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University Physics Volume 1

2.2 Coordinate Systems and Components of a Vector

University Physics Volume 12.2 Coordinate Systems and Components of a Vector
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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:
  • Describe vectors in two and three dimensions in terms of their components, using unit vectors along the axes.
  • Distinguish between the vector components of a vector and the scalar components of a vector.
  • Explain how the magnitude of a vector is defined in terms of the components of a vector.
  • Identify the direction angle of a vector in a plane.
  • Explain the connection between polar coordinates and Cartesian coordinates in a plane.

Vectors are usually described in terms of their components in a coordinate system. Even in everyday life we naturally invoke the concept of orthogonal projections in a rectangular coordinate system. For example, if you ask someone for directions to a particular location, you will more likely be told to go 40 km east and 30 km north than 50 km in the direction 37°37° north of east.

In a rectangular (Cartesian) xy-coordinate system in a plane, a point in a plane is described by a pair of coordinates (x, y). In a similar fashion, a vector AA in a plane is described by a pair of its vector coordinates. The x-coordinate of vector AA is called its x-component and the y-coordinate of vector AA is called its y-component. The vector x-component is a vector denoted by AxAx. The vector y-component is a vector denoted by AyAy. In the Cartesian system, the x and y vector components of a vector are the orthogonal projections of this vector onto the x- and y-axes, respectively. In this way, following the parallelogram rule for vector addition, each vector on a Cartesian plane can be expressed as the vector sum of its vector components:

A=Ax+Ay.A=Ax+Ay.
(2.10)

As illustrated in Figure 2.16, vector AA is the diagonal of the rectangle where the x-component AxAx is the side parallel to the x-axis and the y-component AyAy is the side parallel to the y-axis. Vector component AxAx is orthogonal to vector component AyAy.

Vector A is shown in the x y coordinate system and extends from point b at A’s tail to point e and its head. Vector A points up and to the right. Unit vectors I hat and j hat are small vectors pointing in the x and y directions, respectively, and are at right angles to each other. The x component of vector A is a vector pointing horizontally from the point b to a point directly below point e at the tip of vector A. On the x axis, we see that the vector A sub x extends from x sub b to x sub e and is equal to magnitude A sub x times I hat. The magnitude A sub x equals x sub e minus x sub b. The y component of vector A is a vector pointing vertically from point b to a point directly to the left of point e at the tip of vector A. On the y axis, we see that the vector A sub y extends from y sub b to y sub e and is equal to magnitude A sub y times j hat. The magnitude A sub y equals y sub e minus y sub b.
Figure 2.16 Vector AA in a plane in the Cartesian coordinate system is the vector sum of its vector x- and y-components. The x-vector component AxAx is the orthogonal projection of vector AA onto the x-axis. The y-vector component AyAy is the orthogonal projection of vector AA onto the y-axis. The numbers AxAx and AyAy that multiply the unit vectors are the scalar components of the vector.

It is customary to denote the positive direction on the x-axis by the unit vector i^i^ and the positive direction on the y-axis by the unit vector j^j^. Unit vectors of the axes, i^i^ and j^j^, define two orthogonal directions in the plane. As shown in Figure 2.16, the x- and y- components of a vector can now be written in terms of the unit vectors of the axes:

{Ax=Axi^Ay=Ayj^.{Ax=Axi^Ay=Ayj^.
(2.11)

The vectors AxAx and AyAy defined by Equation 2.11 are the vector components of vector AA. The numbers AxAx and AyAy that define the vector components in Equation 2.11 are the scalar components of vector AA. Combining Equation 2.10 with Equation 2.11, we obtain the component form of a vector:

A=Axi^+Ayj^.A=Axi^+Ayj^.
(2.12)

If we know the coordinates b(xb,yb)b(xb,yb) of the origin point of a vector (where b stands for “beginning”) and the coordinates e(xe,ye)e(xe,ye) of the end point of a vector (where e stands for “end”), we can obtain the scalar components of a vector simply by subtracting the origin point coordinates from the end point coordinates:

{Ax=xexbAy=yeyb.{Ax=xexbAy=yeyb.
(2.13)

Example 2.3

Displacement of a Mouse Pointer A mouse pointer on the display monitor of a computer at its initial position is at point (6.0 cm, 1.6 cm) with respect to the lower left-side corner. If you move the pointer to an icon located at point (2.0 cm, 4.5 cm), what is the displacement vector of the pointer?

Strategy The origin of the xy-coordinate system is the lower left-side corner of the computer monitor. Therefore, the unit vector i^i^ on the x-axis points horizontally to the right and the unit vector j^j^ on the y-axis points vertically upward. The origin of the displacement vector is located at point b(6.0, 1.6) and the end of the displacement vector is located at point e(2.0, 4.5). Substitute the coordinates of these points into Equation 2.13 to find the scalar components DxDx and DyDy of the displacement vector DD. Finally, substitute the coordinates into Equation 2.12 to write the displacement vector in the vector component form.

Solution We identify xb=6.0xb=6.0, xe=2.0xe=2.0, yb=1.6yb=1.6, and ye=4.5ye=4.5, where the physical unit is 1 cm. The scalar x- and y-components of the displacement vector are

Dx=xexb=(2.06.0)cm=−4.0cm,Dy=yeyb=(4.51.6)cm=+2.9cm.Dx=xexb=(2.06.0)cm=−4.0cm,Dy=yeyb=(4.51.6)cm=+2.9cm.

The vector component form of the displacement vector is

D=Dxi^+Dyj^=(−4.0cm)i^+(2.9cm)j^=(−4.0i^+2.9j^)cm.D=Dxi^+Dyj^=(−4.0cm)i^+(2.9cm)j^=(−4.0i^+2.9j^)cm.
(2.14)

This solution is shown in Figure 2.17.

Vector D extends from coordinates 6.0, 1.6 to coordinates 2.0, 4.5. Vector D equals vector D sub x plus vector D sub y. D sub x equals minus 4.0 I hat, and extends from x=6.0 to x =2.0. The magnitude D sub x equals 2.0-6.0 = -4.0. D sub y equals plus 2.9 j hat, and extends from y=1.6 to y=4.5. The magnitude D sub y equals 4.5 − 1.6.
Figure 2.17 The graph of the displacement vector. The vector points from the origin point at b to the end point at e.

Significance Notice that the physical unit—here, 1 cm—can be placed either with each component immediately before the unit vector or globally for both components, as in Equation 2.14. Often, the latter way is more convenient because it is simpler.

The vector x-component Dx=−4.0i^=4.0(i^)Dx=−4.0i^=4.0(i^) of the displacement vector has the magnitude |Dx|=|4.0||i^|=4.0|Dx|=|4.0||i^|=4.0 because the magnitude of the unit vector is |i^|=1|i^|=1. Notice, too, that the direction of the x-component is i^i^, which is antiparallel to the direction of the +x-axis; hence, the x-component vector DxDx points to the left, as shown in Figure 2.17. The scalar x-component of vector DD is Dx=−4.0Dx=−4.0.

Similarly, the vector y-component Dy=+2.9j^Dy=+2.9j^ of the displacement vector has magnitude |Dy|=|2.9||j^|=2.9|Dy|=|2.9||j^|=2.9 because the magnitude of the unit vector is |j^|=1|j^|=1. The direction of the y-component is +j^+j^, which is parallel to the direction of the +y-axis. Therefore, the y-component vector DyDy points up, as seen in Figure 2.17. The scalar y-component of vector DD is Dy=+2.9Dy=+2.9. The displacement vector DD is the resultant of its two vector components.

The vector component form of the displacement vector Equation 2.14 tells us that the mouse pointer has been moved on the monitor 4.0 cm to the left and 2.9 cm upward from its initial position.

Check Your Understanding 2.4

A blue fly lands on a sheet of graph paper at a point located 10.0 cm to the right of its left edge and 8.0 cm above its bottom edge and walks slowly to a point located 5.0 cm from the left edge and 5.0 cm from the bottom edge. Choose the rectangular coordinate system with the origin at the lower left-side corner of the paper and find the displacement vector of the fly. Illustrate your solution by graphing.

When we know the scalar components AxAx and AyAy of a vector AA, we can find its magnitude A and its direction angle θAθA. The direction angle—or direction, for short—is the angle the vector forms with the positive direction on the x-axis. The angle θAθA is measured in the counterclockwise direction from the +x-axis to the vector (Figure 2.18). Because the lengths A, AxAx, and AyAy form a right triangle, they are related by the Pythagorean theorem:

A2=Ax2+Ay2A=Ax2+Ay2.A2=Ax2+Ay2A=Ax2+Ay2.
(2.15)

This equation works even if the scalar components of a vector are negative. The direction angle θAθA of a vector is defined via the tangent function of angle θAθA in the triangle shown in Figure 2.18:

tanθ=AyAxtanθ=AyAx
(2.16)
Vector A has horizontal x component A sub x equal to magnitude A sub x I hat and vertical y component A sub y equal to magnitude A sub y j hat. Vector A and the components form a right triangle with sides length magnitude A sub x and magnitude A sub y and hypotenuse magnitude A equal to the square root of A sub x squared plus A sub y squared. The angle between the horizontal side A sub x and the hypotenuse A is theta sub A.
Figure 2.18 When the vector lies either in the first quadrant or in the fourth quadrant, where component AxAxis positive (Figure 2.19), the direction angle θAθA in Equation 2.16) is identical to the angle θθ.

When the vector lies either in the first quadrant or in the fourth quadrant, where component AxAx is positive (Figure 2.19), the angle θθ in Equation 2.16 is identical to the direction angle θAθA. For vectors in the fourth quadrant, angle θθ is negative, which means that for these vectors, direction angle θAθA is measured clockwise from the positive x-axis. Similarly, for vectors in the second quadrant, angle θθ is negative. When the vector lies in either the second or third quadrant, where component AxAx is negative, the direction angle is θA=θ+180°θA=θ+180° (Figure 2.19).

Figure I shows vector A in the first quadrant (pointing up and right.) It has positive x and y components A sub x and A sub y, and the angle theta sub A measured counterclockwise from the positive x axis is smaller than 90 degrees. Figure II shows vector A in the first second (pointing up and left.) It has negative x and positive y components A sub x and A sub y. The angle theta sub A measured counterclockwise from the positive x axis is larger than 90 degrees but less than 180 degrees. The angle theta, measured clockwise from the negative x axis, is smaller than 90 degrees. Figure III shows vector A in the third quadrant (pointing down and left.) It has negative x and y components A sub x and A sub y, and the angle theta sub A measured counterclockwise from the positive x axis is larger than 180 degrees and smaller than 270 degrees. The angle theta, measured counterclockwise from the negative x axis, is smaller than 90 degrees. Figure IV shows vector A in the fourth quadrant (pointing down and right.) It has positive x and negative y components A sub x and A sub y, and the angle theta sub A measured clockwise from the positive x axis is smaller than 90 degrees.
Figure 2.19 Scalar components of a vector may be positive or negative. Vectors in the first quadrant (I) have both scalar components positive and vectors in the third quadrant have both scalar components negative. For vectors in quadrants II and III, the direction angle of a vector is θA=θ+180°θA=θ+180°.

Example 2.4

Magnitude and Direction of the Displacement VectorYou move a mouse pointer on the display monitor from its initial position at point (6.0 cm, 1.6 cm) to an icon located at point (2.0 cm, 4.5 cm). What are the magnitude and direction of the displacement vector of the pointer?

Strategy In Example 2.3, we found the displacement vector DD of the mouse pointer (see Equation 2.14). We identify its scalar components Dx=−4.0cmDx=−4.0cm and Dy=+2.9cmDy=+2.9cm and substitute into Equation 2.15 and Equation 2.16 to find the magnitude D and direction θDθD, respectively.

Solution The magnitude of vector DD is

D=Dx2+Dy2=(−4.0cm)2+(2.9cm)2=(4.0)2+(2.9)2cm=4.9cm.D=Dx2+Dy2=(−4.0cm)2+(2.9cm)2=(4.0)2+(2.9)2cm=4.9cm.

The direction angle is

tanθ=DyDx=+2.9cm−4.0cm=−0.725θ=tan−1(−0.725)=−35.9°.tanθ=DyDx=+2.9cm−4.0cm=−0.725θ=tan−1(−0.725)=−35.9°.

Vector DD lies in the second quadrant, so its direction angle is

θD=θ+180°=−35.9°+180°=144.1°.θD=θ+180°=−35.9°+180°=144.1°.
Check Your Understanding 2.5

If the displacement vector of a blue fly walking on a sheet of graph paper is D=(−5.00i^3.00j^)cmD=(−5.00i^3.00j^)cm, find its magnitude and direction.

In many applications, the magnitudes and directions of vector quantities are known and we need to find the resultant of many vectors. For example, imagine 400 cars moving on the Golden Gate Bridge in San Francisco in a strong wind. Each car gives the bridge a different push in various directions and we would like to know how big the resultant push can possibly be. We have already gained some experience with the geometric construction of vector sums, so we know the task of finding the resultant by drawing the vectors and measuring their lengths and angles may become intractable pretty quickly, leading to huge errors. Worries like this do not appear when we use analytical methods. The very first step in an analytical approach is to find vector components when the direction and magnitude of a vector are known.

Let us return to the right triangle in Figure 2.18. The quotient of the adjacent side AxAx to the hypotenuse A is the cosine function of direction angle θAθA, Ax/A=cosθAAx/A=cosθA, and the quotient of the opposite side AyAy to the hypotenuse A is the sine function of θAθA, Ay/A=sinθAAy/A=sinθA. When magnitude A and direction θAθA are known, we can solve these relations for the scalar components:

{Ax=AcosθAAy=AsinθA.{Ax=AcosθAAy=AsinθA.
(2.17)

When calculating vector components with Equation 2.17, care must be taken with the angle. The direction angle θAθA of a vector is the angle measured counterclockwise from the positive direction on the x-axis to the vector. The clockwise measurement gives a negative angle.

Example 2.5

Components of Displacement Vectors A rescue party for a missing child follows a search dog named Trooper. Trooper wanders a lot and makes many trial sniffs along many different paths. Trooper eventually finds the child and the story has a happy ending, but his displacements on various legs seem to be truly convoluted. On one of the legs he walks 200.0 m southeast, then he runs north some 300.0 m. On the third leg, he examines the scents carefully for 50.0 m in the direction 30°30° west of north. On the fourth leg, Trooper goes directly south for 80.0 m, picks up a fresh scent and turns 23°23° west of south for 150.0 m. Find the scalar components of Trooper’s displacement vectors and his displacement vectors in vector component form for each leg.

Strategy Let’s adopt a rectangular coordinate system with the positive x-axis in the direction of geographic east, with the positive y-direction pointed to geographic north. Explicitly, the unit vector i^i^ of the x-axis points east and the unit vector j^j^ of the y-axis points north. Trooper makes five legs, so there are five displacement vectors. We start by identifying their magnitudes and direction angles, then we use Equation 2.17 to find the scalar components of the displacements and Equation 2.12 for the displacement vectors.

Solution On the first leg, the displacement magnitude is L1=200.0mL1=200.0m and the direction is southeast. For direction angle θ1θ1 we can take either 45°45° measured clockwise from the east direction or 45°+270°45°+270° measured counterclockwise from the east direction. With the first choice, θ1=−45°θ1=−45°. With the second choice, θ1=+315°θ1=+315°. We can use either one of these two angles. The components are

L1x=L1cosθ1=(200.0m)cos315°=141.4m,L1y=L1sinθ1=(200.0m)sin315°=−141.4m.L1x=L1cosθ1=(200.0m)cos315°=141.4m,L1y=L1sinθ1=(200.0m)sin315°=−141.4m.

The displacement vector of the first leg is

L1=L1xi^+L1yj^=(141.4i^141.4j^)m.L1=L1xi^+L1yj^=(141.4i^141.4j^)m.

On the second leg of Trooper’s wanderings, the magnitude of the displacement is L2=300.0mL2=300.0m and the direction is north. The direction angle is θ2=+90°θ2=+90°. We obtain the following results:

L2x=L2cosθ2=(300.0m)cos90°=0.0,L2y=L2sinθ2=(300.0m)sin90°=300.0m,L2=L2xi^+L2yj^=(300.0m)j^.L2x=L2cosθ2=(300.0m)cos90°=0.0,L2y=L2sinθ2=(300.0m)sin90°=300.0m,L2=L2xi^+L2yj^=(300.0m)j^.

On the third leg, the displacement magnitude is L3=50.0mL3=50.0m and the direction is 30°30° west of north. The direction angle measured counterclockwise from the eastern direction is θ3=30°+90°=+120°θ3=30°+90°=+120°. This gives the following answers:

L3x=L3cosθ3=(50.0m)cos120°=−25.0m,L3y=L3sinθ3=(50.0m)sin120°=+43.3m,L3=L3xi^+L3yj^=(−25.0i^+43.3j^)m.L3x=L3cosθ3=(50.0m)cos120°=−25.0m,L3y=L3sinθ3=(50.0m)sin120°=+43.3m,L3=L3xi^+L3yj^=(−25.0i^+43.3j^)m.

On the fourth leg of the excursion, the displacement magnitude is L4=80.0mL4=80.0m and the direction is south. The direction angle can be taken as either θ4=−90°θ4=−90° or θ4=+270°θ4=+270°. We obtain

L4x=L4cosθ4=(80.0m)cos(−90°)=0,L4y=L4sinθ4=(80.0m)sin(−90°)=−80.0m,L4=L4xi^+L4yj^=(−80.0m)j^.L4x=L4cosθ4=(80.0m)cos(−90°)=0,L4y=L4sinθ4=(80.0m)sin(−90°)=−80.0m,L4=L4xi^+L4yj^=(−80.0m)j^.

On the last leg, the magnitude is L5=150.0mL5=150.0m and the angle is θ5=−23°+270°=+247°θ5=−23°+270°=+247° (23°(23° west of south), which gives

L5x=L5cosθ5=(150.0m)cos247°=−58.6m,L5y=L5sinθ5=(150.0m)sin247°=−138.1m,L5=L5xi^+L5yj^=(−58.6i^138.1j^)m.L5x=L5cosθ5=(150.0m)cos247°=−58.6m,L5y=L5sinθ5=(150.0m)sin247°=−138.1m,L5=L5xi^+L5yj^=(−58.6i^138.1j^)m.
Check Your Understanding 2.6

If Trooper runs 20 m west before taking a rest, what is his displacement vector?

Polar Coordinates

To describe locations of points or vectors in a plane, we need two orthogonal directions. In the Cartesian coordinate system these directions are given by unit vectors i^i^ and j^j^ along the x-axis and the y-axis, respectively. The Cartesian coordinate system is very convenient to use in describing displacements and velocities of objects and the forces acting on them. However, it becomes cumbersome when we need to describe the rotation of objects. When describing rotation, we usually work in the polar coordinate system.

In the polar coordinate system, the location of point P in a plane is given by two polar coordinates (Figure 2.20). The first polar coordinate is the radial coordinate r, which is the distance of point P from the origin. The second polar coordinate is an angle φφ that the radial vector makes with some chosen direction, usually the positive x-direction. In polar coordinates, angles are measured in radians, or rads. The radial vector is attached at the origin and points away from the origin to point P. This radial direction is described by a unit radial vector r^r^. The second unit vector t^t^ is a vector orthogonal to the radial direction r^r^. The positive +t^+t^ direction indicates how the angle φφ changes in the counterclockwise direction. In this way, a point P that has coordinates (x, y) in the rectangular system can be described equivalently in the polar coordinate system by the two polar coordinates (r,φ)(r,φ). Equation 2.17 is valid for any vector, so we can use it to express the x- and y-coordinates of vector rr. In this way, we obtain the connection between the polar coordinates and rectangular coordinates of point P:

{x=rcosφy=rsinφ.{x=rcosφy=rsinφ.
(2.18)
Vector r points from the origin of the x y coordinate system to point P. The angle between the vector r and the positive x direction is phi. X equals r cosine phi and y equals r sine phi. Extending a line in the direction of r vector past point P, a unit vector r hat is drawn in the same direction as r. A unit vector t hat is perpendicular to r hat, pointing 90 degrees counterclockwise to r hat.
Figure 2.20 Using polar coordinates, the unit vector r^r^ defines the positive direction along the radius r (radial direction) and, orthogonal to it, the unit vector t^t^ defines the positive direction of rotation by the angle φφ.

Example 2.6

Polar Coordinates A treasure hunter finds one silver coin at a location 20.0 m away from a dry well in the direction 20°20° north of east and finds one gold coin at a location 10.0 m away from the well in the direction 20°20° north of west. What are the polar and rectangular coordinates of these findings with respect to the well?

Strategy The well marks the origin of the coordinate system and east is the +x-direction. We identify radial distances from the locations to the origin, which are rS=20.0mrS=20.0m (for the silver coin) and rG=10.0mrG=10.0m (for the gold coin). To find the angular coordinates, we convert 20°20° to radians: 20°=π20/180=π/920°=π20/180=π/9. We use Equation 2.18 to find the x- and y-coordinates of the coins.

Solution The angular coordinate of the silver coin is φS=π/9φS=π/9, whereas the angular coordinate of the gold coin is φG=ππ/9=8π/9φG=ππ/9=8π/9. Hence, the polar coordinates of the silver coin are (rS,φS)=(20.0m,π/9)(rS,φS)=(20.0m,π/9) and those of the gold coin are (rG,φG)=(10.0m,8π/9)(rG,φG)=(10.0m,8π/9). We substitute these coordinates into Equation 2.18 to obtain rectangular coordinates. For the gold coin, the coordinates are

{xG=rGcosφG=(10.0m)cos8π/9=−9.4myG=rGsinφG=(10.0m)sin8π/9=3.4m(xG,yG)=(−9.4m,3.4m).{xG=rGcosφG=(10.0m)cos8π/9=−9.4myG=rGsinφG=(10.0m)sin8π/9=3.4m(xG,yG)=(−9.4m,3.4m).

For the silver coin, the coordinates are

{xS=rScosφS=(20.0m)cosπ/9=18.9myS=rSsinφS=(20.0m)sinπ/9=6.8m(xS,yS)=(18.9m,6.8m).{xS=rScosφS=(20.0m)cosπ/9=18.9myS=rSsinφS=(20.0m)sinπ/9=6.8m(xS,yS)=(18.9m,6.8m).

Vectors in Three Dimensions

To specify the location of a point in space, we need three coordinates (x, y, z), where coordinates x and y specify locations in a plane, and coordinate z gives a vertical position above or below the plane. Three-dimensional space has three orthogonal directions, so we need not two but three unit vectors to define a three-dimensional coordinate system. In the Cartesian coordinate system, the first two unit vectors are the unit vector of the x-axis i^i^ and the unit vector of the y-axis j^j^. The third unit vector k^k^ is the direction of the z-axis (Figure 2.21). The order in which the axes are labeled, which is the order in which the three unit vectors appear, is important because it defines the orientation of the coordinate system. The order x-y-z, which is equivalent to the order i^i^ - j^j^ - k^k^, defines the standard right-handed coordinate system (positive orientation).

The x y z coordinate system, with unit vectors I hat, j hat and k hat respectively. I hat points out at us, j hat points to the right, and k hat points up the page. The unit vectors form the sides of a cube.
Figure 2.21 Three unit vectors define a Cartesian system in three-dimensional space. The order in which these unit vectors appear defines the orientation of the coordinate system. The order shown here defines the right-handed orientation.

In three-dimensional space, vector AA has three vector components: the x-component Ax=Axi^Ax=Axi^, which is the part of vector AA along the x-axis; the y-component Ay=Ayj^Ay=Ayj^, which is the part of AA along the y-axis; and the z-component Az=Azk^Az=Azk^, which is the part of the vector along the z-axis. A vector in three-dimensional space is the vector sum of its three vector components (Figure 2.22):

A=Axi^+Ayj^+Azk^.A=Axi^+Ayj^+Azk^.
(2.19)

If we know the coordinates of its origin b(xb,yb,zb)b(xb,yb,zb) and of its end e(xe,ye,ze)e(xe,ye,ze), its scalar components are obtained by taking their differences: AxAx and AyAy are given by Equation 2.13 and the z-component is given by

Az=zezb.Az=zezb.
(2.20)

Magnitude A is obtained by generalizing Equation 2.15 to three dimensions:

A=Ax2+Ay2+Az2.A=Ax2+Ay2+Az2.
(2.21)

This expression for the vector magnitude comes from applying the Pythagorean theorem twice. As seen in Figure 2.22, the diagonal in the xy-plane has length Ax2+Ay2Ax2+Ay2 and its square adds to the square Az2Az2 to give A2A2. Note that when the z-component is zero, the vector lies entirely in the xy-plane and its description is reduced to two dimensions.

Vector A in the x y z coordinate system extends from the origin. Vector A equals the sum of vectors A sub x, A sub y and A sub z. Vector A sub x is the x component along the x axis and has length A sub x I hat. Vector A sub y is the y component along the y axis and has length A sub y j hat. Vector A sub z is the z component along the z axis and has length A sub x k hat. The components form the sides of a rectangular box with sides length A sub x, A sub y, and A sub z.
Figure 2.22 A vector in three-dimensional space is the vector sum of its three vector components.

Example 2.7

Takeoff of a Drone During a takeoff of IAI Heron (Figure 2.23), its position with respect to a control tower is 100 m above the ground, 300 m to the east, and 200 m to the north. One minute later, its position is 250 m above the ground, 1200 m to the east, and 2100 m to the north. What is the drone’s displacement vector with respect to the control tower? What is the magnitude of its displacement vector?

A photo of a drone plane.
Figure 2.23 The drone IAI Heron in flight. (credit: SSgt Reynaldo Ramon, USAF)

Strategy We take the origin of the Cartesian coordinate system as the control tower. The direction of the +x-axis is given by unit vector i^i^ to the east, the direction of the +y-axis is given by unit vector j^j^ to the north, and the direction of the +z-axis is given by unit vector k^k^, which points up from the ground. The drone’s first position is the origin (or, equivalently, the beginning) of the displacement vector and its second position is the end of the displacement vector.

SolutionWe identify b(300.0 m, 200.0 m, 100.0 m) and e(1200 m, 2100 m, 250 m), and use Equation 2.13 and Equation 2.20 to find the scalar components of the drone’s displacement vector:

{Dx=xexb=1200.0m300.0m=900.0m,Dy=yeyb=2100.0m200.0m=1900.0m,Dz=zezb=250.0m100.0m=150.0m.{Dx=xexb=1200.0m300.0m=900.0m,Dy=yeyb=2100.0m200.0m=1900.0m,Dz=zezb=250.0m100.0m=150.0m.

We substitute these components into Equation 2.19 to find the displacement vector:

D=Dxi^+Dyj^+Dzk^=900.0mi^+1900.0mj^+150.0mk^=(0.90i^+1.90j^+0.15k^)km.D=Dxi^+Dyj^+Dzk^=900.0mi^+1900.0mj^+150.0mk^=(0.90i^+1.90j^+0.15k^)km.

We substitute into Equation 2.21 to find the magnitude of the displacement:

D=Dx2+Dy2+Dz2=(0.90km)2+(1.90km)2+(0.15km)2=2.11km.D=Dx2+Dy2+Dz2=(0.90km)2+(1.90km)2+(0.15km)2=2.11km.
Check Your Understanding 2.7

If the average velocity vector of the drone in the displacement in Example 2.7 is u=(15.0i^+31.7j^+2.5k^)m/su=(15.0i^+31.7j^+2.5k^)m/s, what is the magnitude of the drone’s velocity vector?

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