Pulley with 4 rope segments: least force

A pulley system lifts a 30 kg object at constant speed, with 4 rope segments supporting the load and a lifting height of 0.5 m. Given g = 10 N/kg, find the minimum pulling force and the distance moved by the free end of the rope.

Step-by-step

1. Weight G = mg = 30×10 = 300 N
   Find the weight of the object being lifted first
2. F = G/n = 300/4 = 75 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

The principle of force reduction in a pulley system is 'tension is uniform throughout the same rope', so if there are n segments supporting the load, the pulling force is 1/n of the object's weight. The number of rope segments is determined by the movable pulleys; fixed pulleys only change the direction of force without reducing it. The distance moved by the free end of the rope is n times the lifting height. These two rules should be memorized together as a pair.

Common mistakes

• 1) Counting fixed pulleys in the force-reducing segments, leading to an overestimated number of segments.
• 2) Only answering the pulling force while omitting the distance moved by the free end of the rope.
• 3) Reversing the force-reduction relationship, multiplying the pulling force by the number of segments to get the object's weight.

Related

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