Pulley Systems & Mechanical Advantage

how many strands hold the load, and what a real rope charges for it

Worked example — Pulley Systems & Mechanical Advantage

Ideal MA (n)	4.0
Actual MA (n · η)	3.40
Effort Force (E)	117.6 N
Rope Pulled per 1 m Lift	4.0 m
Supporting Ropes (n)	4
Load Force (L)	400 N
Efficiency (η)	85%

These are the values the calculator shows when the page opens. Change any input above to recompute them.

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Count the ropes; that is the whole calculation 🖖

Ideal mechanical advantage is the number of rope sections supporting the moving block, and nothing else about the arrangement matters. One section for a fixed pulley, two for a movable one, four for a block and tackle. The rope you have to pull is the same number of metres per metre of lift, so a four-fold advantage costs four metres of hauling. A fixed pulley has an advantage of exactly 1: it buys you a direction, not a force.

Friction is charged to the force, not to the distance 🖖

Set the block and tackle to 85% and the rope you pull is still exactly 4 m per metre of lift, because that number is fixed by the geometry of how it is threaded. What changes is the pull: 235.3 N instead of 200 N, so 941.2 J go in to deliver 800 J and 141.2 J are lost in the sheaves. Distance is set by the rigging; efficiency is paid in force.

A big advantage makes losses expensive 🖖

The heavy crane preset threads six sections at 80%, lifting 1,500 N with 312.5 N of pull. Every metre of lift costs 1,875 J and wastes 375 J. That 20% comes off the force and never off the rope: six metres still run through your hands for every metre the hook rises, while the hand feels 312.5 N where a frictionless six-rope tackle would ask 250. Each sheave you add takes another cut of the same kind, so the efficiency reading falls as the count rises, and the second worked problem below finds where the two balance.

Problems solved in full

Four ropes, eight hundred newtons, and where the missing 141 joules go 6 steps
Load Block & Tackle: four rope sections, an 800 N load, 85% efficiency. The panel says 235.3 N of pull and 4 m of rope per metre of lift. Work out both, then say exactly what the 15% was spent on.
1. Start with the ideal machine. Four sections of rope share the load, so each carries a quarter of it, and the hand holding one of them feels a quarter. That is the mechanical advantage: it is the count, not a formula.
  
  $MA_{ideal} = n = 4, E_{ideal} = \frac{800}{4} = 200 N$
2. The rope has to come from somewhere. Raising the block by 1 m shortens all four sections by 1 m each, so 4 m has to run through your hands. The advantage and the rope length are the same number for the same reason.
  
  $ℓ = n \times h = 4 m of rope per 1 m of lift$
3. Now add the real machine. Efficiency multiplies the advantage, not the rope: the sheaves turn against their bearings and the rope resists being bent round them, and both cost force.
  
  $MA_{actual} = n η = 4 \times 0.85 = 3.40$
4. So the pull rises. Note the second form: dividing the ideal 200 N by 0.85 gives the same answer, which is what "85% efficient" is saying.
  
  $E = \frac{L}{n η} = \frac{800}{3.40} = 235.3 N = \frac{200}{0.85}$
5. Count the joules on both sides. You move 4 m against 235.3 N; the load moves 1 m against 800 N.
  
  $W_{in} = 235.3 \times 4 = 941.2 J, W_{out} = 800 \times 1 = 800 J$
6. The difference is the loss, and it is exactly the fraction the efficiency named.
  
  $W_{lost} = 141.2 J = (1 - η) W_{in}$
Answer

235.3 N, over 4 m of rope, and the 141.2 J went into the sheaves as heat.

The part worth carrying away is which of the two numbers moved. The rope stayed at exactly 4 m, because that is set by how the rope is threaded and no amount of friction can change how many strands hold the block. Everything the inefficiency cost was charged to the force. That is why a stiff rope or a dry bearing makes a tackle feel heavy rather than slow, and it is why the count and the efficiency have to be read together: each sheave adds a bend and a bearing, so the advantage climbs by one while the efficiency falls. Past some count the extra strand costs more than it earns, and the second problem below finds that count. On the heavy crane's sheaves it sits at twenty-seven sections, well beyond anything this tool will let you thread.
The seventh rope, and when it starts costing you 6 steps
Load Heavy Crane (6×): six rope sections, 1,500 N of load, 80% efficiency, 312.5 N of pull. The page says three separate times that adding sections eventually stops helping. Find where it stops, then decide whether a seventh section on this crane is worth threading.
1. The panel first. Six sections at 80% give an advantage of 4.80, so the pull is 1,500 divided by that, and six metres of rope run through your hands for every metre the hook rises.
  
  $MA = n η = 6 \times 0.80 = 4.80, E = \frac{1500}{4.80} = 312.5 N, ℓ = 6.0 m$
2. The tool lets you set the count and the efficiency separately; rigging does not. Each sheave takes its own cut — one bearing, one bend of the rope — so the losses compound section by section. The tool’s own presets agree on the size of the per-section cut: 80% over six sections and 85% over four both come out at about 96%.
  
  $η = η_{s}^{n} ⟹ η_{s} = 0.8 0^{1/6} = 0.9635; block and tackle: 0.8 5^{1/4} = 0.9602$
3. Substitute that in and the pull stops being a straight division. Differentiate it and the count where the two effects balance falls out with no load and no rigging left in it.
  
  $E (n) = \frac{L}{n η _{s}^{n}}, \frac{d}{d n} ln E = - \frac{1}{n} - ln η_{s} = 0 ⟹ n^{*} = \frac{1}{ln ( 1/ η _{s} )}$
4. For 96.3% sheaves the balance sits at twenty-seven sections. Nothing in the tool’s range comes near it — at ten sections the pull is still falling steeply.
  
  $η_{s} = 0.9635 : n^{*} = 26.9; E (6) = 312.5, E (7) = 278.0, E (10) = 217.6 N$
5. Make the sheaves worse and the answer moves a long way. At 85% each, six sections is the minimum, and the seventh is already heavier than the sixth.
  
  $η_{s} = 0.85 : n^{*} = 6.15; E (5) = 676.1, E (6) = 662.9, E (7) = 668.4 N$
6. The efficiency card is not the thing to watch while you do this. Ten sections on this crane’s own sheaves is 68.9% efficient, well under the 80% showing now, and the pull is a third lighter.
  
  $η (10) = 0.963 5^{10} = 0.689 < 0.80, E (10) = \frac{1500}{10 \times 0.689} = 217.6 N$
Answer

Thread it. On these sheaves the turning point is at twenty-seven sections, and the seventh drops the pull from 312.5 N to 278.0 N.

The formula is the part worth keeping, because the load and the rigging cancel out of it: n* = 1/ln(1/η_s). Good sheaves at 99% put the crossover near a hundred sections, 96% puts it at twenty-four, 85% at six, and 70% at three. Whether another rope helps is a question about the sheave you are about to add, and never about how many you already have. Which leaves the eighth rope needing a different explanation. Eight sections means eight metres of rope through the drum for every metre of lift, and a hook that rises at an eighth of the rope speed. Drum capacity and lifting speed run out long before the arithmetic does.

$n^{*} = \frac{1}{ln ( 1/ η _{s} )} : 0.99 \to 99.5, 0.96 \to 24.5, 0.85 \to 6.2, 0.70 \to 2.8$

Learning path

Simple machines: trading force for distance

Leads to Inclined plane, screw and wedge an advantage you count rather than calculate, and the chapter’s first real loss.

By the 42Math team · Updated 2026-08-31

Reviewed against published sources. Corrections welcome. Editorial policy · Corrections

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Example problems

Single Fixed (1×) - One fixed pulley, no advantage: 200 N of load takes 200 N of pull and 1 m of rope. You have changed the direction and nothing else.
Single Movable (2×) - A movable pulley doubles the force and doubles the rope: 400 N held by 210.5 N at 95%, over 2 m of hauling.
Block & Tackle (4×) - Four sections at 85%: 800 N lifted with 235.3 N. Still 4 m of rope, but 141.2 J lost per metre of lift.
Heavy Crane (6×) - Six sections at 80%: 1,500 N on a 312.5 N pull, and 375 J of every 1,875 J goes into the sheaves.