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 m 1 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 m 2 , (c) the acceleration of M , and (d) the acceleration of m 1 . ( Note: The pulley accelerates along with the cart.) Figure P5.49 Problems 49 and 53
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 m 1 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 m 2 , (c) the acceleration of M , and (d) the acceleration of m 1 . ( Note: The pulley accelerates along with the cart.) Figure P5.49 Problems 49 and 53
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
(a)
Expert Solution
To determine
The tension in the string.
Answer to Problem 5.98CP
The tension in the string is m2g(m1Mm2M+m1(m2+M)).
Explanation of Solution
Consider the free body diagram given below,
Figure I
Here, a is the acceleration of hanging block having mass m1, A is the acceleration of large block having mass M and a−A is the acceleration of top block having mass m2.
Write the expression for the equilibrium condition for hanging block
m1g−T=m1aT=m1(g−a) (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(a−A)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.
Therefore, the tension in the string is m2g(m1Mm2M+m1(m2+M)).
(b)
Expert Solution
To determine
The acceleration of m2.
Answer to Problem 5.98CP
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.
Therefore, the acceleration of m2 is m1g(M+m2)Mm2+m1(M+m2).
(c)
Expert Solution
To determine
The acceleration of M.
Answer to Problem 5.98CP
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
To determine
The acceleration of m1.
Answer to Problem 5.98CP
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=a−A (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 a−A.
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