Physics For Scientists And Engineers With Modern Physics, 9th Edition, The Ohio State University
Physics For Scientists And Engineers With Modern Physics, 9th Edition, The Ohio State University
9th Edition
ISBN: 9781305372337
Author: Raymond A. Serway | John W. Jewett
Publisher: Cengage Learning
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Chapter 5, Problem 98CP

Initially, the system of objects shown in Figure P5.49 is held motionless. The pulley and all surfaces and wheels are frictionless. Let the force F be zero and assume that m1 can move only vertically. At the instant after the system of objects is released, Find (a) the tension T in the string, (b) the acceleration of m2, (c) the acceleration of M, and (d) the acceleration of m1. (Note: The pulley accelerates along with the cart.)

Figure P5.49 Problems 49 and 53

Chapter 5, Problem 98CP, Initially, the system of objects shown in Figure P5.49 is held motionless. The pulley and all

(a)

Expert Solution
Check Mark
To determine

The tension in the string.

Answer to Problem 98CP

Solution: The tension in the string is m2g(m1Mm2M+m1(m2+M)) .

Explanation:

Explanation of Solution

Given info: The mass of large block is M , the mass of top block is m2 , the mass of hanging block is m1 , the tension in string is T and the force F is zero.

Consider the free body diagram given below,

Physics For Scientists And Engineers With Modern Physics, 9th Edition, The Ohio State University, Chapter 5, Problem 98CP

Figure I

In above figure,

  • a is the acceleration of hanging block having mass m1 .
  • A is the acceleration of large block having mass M .
  • aA is the acceleration of top block having mass m2 .

The equilibrium condition for hanging block is,

m1gT=m1aT=m1(ga) (I)

  • g is the acceleration due to gravity.

The equilibrium condition for top block is,

T=m2(aA)a=Tm2+A (II)

The equilibrium condition for large block is,

MA=TA=TM (III)

Substitute (Tm2+A) for a and TM for A in equation (I) to find T .

T=m1(g(Tm2+A))=m1(g(Tm2+TM))=m1g(m1Tm2+m1TM)=m1gm1T(M+m2Mm2)

Further, solve for T .

T=m1gm1T(M+m2Mm2)T+m1T(M+m2Mm2)=m1gT=m2g(m1Mm2M+m1(m2+M))

Conclusion:

Therefore, the tension in the string is m2g(m1Mm2M+m1(m2+M)) .

(b)

Expert Solution
Check Mark
To determine

The acceleration of m2 .

Answer to Problem 98CP

Solution: The acceleration of m2 is m1g(M+m2)Mm2+m1(M+m2) .

Explanation of Solution

Given info: The mass of large block is M , the mass of top block is m2 , the mass of hanging block is m1 , the tension in string is T and the force F is zero.

The force applied on the block of mass M is zero initially and the block of mass m2 has acceleration in synchronization with the big block so the net acceleration on the block is a .

Substitute TM for A in equation (II).

a=Tm2+TM=T(M+m2Mm2)

Substitute m1g(Mm2Mm2+m1(M+m2)) for T in above equation to find a .

a=(m1g(Mm2Mm2+m1(M+m2)))(M+m2Mm2)=m1g(M+m2)Mm2+m1(M+m2)

Conclusion:

Therefore, the acceleration of m2 is m1g(M+m2)Mm2+m1(M+m2) .

(c)

Expert Solution
Check Mark
To determine

The acceleration of M .

Answer to Problem 98CP

Solution: The acceleration of M is m1m2gm2M+m1(m2+M) .

Explanation of Solution

Given info: The mass of large block is M , the mass of top block is m2 , the mass of hanging block is m1 , the tension in string is T and the force F is zero.

The acceleration of M is A .

Substitute m1g(Mm2Mm2+m1(M+m2)) for T in equation (II).

A=m1g(Mm2Mm2+m1(M+m2))M=m1m2gm2M+m1(m2+M)

Conclusion:

Therefore, the acceleration of M is m1m2gm2M+m1(m2+M) .

(d)

Expert Solution
Check Mark
To determine

The acceleration of m1 .

Answer to Problem 98CP

Solution: The acceleration of m1 is Mm1gMm2+m1(M+m2) .

Explanation of Solution

Given info: The mass of large block is M , the mass of top block is m2 , the mass of hanging block is m1 , the tension in string is T and the force F is zero.

The block of mass m1 moves in vertical direction only but the net acceleration is the difference between the acceleration of the big block of mass M and the acceleration a of m2 .

The formula to calculate the acceleration of m1 is,

am1=aA . (4)

Substitute m1g(Mm2Mm2+m1(M+m2)) for T in equation (II).

A=m1g(Mm2Mm2+m1(M+m2))M=m1m2gm2M+m1(m2+M)

Substitute (m1g(M+m2)Mm2+m1(M+m2)) for a and m1m2gm2M+m1(m2+M) for A in equation (4) to find the value of aA .

aA=(m1g(M+m2)Mm2+m1(M+m2))(m1m2gMm2+m1(M+m2))=(Mm1gMm2+m1(M+m2))

Conclusion:

Therefore, the acceleration of m1 is Mm1gMm2+m1(M+m2) .

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Chapter 5 Solutions

Physics For Scientists And Engineers With Modern Physics, 9th Edition, The Ohio State University

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