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Question: A man of mass m stands on a crate of mass M. He pulls on a light rope passing over a smooth light pu...

A man of mass m stands on a crate of mass M. He pulls on a light rope passing over a smooth light pulley. The other end of the rope is attached to the crate. For the system to be in equilibrium, the force exerted by the men on the rope will be

A

(M + m)g

B

12(M+m)g\frac { 1 } { 2 } ( M + m ) g

C

Mg

D

mg

Answer

12(M+m)g\frac { 1 } { 2 } ( M + m ) g

Explanation

Solution

From the free body diagram of man and crate system:

For vertical equilibrium

2T=(M+m)g2 T = ( M + m ) g

T=(M+m)g2\therefore T = \frac { ( M + m ) g } { 2 }