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

To find the force exerted by the man on the rope for the system to remain in equilibrium, let us analyze the forces acting on the man and the crate.

Let:
$m$ be the mass of the man.
$M$ be the mass of the crate.
$T$ be the tension in the rope, which is equal to the force exerted by the man on the rope.
$N$ be the normal force acting between the man and the crate.
$g$ be the acceleration due to gravity.

Since the pulley is smooth and light, the tension $T$ is uniform throughout the rope. The man pulls the rope downward with a force $T$, and by Newton's third law, the rope pulls the man upward with an equal tension force $T$. The other end of the rope is attached to the crate, pulling the crate upward with a tension force $T$.

Let us analyze the forces on the man-crate system as a combined single system of total mass $(M+m)$:
1. The total downward gravitational force acting on the system is:
$F_g=(M+m)g$
2. The upward forces acting on this combined system are:
- The tension $T$ acting upward on the man (who is holding the rope).
- The tension $T$ acting upward on the crate (where the rope is tied).
Thus, the total upward force is:
$F_{\mathrm{up}}=T+T=2T$

For the system to be in equilibrium, the net force must be zero. Therefore, the total upward force must balance the total downward gravitational force:
$2T=(M+m)g$

Solving for $T$:
$T=\frac{(M+m)g}{2}$

Thus, the force exerted by the man on the rope is $(M+m)g/2$.