Physics for Scientists and Engineers
Physics for Scientists and Engineers
6th Edition
ISBN: 9781429281843
Author: Tipler
Publisher: MAC HIGHER
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Chapter 4, Problem 75P

(a)

To determine

The acceleration of the objects and the tension in the string in terms of θ , m1 and m2 .

(a)

Expert Solution
Check Mark

Answer to Problem 75P

The expression for the tension in the string is m2g(m1+m1sinθ)m1+m2 and the acceleration of the block is g(m2m1sinθ)m1+m2 .

Explanation of Solution

Calculation:

The free body diagram for the blocks is shown in Figure 1

  Physics for Scientists and Engineers, Chapter 4, Problem 75P

Figure 1

Consider the tension in the string is T and the acceleration of the system is a . Then consider that the block of mass m1 is accelerating to the right and the block of mass m2 is accelerating downwards.

The expression for the tension T for block m1 is given by,

  T=m1gsinθ

The expression for the tension in terms of the acceleration for mass m1 is given by,

  T=m1am1gsinθ=m1a

The expression for the tension T for block m2 is given by,

  m2gT=m2am2(g+a)=T

The expression for the tension in terms of the acceleration for mass m2 is given by,

  T=m2(g+a)m2(g+a)=m2a

The expression for the acceleration is evaluated as,

  m2(ga)m1gsinθ=m1aa=g( m 2 m 1 sinθ)m1+m2

The expression for the tension in terms of acceleration is given by,

  T=m2(ga)=m2(g( g( m 2 m 1 sinθ ) m 1 + m 2 ))=m2g( m 1 + m 1 sinθ)m1+m2

Conclusion:

Therefore, the expression for the tension in the string is m2g(m1+m1sinθ)m1+m2 and the acceleration of the block is g(m2m1sinθ)m1+m2 .

(b)

To determine

The acceleration and the tension of the string.

(b)

Expert Solution
Check Mark

Answer to Problem 75P

The tension of the string is 36.75N and the acceleration of the string is 2.45m/s2 .

Explanation of Solution

Given:

The angle of inclination is θ=30° .

The mass of first block is m1=5.0kg .

The mass of the second block is m2=5.0kg .

Calculation:

The acceleration of the string is calculated as,

  a=g( m 2 m 1 sinθ)m1+m2=( 9.81m/ s 2 )( 5.00kg( 5.00kg )sin30°)5.00kg+5.00kg=2.45m/s2

The tension in the string is calculated as,

  T=m2g( m 1 + m 1 sinθ)m1+m2=( 5.00kg)( 9.81m/ s 2 )( 5.00kg+( 5.00kg )sin30°)5.00kg+5.00kg=36.75N

Conclusion:

Therefore, the tension of the string is 36.75N and the acceleration of the string is 2.45m/s2 .

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