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University Physics Volume 1

2.3 Algebra of Vectors

University Physics Volume 12.3 Algebra of Vectors
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  1. Preface
  2. Unit 1. Mechanics
    1. 1 Units and Measurement
      1. Introduction
      2. 1.1 The Scope and Scale of Physics
      3. 1.2 Units and Standards
      4. 1.3 Unit Conversion
      5. 1.4 Dimensional Analysis
      6. 1.5 Estimates and Fermi Calculations
      7. 1.6 Significant Figures
      8. 1.7 Solving Problems in Physics
      9. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    2. 2 Vectors
      1. Introduction
      2. 2.1 Scalars and Vectors
      3. 2.2 Coordinate Systems and Components of a Vector
      4. 2.3 Algebra of Vectors
      5. 2.4 Products of Vectors
      6. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    3. 3 Motion Along a Straight Line
      1. Introduction
      2. 3.1 Position, Displacement, and Average Velocity
      3. 3.2 Instantaneous Velocity and Speed
      4. 3.3 Average and Instantaneous Acceleration
      5. 3.4 Motion with Constant Acceleration
      6. 3.5 Free Fall
      7. 3.6 Finding Velocity and Displacement from Acceleration
      8. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    4. 4 Motion in Two and Three Dimensions
      1. Introduction
      2. 4.1 Displacement and Velocity Vectors
      3. 4.2 Acceleration Vector
      4. 4.3 Projectile Motion
      5. 4.4 Uniform Circular Motion
      6. 4.5 Relative Motion in One and Two Dimensions
      7. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    5. 5 Newton's Laws of Motion
      1. Introduction
      2. 5.1 Forces
      3. 5.2 Newton's First Law
      4. 5.3 Newton's Second Law
      5. 5.4 Mass and Weight
      6. 5.5 Newton’s Third Law
      7. 5.6 Common Forces
      8. 5.7 Drawing Free-Body Diagrams
      9. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    6. 6 Applications of Newton's Laws
      1. Introduction
      2. 6.1 Solving Problems with Newton’s Laws
      3. 6.2 Friction
      4. 6.3 Centripetal Force
      5. 6.4 Drag Force and Terminal Speed
      6. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    7. 7 Work and Kinetic Energy
      1. Introduction
      2. 7.1 Work
      3. 7.2 Kinetic Energy
      4. 7.3 Work-Energy Theorem
      5. 7.4 Power
      6. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    8. 8 Potential Energy and Conservation of Energy
      1. Introduction
      2. 8.1 Potential Energy of a System
      3. 8.2 Conservative and Non-Conservative Forces
      4. 8.3 Conservation of Energy
      5. 8.4 Potential Energy Diagrams and Stability
      6. 8.5 Sources of Energy
      7. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
    9. 9 Linear Momentum and Collisions
      1. Introduction
      2. 9.1 Linear Momentum
      3. 9.2 Impulse and Collisions
      4. 9.3 Conservation of Linear Momentum
      5. 9.4 Types of Collisions
      6. 9.5 Collisions in Multiple Dimensions
      7. 9.6 Center of Mass
      8. 9.7 Rocket Propulsion
      9. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    10. 10 Fixed-Axis Rotation
      1. Introduction
      2. 10.1 Rotational Variables
      3. 10.2 Rotation with Constant Angular Acceleration
      4. 10.3 Relating Angular and Translational Quantities
      5. 10.4 Moment of Inertia and Rotational Kinetic Energy
      6. 10.5 Calculating Moments of Inertia
      7. 10.6 Torque
      8. 10.7 Newton’s Second Law for Rotation
      9. 10.8 Work and Power for Rotational Motion
      10. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    11. 11 Angular Momentum
      1. Introduction
      2. 11.1 Rolling Motion
      3. 11.2 Angular Momentum
      4. 11.3 Conservation of Angular Momentum
      5. 11.4 Precession of a Gyroscope
      6. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    12. 12 Static Equilibrium and Elasticity
      1. Introduction
      2. 12.1 Conditions for Static Equilibrium
      3. 12.2 Examples of Static Equilibrium
      4. 12.3 Stress, Strain, and Elastic Modulus
      5. 12.4 Elasticity and Plasticity
      6. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    13. 13 Gravitation
      1. Introduction
      2. 13.1 Newton's Law of Universal Gravitation
      3. 13.2 Gravitation Near Earth's Surface
      4. 13.3 Gravitational Potential Energy and Total Energy
      5. 13.4 Satellite Orbits and Energy
      6. 13.5 Kepler's Laws of Planetary Motion
      7. 13.6 Tidal Forces
      8. 13.7 Einstein's Theory of Gravity
      9. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    14. 14 Fluid Mechanics
      1. Introduction
      2. 14.1 Fluids, Density, and Pressure
      3. 14.2 Measuring Pressure
      4. 14.3 Pascal's Principle and Hydraulics
      5. 14.4 Archimedes’ Principle and Buoyancy
      6. 14.5 Fluid Dynamics
      7. 14.6 Bernoulli’s Equation
      8. 14.7 Viscosity and Turbulence
      9. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
  3. Unit 2. Waves and Acoustics
    1. 15 Oscillations
      1. Introduction
      2. 15.1 Simple Harmonic Motion
      3. 15.2 Energy in Simple Harmonic Motion
      4. 15.3 Comparing Simple Harmonic Motion and Circular Motion
      5. 15.4 Pendulums
      6. 15.5 Damped Oscillations
      7. 15.6 Forced Oscillations
      8. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    2. 16 Waves
      1. Introduction
      2. 16.1 Traveling Waves
      3. 16.2 Mathematics of Waves
      4. 16.3 Wave Speed on a Stretched String
      5. 16.4 Energy and Power of a Wave
      6. 16.5 Interference of Waves
      7. 16.6 Standing Waves and Resonance
      8. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
    3. 17 Sound
      1. Introduction
      2. 17.1 Sound Waves
      3. 17.2 Speed of Sound
      4. 17.3 Sound Intensity
      5. 17.4 Normal Modes of a Standing Sound Wave
      6. 17.5 Sources of Musical Sound
      7. 17.6 Beats
      8. 17.7 The Doppler Effect
      9. 17.8 Shock Waves
      10. Chapter Review
        1. Key Terms
        2. Key Equations
        3. Summary
        4. Conceptual Questions
        5. Problems
        6. Additional Problems
        7. Challenge Problems
  4. A | Units
  5. B | Conversion Factors
  6. C | Fundamental Constants
  7. D | Astronomical Data
  8. E | Mathematical Formulas
  9. F | Chemistry
  10. G | The Greek Alphabet
  11. Answer Key
    1. Chapter 1
    2. Chapter 2
    3. Chapter 3
    4. Chapter 4
    5. Chapter 5
    6. Chapter 6
    7. Chapter 7
    8. Chapter 8
    9. Chapter 9
    10. Chapter 10
    11. Chapter 11
    12. Chapter 12
    13. Chapter 13
    14. Chapter 14
    15. Chapter 15
    16. Chapter 16
    17. Chapter 17
  12. Index

Learning Objectives

By the end of this section, you will be able to:
  • Apply analytical methods of vector algebra to find resultant vectors and to solve vector equations for unknown vectors.
  • Interpret physical situations in terms of vector expressions.

Vectors can be added together and multiplied by scalars. Vector addition is associative (Equation 2.8) and commutative (Equation 2.7), and vector multiplication by a sum of scalars is distributive (Equation 2.9). Also, scalar multiplication by a sum of vectors is distributive:

α(A+B)=αA+αB.α(A+B)=αA+αB.
(2.22)

In this equation, αα is any number (a scalar). For example, a vector antiparallel to vector A=Axi^+Ayj^+Azk^A=Axi^+Ayj^+Azk^ can be expressed simply by multiplying AA by the scalar α=−1α=−1:

A=Axi^Ayj^Azk^.A=Axi^Ayj^Azk^.
(2.23)

Example 2.8

Direction of Motion In a Cartesian coordinate system where i^i^ denotes geographic east, j^j^ denotes geographic north, and k^k^ denotes altitude above sea level, a military convoy advances its position through unknown territory with velocity v=(4.0i^+3.0j^+0.1k^)km/hv=(4.0i^+3.0j^+0.1k^)km/h. If the convoy had to retreat, in what geographic direction would it be moving?

Solution The velocity vector has the third component vz=(+0.1km/h)k^vz=(+0.1km/h)k^, which says the convoy is climbing at a rate of 100 m/h through mountainous terrain. At the same time, its velocity is 4.0 km/h to the east and 3.0 km/h to the north, so it moves on the ground in direction tan−1(3/4)37°tan−1(3/4)37° north of east. If the convoy had to retreat, its new velocity vector uu would have to be antiparallel to vv and be in the form u=αvu=αv, where αα is a positive number. Thus, the velocity of the retreat would be u=α(−4.0i^3.0j^0.1k^)km/hu=α(−4.0i^3.0j^0.1k^)km/h. The negative sign of the third component indicates the convoy would be descending. The direction angle of the retreat velocity is tan−1(−3α/4α)37°tan−1(−3α/4α)37° south of west. Therefore, the convoy would be moving on the ground in direction 37°37° south of west while descending on its way back.

The generalization of the number zero to vector algebra is called the null vector, denoted by 00. All components of the null vector are zero, 0=0i^+0j^+0k^0=0i^+0j^+0k^, so the null vector has no length and no direction.

Two vectors AA and BB are equal vectors if and only if their difference is the null vector:

0=AB=(Axi^+Ayj^+Azk^)(Bxi^+Byj^+Bzk^)=(AxBx)i^+(AyBy)j^+(AzBz)k^.0=AB=(Axi^+Ayj^+Azk^)(Bxi^+Byj^+Bzk^)=(AxBx)i^+(AyBy)j^+(AzBz)k^.

This vector equation means we must have simultaneously AxBx=0AxBx=0, AyBy=0AyBy=0, and AzBz=0AzBz=0. Hence, we can write A=BA=B if and only if the corresponding components of vectors AA and BB are equal:

A=B{Ax=BxAy=ByAz=Bz.A=B{Ax=BxAy=ByAz=Bz.
(2.24)

Two vectors are equal when their corresponding scalar components are equal.

Resolving vectors into their scalar components (i.e., finding their scalar components) and expressing them analytically in vector component form (given by Equation 2.19) allows us to use vector algebra to find sums or differences of many vectors analytically (i.e., without using graphical methods). For example, to find the resultant of two vectors AA and BB, we simply add them component by component, as follows:

R=A+B=(Axi^+Ayj^+Azk^)+(Bxi^+Byj^+Bzk^)=(Ax+Bx)i^+(Ay+By)j^+(Az+Bz)k^.R=A+B=(Axi^+Ayj^+Azk^)+(Bxi^+Byj^+Bzk^)=(Ax+Bx)i^+(Ay+By)j^+(Az+Bz)k^.

In this way, using Equation 2.24, scalar components of the resultant vector R=Rxi^+Ryj^+Rzk^R=Rxi^+Ryj^+Rzk^ are the sums of corresponding scalar components of vectors AA and BB:

{Rx=Ax+Bx,Ry=Ay+By,Rz=Az+Bz.{Rx=Ax+Bx,Ry=Ay+By,Rz=Az+Bz.

Analytical methods can be used to find components of a resultant of many vectors. For example, if we are to sum up NN vectors F1,F2,F3,,FNF1,F2,F3,,FN, where each vector is Fk=Fkxi^+Fkyj^+Fkzk^Fk=Fkxi^+Fkyj^+Fkzk^, the resultant vector FRFR is

FR=F1+F2+F3++FN=k=1NFk=k=1N(Fkxi^+Fkyj^+Fkzk^)=(k=1NFkx)i^+(k=1NFky)j^+(k=1NFkz)k^.FR=F1+F2+F3++FN=k=1NFk=k=1N(Fkxi^+Fkyj^+Fkzk^)=(k=1NFkx)i^+(k=1NFky)j^+(k=1NFkz)k^.

Therefore, scalar components of the resultant vector are

{FRx=k=1NFkx=F1x+F2x++FNxFRy=k=1NFky=F1y+F2y++FNyFRz=k=1NFkz=F1z+F2z++FNz.{FRx=k=1NFkx=F1x+F2x++FNxFRy=k=1NFky=F1y+F2y++FNyFRz=k=1NFkz=F1z+F2z++FNz.
(2.25)

Having found the scalar components, we can write the resultant in vector component form:

FR=FRxi^+FRyj^+FRzk^.FR=FRxi^+FRyj^+FRzk^.

Analytical methods for finding the resultant and, in general, for solving vector equations are very important in physics because many physical quantities are vectors. For example, we use this method in kinematics to find resultant displacement vectors and resultant velocity vectors, in mechanics to find resultant force vectors and the resultants of many derived vector quantities, and in electricity and magnetism to find resultant electric or magnetic vector fields.

Example 2.9

Analytical Computation of a Resultant Three displacement vectors AA, BB, and CC in a plane (Figure 2.13) are specified by their magnitudes A = 10.0, B = 7.0, and C = 8.0, respectively, and by their respective direction angles with the horizontal direction α=35°,α=35°, β=−110°β=−110°, and γ=30°γ=30°. The physical units of the magnitudes are centimeters. Resolve the vectors to their scalar components and find the following vector sums: (a) R=A+B+CR=A+B+C, (b) D=ABD=AB, and (c) S=A3B+CS=A3B+C.

Strategy First, we use Equation 2.17 to find the scalar components of each vector and then we express each vector in its vector component form given by Equation 2.12. Then, we use analytical methods of vector algebra to find the resultants.

Solution We resolve the given vectors to their scalar components:

{Ax=Acosα=(10.0cm)cos35°=8.19cmAy=Asinα=(10.0cm)sin35°=5.73cm{Bx=Bcosβ=(7.0cm)cos(−110°)=−2.39cmBy=Bsinβ=(7.0cm)sin(−110°)=−6.58cm{Cx=Ccosγ=(8.0cm)cos30°=6.93cmCy=Csinγ=(8.0cm)sin30°=4.00cm.{Ax=Acosα=(10.0cm)cos35°=8.19cmAy=Asinα=(10.0cm)sin35°=5.73cm{Bx=Bcosβ=(7.0cm)cos(−110°)=−2.39cmBy=Bsinβ=(7.0cm)sin(−110°)=−6.58cm{Cx=Ccosγ=(8.0cm)cos30°=6.93cmCy=Csinγ=(8.0cm)sin30°=4.00cm.

For (a) we may substitute directly into Equation 2.25 to find the scalar components of the resultant:

{Rx=Ax+Bx+Cx=8.19cm2.39cm+6.93cm=12.73cmRy=Ay+By+Cy=5.73cm6.58cm+4.00cm=3.15cm.{Rx=Ax+Bx+Cx=8.19cm2.39cm+6.93cm=12.73cmRy=Ay+By+Cy=5.73cm6.58cm+4.00cm=3.15cm.

Therefore, the resultant vector is R=Rxi^+Ryj^=(12.7i^+3.1j^)cmR=Rxi^+Ryj^=(12.7i^+3.1j^)cm.

For (b), we may want to write the vector difference as

D=AB=(Axi^+Ayj^)(Bxi^+Byj^)=(AxBx)i^+(AyBy)j^.D=AB=(Axi^+Ayj^)(Bxi^+Byj^)=(AxBx)i^+(AyBy)j^.

Then, the scalar components of the vector difference are

{Dx=AxBx=8.19cm(−2.39cm)=10.58cmDy=AyBy=5.73cm(−6.58cm)=12.31cm.{Dx=AxBx=8.19cm(−2.39cm)=10.58cmDy=AyBy=5.73cm(−6.58cm)=12.31cm.

Hence, the difference vector is D=Dxi^+Dyj^=(10.6i^+12.3j^)cmD=Dxi^+Dyj^=(10.6i^+12.3j^)cm.

For (c), we can write vector SS in the following explicit form:

S=A3B+C=(Axi^+Ayj^)3(Bxi^+Byj^)+(Cxi^+Cyj^)=(Ax3Bx+Cx)i^+(Ay3By+Cy)j^.S=A3B+C=(Axi^+Ayj^)3(Bxi^+Byj^)+(Cxi^+Cyj^)=(Ax3Bx+Cx)i^+(Ay3By+Cy)j^.

Then, the scalar components of SS are

{Sx=Ax3Bx+Cx=8.19cm3(−2.39cm)+6.93cm=22.29cmSy=Ay3By+Cy=5.73cm3(−6.58cm)+4.00cm=29.47cm.{Sx=Ax3Bx+Cx=8.19cm3(−2.39cm)+6.93cm=22.29cmSy=Ay3By+Cy=5.73cm3(−6.58cm)+4.00cm=29.47cm.

The vector is S=Sxi^+Syj^=(22.3i^+29.5j^)cmS=Sxi^+Syj^=(22.3i^+29.5j^)cm.

Significance Having found the vector components, we can illustrate the vectors by graphing or we can compute magnitudes and direction angles, as shown in Figure 2.24. Results for the magnitudes in (b) and (c) can be compared with results for the same problems obtained with the graphical method, shown in Figure 2.14 and Figure 2.15. Notice that the analytical method produces exact results and its accuracy is not limited by the resolution of a ruler or a protractor, as it was with the graphical method used in Example 2.2 for finding this same resultant.

Vector R has magnitude 13.11. The angle between R and the positive x direction is theta sub R equals 13.9 degrees. The components of R are R sub x on the x axis and R sub y on the y axis. Vector D has magnitude 16.23. The angle between D and the positive x direction is theta sub D equals 49.3 degrees. The components of D are D sub x on the x axis and D sub y on the y axis. Vector S has magnitude 36.95. The angle between S and the positive x direction is theta sub S equals 52.9 degrees. The components of S are S sub x on the x axis and S sub y on the y axis.
Figure 2.24 Graphical illustration of the solutions obtained analytically in Example 2.9.
Check Your Understanding 2.8

Three displacement vectors AA, BB, and FF (Figure 2.13) are specified by their magnitudes A = 10.00, B = 7.00, and F = 20.00, respectively, and by their respective direction angles with the horizontal direction α=35°α=35°, β=−110°β=−110°, and φ=110°φ=110°. The physical units of the magnitudes are centimeters. Use the analytical method to find vector G=A+2BFG=A+2BF. Verify that
G = 28.15 cm and that θG=−68.65°θG=−68.65°.

Example 2.10

The Tug-of-War GameFour dogs named Astro, Balto, Clifford, and Dug play a tug-of-war game with a toy (Figure 2.25). Astro pulls on the toy in direction α=55°α=55° south of east, Balto pulls in direction β=60°β=60° east of north, and Clifford pulls in direction γ=55°γ=55° west of north. Astro pulls strongly with 160.0 units of force (N), which we abbreviate as A = 160.0 N. Balto pulls even stronger than Astro with a force of magnitude B = 200.0 N, and Clifford pulls with a force of magnitude C = 140.0 N. When Dug pulls on the toy in such a way that his force balances out the resultant of the other three forces, the toy does not move in any direction. With how big a force and in what direction must Dug pull on the toy for this to happen?

Illustration of 4 dogs pulling on a toy. The toy is at the origin of a coordinate system, with plus x aligned with east and plus y with north. Ang is pulling at an angle alpha which is 55 degrees clockwise from the plus x (east) direction. Bing is pulling at an angle beta which is 60 degrees clockwise from the plus y (north) direction. Chang is pulling at an angle gamma which is 55 degrees counterclockwise from the plus y (north) direction. Dong is pulling in an unspecified direction in the third quadrant.
Figure 2.25 Four dogs play a tug-of-war game with a toy.

StrategyWe assume that east is the direction of the positive x-axis and north is the direction of the positive y-axis. As in Example 2.9, we have to resolve the three given forces— AA (the pull from Astro), BB (the pull from Balto), and CC (the pull from Clifford)—into their scalar components and then find the scalar components of the resultant vector R=A+B+CR=A+B+C. When the pulling force DD from Dug balances out this resultant, the sum of DD and RR must give the null vector D+R=0D+R=0. This means that D=RD=R, so the pull from Dug must be antiparallel to RR.

Solution The direction angles are θA=α=−55°θA=α=−55°, θB=90°β=30°θB=90°β=30°, and θC=90°+γ=145°θC=90°+γ=145°, and substituting them into Equation 2.17 gives the scalar components of the three given forces:

{Ax=AcosθA=(160.0N)cos(−55°)=+91.8NAy=AsinθA=(160.0N)sin(−55°)=−131.1N{Bx=BcosθB=(200.0N)cos30°=+173.2NBy=BsinθB=(200.0N)sin30°=+100.0N{Cx=CcosθC=(140.0N)cos145°=−114.7NCy=CsinθC=(140.0N)sin145°=+80.3N.{Ax=AcosθA=(160.0N)cos(−55°)=+91.8NAy=AsinθA=(160.0N)sin(−55°)=−131.1N{Bx=BcosθB=(200.0N)cos30°=+173.2NBy=BsinθB=(200.0N)sin30°=+100.0N{Cx=CcosθC=(140.0N)cos145°=−114.7NCy=CsinθC=(140.0N)sin145°=+80.3N.

Now we compute scalar components of the resultant vector R=A+B+CR=A+B+C:

{Rx=Ax+Bx+Cx=+91.8N+173.2N114.7N=+150.3NRy=Ay+By+Cy=−131.1N+100.0N+80.3N=+49.2N.{Rx=Ax+Bx+Cx=+91.8N+173.2N114.7N=+150.3NRy=Ay+By+Cy=−131.1N+100.0N+80.3N=+49.2N.

The antiparallel vector to the resultant RR is

D=R=Rxi^Ryj^=(−150.3i^49.2j^)N.D=R=Rxi^Ryj^=(−150.3i^49.2j^)N.

The magnitude of Dug’s pulling force is

D=Dx2+Dy2=(−150.3)2+(−49.2)2N=158.1N.D=Dx2+Dy2=(−150.3)2+(−49.2)2N=158.1N.

The direction of Dug’s pulling force is

θ=tan−1(DyDx)=tan−1(−49.2N−150.3N)=tan−1(49.2150.3)=18.1°.θ=tan−1(DyDx)=tan−1(−49.2N−150.3N)=tan−1(49.2150.3)=18.1°.

Dug pulls in the direction 18.1°18.1° south of west because both components are negative, which means the pull vector lies in the third quadrant (Figure 2.19).

Check Your Understanding 2.9

Suppose that Balto in Example 2.10 leaves the game to attend to more important matters, but Astro, Clifford, and Dug continue playing. Astro and Clifford's pull on the toy does not change, but Dug runs around and bites on the toy in a different place. With how big a force and in what direction must Dug pull on the toy now to balance out the combined pulls from Clifford and Astro? Illustrate this situation by drawing a vector diagram indicating all forces involved.

Example 2.11

Vector Algebra Find the magnitude of the vector CC that satisfies the equation 2A6B+3C=2j^2A6B+3C=2j^, where A=i^2k^A=i^2k^ and B=j^+k^/2B=j^+k^/2.

Strategy We first solve the given equation for the unknown vector CC. Then we substitute AA and BB; group the terms along each of the three directions i^i^, j^j^, and k^k^; and identify the scalar components CxCx, CyCy, and CzCz. Finally, we substitute into Equation 2.21 to find magnitude C.

Solution

2A6B+3C=2j^3C=2j^2A+6BC=23j^23A+2B=23j^23(i^2k^)+2(j^+k^2)=23j^23i^+43k^2j^+k^=23i^+(232)j^+(43+1)k^=23i^43j^+73k^.2A6B+3C=2j^3C=2j^2A+6BC=23j^23A+2B=23j^23(i^2k^)+2(j^+k^2)=23j^23i^+43k^2j^+k^=23i^+(232)j^+(43+1)k^=23i^43j^+73k^.

The components are Cx=2/3Cx=2/3, Cy=−4/3Cy=−4/3, and Cz=7/3Cz=7/3, and substituting into Equation 2.21 gives

C=Cx2+Cy2+Cz2=(−2/3)2+(4/3)2+(7/3)2=23/3.C=Cx2+Cy2+Cz2=(−2/3)2+(4/3)2+(7/3)2=23/3.

Example 2.12

Displacement of a Skier Starting at a ski lodge, a cross-country skier goes 5.0 km north, then 3.0 km west, and finally 4.0 km southwest before taking a rest. Find his total displacement vector relative to the lodge when he is at the rest point. How far and in what direction must he ski from the rest point to return directly to the lodge?

StrategyWe assume a rectangular coordinate system with the origin at the ski lodge and with the unit vector i^i^ pointing east and the unit vector j^j^ pointing north. There are three displacements: D1D1, D2D2, and D3D3. We identify their magnitudes as D1=5.0kmD1=5.0km, D2=3.0kmD2=3.0km, and D3=4.0kmD3=4.0km. We identify their directions are the angles θ1=90°θ1=90°, θ2=180°θ2=180°, and θ3=180°+45°=225°θ3=180°+45°=225°. We resolve each displacement vector to its scalar components and substitute the components into Equation 2.25 to obtain the scalar components of the resultant displacement DD from the lodge to the rest point. On the way back from the rest point to the lodge, the displacement is B=DB=D. Finally, we find the magnitude and direction of BB.

Solution Scalar components of the displacement vectors are

{D1x=D1cosθ1=(5.0km)cos90°=0D1y=D1sinθ1=(5.0km)sin90°=5.0km{D2x=D2cosθ2=(3.0km)cos180°=−3.0kmD2y=D2sinθ2=(3.0km)sin180°=0{D3x=D3cosθ3=(4.0km)cos225°=−2.8kmD3y=D3sinθ3=(4.0km)sin225°=−2.8km.{D1x=D1cosθ1=(5.0km)cos90°=0D1y=D1sinθ1=(5.0km)sin90°=5.0km{D2x=D2cosθ2=(3.0km)cos180°=−3.0kmD2y=D2sinθ2=(3.0km)sin180°=0{D3x=D3cosθ3=(4.0km)cos225°=−2.8kmD3y=D3sinθ3=(4.0km)sin225°=−2.8km.

Scalar components of the net displacement vector are

{Dx=D1x+D2x+D3x=(03.02.8)km=−5.8kmDy=D1y+D2y+D3y=(5.0+02.8)km=+2.2km.{Dx=D1x+D2x+D3x=(03.02.8)km=−5.8kmDy=D1y+D2y+D3y=(5.0+02.8)km=+2.2km.

Hence, the skier’s net displacement vector is D=Dxi^+Dyj^=(−5.8i^+2.2j^)kmD=Dxi^+Dyj^=(−5.8i^+2.2j^)km. On the way back to the lodge, his displacement is B=D=(−5.8i^+2.2j^)km=(5.8i^2.2j^)kmB=D=(−5.8i^+2.2j^)km=(5.8i^2.2j^)km. Its magnitude is B=Bx2+By2=(5.8)2+(−2.2)2km=6.2kmB=Bx2+By2=(5.8)2+(−2.2)2km=6.2km and its direction angle is θ=tan−1(−2.2/5.8)=−20.8°θ=tan−1(−2.2/5.8)=−20.8°. Therefore, to return to the lodge, he must go 6.2 km in a direction about 21°21° south of east.

Significance Notice that no figure is needed to solve this problem by the analytical method. Figures are required when using a graphical method; however, we can check if our solution makes sense by sketching it, which is a useful final step in solving any vector problem.

Example 2.13

Displacement of a Jogger A jogger runs up a flight of 200 identical steps to the top of a hill and then runs along the top of the hill 50.0 m before he stops at a drinking fountain (Figure 2.26). His displacement vector from point A at the bottom of the steps to point B at the fountain is DAB=(−90.0i^+30.0j^)mDAB=(−90.0i^+30.0j^)m. What is the height and width of each step in the flight? What is the actual distance the jogger covers? If he makes a loop and returns to point A, what is his net displacement vector?

A coordinate system is shown with positive x to the right and positive y up. A jogger is at point A at the bottom of steps which lead up and to the left. The top of the steps is labeled as point T. At the top of the steps is a flat section extending from point T to the fountain at point B. The distance between T and B is 50 meters.
Figure 2.26 A jogger runs up a flight of steps.

StrategyThe displacement vector DABDAB is the vector sum of the jogger’s displacement vector DATDAT along the stairs (from point A at the bottom of the stairs to point T at the top of the stairs) and his displacement vector DTBDTB on the top of the hill (from point A at the top of the stairs to the fountain at point T). We must find the horizontal and the vertical components of DATDAT. If each step has width w and height h, the horizontal component of DATDAT must have a length of 200w and the vertical component must have a length of 200h. The actual distance the jogger covers is the sum of the distance he runs up the stairs and the distance of 50.0 m that he runs along the top of the hill.

Solution In the coordinate system indicated in Figure 2.26, the jogger’s displacement vector on the top of the hill is DTB=(−50.0m)i^DTB=(−50.0m)i^. His net displacement vector is

DAB=DAT+DTB.DAB=DAT+DTB.

Therefore, his displacement vector DTBDTB along the stairs is

DAT=DABDTB=(−90.0i^+30.0j^)m(−50.0m)i^=[(−90.0+50.0)i^+30.0j^)]m=(−40.0i^+30.0j^)m.DAT=DABDTB=(−90.0i^+30.0j^)m(−50.0m)i^=[(−90.0+50.0)i^+30.0j^)]m=(−40.0i^+30.0j^)m.

Its scalar components are DATx=−40.0mDATx=−40.0m and DATy=30.0mDATy=30.0m. Therefore, we must have

200w=|40.0|mand200h=30.0m.200w=|40.0|mand200h=30.0m.

Hence, the step width is w = 40.0 m/200 = 0.2 m = 20 cm, and the step height is h = 30.0 m/200 = 0.15 m = 15 cm. The distance that the jogger covers along the stairs is

DAT=DATx2+DATy2=(−40.0)2+(30.0)2m=50.0m.DAT=DATx2+DATy2=(−40.0)2+(30.0)2m=50.0m.

Thus, the actual distance he runs is DAT+DTB=50.0m+50.0m=100.0mDAT+DTB=50.0m+50.0m=100.0m. When he makes a loop and comes back from the fountain to his initial position at point A, the total distance he covers is twice this distance, or 200.0 m. However, his net displacement vector is zero, because when his final position is the same as his initial position, the scalar components of his net displacement vector are zero (Equation 2.13).

In many physical situations, we often need to know the direction of a vector. For example, we may want to know the direction of a magnetic field vector at some point or the direction of motion of an object. We have already said direction is given by a unit vector, which is a dimensionless entity—that is, it has no physical units associated with it. When the vector in question lies along one of the axes in a Cartesian system of coordinates, the answer is simple, because then its unit vector of direction is either parallel or antiparallel to the direction of the unit vector of an axis. For example, the direction of vector d=−5mi^d=−5mi^ is unit vector d^=i^d^=i^. The general rule of finding the unit vector V^V^ of direction for any vector VV is to divide it by its magnitude V:

V^=VV.V^=VV.
(2.26)

We see from this expression that the unit vector of direction is indeed dimensionless because the numerator and the denominator in Equation 2.26 have the same physical unit. In this way, Equation 2.26 allows us to express the unit vector of direction in terms of unit vectors of the axes. The following example illustrates this principle.

Example 2.14

The Unit Vector of Direction If the velocity vector of the military convoy in Example 2.8 is v=(4.000i^+3.000j^+0.100k^)km/hv=(4.000i^+3.000j^+0.100k^)km/h, what is the unit vector of its direction of motion?

Strategy The unit vector of the convoy’s direction of motion is the unit vector v^v^ that is parallel to the velocity vector. The unit vector is obtained by dividing a vector by its magnitude, in accordance with Equation 2.26.

Solution The magnitude of the vector vv is

v=vx2+vy2+vz2=4.0002+3.0002+0.1002km/h=5.001km/h.v=vx2+vy2+vz2=4.0002+3.0002+0.1002km/h=5.001km/h.

To obtain the unit vector v^v^, divide vv by its magnitude:

v^=vv=(4.000i^+3.000j^+0.100k^)km/h5.001km/h=(4.000i^+3.000j^+0.100k^)5.001=4.0005.001i^+3.0005.001j^+0.1005.001k^=(79.98i^+59.99j^+2.00k^)×10−2.v^=vv=(4.000i^+3.000j^+0.100k^)km/h5.001km/h=(4.000i^+3.000j^+0.100k^)5.001=4.0005.001i^+3.0005.001j^+0.1005.001k^=(79.98i^+59.99j^+2.00k^)×10−2.

Significance Note that when using the analytical method with a calculator, it is advisable to carry out your calculations to at least three decimal places and then round off the final answer to the required number of significant figures, which is the way we performed calculations in this example. If you round off your partial answer too early, you risk your final answer having a huge numerical error, and it may be far off from the exact answer or from a value measured in an experiment.

Check Your Understanding 2.10

Verify that vector v^v^ obtained in Example 2.14 is indeed a unit vector by computing its magnitude. If the convoy in Example 2.8 was moving across a desert flatland—that is, if the third component of its velocity was zero—what is the unit vector of its direction of motion? Which geographic direction does it represent?

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