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Name: Julie
Status: student
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I know that torque is produced when a force is applied a certain distance from the pivot point of the lever arm (unless it is applied at 0 degrees or 180 degrees). I also know that the further away from the fulcrum the force is applied then the greater the torque resulting. My question is: WHY? Why does the distance the force is applied from the fulcrum heighten the torque? What causes this to be the case?


The effect can be explained with either acceleration or work. I prefer the second because it does not require worrying about NET force. It can be applied to a single force.

A force is applied at a certain distance from the pivot. The force pushes on a portion of the object that moves along the tangent of a circle, so the component directed toward the pivot does no work. Work done, over a very short distance, is tangential force times distance along the circumference. Tangential force is force perpendicular to the radius. Distance is radius times a small angle of rotation (in radians). For truly rotational motion, the radius is constant. For truly rotational motion, work equals tangential force times radius times angle of rotation. This can then be expressed as (perpendicular force) times angle change, or torque times angle change.

Perpendicular force is force magnitude times a sine function: torque equals force times radius times a sine function. If you wish to use lever arm, radius at which the force makes contact times the same sine function yields this quantity called lever arm (portion of radius perpendicular to the force). This allows you to write torque times lever arm.

Dr. Ken Mellendorf
Physics Instructor
Illinois Central College

Hi, Julie.

The simple answer is that Torque is a derived unit. It is defined based on the other two (force and distance). But that kind of avoids the question.

A torsional system consists of a force applied at some lever arm that is reacted by another force at another lever arm. If you "unwrap" a torsional system, you will have a lever. Exactly the same principle is acting there.

Now let us consider the "work" done at each end. Work is force times distance traveled. If I have a lever with a 2:1 length ratio, and I push down with a force of 100 lb., I can lift 200 lb. on the other end. But look at the work done: I have to push the 100 lb. end down twice as far as the 200 lb. end comes up. the work done on one end is 100 lb * 2 feet at one end, and 200 lb. * 1 foot on the other. The work done on either side is equal at 200 foot-pounds. Work and energy are different terms for one another, and can be equated (search for "energy" in Wikipedia, and scroll down to the section on forms of energy and transformation of energy for the theory behind what engineers call the "work-energy equation"), so I have not broken any of the laws of thermodynamics by having more force coming out than went in. It is just that something else must be less, such that the energy or work on each end is the same (less losses, of course). Of course, I have not included any losses, but this simplistic picture gives you the idea.

David Brandt

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