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 82P

(a)

To determine

The acceleration of each of the blocks.

(a)

Expert Solution
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Explanation of Solution

Introduction:

Newton’s third law states that for every action, there is an equal and opposite reaction. It means for every interaction between two objects, there is a pair of forces that act on both the objects.

The free-body diagram for both the block is shown in Figure 1

  EBK PHYSICS FOR SCIENTISTS AND ENGINEER, Chapter 4, Problem 82P

Figure 1

In the above diagram is shown that the force on the block m1 is the gravitational force and the force due to tension on the string. In the same way, the force on the block of mass m2 is the gravitational force and the force due to tension on the string.

Conclusion:

Therefore, the free body diagram for both blocks is shown in Figure 1

(b)

To determine

The expression for the acceleration of the block and the tension in the string.

(b)

Expert Solution
Check Mark

Answer to Problem 82P

The expression for the acceleration is (m1m2)gm1+m2 and the expression for the tension in the string is 2m1m2m1+m2g .

Explanation of Solution

Calculation:

The equation of motion for the block m1 is given by,

  m1a=m1gTT=m1gm1a

The equation of motion for the block m2 is given by,

  m2a=Tm2gT=m2a+m2g

The expression for the acceleration of the block is evaluated as,

  m1gm1a=m2am2gm1gm2g=m1a+m2aa=( m 1 m 2 )gm1+m2

The expression for the tension in the string is evaluated a,

  T=m1gm1a=m1gm1( m 1 m 2 )gm1+m2=m1g[ m 1+ m 2 m 1+ m 2 m 1+ m 2]=2m1m2m1+m2g

Conclusion:

Therefore, the expression for the acceleration is (m1m2)gm1+m2 and the expression for the tension in the string is 2m1m2m1+m2g .

(c)

To determine

Whether the expressions for the tension and the acceleration gives the plausible results if m1=m2 in the limit that m1m2 and in the limit that m1m2 .

(c)

Expert Solution
Check Mark

Answer to Problem 82P

The value of tension for the condition m1=m2 is m1g and the value of acceleration is 0 . For the value of m1m2 the plausible value of the tension is 2m2g and the value of acceleration is g . For the value of m1m2 , the plausible value of the tension is 2m1g and the value of acceleration is g .

Explanation of Solution

Calculation:

For the value of m1=m2 the value of m2 is negligible and the expression for the acceleration can be evaluated as,

  a=( m 1 m 2 )m1+m2g=0

For the value of m1=m2 the value of m2 is negligible and the expression for the tension can be evaluated as,

  T=2m1m2m1+m2g=m1g

For the value of m1m2 the value of m2 is negligible and the expression for the acceleration can be evaluated as,

  a=( m 1 m 2 )m1+m2g=m1m2g=g

For the value of m1m2 the value of m2 is negligible and the expression for the tension can be evaluated as,

  T=2m1m2m1+m2g=2m2g

For the value of m1m2 the value of m1 is negligible and the expression for the acceleration can be evaluated as,

  a=( m 1 m 2 )m1+m2g=m2m2g=g

For the value of m1m2 the value of m1 is negligible and the expression for the acceleration can be evaluated as,

  T=2m1m2m1+m2g=2m1g

Conclusion:

Therefore, the value of tension for the condition m1=m2 is m1g and the value of acceleration is 0 . For the value of m1m2 the plausible value of the tension is 2m2g and the value of acceleration is g . For the value of m1m2 , the plausible value of the tension is 2m1g and the value of acceleration is g .

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