How to Calculate Tension in Pulleys

Written by paul dohrman
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How to Calculate Tension in Pulleys
(Hemera Technologies/ Images)

Multiple pulleys working together can provide a mechanical advantage, reducing the force required to lift a load. This force reduction corresponds to the tension reduction in the line through the pulleys. Therefore, if you can calculate one, then you can calculate the other. Note that the applicable formula for determining mechanical advantage is work = force x distance. So if a line lifts the same load in the air as before adding the pulley system, and the amount of distance of rope for which the input force must be applied has increased X-fold, then the input force required to lift the load has decreased X times. The mechanical advantage is then X-to-1.

Skill level:


  1. 1

    Draw a diagram of a line going over a pulley hanging from a ceiling support, then under a pulley attached to the load to be lifted, then going back up to the hanging pulley again to attach to an eyelet at the bottom of the pulley casing. The line is then doubled up between the stationary pulley affixed above and the pulley attached to the load.

  2. 2

    Determine how much line a workman will have to be pull to lift the load 1 foot.

    Continuing with the example above, you’d have to pull the line out 2 feet to get the two lengths of line between the two pulleys to shorten 1 foot each. In other words, you need to bring the two pulleys together by 1 foot to lift the load 1 foot. The two lengths between the two pulleys must therefore both shorten 1 foot each.

  3. 3

    Divide the distance of line over which force is put in by the distance over which work is done on the output end. This gives your mechanical advantage.

    The result for the example above is 2-to-1.

  4. 4

    Take the reciprocal of your result in step 3 and multiply it by your load weight. This is the tension through the line if the load is held static off the ground. Pulling on the load will make the tension go above this amount, of course, but the static case gives a lower limit.

    Continuing with the above example, suppose the load is 4.54kg. Then (1/2)x4.54kg = 2.27kg of force. (Note that the pound is a unit of force, not mass, in the Imperial measurement system.)

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