A man of mass 50 kg stands on a crate of mass 200 kg. The rope attached to the crate passes over a pulley. What force must the man apply on the rope for the system to remain in equilibrium?

Step 1: Define variables and constants.
Man's mass: $m = 50$ kg
Crate's mass: $M = 200$ kg
Acceleration due to gravity: $g = 10$ m/s$^2$

Step 2: Free-body diagram for the man (vertical equilibrium).
Forces: weight $mg$ downward, normal force $N$ upward from crate, tension $T$ upward from rope.
$N + T - mg = 0$
Substitute $m = 50$ kg, $g = 10$ m/s$^2$:
$N + T - (50)(10) = 0$
$N + T = 500$ N
$N = 500 - T$ N

Step 3: Free-body diagram for the crate (vertical equilibrium).
Forces: weight $Mg$ downward, normal force $N$ downward from man, tension $T$ upward from rope.
$T - Mg - N = 0$
Substitute $M = 200$ kg, $g = 10$ m/s$^2$:
$T - (200)(10) - N = 0$
$T - 2000 - N = 0$
$T = 2000 + N$ N

Step 4: Substitute $N$ from Step 2 into Step 3.
$T = 2000 + (500 - T)$
$T = 2500 - T$
$2T = 2500$ N
$T = 1250$ N

Step 5: The force the man must apply to the rope is the tension $T$.
The required force is $\boxed{1250 N}$.