Bundle: Physics for Scientists and Engineers, Volume 2, Loose-leaf Version, 10th + WebAssign Printed Access Card, Single-Term
Bundle: Physics for Scientists and Engineers, Volume 2, Loose-leaf Version, 10th + WebAssign Printed Access Card, Single-Term
10th Edition
ISBN: 9781337888745
Author: SERWAY, Raymond A., Jewett, John W.
Publisher: Cengage Learning
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Chapter 5, Problem 53CP

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 53CP, 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 53CP

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

Explanation of Solution

Consider the free body diagram given below,

Bundle: Physics for Scientists and Engineers, Volume 2, Loose-leaf Version, 10th + WebAssign Printed Access Card, Single-Term, Chapter 5, Problem 53CP

Figure I

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

Write the expression for the equilibrium condition for hanging block

    m1gT=m1aT=m1(ga)                                                                         (I)

Here, m1 is the mass of the hanging block, a is the acceleration of the hanging block, g is the acceleration due to gravity and T is the tension of the cord.

Write the expression for the equilibrium condition for top block

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

Here, m2 is the mass of the top block and A is the acceleration of the top block

Write the expression for the equilibrium condition for large block

    MA=TA=TM                                                                                           (III)

Here, M is the acceleration of the large mass.

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 53CP

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

Explanation of Solution

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 53CP

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

Explanation of Solution

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 53CP

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

Explanation of Solution

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.

Write the formula to calculate the acceleration of m1

    am1=aA                                                                                   (IV)

Here, am1 is the acceleration of mass m1.

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

Bundle: Physics for Scientists and Engineers, Volume 2, Loose-leaf Version, 10th + WebAssign Printed Access Card, Single-Term

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