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

Question

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) (1/2)(M+ m) g (C) M g (D) m g

Tardigrade Admin · Accepted Answer

Considering free body diagram of man and crate system in vertical direction. 2 T =(M+ m) g ∴ T =((M+ m) g/2)

Q. 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

$(M + m) g$

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

$M g$

$m g$

Considering free body diagram of man and crate system in vertical direction.
$2 T = (M + m) g$
$∴ T = \frac{( M + m ) g}{2}$

Q. 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

(M+m)g

21​(M+m)g

Mg

mg