Two movable pulleys: minimum force

A pulley block with 2 movable pulleys lifts a 24 kg load at constant speed; 4 rope segments support the load. The load rises 0.5 m. Find the minimum pulling force and the distance the rope end moves. (Take g = 10 N/kg.)

Step-by-step

1. Weight G = mg = 24×10 = 240 N
   Find the weight of the object being lifted first
2. F = G/n = 240/4 = 60 N
   n rope segments share the load, so each needs only G/n (n = 4 ✓)
3. s = n·h = 4×0.5 = 2 m
   Less force means more distance: the rope end moves n times the lift height (work is not saved ✓)

Why it works

With n rope segments sharing the load, the minimum pull is the weight divided by n, while the rope end moves n times the load's rise. The work done is the same either way.

Common mistakes

• Counting only one movable pulley and using 2 rope segments.
• Forgetting that the rope end moves 4 times the load's rise.

Related

conveyor-belt incline-friction pulley-two-movable-30kg-lift-1m