 256

10 Simple Harmonic Motion and Elasticity 257

10.1 The Ideal Spring and Simple Harmonic Motion 257 10.2 Simple Harmonic Motion and the Reference Circle 261 10.3 Energy and Simple Harmonic Motion 267 10.4 The Pendulum 270 10.5 Damped Harmonic Motion 273 10.6 Driven Harmonic Motion and Resonance 274 10.7 Elastic Deformation 275 10.8 Stress, Strain, and Hooke’s Law 279 Concept Summary 280 Focus on Concepts 281 Problems 282 Additional Problems 287 Concepts and Calculations Problems 288 Team Problems 288

11 Fluids 289 11.1 Mass Density 289 11.2 Pressure 291 11.3 Pressure and Depth in a Static Fluid 293 11.4 Pressure Gauges 297 11.5 Pascal’s Principle 298 11.6 Archimedes’ Principle 300 11.7 Fluids in Motion 305 11.8 The Equation of Continuity 307 11.9 Bernoulli’s Equation 309 11.10 Applications of Bernoulli’s Equation 311 11.11 *Viscous Flow 314 Concept Summary 317 Focus on Concepts 318 Problems 319 Additional Problems 323 Concepts and Calculations Problems 324 Team Problems 325

12 Temperature and Heat 326 12.1 Common Temperature Scales 326 12.2 The Kelvin Temperature Scale 328 12.3 Thermometers 329 12.4 Linear Thermal Expansion 330 12.5 Volume Thermal Expansion 337 12.6 Heat and Internal Energy 339

12.7 Heat and Temperature Change: Specific Heat Capacity 340

12.8 Heat and Phase Change: Latent Heat 343 12.9 *Equilibrium Between Phases of Matter 347 12.10 *Humidity 350 Concept Summary 352 Focus on Concepts 352 Problems 353 Additional Problems 358 Concepts and Calculations Problems 358 Team Problems 359

13 The Transfer of Heat 360 13.1 Convection 360 13.2 Conduction 363 13.3 Radiation 370 13.4 Applications 373 Concept Summary 375 Focus on Concepts 375 Problems 376 Additional Problems 378 Concepts and Calculations Problems 379 Team Problems 379

14 The Ideal Gas Law and Kinetic Theory 380

14.1 Molecular Mass, the Mole, and Avogadro’s Number 380 14.2 The Ideal Gas Law 383 14.3 Kinetic Theory of Gases 388 14.4 *Diff usion 392 Concept Summary 395 Focus on Concepts 396 Problems 397 Additional Problems 399 Concepts and Calculations Problems 400 Team Problems 400

15 Thermodynamics 401 15.1 Thermodynamic Systems and Their Surroundings 401 15.2 The Zeroth Law of Thermodynamics 402 15.3 The First Law of Thermodynamics 402 15.4 Thermal Processes 404 15.5 Thermal Processes Using an Ideal Gas 408 15.6 Specific Heat Capacities 411 15.7 The Second Law of Thermodynamics 412 15.8 Heat Engines 413 15.9 Carnot’s Principle and the Carnot Engine 414 15.10 Refrigerators, Air Conditioners, and Heat Pumps 417 15.11 Entropy 420 15.12 The Third Law of Thermodynamics 425

Concept Summary 425 Focus on Concepts 426 Problems 427 Additional Problems 431 Concepts and Calculations Problems 432 Team Problems 432

16 Waves and Sound 433 16.1 The Nature of Waves 433 16.2 Periodic Waves 435 16.3 The Speed of a Wave on a String 436 16.4 *The Mathematical Description of a Wave 439 16.5 The Nature of Sound 439 16.6 The Speed of Sound 442 16.7 Sound Intensity 446 16.8 Decibels 448 16.9 The Doppler Eff ect 450 16.10 Applications of Sound in Medicine 454 16.11 *The Sensitivity of the Human Ear 455 Concept Summary 456 Focus on Concepts 457 Problems 458 Additional Problems 463 Concepts and Calculations Problems 464 Team Problems 464

17 The Principle of Linear Superposition and Interference Phenomena 465

17.1 The Principle of Linear Superposition 465 17.2 Constructive and Destructive Interference of

Sound Waves 466 17.3 Diff raction 470 17.4 Beats 473 17.5 Transverse Standing Waves 474 17.6 Longitudinal Standing Waves 478 17.7 *Complex Sound Waves 481 Concept Summary 482 Focus on Concepts 483 Problems 484 Additional Problems 487 Concepts and Calculations Problems 488 Team Problems 488

18 Electric Forces and Electric Fields 489 18.1 The Origin of Electricity 489 18.2 Charged Objects and the Electric Force 490 18.3 Conductors and Insulators 493 18.4 Charging by Contact and by Induction 493 18.5 Coulomb’s Law 495 18.6 The Electric Field 500

viii Contents

Contents ix

18.7 Electric Field Lines 505 18.8 The Electric Field Inside a Conductor: Shielding 508 18.9 Gauss’ Law 510 18.10 *Copiers and Computer Printers 513 Concept Summary 516 Focus on Concepts 516 Problems 517 Additional Problems 521 Concepts and Calculations Problems 521 Team Problems 522

19 Electric Potential Energy and the Electric Potential 523

19.1 Potential Energy 523 19.2 The Electric Potential Diff erence 524 19.3 The Electric Potential Diff erence Created by Point

Charges 530 19.4 Equipotential Surfaces and Their Relation to the

Electric Field 534 19.5 Capacitors and Dielectrics 537 19.6 *Biomedical Applications of Electric Potential

Diff erences 541 Concept Summary 544 Focus on Concepts 544 Problems 546 Additional Problems 548 Concepts and Calculations Problems 549 Team Problems 550

20 Electric Circuits 551 20.1 Electromotive Force and Current 551 20.2 Ohm’s Law 553 20.3 Resistance and Resistivity 554 20.4 Electric Power 557 20.5 Alternating Current 559 20.6 Series Wiring 562 20.7 Parallel Wiring 565 20.8 Circuits Wired Partially in Series and Partially in

Parallel 569 20.9 Internal Resistance 570 20.10 Kirchhoff ’s Rules 571 20.11 The Measurement of Current and Voltage 574 20.12 Capacitors in Series and in Parallel 575 20.13 RC Circuits 577 20.14 Safety and the Physiological Eff ects of Current 579 Concept Summary 580 Focus on Concepts 581 Problems 582 Additional Problems 588 Concepts and Calculations Problems 589 Team Problems 589

21 Magnetic Forces and Magnetic Fields 590

21.1 Magnetic Fields 590 21.2 The Force That a Magnetic Field Exerts on a Moving

Charge 592 21.3 The Motion of a Charged Particle in a Magnetic Field 595 21.4 The Mass Spectrometer 599 21.5 The Force on a Current in a Magnetic Field 600 21.6 The Torque on a Current-Carrying Coil 602 21.7 Magnetic Fields Produced by Currents 605 21.8 Ampère’s Law 612 21.9 Magnetic Materials 613 Concept Summary 616 Focus on Concepts 617 Problems 618 Additional Problems 623 Concepts and Calculations Problems 624 Team Problems 624

22 Electromagnetic Induction 625 22.1 Induced Emf and Induced Current 625 22.2 Motional Emf 627 22.3 Magnetic Flux 631 22.4 Faraday’s Law of Electromagnetic Induction 634 22.5 Lenz’s Law 637 22.6 *Applications of Electromagnetic Induction to the

Reproduction of Sound 640 22.7 The Electric Generator 641 22.8 Mutual Inductance and Self-Inductance 646 22.9 Transformers 649 Concept Summary 652 Focus on Concepts 653 Problems 654 Additional Problems 659 Concepts and Calculations Problems 659 Team Problems 660

23 Alternating Current Circuits 661 23.1 Capacitors and Capacitive Reactance 661 23.2 Inductors and Inductive Reactance 664 23.3 Circuits Containing Resistance, Capacitance, and

Inductance 665 23.4 Resonance in Electric Circuits 670 23.5 Semiconductor Devices 672 Concept Summary 678 Focus on Concepts 679 Problems 680 Additional Problems 681 Concepts and Calculations Problems 682 Team Problems 683

24 Electromagnetic Waves 684 24.1 The Nature of Electromagnetic Waves 684 24.2 The Electromagnetic Spectrum 688 24.3 The Speed of Light 690 24.4 The Energy Carried by Electromagnetic Waves 692 24.5 The Doppler Eff ect and Electromagnetic Waves 695 24.6 Polarization 697 Concept Summary 704 Focus on Concepts 704 Problems 705 Additional Problems 708 Concepts and Calculations Problems 709 Team Problems 710

25 The Reflection of Light: Mirrors 711 25.1 Wave Fronts and Rays 711 25.2 The Reflection of Light 712 25.3 The Formation of Images by a Plane Mirror 713 25.4 Spherical Mirrors 716 25.5 The Formation of Images by Spherical Mirrors 718 25.6 The Mirror Equation and the Magnification

Equation 722 Concept Summary 728 Focus on Concepts 728 Problems 729 Additional Problems 731 Concepts and Calculations Problems 731 Team Problems 732

26 The Refraction of Light: Lenses and Optical Instruments 733

26.1 The Index of Refraction 733 26.2 Snell’s Law and the Refraction of Light 734 26.3 Total Internal Reflection 739 26.4 Polarization and the Reflection and Refraction of

Light 745 26.5 The Dispersion of Light: Prisms and Rainbows 746 26.6 Lenses 748 26.7 The Formation of Images by Lenses 749 26.8 The Thin-Lens Equation and the Magnification

Equation 752 26.9 Lenses in Combination 755 26.10 The Human Eye 756 26.11 Angular Magnification and the Magnifying Glass 761 26.12 The Compound Microscope 763 26.13 The Telescope 764 26.14 Lens Aberrations 765 Concept Summary 767 Focus on Concepts 768 Problems 769

Additional Problems 775 Concepts and Calculations Problems 775 Team Problems 776

27 Interference and the Wave Nature of Light 777

27.1 The Principle of Linear Superposition 777 27.2 Young’s Double-Slit Experiment 779 27.3 Thin-Film Interference 782 27.4 The Michelson Interferometer 786 27.5 Diff raction 787 27.6 Resolving Power 791 27.7 The Diff raction Grating 796 27.8 *Compact Discs, Digital Video Discs, and the Use of

Interference 798 27.9 X-Ray Diff raction 799 Concept Summary 801 Focus on Concepts 802 Problems 803 Additional Problems 805 Concepts and Calculations Problems 806 Team Problems 807

28 Special Relativity 808 28.1 Events and Inertial Reference Frames 808 28.2 The Postulates of Special Relativity 809 28.3 The Relativity of Time: Time Dilation 811 28.4 The Relativity of Length: Length Contraction 815 28.5 Relativistic Momentum 817 28.6 The Equivalence of Mass and Energy 819 28.7 The Relativistic Addition of Velocities 824 Concept Summary 827 Focus on Concepts 827 Problems 828 Additional Problems 830 Concepts and Calculations Problems 831 Team Problems 831

29 Particles and Waves 832 29.1 The Wave–Particle Duality 832 29.2 Blackbody Radiation and Planck’s Constant 833 29.3 Photons and the Photoelectric Eff ect 834 29.4 The Momentum of a Photon and the Compton

Eff ect 840 29.5 The De Broglie Wavelength and the Wave Nature

of Matter 843 29.6 The Heisenberg Uncertainty Principle 845 Concept Summary 849 Focus on Concepts 849 Problems 850

x Contents

Contents xi

Additional Problems 852 Concepts and Calculations Problems 852 Team Problems 852

30 The Nature of the Atom 853 30.1 Rutherford Scattering and the Nuclear Atom 853 30.2 Line Spectra 855 30.3 The Bohr Model of the Hydrogen Atom 857 30.4 De Broglie’s Explanation of Bohr’s Assumption About

Angular Momentum 861 30.5 The Quantum Mechanical Picture of the Hydrogen

Atom 862 30.6 The Pauli Exclusion Principle and the Periodic Table of

the Elements 866 30.7 X-Rays 868 30.8 The Laser 872 30.9 *Medical Applications of the Laser 874 30.10 *Holography 876 Concept Summary 878 Focus on Concepts 879 Problems 880 Additional Problems 883 Concepts and Calculations Problems 883 Team Problems 883

31 Nuclear Physics and Radioactivity 885 31.1 Nuclear Structure 885 31.2 The Strong Nuclear Force and the Stability of the

Nucleus 887 31.3 The Mass Defect of the Nucleus and Nuclear Binding

Energy 888 31.4 Radioactivity 890 31.5 The Neutrino 896 31.6 Radioactive Decay and Activity 897 31.7 Radioactive Dating 900 31.8 Radioactive Decay Series 903 31.9 Radiation Detectors 904

Concept Summary 906 Focus on Concepts 907 Problems 908 Additional Problems 910 Concepts and Calculations Problems 910 Team Problems 910

32 Ionizing Radiation, Nuclear Energy, and Elementary Particles 911

32.1 Biological Eff ects of Ionizing Radiation 911 32.2 Induced Nuclear Reactions 915 32.3 Nuclear Fission 916 32.4 Nuclear Reactors 919 32.5 Nuclear Fusion 920 32.6 Elementary Particles 922 32.7 Cosmology 928 Concept Summary 931 Focus on Concepts 932 Problems 932 Additional Problems 934 Concepts and Calculations Problems 935 Team Problems 935

Appendixes A-1

APPENDIX A Powers of Ten and Scientific Notation A-1 APPENDIX B Significant Figures A-1 APPENDIX C Algebra A-2 APPENDIX D Exponents and Logarithms A-3 APPENDIX E Geometry and Trigonometry A-4 APPENDIX F Selected Isotopes A-5

ANSWERS TO CHECK YOUR UNDERSTANDING A-10

ANSWERS TO ODD-NUMBERED PROBLEMS A-18

INDEX I -1

Note: Chapter sections marked with an asterisk (*) can be omitted with little impact to the overall development of the material.

Our Vision Our goal is to provide students with the skills they need to succeed in this course, and instructors with the tools they need to develop those skills.

Skills Development One of the great strengths of this text is the synergistic relationship

between conceptual understanding, problem solving, and establish-

ing relevance. We identify here some of the core features of the text

that support these synergies.

Conceptual Understanding Students often regard physics as a collection of equations that can be used blindly to solve problems. How-

ever, a good problem-solving technique does not begin with equations. It

starts with a fi rm grasp of physics concepts and how they fi t together to

provide a coherent description of natural phenomena. Helping students

develop a conceptual understanding of physics principles is a primary

goal of this text. The features in the text that work toward this goal are:

• Lecture Videos (one for each section of the text) • Conceptual Examples • Concepts & Calculations problems (now with video solutions) • Focus on Concepts homework material • Check Your Understanding questions • Concept Simulations (an online feature)

Problem Solving The ability to reason in an organized and mathematically correct manner is essential to solving problems, and

helping students to improve their reasoning skills is also one of our

primary goals. To this end, we have included the following features:

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Chemical and Physical Foundations of Biological Systems Passages section of the MCAT. All biomedical examples and end-of-chapter

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EXAMPLE 7 BIO The Physics of Hearing Loss— Standing Waves in the Ear

Inner ear

Semicircular canals

Anvil

Hammer

Cochlea

Auditory nerve

Eustachian tubeOval

window Stirrup

Middle ear

Tympanic membrane

Outer ear

Auditory canal

Pinna

2.3 cm

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xvi Our V is ion and the Wi leyPLUS with ORION Advantage

The publishing world is changing rapidly! The digital age is here, and college

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Many of our physics colleagues and their students have generously

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editions, we especially owe a large debt of gratitude. Specifi cally, we thank:

Lai Cao, Baton Rouge Magnet High School

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About the cover: The cover image shows an artist’s rendition of a synaptic gap between an axon and a dendrite of a human nerve cell. Just like

the wires in the electrical system of your home, the nerve cells make connec-

tions in circuits called neural pathways. The transmission of chemical signals

between the axon and dendrite relies on the electrical potential diff erence

across the gap, which is a topic in Volume 2 of the text. Our hope is that this

book and its resources will help you develop some new neural pathways of

your own!

In spite of our best eff orts to produce an error-free book, errors no doubt remain. They are solely our responsibility, and we would appreciate hearing of any that

you fi nd. We hope that this text makes learning and teaching physics easier and more enjoyable, and we look forward to hearing about your experiences with

it. Please feel free to write us care of Physics Editor, Global Education, John Wiley & Sons, Inc., 111 River Street, Hoboken, NJ 07030, or contact the authors

at dyoun14@gmail.com or sstadler23@gmail.com.

LEARNING OBJECTIVES

Aft er reading this module, you should be able to...

1.1 Describe the fundamental nature of physics.

1.2 Describe diff erent systems of units.

1.3 Solve unit conversion problems.

1.4 Solve trigonometry problems.

1.5 Distinguish between vectors and scalars.

1.6 Solve vector addition and subtraction problems by graphical methods.

1.7 Calculate vector components.

1.8 Solve vector addition and subtraction problems using components.

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CHAPTER 1

Introduction and Mathematical Concepts

The animation techniques and special eff ects used in the fi lm The Avengers rely on computers and mathematical concepts such as trigonometry and vectors. Such mathematical concepts will be very useful

throughout this book in our discussion of physics.

1.1 The Nature of Physics Physics is the most basic of the sciences, and it is at the very root of subjects like chem-

istry, engineering, astronomy, and even biology. The discipline of physics has developed

over many centuries, and it continues to evolve. It is a mature science, and its laws en-

compass a wide scope of phenomena that range from the formation of galaxies to the in-

teractions of particles in the nuclei of atoms. Perhaps the most visible evidence of physics

in everyday life is the eruption of new applications that have improved our quality of life,

such as new medical devices, and advances in computers and high-tech communications.

The exciting feature of physics is its capacity for predicting how nature will be-

have in one situation on the basis of experimental data obtained in another situation.

Such predictions place physics at the heart of modern technology and, therefore, can

have a tremendous impact on our lives. Rocketry and the development of space travel

have their roots fi rmly planted in the physical laws of Galileo Galilei (1564–1642) and

Isaac Newton (1642–1727). The transportation industry relies heavily on physics in

the development of engines and the design of aerodynamic vehicles. Entire electronics

and computer industries owe their existence to the invention of the transistor, which

grew directly out of the laws of physics that describe the electrical behavior of solids.

The telecommunications industry depends extensively on electromagnetic waves, 1

2 CHAPTER 1 Introduction and Mathematical Concepts

whose existence was predicted by James Clerk Maxwell (1831–1879) in his theory of electricity and magnetism. The medical profession uses X-ray, ultrasonic, and magnetic resonance methods for obtaining images of the interior of the human body, and physics lies at the core of all these. Perhaps the most widespread impact in modern technology is that due to the laser. Fields ranging from space exploration to medicine benefi t from this incredible device, which is a direct applica- tion of the principles of atomic physics.

Because physics is so fundamental, it is a required course for students in a wide range of major areas. We welcome you to the study of this fascinating topic. You will learn how to see the world through the “eyes” of physics and to reason as a physicist does. In the process, you will learn how to apply physics principles to a wide range of problems. We hope that you will come to recognize that physics has important things to say about your environment.

1.2 Units Physics experiments involve the measurement of a variety of quantities, and a great deal of eff ort goes into making these measurements as accurate and reproducible as possible. The fi rst step toward ensuring accuracy and reproducibility is defi ning the units in which the measurements are made.

In this text, we emphasize the system of units known as SI units, which stands for the French phrase “Le Système International d’Unités.” By international agreement, this system employs the meter (m) as the unit of length, the kilogram (kg) as the unit of mass, and the second (s) as the unit of time. Two other systems of units are also in use, however. The CGS system utilizes the centimeter (cm), the gram (g), and the second for length, mass, and time, respectively, and the BE or British Engineering system (the gravitational version) uses the foot (ft), the slug (sl), and the second. Table 1.1 summarizes the units used for length, mass, and time in the three systems.

Originally, the meter was defi ned in terms of the distance measured along the earth’s surface between the north pole and the equator. Eventually, a more accurate measurement standard was needed, and by international agreement the meter became the distance between two marks on a bar of platinum–iridium alloy (see Figure 1.1) kept at a temperature of 0 °C. Today, to meet further demands for increased accuracy, the meter is defi ned as the distance that light travels in a vacuum in a time of 1/299 792 458 second. This defi nition arises because the speed of light is a universal constant that is defi ned to be 299 792 458 m/s.

The defi nition of a kilogram as a unit of mass has also undergone changes over the years. As Chapter 4 discusses, the mass of an object indicates the tendency of the object to continue in motion with a constant velocity. Originally, the kilogram was expressed in terms of a specifi c amount of water. Today, one kilogram is defi ned to be the mass of a standard cylinder of platinum– iridium alloy, like the one in Figure 1.2.

As with the units for length and mass, the present defi nition of the second as a unit of time is diff erent from the original defi nition. Originally, the second was defi ned according to the average time for the earth to rotate once about its axis, one day being set equal to 86 400 seconds. The earth’s rotational motion was chosen because it is naturally repetitive, occurring over and over again. Today, we still use a naturally occurring repetitive phenomenon to defi ne the second, but of a very diff erent kind. We use the electromagnetic waves emitted by cesium-133 atoms in an atomic clock like that in Figure 1.3. One second is defi ned as the time needed for 9 192 631 770 wave cycles to occur.*

The units for length, mass, and time, along with a few other units that will arise later, are regarded as base SI units. The word “base” refers to the fact that these units are used along with

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FIGURE 1.1 The standard platinum–iridium meter bar.

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FIGURE 1.2 The standard platinum–iridium kilogram is kept at the International Bureau of Weights and Measures in Sèvres, France. This copy of it was assigned to the United States in 1889 and is housed at the National Institute of Standards and Technology.

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FIGURE 1.3 This atomic clock, the NIST-F1, keeps time with an uncertainty of about one second in sixty million years. *See Chapter 16 for a discussion of waves in general and Chapter 24 for a discussion of electromagnetic waves in particular.

TABLE 1.1 Units of Measurement

System SI CGS BE

Length Meter (m) Centimeter (cm) Foot (ft)

Mass Kilogram (kg) Gram (g) Slug (sl)

Time Second (s) Second (s) Second (s)

1.3 The Role of Units in Problem Solving 3

various laws to defi ne additional units for other important physical quantities, such as force and energy. The units for such other physical quantities are referred to as derived units, since they are combinations of the base units. Derived units will be introduced from time to time, as they arise naturally along with the related physical laws.

The value of a quantity in terms of base or derived units is sometimes a very large or very small number. In such cases, it is convenient to introduce larger or smaller units that are related to the normal units by multiples of ten. Table 1.2 summarizes the prefi xes that are used to denote multiples of ten. For example, 1000 or 103 meters are referred to as 1 kilometer (km), and 0.001 or 10 −3 meter is called 1 millimeter (mm). Similarly, 1000 grams and 0.001 gram are referred to as 1 kilogram (kg) and 1 milligram (mg), respectively. Appendix A contains a discussion of scientifi c notation and powers of ten, such as 103 and 10 −3.

1.3 The Role of Units in Problem Solving The Conversion of Units Since any quantity, such as length, can be measured in several diff erent units, it is important to know how to convert from one unit to another. For instance, the foot can be used to express the distance between the two marks on the standard platinum–iridium meter bar. There are 3.281 feet in one meter, and this number can be used to convert from meters to feet, as the following example demonstrates.

TABLE 1.2 Standard Prefixes Used to Denote Multiples of Ten

Prefix Symbol Factora

tera T 1012

giga G 109

mega M 106

kilo k 103

hecto h 102

deka da 101

deci d 10−1

centi c 10−2

milli m 10−3

micro μ 10−6

nano n 10−9

pico p 10−12

femto f 10−15

aAppendix A contains a discussion of powers of ten and

scientific notation.

EXAMPLE 1 The World’s Highest Waterfall

The highest waterfall in the world is Angel Falls in Venezuela, with a total drop of 979.0 m (see Figure 1.4). Express this drop in feet.

Reasoning When converting between units, we write down the units explicitly in the calculations and treat them like any algebraic quantity. In particular, we will take advantage of the following algebraic fact: Multiplying or dividing an equation by a factor of 1 does not alter an equation.

Solution Since 3.281 feet = 1 meter, it follows that (3.281 feet)/ (1 meter) = 1. Using this factor of 1 to multiply the equation “Length = 979.0 meters,” we fi nd that

Length = (979.0 m)(1) = (979.0 meters) (3.281 feet1 meter ) = 3212 feet The colored lines emphasize that the units of meters behave like any al- gebraic quantity and cancel when the multiplication is performed, leaving only the desired unit of feet to describe the answer. In this regard, note that 3.281 feet = 1 meter also implies that (1 meter)/(3.281 feet) = 1. However, we chose not to multiply by a factor of 1 in this form, because the units of meters would not have canceled.

A calculator gives the answer as 3212.099 feet. Standard proced- ures for signifi cant fi gures, however, indicate that the answer should be rounded off to four signifi cant fi gures, since the value of 979.0 meters is accurate to only four signifi cant fi gures. In this regard, the “1 meter” in the denominator does not limit the signifi cant fi gures of the answer, be- cause this number is precisely one meter by defi nition of the conversion factor. Appendix B contains a review of signifi cant fi gures. ©

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FIGURE 1.4 Angel Falls in Venezuela is the highest waterfall in the world.

Problem-Solving Insight In any conversion, if the units do not combine algebraically to give the desired result, the conversion has not been carried out properly.

4 CHAPTER 1 Introduction and Mathematical Concepts

EXAMPLE 2 Interstate Speed Limit

Express the speed limit of 65 miles/hour in terms of meters/second.

Reasoning As in Example 1, it is important to write down the units explicitly in the calculations and treat them like any algebraic quantity. Here, we take advantage of two well-known relationships—namely, 5280 feet = 1 mile and 3600 seconds = 1 hour. As a result, (5280 feet)/ (1 mile) = 1 and (3600 seconds)/(1 hour) = 1. In our solution we will use the fact that multiplying and dividing by these factors of unity does not alter an equation.

Solution Multiplying and dividing by factors of unity, we fi nd the speed limit in feet per second as shown below:

Speed = (65 mileshour ) (1)(1) =

(65 mileshour ) ( 5280 feet

1 mile ) ( 1 hour

3600 seconds) = 95 feet

second

To convert feet into meters, we use the fact that (1 meter)/(3.281 feet) = 1:

Speed = (95 feetsecond) (1) =

(95 feetsecond) ( 1 meter

3.281 feet) = 29 meters second

In addition to their role in guiding the use of conversion factors, units serve a useful purpose in solving problems. They can provide an internal check to eliminate errors, if they are carried along during each step of a calculation and treated like any algebraic factor.

Problem-Solving Insight In particular, remember that only quantities with the same units can be added or subtracted.

Thus, at one point in a calculation, if you fi nd yourself adding 12 miles to 32 kilometers, stop and reconsider. Either miles must be converted into kilometers or kilometers must be converted into miles before the addition can be carried out.

A collection of useful conversion factors is given on the page facing the inside of the front cover. The reasoning strategy that we have followed in Examples 1 and 2 for converting between units is outlined as follows:

REASONING STRATEGY Converting Between Units 1. In all calculations, write down the units explicitly. 2. Treat all units as algebraic quantities. In particular, when identical units are divided,

they are eliminated algebraically. 3. Use the conversion factors located on the page facing the inside of the front cover. Be

guided by the fact that multiplying or dividing an equation by a factor of 1 does not alter the equation. For instance, the conversion factor of 3.281 feet = 1 meter might be applied in the form (3.281 feet)/(1 meter) = 1. This factor of 1 would be used to multiply an equa- tion such as “Length = 5.00 meters” in order to convert meters to feet.

4. Check to see that your calculations are correct by verifying that the units combine algeb- raically to give the desired unit for the answer. Only quantities with the same units can be added or subtracted.

Sometimes an equation is expressed in a way that requires specifi c units to be used for the variables in the equation. In such cases it is important to understand why only certain units can be used in the equation, as the following example illustrates.

EXAMPLE 3 BIO The Physics of the Body Mass Index

The body mass index (BMI) takes into account your mass in kilograms (kg) and your height in meters (m) and is defi ned as follows:

BMI = Mass in kg

(Height in m)2

However, the BMI is often computed using the weight* of a person in pounds (lb) and his or her height in inches (in.). Thus, the expression for the BMI incorporates these quantities, rather than the mass in kilograms and the height in meters. Starting with the defi nition above, determine the expression for the BMI that uses pounds and inches.

*Weight and mass are different concepts, and the relationship between them will be discussed in Section 4.7.

With this in mind, the next example stresses the importance of writing down the units and illustrates a typical situation in which several conversions are required.

1.3 The Role of Units in Problem Solving 5

Dimensional Analysis We have seen that many quantities are denoted by specifying both a number and a unit. For

example, the distance to the nearest telephone may be 8 meters, or the speed of a car might be

25 meters/second. Each quantity, according to its physical nature, requires a certain type of unit. Distance must be measured in a length unit such as meters, feet, or miles, and a time unit will not

do. Likewise, the speed of an object must be specifi ed as a length unit divided by a time unit. In

physics, the term dimension is used to refer to the physical nature of a quantity and the type of unit used to specify it. Distance has the dimension of length, which is symbolized as [L], while

speed has the dimensions of length [L] divided by time [T], or [L/T]. Many physical quantities

can be expressed in terms of a combination of fundamental dimensions such as length [L], time

[T], and mass [M]. Later on, we will encounter certain other quantities, such as temperature,

which are also fundamental. A fundamental quantity like temperature cannot be expressed as a

combination of the dimensions of length, time, mass, or any other fundamental dimension.

Dimensional analysis is used to check mathematical relations for the consistency of their

dimensions. As an illustration, consider a car that starts from rest and accelerates to a speed υ in a time t. Suppose we wish to calculate the distance x traveled by the car but are not sure whether the correct relation is x = 12 𝜐t

2 or x = 12 𝜐t. We can decide by checking the quantities on both sides of the equals sign to see whether they have the same dimensions. If the dimensions are not the

same, the relation is incorrect. For x = 12 𝜐t 2, we use the dimensions for distance [L], time [T], and

speed [L/T] in the following way:

x = 12 𝜐t 2

Dimensions [L] ≟ [ LT ] [T]2 = [L][T] Dimensions cancel just like algebraic quantities, and pure numerical factors like

1

2 have no dimen-

sions, so they can be ignored. The dimension on the left of the equals sign does not match those

on the right, so the relation x = 12 𝜐t 2 cannot be correct. On the other hand, applying dimensional

analysis to x = 12 𝜐t, we fi nd that

x = 12 𝜐t 2

Dimensions [L] ≟ [ LT ] [T] = [L]

Reasoning We will begin with the BMI defi nition and work separately with the numerator and the denominator. We will determine the mass in

kilograms that appears in the numerator from the weight in pounds by using

the fact that 1 kg corresponds to 2.205 lb. Then, we will determine the height

in meters that appears in the denominator from the height in inches with the

aid of the facts that 1 m = 3.281 ft and 1 ft = 12 in. These conversion factors

are located on the page facing the inside of the front cover of the text.

Solution Since 1 kg corresponds to 2.205 lb, the mass in kilograms can be determined from the weight in pounds in the following way:

Mass in kg = (Weight in lb)( 1 kg

2.205 lb) Since 1 ft = 12 in. and 1 m = 3.281 ft, we have

Height in m = (Height in in.)( 1 ft12 in.)( 1 m

3.281 ft) Substituting these results into the numerator and denominator of the BMI

defi nition gives

BMI = Mass in kg

(Height in m)2 =

(Weight in lb)( 1 kg

2.205 lb) (Height in in.)2 ( 1 ft12 in.)

2

( 1 m3.281 ft) 2

= ( 1 kg

2.205 lb)( 12 in.

1 ft ) 2

(3.281 ft1 m ) 2 (Weight in lb)

(Height in in.)2

BMI = (703.0 kg · in.2

lb · m2 ) (Weight in lb)

(Height in in.)2

For example, if your weight and height are 180 lb and 71 in., your body

mass index is 25 kg/m2. The BMI can be used to assess approximately

whether your weight is normal for your height (see Table 1.3).

TABLE 1.3 The Body Mass Index

BMI (kg/m2) Evaluation Below 18.5 Underweight

18.5–24.9 Normal

25.0–29.9 Overweight

30.0–39.9 Obese

40 and above Morbidly obese

6 CHAPTER 1 Introduction and Mathematical Concepts

Problem-Solving Insight You can check for errors that may have arisen during algebraic manipulations by performing a dimensional analysis on the fi nal expression.

The dimension on the left of the equals sign matches that on the right, so this relation is dimen- sionally correct. If we know that one of our two choices is the right one, then x = 12 𝜐t is it. In the absence of such knowledge, however, dimensional analysis cannot identify the correct relation. It can only identify which choices may be correct, since it does not account for numerical factors like 12 or for the manner in which an equation was derived from physics principles.

Check Your Understanding

(The answers are given at the end of the book.) 1. (a) Is it possible for two quantities to have the same dimensions but diff erent units? (b) Is it possible for two quantities to have the same units but diff erent dimensions? 2. You can always add two numbers that have the same units (such as 6 meters + 3 meters). Can you always add

two numbers that have the same dimensions, such as two numbers that have the dimensions of length [L]?

3. The following table lists four variables, along with their units:

Variable Units x Meters (m) υ Meters per second (m/s) t Seconds (s) a Meters per second squared (m/s2)

These variables appear in the following equations, along with a few numbers that have no units. In which of the equations are the units on the left side of the equals sign consistent with the units on the right side?

(a) x = 𝜐t (d) 𝜐 = at + 12 at 3

(b) x = 𝜐t + 12 at 2 (e) 𝜐 3 = 2ax 2

(c) 𝜐 = at (f) t = √2xa 4. In the equation y = cnat2 you wish to determine the integer value (1, 2, etc.) of the exponent n. The di-

mensions of y, a, and t are known. It is also known that c has no dimensions. Can dimensional analysis be used to determine n?

1.4 Trigonometry Scientists use mathematics to help them describe how the physical universe works, and tri- gonometry is an important branch of mathematics. Three trigonometric functions are utilized throughout this text. They are the sine, the cosine, and the tangent of the angle θ (Greek theta), abbreviated as sin θ, cos θ, and tan θ, respectively. These functions are defi ned below in terms of the symbols given along with the right triangle in Interactive Figure 1.5.

DEFINITION OF SIN θ, COS θ, AND TAN θ

sin θ = ho h

(1.1)

cos θ = ha h

(1.2)

tan θ = ho ha

(1.3)

h = length of the hypotenuse of a right triangle ho = length of the side opposite the angle θ ha = length of the side adjacent to the angle θ

INTERACTIVE FIGURE 1.5 A right triangle.

90°

h = hypotenuse

θ

ho = length of side opposite the angle

ha = length of side adjacent to the angle θ

θ

1.4 Trigonometry 7

The sine, cosine, and tangent of an angle are numbers without units, because each is the ratio of the lengths of two sides of a right triangle. Example 4 illustrates a typical application of Equation 1.3.

EXAMPLE 4 Using Trigonometric Functions

On a sunny day, a tall building casts a shadow that is 67.2 m long. The angle between the sun’s rays and the ground is θ = 50.0°, as Figure 1.6 shows. Determine the height of the building.

Reasoning We want to fi nd the height of the building. Therefore, we begin with the colored right triangle in Figure 1.6 and identify the height as the length ho of the side opposite the angle θ. The length of the shadow is the length ha of the side that is adjacent to the angle θ. The ratio of the length of the opposite side to the length of the adjacent side is the tangent of the angle θ, which can be used to fi nd the height of the building.

Solution We use the tangent function in the following way, with θ = 50.0° and ha = 67.2 m:

tan θ = ho ha

(1.3)

ho = ha tan θ = (67.2 m)(tan 50.0°) = (67.2 m)(1.19) = 80.0 m

The value of tan 50.0° is found by using a calculator. FIGURE 1.6 From a value for the angle θ and the length ha of the

shadow, the height ho of the building can be found using trigonometry.

= 50.0°θ

= 67.2 mha

ho

The sine, cosine, or tangent may be used in calculations such as that in Example 4, depend- ing on which side of the triangle has a known value and which side is asked for.

Problem-Solving Insight However, the choice of which side of the triangle to label ho (opposite) and which to label ha (adjacent) can be made only after the angle θ is identifi ed.

Often the values for two sides of the right triangle in Interactive Figure 1.5 are available, and the value of the angle θ is unknown. The concept of inverse trigonometric functions plays an important role in such situations. Equations 1.4–1.6 give the inverse sine, inverse cosine, and inverse tangent in terms of the symbols used in the drawing. For instance, Equation 1.4 is read as “θ equals the angle whose sine is ho/h.”

θ = sin−1 ( ho h ) (1.4)

θ = cos−1 ( ha h ) (1.5)

θ = tan−1 ( ho ha) (1.6)

The use of −1 as an exponent in Equations 1.4–1.6 does not mean “take the reciprocal.” For instance, tan−1 (ho/ha) does not equal 1/tan (ho/ha). Another way to express the inverse trigono- metric functions is to use arc sin, arc cos, and arc tan instead of sin−1, cos−1, and tan−1. Example 5 illustrates the use of an inverse trigonometric function.

EXAMPLE 5 Using Inverse Trigonometric Functions

A lakefront drops off gradually at an angle θ, as Figure 1.7 indicates. For safety reasons, it is necessary to know how deep the lake is at vari- ous distances from the shore. To provide some information about the depth, a lifeguard rows straight out from the shore a distance of 14.0 m

and drops a weighted fi shing line. By measuring the length of the line, the lifeguard determines the depth to be 2.25 m. (a) What is the value of θ? (b) What would be the depth d of the lake at a distance of 22.0 m from the shore?

8 CHAPTER 1 Introduction and Mathematical Concepts

The right triangle in Interactive Figure 1.5 provides the basis for defi ning the various tri- gonometric functions according to Equations 1.1–1.3. These functions always involve an angle and two sides of the triangle. There is also a relationship among the lengths of the three sides of a right triangle. This relationship is known as the Pythagorean theorem and is used often in this text.

PYTHAGOREAN THEOREM The square of the length of the hypotenuse of a right triangle is equal to the sum of the squares of the lengths of the other two sides:

h2 = ho2 + ha2 (1.7)

1.5 Scalars and Vectors The volume of water in a swimming pool might be 50 cubic meters, or the winning time of a race could be 11.3 seconds. In cases like these, only the size of the numbers matters. In other words, how much volume or time is there? The 50 specifi es the amount of water in units of cubic meters, while the 11.3 specifi es the amount of time in seconds. Volume and time are examples of scalar quantities. A scalar quantity is one that can be described with a single number (including any units) giving its size or magnitude. Some other common scalars are temperature (e.g., 20 °C) and mass (e.g., 85 kg).

While many quantities in physics are scalars, there are also many that are not, and for these quantities the magnitude tells only part of the story. Consider Figure 1.8, which depicts a car that has moved 2 km along a straight line from start to fi nish. When describing the motion, it is incomplete to say that “the car moved a distance of 2 km.” This statement would indicate only that the car ends up somewhere on a circle whose center is at the starting point and whose radius

Reasoning Near the shore, the lengths of the opposite and adjacent sides of the right triangle in Figure 1.7 are ho = 2.25 m and ha = 14.0 m, relative to the angle θ. Having made this identifi cation, we can use the inverse tangent to fi nd the angle in part (a). For part (b) the opposite and adjacent sides farther from the shore become ho = d and ha = 22.0 m. With the value for θ obtained in part (a), the tangent function can be used to fi nd the unknown depth. Considering the way in which the lake bottom drops off in Figure 1.7, we expect the unknown depth to be greater than 2.25 m.

Solution (a) Using the inverse tangent given in Equation 1.6, we fi nd that

θ = tan−1 ( ho ha) = tan

−1 (2.25 m14.0 m) = 9.13° (b) With θ = 9.13°, the tangent function given in Equation 1.3 can be used to fi nd the unknown depth farther from the shore, where ho = d and ha = 22.0 m. Since tan θ = ho /ha, it follows that

ho = ha tan θ d = (22.0 m)(tan 9.13°) = 3.54 m

which is greater than 2.25 m, as expected.

FIGURE 1.7 If the distance from the shore and the depth of the water at any one point are known, the angle θ can be found with the aid of trigonometry. Knowing the value of θ is useful, because then the depth d at another point can be determined.

14.0 m

2.25 m d

22.0 m

θ

1.5 Scalars and Vectors 9

is 2 km. A complete description must include the direction along with the distance, as in the state- ment “the car moved a distance of 2 km in a direction 30° north of east.” A quantity that deals inherently with both magnitude and direction is called a vector quantity. Because direction is an important characteristic of vectors, arrows are used to represent them; the direction of the arrow gives the direction of the vector. The colored arrow in Figure 1.8, for example, is called the displacement vector, because it shows how the car is displaced from its starting point. Chapter 2 discusses this particular vector.

The length of the arrow in Figure 1.8 represents the magnitude of the displacement vector. If the car had moved 4 km instead of 2 km from the starting point, the arrow would have been drawn twice as long. By convention, the length of a vector arrow is proportional to the magnitude of the vector.

In physics there are many important kinds of vectors, and the practice of using the length of an arrow to represent the magnitude of a vector applies to each of them. All forces, for instance, are vectors. In common usage a force is a push or a pull, and the direction in which a force acts is just as important as the strength or magnitude of the force. The magnitude of a force is measured in SI units called newtons (N). An arrow representing a force of 20 newtons is drawn twice as long as one representing a force of 10 newtons.

The fundamental distinction between scalars and vectors is the characteristic of direction. Vectors have it, and scalars do not. Conceptual Example 6 helps to clarify this distinction and explains what is meant by the “direction” of a vector.

FIGURE 1.8 A vector quantity has a magnitude and a direction. The colored arrow in this drawing represents a displacement vector.

N

S

W E

30.0°

Start

Finish2 km

CONCEPTUAL EXAMPLE 6 Vectors, Scalars, and the Role of Plus and Minus Signs

There are places where the temperature is +20 °C at one time of the year and −20 °C at another time. Do the plus and minus signs that signify positive and negative temperatures imply that temperature is a vector quantity? (a) Yes (b) No

Reasoning A hallmark of a vector is that there is both a magnitude and a physical direction associated with it, such as 20 meters due east or 20 meters due west.

Answer (a) is incorrect. The plus and minus signs associated with +20 °C and −20 °C do not convey a physical direction, such as due east or due west. Therefore, temperature cannot be a vector quantity.

Answer (b) is correct. On a thermometer, the algebraic signs simply mean that the temperature is a number less than or greater than zero on the temperature scale being used and have nothing to do with east, west, or any other physical direction. Temperature, then, is not a vector. It is a scalar, and scalars can sometimes be negative.

*Vectors are also sometimes written in other texts as boldface symbols without arrows above them.

Often, for the sake of convenience, quantities such as volume, time, displacement, velo- city, and force are represented in physics by symbols. In this text, we write vectors in boldface symbols (this is boldface) with arrows above them* and write scalars in italic symbols (this is italic). Thus, a displacement vector is written as “A→ = 750 m, due east,” where the A→ is a boldface symbol. By itself, however, separated from the direction, the magnitude of this vector is a scalar quantity. Therefore, the magnitude is written as “A = 750 m,” where the A is an italic symbol without an arrow.

Check Your Understanding

(The answer is given at the end of the book.) 5. Which of the following statements, if any, involves a vector? (a) I walked 2 miles along the beach. (b) I

walked 2 miles due north along the beach. (c) I jumped off a cliff and hit the water traveling at 17 miles per hour. (d) I jumped off a cliff and hit the water traveling straight down at a speed of 17 miles per hour. (e) My bank account shows a negative balance of −25 dollars.

10 CHAPTER 1 Introduction and Mathematical Concepts

1.6 Vector Addition and Subtraction Addition Often it is necessary to add one vector to another, and the process of addition must take into ac- count both the magnitude and the direction of the vectors. The simplest situation occurs when the vectors point along the same direction—that is, when they are colinear, as in Figure 1.9. Here, a car fi rst moves along a straight line, with a displacement vector A→ of 275 m, due east. Then the car moves again in the same direction, with a displacement vector B→ of 125 m, due east. These two vectors add to give the total displacement vector R→ , which would apply if the car had moved from start to fi nish in one step. The symbol R→ is used because the total vector is often called the resultant vector. With the tail of the second arrow located at the head of the fi rst arrow, the two lengths simply add to give the length of the total displacement. This kind of vector addition is identical to the familiar addition of two scalar numbers (2 + 3 = 5) and can be carried out here only because the vectors point along the same direction. In such cases we add the individual magnitudes to get the magnitude of the total, knowing in advance what the direction must be. Formally, the addition is written as follows:

R→ = A→ + B→

R→ = 275 m, due east + 125 m, due east = 400 m, due east Perpendicular vectors are frequently encountered, and Figure 1.10 indicates how they can

be added. This fi gure applies to a car that fi rst travels with a displacement vector A→ of 275 m, due east, and then with a displacement vector B→ of 125 m, due north. The two vectors add to give a resultant displacement vector R→ . Once again, the vectors to be added are arranged in a tail-to- head fashion, and the resultant vector points from the tail of the fi rst to the head of the last vector added. The resultant displacement is given by the vector equation

R→ = A→ + B→

The addition in this equation cannot be carried out by writing R = 275 m + 125 m, because the vectors have diff erent directions. Instead, we take advantage of the fact that the triangle in Figure 1.10 is a right triangle and use the Pythagorean theorem (Equation 1.7). According to this theorem, the magnitude of R→ is

R = √(275 m)2 + (125 m)2 = 302 m

The angle θ in Figure 1.10 gives the direction of the resultant vector. Since the lengths of all three sides of the right triangle are now known, sin θ, cos θ, or tan θ can be used to determine θ. Noting that tan θ = B/A and using the inverse trigonometric function, we fi nd that:

θ = tan−1 (BA) = tan−1 ( 125 m 275 m) = 24.4°

Thus, the resultant displacement of the car has a magnitude of 302 m and points north of east at an angle of 24.4°. This displacement would bring the car from the start to the fi nish in Figure 1.10 in a single straight-line step.

When two vectors to be added are not perpendicular, the tail-to-head arrangement does not lead to a right triangle, and the Pythagorean theorem cannot be used. Figure 1.11a illustrates such a case for a car that moves with a displacement A→ of 275 m, due east, and then with a dis- placement B→ of 125 m, in a direction 55.0° north of west. As usual, the resultant displacement vector R→ is directed from the tail of the fi rst to the head of the last vector added. The vector addition is still given according to

R→ = A→ + B→

However, the magnitude of R→ is not R = A + B, because the vectors A→ and B→ do not have the same direction, and neither is it R = √A2 + B2, because the vectors are not perpendicular, so the Pythagorean theorem does not apply. Some other means must be used to fi nd the magnitude and direction of the resultant vector.

N

S

W E

Tail-to-head

Start Finish A

R

B

FIGURE 1.9 Two colinear displacement

vectors A→ and B→ add to give the resultant displacement vector R→ .

FIGURE 1.10 The addition of two perpendicular displacement vectors A→ and B→ gives the resultant vector R→ .

N

S

W E

Tail-to-head

Start

R B

Finish

90°θ

A

FIGURE 1.11 (a) The two displacement vectors A→ and B→ are neither colinear nor perpendicular, but even so they add to give the resultant vector R→. (b) In one method for adding them together, a graphical technique is used.

Finish

Start

Tail-to-head

55.0°

55.0°

( )a

( )b

22. 8 c

m

4 8

12

20

0

24 cm

16

28

N

S

W E

θ

θ

A

R B

A

R B

1.6 Vector Addition and Subtraction 11

One approach uses a graphical technique. In this method, a diagram is constructed in which the arrows are drawn tail to head. The lengths of the vector arrows are drawn to scale, and the angles are drawn accurately (with a protractor, perhaps). Then the length of the arrow representing the resultant vector is measured with a ruler. This length is converted to the mag- nitude of the resultant vector by using the scale factor with which the drawing is constructed. In Figure 1.11b, for example, a scale of one centimeter of arrow length for each 10.0 m of displacement is used, and it can be seen that the length of the arrow representing R→ is 22.8 cm. Since each centimeter corresponds to 10.0 m of displacement, the magnitude of R→ is 228 m. The angle θ, which gives the direction of R→ , can be measured with a protractor to be θ = 26.7° north of east.

Subtraction The subtraction of one vector from another is carried out in a way that depends on the follow- ing fact. When a vector is multiplied by −1, the magnitude of the vector remains the same, but the direction of the vector is reversed. Conceptual Example 7 illustrates the meaning of this statement.

Reasoning A displacement vector of −D→ is (−1) D→ . The presence of the (−1) factor reverses the direction of the vector, but does not change its magnitude. Similarly, a force vector of −F→ has the same magnitude as the vector F→ but has the opposite direction.

Answer (a) and (b) are incorrect. While scalars can sometimes be negative, magnitudes of vectors are never negative.

Answer (c) is correct. The vectors −D→ and −F→ have the same mag- nitudes as D→ and F→, but point in the opposite direction, as indicated in Figures 1.12b and 1.13b.

Related Homework: Problems 67

CONCEPTUAL EXAMPLE 7 Multiplying a Vector by −1

Consider two vectors described as follows:

1. A woman climbs 1.2 m up a ladder, so that her displacement vec- tor D→ is 1.2 m, upward along the ladder, as in Figure 1.12a.

2. A man is pushing with 450 N of force on his stalled car, trying to move it eastward. The force vector F→ that he applies to the car is 450 N, due east, as in Figure 1.13a.

What are the physical meanings of the vectors −D→ and −F→? (a) −D→ points upward along the ladder and has a magnitude of −1.2 m; −F→ points due east and has a magnitude of −450 N. (b) −D→ points downward along the ladder and has a magnitude of −1.2 m; −F→ points due west and has a magnitude of −450 N. (c) −D→ points downward along the ladder and has a magnitude of 1.2 m; −F→ points due west and has a magnitude of 450 N.

FIGURE 1.12 (a) The displacement vector for a woman climbing 1.2 m up a ladder is D→ . (b) The displacement vector for a woman climbing 1.2 m down a ladder is −D→ .

(a) (b)

–DD

FIGURE 1.13 (a) The force vector for a man pushing on a car with 450 N of force in a direction due east is F→ . (b) The force vector for a man pushing on a car with 450 N of force in a direction due west is −F→ .

(a)

(b)

F

–F

12 CHAPTER 1 Introduction and Mathematical Concepts

In practice, vector subtraction is carried out exactly like vector addition, except that one of the vectors added is multiplied by a scalar factor of −1. To see why, look at the two vectors A→ and B→ in Figure 1.14a. These vectors add together to give a third vector C→, according to C→ = A→ + B→. Therefore, we can calculate vector A→ as A→ = C→ − B→, which is an example of vector subtraction. However, we can also write this result as A→ = C→ + (−B→) and treat it as vector addi- tion. Figure 1.14b shows how to calculate vector A→ by adding the vectors C→ and −B→. Notice that vectors C→ and −B→ are arranged tail to head and that any suitable method of vector addition can be employed to determine A→.

Check Your Understanding

(The answers are given at the end of the book.) 6. Two vectors A→ and B→ are added together to give a resultant vector R→: R→ = A→ + B→. The magnitudes

of A→ and B→ are 3 m and 8 m, respectively, but the vectors can have any orientation. What are (a) the maximum possible value and (b) the minimum possible value for the magnitude of R→?

7. Can two nonzero perpendicular vectors be added together so their sum is zero? 8. Can three or more vectors with unequal magnitudes be added together so their sum is zero? 9. In preparation for this question, review Conceptual Example 7. Vectors A→ and B→ satisfy the vector

equation A→ + B→ = 0. (a) How does the magnitude of B→ compare with the magnitude of A→? (b) How does the direction of B→ compare with the direction of A→?

10. Vectors A→, B→, and C→ satisfy the vector equation A→ + B→ = C→, and their magnitudes are related by the scalar equation A2 + B2 = C2. How is vector A→ oriented with respect to vector B→?

11. Vectors A→, B→, and C→ satisfy the vector equation A→ + B→ = C→, and their magnitudes are related by the scalar equation A + B = C. How is vector A→ oriented with respect to vector B→?

1.7 The Components of a Vector Vector Components Suppose a car moves along a straight line from start to fi nish, as in Figure 1.15, the corres- ponding displacement vector being r→. The magnitude and direction of the vector r→ give the distance and direction traveled along the straight line. However, the car could also arrive at the fi nish point by fi rst moving due east, turning through 90°, and then moving due north. This alternative path is shown in the drawing and is associated with the two displacement vectors x→ and y→. The vectors x→ and y→ are called the x vector component and the y vector component of r→.

Vector components are very important in physics and have two basic features that are appar- ent in Figure 1.15. One is that the components add together to equal the original vector:

r→ = x→ + y→

The components x→ and y→, when added vectorially, convey exactly the same meaning as does the original vector r→: they indicate how the fi nish point is displaced relative to the starting point. The other feature of vector components that is apparent in Figure 1.15 is that x→ and y→ are not just any two vectors that add together to give the original vector r→: they are perpendicular vectors. This perpendicularity is a valuable characteristic, as we will soon see.

Any type of vector may be expressed in terms of its components, in a way similar to that illustrated for the displacement vector in Figure 1.15. Interactive Figure 1.16 shows an arbitrary vector A→ and its vector components A→ x and A

→ y. The components are drawn parallel

to convenient x and y axes and are perpendicular. They add vectorially to equal the original vector A→ :

A→ = A→ x + A →

y

FIGURE 1.15 The displacement vector r→ and its vector components x→ and y→.

FIGURE 1.14 (a) Vector addition according to C→ = A→ + B→ . (b) Vector subtraction accord- ing to A→ = C→ − B→ = C→ + (−B→ ).

A

B

C = A + B

( )a

B–

C

Tail-to-head

( )b

A = C – B

Start

Finish

90°

N

S

W E

x

yr

INTERACTIVE FIGURE 1.16 An arbitrary

vector A→ and its vector components A→ x and A →

y.

A

y+

y

x+ Ax

A

θ

1.7 The Components of a Vector 13

There are times when a drawing such as Interactive Figure 1.16 is not the most convenient way to represent vector components, and Figure 1.17 presents an alternative method. The disadvant- age of this alternative is that the tail-to-head arrangement of A→ x and A

→ y is missing, an arrange-

ment that is a nice reminder that A→ x and A →

y add together to equal A →

. The defi nition that follows summarizes the meaning of vector components:

DEFINITION OF VECTOR COMPONENTS In two dimensions, the vector components of a vector A→ are two perpendicular vectors A→ x and A

→ y that are parallel to the x and y axes, respectively, and add together vectorially

according to A→ = A→ x + A →

y.

Problem-Solving Insight In general, the components of any vector can be used in place of the vector itself in any calculation where it is convenient to do so.

The values calculated for vector components depend on the orientation of the vector relative to the axes used as a reference. Figure 1.18 illustrates this fact for a vector A→ by showing two sets of axes, one set being rotated clockwise relative to the other. With respect to the black axes, vector A→ has perpendicular vector components A→ x and A

→ y; with respect to the colored rotated axes,

vector A→ has diff erent vector components A→ x′ and A →

y′. The choice of which set of components to use is purely a matter of convenience.

Scalar Components It is often easier to work with the scalar components, Ax and Ay (note the italic symbols), rather than the vector components A→ x and A

→ y. Scalar components are positive or negative numbers (with

units) that are defi ned as follows: The scalar component Ax has a magnitude equal to that of A →

x and is given a positive sign if A→ x points along the +x axis and a negative sign if A

→ x points along

the −x axis. The scalar component Ay is defi ned in a similar manner. The following table shows an example of vector and scalar components:

In this text, when we use the term “component,” we will be referring to a scalar component, unless otherwise indicated.

Another method of expressing vector components is to use unit vectors. A unit vector is a vector that has a magnitude of 1, but no dimensions. We will use a caret (^) to distinguish it from other vectors. Thus,

x̂ is a dimensionless unit vector of length l that points in the positive x direction, and ŷ is a dimensionless unit vector of length l that points in the positive y direction.

These unit vectors are illustrated in Figure 1.19. With the aid of unit vectors, the vector com- ponents of an arbitrary vector A→ can be written as A→ x = Ax x̂ and A

→ y = Ay ŷ, where Ax and Ay are

its scalar components (see the drawing and the third column of the table above). The vector A→ is then written as A→ = Ax x̂ + Ay ŷ.

Resolving a Vector into Its Components If the magnitude and direction of a vector are known, it is possible to fi nd the components of the vector. The process of fi nding the components is called “resolving the vector into its components.” As Example 8 illustrates, this process can be carried out with the aid of trigo- nometry, because the two perpendicular vector components and the original vector form a right triangle.

Ay

+y

+x Ax

A

θ

FIGURE 1.17 This alternative way of

drawing the vector A→ and its vector components is completely equivalent to that shown in Interactive Figure 1.16.

+y

+x

+y´

+x´

Ax

Áy

Á x

A

Ay

FIGURE 1.18 The vector components of the vector depend on the orientation of the axes used as a reference.

x

y Axx

Ayy

+x

+y

FIGURE 1.19 The dimensionless unit vectors x̂ and ŷ have magnitudes equal to 1, and they point in the +x and +y directions, respectively. Expressed in terms of unit vectors, the vector components of the vector A→ are Ax x̂ and Ay ŷ.

Vector Components Scalar Components Unit Vectors

A→ x = 8 meters, directed along the +x axis Ax = +8 meters A →

x = (+8 meters) x̂

A→ y = 10 meters, directed along the –y axis Ay = −10 meters A →

y = (–10 meters) ŷ

14 CHAPTER 1 Introduction and Mathematical Concepts

EXAMPLE 8 Finding the Components of a Vector

A displacement vector r→ has a magnitude of r = 175 m and points at an angle of 50.0° relative to the x axis in Figure 1.20. Find the x and y com- ponents of this vector.

Reasoning We will base our solution on the fact that the triangle formed in Figure 1.20 by the vector r→ and its components x→ and y→ is a right triangle. This fact enables us to use the trigonometric sine and cosine functions, as defi ned in Equations 1.1 and 1.2.

Problem-Solving Insight You can check to see whether the components of a vector are correct by substituting them into the Pythagorean theorem in order to calculate the magnitude of the original vector.

Solution The y component can be obtained using the 50.0° angle and Equation 1.1, sin θ = y/r:

y = r sin θ = (175 m)(sin 50.0°) = 134 m

In a similar fashion, the x component can be obtained using the 50.0° angle and Equation 1.2, cos θ = x/r:

x = r cos θ = (175 m)(cos 50.0°) = 112 m

Math Skills Either acute angle of a right triangle can be used to determine the components of a vector. The choice of angle is a matter of convenience. For instance, instead of the 50.0° angle, it is also possible to use the angle α in Figure 1.20. Since α + 50.0° = 90.0°, it follows that α = 40.0°. The solution using α yields the same answers as the solution using the 50.0° angle:

cos α = y r

y = r cos α = (175 m)(cos 40.0°) = 134 m

sin α = x r

x = r sin α = (175 m)(sin 40.0°) = 112 m90.0°50.0°

x

yr

α

FIGURE 1.20 The x and y components of the displacement vector r→ can be found using trigonometry.

Since the vector components and the original vector form a right triangle, the Pythagorean theorem can be applied to check the validity of calculations such as those in Example 8. Thus, with the components obtained in Example 8, the theorem can be used to verify that the magnitude of the original vector is indeed 175 m, as given initially:

r = √(112 m) 2 + (134 m) 2 = 175 m

It is possible for one of the components of a vector to be zero. This does not mean that the vector itself is zero, however.

Problem-Solving Insight For a vector to be zero, every vector component must individually be zero.

Thus, in two dimensions, saying that A→ = 0 is equivalent to saying that A→ x = 0 and A →

y = 0. Or, stated in terms of scalar components, if A→ = 0, then Ax = 0 and Ay = 0.

Problem-Solving Insight Two vectors are equal if, and only if, they have the same magnitude and direction.

Thus, if one displacement vector points east and another points north, they are not equal, even if each has the same magnitude of 480 m. In terms of vector components, two vectors A→ and B →

are equal if, and only if, each vector component of one is equal to the corresponding vector component of the other. In two dimensions, if A→ = B

→ , then A→ x = B

→ x and A

→ y = A

→ y. Alternatively,

using scalar components, we write that Ax = Bx and Ay = By.

Check Your Understanding

(The answers are given at the end of the book.) 12. Which of the following displacement vectors

(if any) are equal?

Variable Magnitude Direction A→ 100 m 30° north of east

B→ 100 m 30° south of west

C→ 50 m 30° south of west

D→ 100 m 60° east of north

1.8 Addition of Vectors by Means of Components 15

13. Two vectors, A→ and B→, are shown in CYU Figure 1.1. (a) What are the signs (+ or −) of the scalar components, Ax and Ay, of the vector A→? (b) What are the signs of the scalar components, Bx and By, of the vector B

→ ? (c) What are the signs of the scalar

components, Rx and Ry, of the vector R →

, where R→ = A→ + B→? 14. Are two vectors with the same magnitude necessarily equal? 15. The magnitude of a vector has doubled, its direction remaining

the same. Can you conclude that the magnitude of each com- ponent of the vector has doubled?

16. The tail of a vector is fi xed to the origin of an x, y axis sys- tem. Originally the vector points along the +x axis and has a magnitude of 12 units. As time passes, the vector rotates coun- terclockwise. What are the sizes of the x and y components of the vector for the following rotational angles? (a) 90° (b) 180° (c) 270° (d) 360°

17. A vector has a component of zero along the x axis of a certain axis system. Does this vector necessarily have a component of zero along the x axis of another (rotated) axis system?

1.8 Addition of Vectors by Means of Components The components of a vector provide the most convenient and accurate way of adding (or sub- tracting) any number of vectors. For example, suppose that vector A→ is added to vector B→. The resultant vector is C→, where C→ = A→ + B→. Interactive Figure 1.21a illustrates this vector addition, along with the x and y vector components of A→ and B→. In part b of the drawing, the vectors A→ and B→ have been removed, because we can use the vector components of these vectors in place of them. The vector component B→x has been shifted downward and arranged tail to head with vector component A→ x. Similarly, the vector component A

→ y has been shifted to the right and arranged tail

to head with the vector component B→y. The x components are colinear and add together to give the x component of the resultant vector C→. In like fashion, the y components are colinear and add together to give the y component of C→. In terms of scalar components, we can write

Cx = Ax + Bx and Cy = Ay + By

The vector components C→x and C →

y of the resultant vector form the sides of the right triangle shown in Interactive Figure 1.21c. Thus, we can fi nd the magnitude of C→ by using the Pythagorean theorem:

C = √C 2x + C 2y The angle θ that C

→ makes with the x axis is given by θ = tan−1 (Cy/Cx). Example 9 illustrates how

to add several vectors using the component method.

C

A

B

Bx

By

Ax

Ay +x

+y

(a) (c)

Cy

Cx

C

C

Bx

By

Ax

Ay

(b)

A

θ

INTERACTIVE FIGURE 1.21 (a) The vectors A→ and B→ add together to give the resultant vector C→. The x and y components of A→ and B→ are also shown. (b) The drawing illustrates that C

→ x = A

→ x + B

→ x and C

→ y = A

→ y +

B→ y. (c) Vector C →

and its components form a right triangle.

+x

+y

A B

CYU FIGURE 1.1

16 CHAPTER 1 Introduction and Mathematical Concepts

Analyzing Multiple -Concept Problems

EXAMPLE 9 The Component Method of Vector Addition

A jogger runs 145 m in a direction 20.0° east of north (displacement vector A→) and then 105 m in a direction 35.0° south of east (displacement vector B→). Using components, determine the magnitude and direction of the resultant vector C→ for these two displacements.

Reasoning Figure 1.22 shows the vectors A→ and B→, assuming that the y axis corresponds to the dir- ection due north. The vectors are arranged in a tail-to-head fashion, with the resultant vector C→ drawn from the tail of A→ to the head of B→. The components of the vectors are also shown in the fi gure. Since C→ and its components form a right triangle (red in the drawing), we will use the Pythagorean theorem and trigonometry to express the magnitude and directional angle θ for C→ in terms of its components. The components of C→ will then be obtained from the components of A→ and B→ and the data given for these two vectors.

Knowns and Unknowns The data for this problem are listed in the table that follows:

Description Symbol Value Comment

Magnitude of vector A→ 145 m

Direction of vector A→ 20.0° east of north See Figure 1.22.

Magnitude of vector B→ 105 m

Direction of vector B→ 35.0° south of east See Figure 1.22.

Unknown Variable Magnitude of resultant vector C ?

Direction of resultant vector θ ?

FIGURE 1.22 The vectors A→ and B→ add together to give the resultant vector C→. The vector components of A→ and B→ are also shown. The resultant vector C→ can be obtained once its components have been found.

N

S

W E

20.0°

+x

35.0°

+y Bx

By

Ax

Ay

CyC

Cx

B A

θ

Modeling the Problem

STEP 1 Magnitude and Direction of C→ In Figure 1.22 the vector C→ and its components C→x and C→y form a right triangle, as the red arrows show. Applying the Pythagorean theorem to this right triangle shows that the magnitude of C→ is given by Equation 1a at the right. From the red triangle it also follows that the directional angle θ for the vector C→ is given by Equation 1b at the right.

STEP 2 Components of C→ Since vector C→ is the resultant of vectors A→ and B→, we have C→ = A→ + B→ and can write the scalar components of C→ as the sum of the scalar components of A→ and B→:

Cx = Ax + Bx and Cy = Ay + By

These expressions can be substituted into Equations 1a and 1b for the magnitude and direction of C→, as shown at the right.

Solution Algebraically combining the results of each step, we fi nd that

C = √C 2x + C 2y = √(Ax + Bx)2 + (Ay + By )2

θ = tan−1 ( Cy Cx) = tan−1(

Ay + By Ax + Bx)

To use these results we need values for the individual components of A→ and B→.

C = √Cx 2 + Cy 2 (1a)

θ = tan−1 ( Cy Cx) (1b)

C = √Cx 2 + Cy 2 (1a)

Cx = Ax + Bx Cy = Ay + By (2)

θ = tan−1 ( Cy Cx) (1b)

Cx = Ax + Bx Cy = Ay + By (2)

STEP 1 STEP 2

STEP 1 STEP 2

1.8 Addition of Vectors by Means of Components 17

Referring to Figure 1.22, we fi nd these values to be

Ax = (145 m) sin 20.0° = 49.6 m and Ay = (145 m) cos 20.0° = 136 m

Bx = (105 m) cos 35.0° = 86.0 m and By = −(105 m) sin 35.0° = −60.2 m

Note that the component By is negative, because B →

y points downward, in the negative y direction in the drawing. Substituting these values into the results for C and θ gives

C = √(Ax + Bx)2 + (Ay + By)2

= √(49.6 m + 86.0 m)2 + (136 m − 60.2 m)2 = 155 m

θ = tan−1( Ay + By Ax + Bx)

= tan−1( 136 m − 60.2 m49.6 m + 86.0 m) = 29°

Math Skills According to the defi nitions given in Equations 1.1 and 1.2, the sine and

cosine functions are sin ϕ = ho h

and cos ϕ = ha h

, where ho is the length of the side of a

right triangle that is opposite the angle ϕ, ha is the length of the side adjacent to the angle ϕ, and h is the length of the hypotenuse (see Figure 1.23a). Applications of the sine and cosine functions to determine the scalar components of a vector occur frequently. In such applications we begin by identifying the angle ϕ. Figure 1.23b shows the relevant portion of Figure 1.22 and indicates that ϕ = 20.0° for the vector A→. In this case we have ho = Ax, ha = Ay, and h = A = 145 m; it follows that

sin 20.0° = ho h

= Ax A

or Ax = A sin 20.0° = (145 m) sin 20.0° = 49.6 m

cos 20.0° = ha h

= Ay A

or Ay = A cos 20.0° = (145 m) cos 20.0° = 136 m

(a) (b)

Ax

+y

+x

Ay ha

ho

h

ha

ho

h A

ϕ

20.0°

90°

FIGURE 1.23 Math Skills drawing.

Related Homework: Problems 45, 47, 50, 54

18 CHAPTER 1 Introduction and Mathematical Concepts

In later chapters we will often use the component method for vector addition. For future reference, the main features of the reasoning strategy used in this technique are summarized below.

REASONING STRATEGY The Component Method of Vector Addition 1. For each vector to be added, determine the x and y components relative to a conveniently

chosen x, y coordinate system. Be sure to take into account the directions of the compon- ents by using plus and minus signs to denote whether the components point along the positive or negative axes.

2. Find the algebraic sum of the x components, which is the x component of the resultant vector. Similarly, fi nd the algebraic sum of the y components, which is the y component of the resultant vector.

3. Use the x and y components of the resultant vector and the Pythagorean theorem to de- termine the magnitude of the resultant vector.

4. Use the inverse sine, inverse cosine, or inverse tangent function to fi nd the angle that specifi es the direction of the resultant vector.

Check Your Understanding

(1) (2) (3) (4)

Ax

Ay

R

R

R

RBy Bx

(The answer is given at the end of the book.) 18. Two vectors, A→ and B→, have vector components that are shown (to the same scale) in CYU Figure 1.2.

The resultant vector is labeled R→. Which drawing shows the correct vector sum of A→ + B→? (a) 1, (b) 2, (c) 3, (d) 4

joint, and vector B→ represents the position of the ball relative to the elbow joint. Use the component method of vector addition and the angles given in the fi gure to fi nd the magnitude and direction (θ) of vector C→, which represents the position of the ball relative to the shoulder joint. The mag- nitude of vector A→ is 35.6 cm, and the magnitude of vector B→ is 31.2 cm. The angle θ is measured relative to a vertical anatomical plane known as the frontal or coronal plane.

Reasoning Similar to Example 9, the vectors A→ and B→ in Figure 1.24 are drawn tail-to-head. Thus, the resultant vector C→ = A→ + B→. The com- ponents of vector C→ will be obtained from the components of vectors A→ and B→. Once Cx and Cy are known, we can calculate the magnitude of vector C→ using the Pythagorean theorem. The directional angle θ will be determined from the components of vector C→.

Solution Applying the component method of vector addition, we know that Cx = Ax + Bx, and Cy = Ay + By. From Figure 1.24, we see that Ax = (35.6 cm) cos 45° = 25.2 cm, Bx = – (31.2) cos 80° = – 5.42 cm, Ay = (35.6 cm) sin 45° = 25.2 cm, and By = (31.2 cm) sin 80° = 30.7 cm. Therefore, Cx = 25.2 cm + (– 5.42 cm) = 19.8 cm, and Cy = 25.2 cm + 30.7 cm = 55.9 cm. The magnitude of vector C→ is now found by the Pythagorean theorem: C = √C2x + C2y = √(19.8 cm)2 + (55.9 cm)2 = 59.3 cm The directional angle θ is found by using the tangent function:

θ = tan−1( Cy Cx) = tan−1(

55.9 cm 19.8 cm) = 70.5° .

EXAMPLE 10 BIO Multi-joint Movements

Figure 1.24 shows an example of a multi-joint movement involving the shoulder and elbow joints. The view from above shows a person holding a ball in a position that involves both shoulder fl exion and elbow extension. Vector A→ represents the position of the elbow joint relative to the shoulder

FIGURE 1.24 Top view of a multi-joint movement involving the shoulder and elbow joints. The vectors A→ and B→ add together to give the resultant vector C→, which represents the position vector of the ball with respect to the shoulder joint. The resultant vector C→ can be obtained, once its components are determined.

Shoulder joint

Elbow joint

Ball

Frontal plane

45.0°

80.0°

+y

B

A

C

+x θ

CYU FIGURE 1.2

Focus on Concepts 19

Concept Summary 1.2 Units The SI system of units includes the meter (m), the kilogram (kg), and the second (s) as the base units for length, mass, and time, respectively. One meter is the distance that light travels in a vacuum in a time of 1/299 792 458 second. One kilogram is the mass of a standard cylinder of platinum–iridium alloy kept at the International Bureau of Weights and Measures. One second is the time for a certain type of electromagnetic wave emitted by cesium-133 atoms to undergo 9 192 631 770 wave cycles.

1.3 The Role of Units in Problem Solving To convert a number from one unit to another, multiply the number by the ratio of the two units. For instance, to convert 979 meters to feet, multiply 979 meters by the factor (3.281 foot/1 meter).

The dimension of a quantity represents its physical nature and the type of unit used to specify it. Three such dimensions are length [L], mass [M], time [T]. Dimensional analysis is a method for checking mathematical rela- tions for the consistency of their dimensions.

1.4 Trigonometry The sine, cosine, and tangent functions of an angle θ are defi ned in terms of a right triangle that contains θ, as in Equations 1.1–1.3, where ho and ha are, respectively, the lengths of the sides opposite and adja- cent to the angle θ, and h is the length of the hypotenuse.

sin θ = ho h

(1.1)

cos θ = ha h

(1.2)

tan θ = ho ha

(1.3)

The inverse sine, inverse cosine, and inverse tangent functions are given in Equations 1.4–1.6.

θ = sin−1 ( ho h ) (1.4)

θ = cos−1 ( ha h ) (1.5)

θ = tan−1 ( ho ha) (1.6)

The Pythagorean theorem states that the square of the length of the hypot- enuse of a right triangle is equal to the sum of the squares of the lengths of the other two sides, according to Equation 1.7.

h2 = ho2 + ha2 (1.7)

1.5 Scalars and Vectors A scalar quantity is described by its size, which is also called its magnitude. A vector quantity has both a magnitude and a direction. Vectors are often represented by arrows, the length of the arrow being proportional to the magnitude of the vector and the direction of the arrow indicating the direction of the vector.

1.6 Vector Addition and Subtraction One procedure for adding vectors utilizes a graphical technique, in which the vectors to be added are ar- ranged in a tail-to-head fashion. The resultant vector is drawn from the tail of the fi rst vector to the head of the last vector. The subtraction of a vector is treated as the addition of a vector that has been multiplied by a scalar factor of −1. Multiplying a vector by −1 reverses the direction of the vector.

1.7 The Components of a Vector In two dimensions, the vector compon- ents of a vector A→ are two perpendicular vectors A→ x and A

→ y that are parallel

to the x and y axes, respectively, and that add together vectorially so that A→ = A→ x + A

→ y. The scalar component Ax has a magnitude that is equal to that

of A→ x and is given a positive sign if A →

x points along the +x axis and a negative sign if A→ x points along the −x axis. The scalar component Ay is defi ned in a similar manner.

Two vectors are equal if, and only if, they have the same magnitude and direction. Alternatively, two vectors are equal in two dimensions if the x vector components of each are equal and the y vector components of each are equal. A vector is zero if, and only if, each of its vector components is zero.

1.8 Addition of Vectors by Means of Components If two vec- tors A→ and B→ are added to give a resultant C→ such that C→ = A→ + B→, then Cx = Ax + Bx and Cy = Ay + By, where Cx, Ax, and Bx are the scalar compon- ents of the vectors along the x direction, and Cy, Ay, and By are the scalar components of the vectors along the y direction.

Note to Instructors: The numbering of the questions shown here refl ects the fact that they are only a representative subset of the total number that are available online. However, all of the questions are available for assignment via WileyPLUS.

Section 1.6 Vector Addition and Subtraction 1. During a relay race, runner A runs a certain distance due north and then hands off the baton to runner B, who runs for the same distance in a direction south of east. The two displacement vectors A→ and B→ can be added together to give a resultant vector R→. Which drawing correctly shows the resultant vector? (a) 1 (b) 2 (c) 3 (d) 4

North

East (1)

North

East (2)

North

East (3)

North

East (4)

A A A

A

BBBB

RRRR

QUESTION 1

2. How is the magnitude R of the resultant vector R→ in the drawing related to the magnitudes A and B of the vectors A→ and B→? (a) The magnitude of the resultant vector R→ is equal to the sum of the magnitudes of A→ and B→, or R = A + B. (b) The magnitude of the resultant vector R→ is greater than the

Focus on Concepts

20 CHAPTER 1 Introduction and Mathematical Concepts

sum of the magnitudes of A→ and B→, or R > A + B. (c) The magnitude of the resultant vector R→ is less than the sum of the magnitudes of A→ and B→, or R < A + B. 5. The fi rst drawing shows three displacement vectors, A→, B→, and C→, which are added in a tail-to-head fashion. The resultant vector is labeled R→. Which of the following drawings shows the correct resultant vector for A→ + B→ − C→? (a) 1 (b) 2 (c) 3

(1) (2) (3)

A

A A

A B

C

R R R

R

QUESTION 5

6. The fi rst drawing shows the sum of three displacement vectors, A→, B→, and C→. The resultant vector is labeled R→. Which of the following drawings shows the correct resultant vector for A→ − B→ − C→? (a) 1 (b) 2 (c) 3

(1) (2) (3)

A

A A

A B

C

R R R

R

QUESTION 6

Section 1.7 The Components of a Vector 8. A person is jogging along a straight line, and her displacement is denoted by the vector A→ in the drawings. Which drawing represents the correct vector components, A→ x and A

→ y, for the vector A

→ ? (a) 1 (b) 2 (c) 3 (d) 4

+y

+x

+x

+x

+x

+y

+y +y

(1)

Ax

Ay Ay

Ay Ay

Ax

Ax Ax

A

(2)

A

(3)

A

(4)

A

QUESTION 8

11. A person drives a car for a distance of 450.0 m. The displacement A→ of the car is illustrated in the drawing. What are the scalar components of this displacement vector?

(a) Ax = 0 m and Ay = +450.0 m (b) Ax = 0 m and Ay = −450.0 m (c) Ax = +450.0 m and Ay = +450.0 m (d) Ax = −450.0 m and Ay = 0 m (e) Ax = −450.0 m and Ay = +450.0 m

12. Drawing a shows a displacement vector A→ (450.0 m along the −y axis). In this x, y coordinate system the scalar components are Ax = 0 m and Ay = −450.0 m. Suppose that the coordinate system is rotated counterclockwise by 35.0°, but the magnitude (450.0 m) and direction of vector A→ remain unchanged, as in drawing b. What are the scalar components, Ax′ and Ay′, of the vector A

→ in the rotated x′, y′ coordinate system?

(a) Ax′ = −369 m and Ay′ = −258 m (b) Ax′ = +369 m and Ay′ = −258 m (c) Ax′ = +258 m and Ay′ = +369 m (d) Ax′ = +258 m and Ay′ = −369 m (e) Ax′ = −258 m and Ay′ = −369 m

QUESTION 12

+y

+x

A

(a)

A

+y´

+x´

35.0°

(b)

15. Suppose the vectors A→ and B→ in the drawing have magnitudes of 6.0 m and are directed as shown. What are Ax and Bx, the scalar components of A

→

and B→ along the x axis?

Ax Bx (a) +(6.0 m) cos 35° = +4.9 m −(6.0 m) cos 35° = −4.9 m

(b) +(6.0 m) sin 35° = +3.4 m −(6.0 m) cos 35° = −4.9 m

(c) −(6.0 m) cos 35° = −4.9 m +(6.0 m) sin 35° = +3.4 m

(d) −(6.0 m) cos 35° = −4.9 m +(6.0 m) cos 35° = +4.9 m

(e) −(6.0 m) sin 35° = −3.4 m +(6.0 m) sin 35° = +3.4 m

QUESTION 15

A B

35° 35°

6.0 m 6.0 m

+y

+x

Section 1.8 Addition of Vectors by Means of Components 17. Drawing a shows two vectors A→ and B→, and drawing b shows their com- ponents. The scalar components of these vectors are as follows:

Ax = −4.9 m Ay = +3.4 m Bx = +4.9 m By = +3.4 m

When the vectors A→ and B→ are added, the resultant vector is R→, so that R→ = A→ + B→. What are the values of Rx and Ry, the x and y components of R

→ ?

QUESTION 17

+y

+x

+x

+y

A B

35° 35°

6.0 m 6.0 m

A Ay By

BxAx

B

(a)

(b)

θ θ

A

B

R

QUESTION 2

QUESTION 11

+y

+x

A

Problems 21

18. The displacement vectors A→ and B→, when added together, give the res- ultant vector R→, so that R→ = A→ + B→. Use the data in the drawing to fi nd the magnitude R of the resultant vector and the angle θ that it makes with the +x axis.

QUESTION 18

A B

R

+y

6.0 m 8.0 m

23°

+xθ

Note to Instructors: Most of the homework problems in this chapter are available for assignment via WileyPLUS. See the Preface for additional details.

SSM Student Solutions Manual MMH Problem-solving help GO Guided Online Tutorial V-HINT Video Hints CHALK Chalkboard Videos

BIO Biomedical application E Easy M Medium H Hard

Section 1.2 Units

Section 1.3 The Role of Units in Problem Solving 1. E GO A student sees a newspaper ad for an apartment that has 1330 square feet (ft2) of fl oor space. How many square meters of area are there?

2. E Bicyclists in the Tour de France reach speeds of 34.0 miles per hour (mi/h) on fl at sections of the road. What is this speed in (a) kilometers per hour (km/h) and (b) meters per second (m/s)? 3. E SSM Vesna Vulovic survived the longest fall on record without a para- chute when her plane exploded and she fell 6 miles, 551 yards. What is this distance in meters?

4. E Suppose a man’s scalp hair grows at a rate of 0.35 mm per day. What is this growth rate in feet per century?

5. E Given the quantities a = 9.7 m, b = 4.2 s, c = 69 m/s, what is the value of the quantity d = a3/(cb2)?

6. E Consider the equation 𝜐 = 13 zxt 2. The dimensions of the variables υ, x, and t are [L]/[T], [L], and [T], respectively. The numerical factor 3 is di- mensionless. What must be the dimensions of the variable z, such that both sides of the equation have the same dimensions? Show how you determined your answer.

7. E SSM A bottle of wine known as a magnum contains a volume of 1.5 liters. A bottle known as a jeroboam contains 0.792 U.S. gallons. How many magnums are there in one jeroboam?

8. E The CGS unit for measuring the viscosity of a liquid is the poise (P): 1 P = 1 g/(s · cm). The SI unit for viscosity is the kg/(s · m). The viscosity of water at 0 °C is 1.78 × 10−3 kg/(s · m). Express this viscosity in poise. 9. E BIO Azelastine hydrochloride is an antihistamine nasal spray. A standard size container holds one fl uid ounce (oz) of the liquid. You are searching for this medication in a European drugstore and are asked how many milliliters (mL) there are in one fl uid ounce. Using the following con- version factors, determine the number of milliliters in a volume of one fl uid ounce: 1 gallon (gal) = 128 oz, 3.785 × 10−3 cubic meters (m3) = 1 gal, and 1 mL = 10−6 m3.

10. M GO A partly full paint can has 0.67 U.S. gallons of paint left in it. (a) What is the volume of the paint in cubic meters? (b) If all the remaining paint is used to coat a wall evenly (wall area = 13 m2), how thick is the layer of wet paint? Give your answer in meters.

11. M SSM A spring is hanging down from the ceiling, and an object of mass m is attached to the free end. The object is pulled down, thereby stretch- ing the spring, and then released. The object oscillates up and down, and the time T required for one complete up-and-down oscillation is given by the equation T = 2π√m/k, where k is known as the spring constant. What must be the dimension of k for this equation to be dimensionally correct?

Section 1.4 Trigonometry 12. E You are driving into St. Louis, Missouri, and in the distance you see the famous Gateway to the West arch. This monument rises to a height of 192 m. You estimate your line of sight with the top of the arch to be 2.0° above the horizontal. Approximately how far (in kilometers) are you from the base of the arch?

13. E SSM A highway is to be built between two towns, one of which lies 35.0 km south and 72.0 km west of the other. What is the shortest length of highway that can be built between the two towns, and at what angle would this highway be directed with respect to due west?

14. E GO A hill that has a 12.0% grade is one that rises 12.0 m vertically for every 100.0 m of distance in the horizontal direction. At what angle is such a hill inclined above the horizontal?

15. E GO The corners of a square lie on a circle of diameter D = 0.35 m. Each side of the square has a length L. Find L.

16. E GO The drawing shows a person looking at a building on top of which an antenna is mounted. The horizontal distance between the per- son’s eyes and the building is 85.0 m. In part a the person is looking at the base of the antenna, and his line of sight makes an angle of 35.0° with the horizontal. In part b the person is looking at the top of the antenna, and his line of sight makes an angle of 38.0° with the horizontal. How tall is the antenna?

(a) (b)

35.0° 38.0°

85.0 m85.0 m

PROBLEM 16

Problems

22 CHAPTER 1 Introduction and Mathematical Concepts

Section 1.6 Vector Addition and Subtraction 23. E SSM (a) Two workers are trying to move a heavy crate. One pushes on the crate with a force A→, which has a magnitude of 445 newtons and is directed due west. The other pushes with a force B→, which has a magnitude of 325 newtons and is directed due north. What are the magnitude and direc- tion of the resultant force A→ + B→ applied to the crate? (b) Suppose that the second worker applies a force −B→ instead of B→. What then are the magnitude and direction of the resultant force A→ − B→ applied to the crate? In both cases express the direction relative to due west.

24. E A force vector F1 →

points due east and has a magnitude of 200 new- tons. A second force F2

→ is added to F1

→ . The resultant of the two vectors has

a magnitude of 400 newtons and points along the east/west line. Find the magnitude and direction of F2

→ . Note that there are two answers.

25. E SSM Consider the following four force vectors:

F1 →

= 50.0 newtons, due east

F2 →

= 10.0 newtons, due east

F3 →

= 40.0 newtons, due west

F4 →

= 30.0 newtons, due west

Which two vectors add together to give a resultant with the smallest mag- nitude, and which two vectors add to give a resultant with the largest mag- nitude? In each case specify the magnitude and direction of the resultant.

26. E GO Vector A→ has a magnitude of 63 units and points due west, while vector B→ has the same magnitude and points due south. Find the magnitude and direction of (a) A→ + B→ and (b) A→ − B→. Specify the directions relative to due west. 27. E Two bicyclists, starting at the same place, are riding toward the same campground by two diff erent routes. One cyclist rides 1080 m due east and then turns due north and travels another 1430 m before reaching the camp- ground. The second cyclist starts out by heading due north for 1950 m and then turns and heads directly toward the campground. (a) At the turning point, how far is the second cyclist from the campground? (b) In what direc- tion (measured relative to due east) must the second cyclist head during the last part of the trip?

28. E GO The drawing shows a triple jump on a checkerboard, starting at the center of square A and ending on the center of square B. Each side of a square measures 4.0 cm. What is the magnitude of the displacement of the colored checker during the triple jump?

B

A PROBLEM 28

29. E Given the vectors P→ and Q→ shown on the grid, sketch and calculate the magnitudes of the vectors (a) M

→ = P→ + Q→ and (b) K→ = 2P→ − Q→.

Use the tail-to-head method and express the mag- nitudes in centimeters with the aid of the grid scale shown in the drawing.

17. E CHALK The two hot-air balloons in the drawing are 48.2 and 61.0 m above the ground. A person in the left balloon observes that the right balloon is 13.3° above the horizontal. What is the horizontal distance x between the two balloons?

PROBLEM 17

48.2 m

13.3°

61.0 m

x

18. M Available on WileyPLUS. 19. M MMH The drawing shows sodium and chloride ions positioned at the corners of a cube that is part of the crystal structure of sodium chloride (common table salt). The edges of the cube are each 0.281 nm (1 nm = 1 nanometer = 10−9 m) in length. What is the value of the angle θ in the drawing? 20. M GO A person is standing at the edge of the water and looking out at the ocean (see the drawing). The height of the person’s eyes above the water is h = 1.6 m, and the radius of the earth is R = 6.38 × 106 m. (a) How far is it to the horizon? In other words, what is the distance d from the person’s eyes to the horizon? (Note: At the horizon the angle between the line of sight and the radius of the earth is 90°.) (b) Express this distance in miles.

PROBLEM 20

Horizon

h

R

d

R

90°

21. M SSM Three deer, A, B, and C, are grazing in a fi eld. Deer B is located 62 m from deer A at an angle of 51° north of west. Deer C is located 77° north of east relative to deer A. The distance between deer B and C is 95 m. What is the distance between deer A and C? (Hint: Consider the law of cosines given in Appendix E.) 22. H An aerialist on a high platform holds on to a trapeze attached to a support by an 8.0-m cord. (See the drawing.) Just before he jumps off the platform, the cord makes an angle of 41° with the vertical. He jumps, swings down, then back up, releasing the trapeze at the instant it is 0.75 m below its initial height. Calculate the angle θ that the trapeze cord makes with the vertical at this instant.

8.0 m

0.75 m

41°θ

PROBLEM 22

8.00 cm P

Q

PROBLEM 29

PROBLEM 19

Chloride ion Sodium ion

0.281 nanometers

θ

Problems 23

30. M MMH Vector A→ has a magnitude of 12.3 units and points due west. Vector B→ points due north. (a) What is the magnitude of B→ if A→ + B→ has a magnitude of 15.0 units? (b) What is the direction of A→ + B→ relative to due west? (c) What is the magnitude of B→ if A→ − B→ has a magnitude of 15.0 units? (d) What is the direction of A→ − B→ relative to due west? 31. M SSM A car is being pulled out of the mud by two forces that are ap- plied by the two ropes shown in the drawing. The dashed line in the drawing

bisects the 30.0° angle. The magnitude of the force applied by each rope is

2900 newtons. Arrange the force vectors tail to head and use the graphical

technique to answer the following questions. (a) How much force would a single rope need to apply to accomplish the same eff ect as the two forces

added together? (b) How would the single rope be directed relative to the dashed line?

30.0°