Wheel & Axle Mechanical Advantage

a doorknob, a steering wheel and a screwdriver are one machine with different numbers

Worked example — Wheel & Axle Mechanical Advantage

Ideal MA (R / r)	6.00
Actual MA	5.40
Effort Force (E)	92.6 N
Wheel Rope per Turn	3.77 m
Wheel Radius (R)	60 cm
Axle Radius (r)	10 cm
Load Force (L)	500 N
Efficiency (η)	90%

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

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The ratio of the radii is the ratio of the distances 🖖

Mechanical advantage is R/r, and one full turn winds 2πR at the rim against 2πr at the axle. The 2π cancels, so the distance ratio is the radius ratio: the well winch turns 3.770 m of handle travel into 0.628 m of rope, exactly six to one, which is the same six as its advantage. Every wheel-and-axle in this tool obeys it, because it is the same statement written twice.

It is a lever that never runs out of stroke 🖖

Put the fulcrum at the centre of the shaft and the two radii are the two arms of a first-class lever: R/r is a/b under a different name. What the round shape adds is that the stroke never ends. A crowbar gives out when the beam reaches the ground; a winch keeps turning, which is why every machine that has to lift a long way — a well, an anchor, a lift — is built on a shaft rather than a bar.

Where the missing joules go 🖖

The doorknob preset runs at 95%: 33.1 J of turning delivers 31.4 J at the latch, and 1.7 J warms the spindle. That is a good machine. The well winch at 85% loses 66.5 J of every 443.5 J, and you feel it as a handle that is heavier than the arithmetic promised — 117.6 N rather than the ideal 100 N. Efficiency is the gap between the advantage the geometry offers and the one your hands get.

Problems solved in full

Why the winch handle travels 3.77 metres to raise 63 centimetres of rope 6 steps
Load Well Winch: a 60 cm handle radius on a 10 cm axle, 600 N of bucket, 85% efficiency. The panel says an ideal advantage of 6.00 and an effort of 117.6 N. Work out both, and then work out how far your hand goes.
1. The advantage is the ratio of the two radii, and it comes straight from moments: the same shaft, so the same torque, and a force six times smaller at six times the radius balances it.
  
  $MA_{ideal} = \frac{R}{r} = \frac{60}{10} = 6.00$
2. Now the distances, which is where the ratio earns its keep. One complete turn winds the circumference at each radius.
  
  $one turn: s_{R} = 2 π R = 3.770 m, s_{r} = 2 π r = 0.628 m$
3. Divide one by the other and the 2π cancels. The distance ratio is the radius ratio, exactly — the same 6 that the force ratio gave, arrived at from geometry rather than from moments.
  
  $\frac{s _{R}}{s _{r}} = \frac{2 π R}{2 π r} = \frac{R}{r} = 6.0000$
4. Efficiency scales the advantage down. At 85% the six becomes 5.10.
  
  $MA_{actual} = \frac{R}{r} η = 6.00 \times 0.85 = 5.10$
5. So the handle is heavier than the ideal machine promised, by exactly the same 15%.
  
  $E = \frac{600}{5.10} = 117.6 N (ideal would be 100.0 N)$
6. Count the work and the missing joules appear on their own: 66.5 J of the 443.5 you put in, warming the bearing.
  
  $W_{in} = 117.65 \times 3.7699 = 443.5 J, W_{out} = 600 \times 0.6283 = 377.0 J$
Answer

Six to one, so 117.6 N on the handle rather than the 100.0 N an ideal winch would need, and 3.770 m of hand travel per 0.628 m of rope.

A crowbar stops when the beam hits the ground, and a lever long enough to raise a bucket from a well would have to be as long as the well is deep. A shaft simply keeps turning, so the same six-to-one advantage runs for as many metres as you like. Every machine that has to move something a long way with a small force is built on a shaft for that one reason: the anchor winch, the lift, the tap you turned this morning.
What the screwdriver does when you let go 6 steps
Load Screwdriver (5:1): a 15 cm handle radius on a 3 cm shaft, 50 N of load, 90% efficient, 11.1 N at the rim. Work out what the load does the moment your hand comes off, then decide how bad a winch has to be before it will hold a bucket on its own.
1. The panel: five to one from the radii, 4.50 once the 90% is applied, so eleven newtons at the rim balance fifty at the shaft.
  
  $MA = \frac{R}{r} η = 5.00 \times 0.90 = 4.50, E = \frac{50}{4.50} = 11.11 N$
2. Count the joules for one complete turn. The rim travels 0.94 m, the shaft 0.19 m, and one joule of the ten and a half goes into the bearing.
  
  $one turn: W_{in} = 11.11 \times 0.9425 = 10.472 J, W_{out} = 50 \times 0.1885 = 9.425 J, W_{loss} = 1.047 J$
3. Now let the load drive it instead. The same surfaces slide the same distance under the same forces, so the bearing takes the same joule — except that this time it comes out of the load’s work rather than being added to yours. That equality is the only assumption in the calculation, and it is what makes the reverse case computable at all.
  
  $driven from the axle, same surfaces and same travel ⟹ W_{loss}^{'} = W_{loss} = 1.047 J$
4. So the reverse efficiency is whatever is left of the load’s own work, and it collapses to an expression with nothing in it but the forward efficiency.
  
  $η_{rev} = \frac{9.425 - 1.047}{9.425} = \frac{8.378}{9.425} = 0.8889 = 2 - \frac{1}{0.90}$
5. In force: the load pushes back at the rim with 8.89 N. You have to hold it almost as hard as you pulled it.
  
  $F_{hold} = \frac{L r}{R} η_{rev} = 50 \times 0.200 \times 0.8889 = 8.89 N (against 11.11 N to raise it)$
6. The expression runs out at one half. Below that the load cannot move the machine at all — and the tool shows what that costs, because raising the same load at 50% needs twice the ideal effort.
  
  $η_{rev} \leq 0 ⟺ η \leq 0.50; η = 0.50 ⟹ E = 20.0 N = 2 \times \frac{50}{5.00}$
Answer

It unwinds. 8.89 N at the rim, against the 11.11 N you needed to raise it.

A machine holds its own load only when it is worse than 50% efficient, and none of this tool’s four presets is close: the doorknob at 95% gives back 94.7%, the steering wheel and screwdriver at 90% give back 88.9%, the well winch at 85% gives back 82.4%. Drag the efficiency slider to its floor of 10% and the reverse figure is −8, meaning the load cannot budge it however heavy it gets. So a well winch is built with a pawl and a screwdriver is built with your hand around it. A hoist is the one machine where you pay for efficiency twice — once when you buy it, and again in the ratchet you have to bolt on because of it.

$η_{rev} = 2 - \frac{1}{η} : 0.95 \to 0.947, 0.90 \to 0.889, 0.85 \to 0.824, 0.50 \to 0, 0.10 \to - 8$

Learning path

Simple machines: trading force for distance

Leads to Pulley systems the same lever with the fulcrum at the centre of a shaft, which removes the one limit a beam has: the stroke never ends.

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

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

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Example problems

Doorknob (6:1) - A 30 cm knob on a 5 cm spindle: 100 N of latch resistance yields to 17.5 N of grip, at 95% efficiency.
Steering Wheel (4:1) - Four to one, at 90%: 400 N of steering resistance takes 111.1 N at the rim, and the rim travels 2.513 m per turn against 0.628 m at the column.
Well Winch (6:1) - Six to one at 85%: 600 N of bucket on a 117.6 N handle, winding 3.770 m of handle for 0.628 m of rope.
Screwdriver (5:1) - The smallest one here and the same machine: a 15 cm handle on a 3 cm shaft turns 11.1 N of grip into 50 N at the tip.