Pulley System Effort Force

F=WnF = \frac{W}{n}
sig figs

Worked example: 800 N load on a 4-part tackle → 200 N. Press Try an example to run it live, then adjust anything.

Enter your known values, leave one input blank, and solves for the missing one. Tap a variable’s symbol to see what it means, with a typical value. Change any unit and the answer follows.

Here the solver did the work. Could you?

Levers and pulleys →

UniversityEngineering Mechanics

Test your skills in the Exam Room: new numbers every attempt. Free lessons for students, no sign-up, just pure learning.

See your Report Card
Learning zone

Pulley System Effort Force explained

WFn

In a block and tackle the load hangs from several sections of the same rope, and since the tension is identical everywhere along an ideal rope, each section carries an equal share: F = W ⁄ n, where n counts only the sections actually pulling up on the moving block. An 800 N load on a four-part tackle needs just 200 N of pull. Plutarch tells of Archimedes using a compound pulley to draw a fully laden ship out of Syracuse harbour single-handed, to King Hiero's astonishment.

The price, again, is rope: to raise the load 1 m you must haul n metres through your hands, so the work is unchanged. Count carefully: the classic error is including the rope section you are pulling on when it runs off a fixed pulley (a fixed pulley only changes direction, adding nothing to n) and forgetting to include it when it leaves the moving block. Real tackle loses roughly 5–10% per sheave to friction, so a nominal 4:1 rigging typically pulls closer to 3.3:1.

Pulley System Effort Force formula

F=WnF = \frac{W}{n}
Where
  • FF= Effort force (N)
  • WW= Load (N)
  • nn= Supporting rope sections
Revised 16 October 2026
Cite this page

formula.expert. (2026, October 16). Pulley System Effort Force Formula & Calculator. https://formula.expert/solvers/pulley-system-effort-force