Physics: Principles with Applications
Physics: Principles with Applications
6th Edition
ISBN: 9780130606204
Author: Douglas C. Giancoli
Publisher: Prentice Hall
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Chapter 3, Problem 13P
Solution

  1. AB+C
  2. A+BC
  3. CAB

Solution:

  1. The magnitude AB+C is 62.6 with a direction of 328.9o counterclockwise from + x -axis.
  2. The magnitude A+BC is 77.4 with a direction of 71.9o counterclockwise from +x-axis.
  3. The magnitude CAB of resultant vector is 77.4 with a direction of 251.9o counterclockwise from +x-axis.

Explanation:

Physics: Principles with Applications, Chapter 3, Problem 13P

Formula used:

Given: The magnitude |A| is 44.0 with a direction θ'A of 28.0˚ counterclockwise from the positive x -axis. For |B| the magnitude is 26.5with a direction θ'B of 56.0˚ clockwise from the negative x -axis and the magnitude of |C| is 31.0 with a direction of θ'C of 270˚ on the negative y-axis.

Formula used:

The resultant R in terms of its components can be defined as:

The x-component of the resultant Rx can be obtained using:

  Rx=Ax+Bx+Cx=|A|cos(θA)+|B|cos(θB)+|C|cos(θC)

The y-component of the resultant Ry can be obtained using:

  Ry=Ay+By+Cy=|A|sin(θA)+|B|sin(θB)+|C|sin(θC)

Where:

  • θA , θB and θC are the directions for vectors A,BandC respectively measured counterclockwise from the + x -axis.
  • Ay , By and Cy are the y-components for vectors A,BandC respectively.
  • Ax , Bx and Cx are the x-components for vectors A,BandC respectively.
  • |A| , A,BandC |B| and |C| are the magnitudes for vectors respectively.

The resultant |R| of the sum of three vectors A+B+C in terms of its magnitude |R| can be obtained using:

  |R|=Rx2+Ry2

Similarly, the direction (angle) θR is defined as:

  θR=tan1(RyRx)

Where:

  • Rx is the x -component of the resultant vector R .
  • Ry is the y -component of the resultant vector R .

Calculation:

x -component of A

  Ax=44cos28o=38.8

y -component of A

  Ay=44sin28o=20.6

x -component of B

  Bx=26.5cos( 180o 56o)=14.8

y -component of B

  By=26.5sin( 180o 56o)=22

x -component of C

  Cx=31cos( 270o)=0

y -component of C

  Cy=31sin( 270o)=31

a. R=AB+C

  Rx=AxBx+CxRx=|A|cos(θA)|B|cos(θB)+|C|cos(θC)Rx=(44.0)cos(28o)(26.5)cos( 180o 56o)+(31.0)cos(270o)Rx=53.6

  Ry=AyBy+CyRy=|A|sin(θA)|B|sin(θB)+|C|sin(θC)Ry=44.0sin(28.0)(26.5)sin(18056.0)+(31.0)sin(270)Ry=32.3

The magnitude can be calculated as

  |R|=Rx2+Ry2= ( 53.6 )2+ ( 32.3 )2=62.6

  θR=tan1( 32.3 53.6)θR=31.1oθR=360o31.1oθR=328.9o

b. A+BC

  Rx=Ax+BxCxRx=|A|cos(θA)+|B|cos(θB)|C|cos(θC)Rx=(44.0)cos(28o)+(26.5)cos( 180o 56o)(31.0)cos(270o)Rx=24

  Ry=Ay+ByCyRy=|A|sin(θA)+|B|sin(θB)|C|sin(θC)Ry=44.0sin(28.0)+(26.5)sin(18056.0)(31.0)sin(270)Ry=73.6

The magnitude can be calculated as

  |R|=Rx2+Ry2= ( 24 )2+ ( 73.6 )2=77.4

  θR=tan1( 73.6 24)θR=71.9o

c. CAB

This is equal to the negative of previous part.

  CAB=(A+BC)

Thus, there will be no change in the resultant but its direction would be opposite.

  |R|=|CAB|=77.4θ=180o+71.9o=251.9o

Conclusion:

  1. The magnitude AB+C is 62.6 with a direction of 328.9o counterclockwise from + x -axis.
  2. The magnitude A+BC is 77.4 with a direction of 71.9o counterclockwise from +x-axis.
  3. The magnitude CAB of resultant vector is 77.4 with a direction of 251.9o counterclockwise from +x-axis.

Chapter 3 Solutions

Physics: Principles with Applications

Ch. 3 - Prob. 11QCh. 3 - Prob. 12QCh. 3 - Prob. 13QCh. 3 - A projectile is launched at an upward angle of 300...Ch. 3 - Prob. 15QCh. 3 - Two cannonballs, A and B, are fired from the...Ch. 3 - 18. A person sitting in an enclosed train car,...Ch. 3 - Prob. 18QCh. 3 - Prob. 19QCh. 3 - Prob. 20QCh. 3 - A car is driven 225 km west and then 98 km...Ch. 3 - A delivery truck travels 21 blocks north, 16...Ch. 3 - If Vx=9.80 units and Vy=6.40 units, determine the...Ch. 3 - Graphically determine the resultant of the...Ch. 3 - V is a vector 24.8 units in magnitude and points...Ch. 3 - Vector V is 6.6 using long and points along the...Ch. 3 - Figure 3-33 shows two vectors, A and B , whose...Ch. 3 - Prob. 8PCh. 3 - Three vectors are shown in Fig. 3-35 Q. Their...Ch. 3 - (a) given the vectors A and B shown in Fig. 3-35,...Ch. 3 - Determine the vector AC , given the vectors A and...Ch. 3 - For the vectors shown in Fig. 3—35, determine (a)...Ch. 3 - Prob. 13PCh. 3 - Prob. 14PCh. 3 - Prob. 15PCh. 3 - Prob. 16PCh. 3 - Prob. 17PCh. 3 - 17. (l) A tiger leaps horizontally from a...Ch. 3 - 18. (l) A diver running 2.5 m/s dives out...Ch. 3 - Prob. 20PCh. 3 - Prob. 21PCh. 3 - Prob. 22PCh. 3 - Prob. 23PCh. 3 - Prob. 24PCh. 3 - Prob. 25PCh. 3 - Prob. 26PCh. 3 - Prob. 27PCh. 3 - Prob. 28PCh. 3 - Prob. 29PCh. 3 - Prob. 30PCh. 3 - Prob. 31PCh. 3 - Prob. 32PCh. 3 - Prob. 33PCh. 3 - Prob. 34PCh. 3 - Prob. 35PCh. 3 - Prob. 36PCh. 3 - Prob. 37PCh. 3 - Prob. 38PCh. 3 - Prob. 39PCh. 3 - Prob. 40PCh. 3 - Prob. 41PCh. 3 - Prob. 42PCh. 3 - Prob. 43PCh. 3 - Prob. 44PCh. 3 - Prob. 45PCh. 3 - Prob. 46PCh. 3 - Prob. 47PCh. 3 - Prob. 48PCh. 3 - Prob. 49PCh. 3 - Prob. 50PCh. 3 - Prob. 51PCh. 3 - Prob. 52PCh. 3 - Prob. 53GPCh. 3 - Prob. 54GPCh. 3 - Prob. 55GPCh. 3 - Prob. 56GPCh. 3 - Prob. 57GPCh. 3 - Prob. 58GPCh. 3 - Prob. 59GPCh. 3 - Prob. 60GPCh. 3 - Prob. 61GPCh. 3 - Prob. 62GPCh. 3 - Prob. 63GPCh. 3 - Prob. 64GPCh. 3 - Prob. 65GPCh. 3 - Prob. 66GPCh. 3 - Prob. 67GPCh. 3 - Prob. 68GPCh. 3 - Prob. 69GPCh. 3 - Prob. 70GPCh. 3 - Prob. 71GPCh. 3 - Prob. 72GPCh. 3 - Prob. 73GPCh. 3 - Prob. 74GPCh. 3 - Prob. 75GP
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