EBK PHYSICS FOR SCIENTISTS AND ENGINEER
EBK PHYSICS FOR SCIENTISTS AND ENGINEER
6th Edition
ISBN: 9781319321710
Author: Mosca
Publisher: VST
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Chapter 4, Problem 37P
To determine

The magnitude of the acceleration of object for the given figures.

Expert Solution & Answer
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Answer to Problem 37P

The magnitude of the acceleration of object for case (a) is 4.24m/s2 and for case (b) is 8.39m/s2 .

Explanation of Solution

Given:

The given diagram is shown in Figure 1.

  EBK PHYSICS FOR SCIENTISTS AND ENGINEER, Chapter 4, Problem 37P

Figure 1

The acceleration when it acts on object on frictionless surface is, a=3.0m/s2

Formula used:

The expression for the force is given by,

  F0=ma

Here, F0 is the required force.

The net force acting on object due to F0 and F0 which are 90° apart is given by,

  Fnet=F0+F0|Fnet|= | F 0 |2+ | F 0 |2+2 | F 0 |2cosθ

The net force acting on object due to F0 and 2F0 which are 45° apart is given by,

  Fnet=F0+2F0|Fnet|= | 2 F 0 |2+ | F 0 |2+2×2× | F 0 |2cosθ

Calculation:

The value of the force is calculated as,

  F0=ma=m(3.0m/ s 2)=3mN

For the case (a),

The net force acting on object due to F0 and F0 which are 90° apart is calculated as,

  |Fnet|= | F 0 |2+ | F 0 |2+2 | F 0 |2cosθ= ( 3m )2+ ( 3m )2+2 ( 3m )2cos90°=18m2=32mN

The value of the acceleration is calculated as,

  Fnet=m|a1|32mN=m|a1||a1|=32m/s2|a1|=4.24m/s2

For the case (b),

The net force acting on object due to F0 and 2F0 which are 45° apart is calculated as,

  |Fnet|= | 2 F 0 |2+ | F 0 |2+2×2× | F 0 |2cosθ= ( ( 2 )3m )2+ ( 3m )2+2×2× ( 3m )2cos45°=( 36+9+ 36 2 )mN=8.39mN

The value of the acceleration is calculated as,

  Fnet=m|a2|8.39mN=m|a2||a2|=8.39m/s2

Conclusion:

Therefore, the magnitude of the acceleration of object for case (a) is 4.24m/s2 and for case (b) is 8.39m/s2 .

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