What factors affect the mechanical advantage of a lever?

1 Answer
Jul 20, 2015

If on one end of a class 1 lever in equilibrium force F is applied on a distance a from a fulcrum and another force f is applied on the other end of a lever on distance b from a fulcrum, then
Ff=ba

Explanation:

Consider a lever of the 1st class that consists of a rigid rod that can rotate around a fulcrum. When one end of a rod goes up, another goes down.

This lever can be used to lift up a heavy object with significantly weaker than its weight force. It all depends on the lengths of points of application of forces from the fulcrum of the lever.

Assume that a heavy load is positioned at a length a from the fulcrum, the force it pushes down on a rod is F.
On the opposite side of a rod at a distance b from the fulcrum we apply a force f down such that two a lever is in equilibrium.

The fact that a lever is in equilibrium means that the work performed by forces F and f when a lever is pushed on either side by a small distance d must be the same - whatever work we, using force f, perform to push down our end of a lever on a distance b from the fulcrum should be equal to work to lift a heavy object on a distance a on the other end of a lever.

Rigidity of a rod that serves as a lever means that the angle a lever turns around a fulcrum is the same on both ends of a lever.

Assume that a lever turned by a small angle ϕ around a fulcrum slightly lifting a heavy weight. Then this heavy weight that exhorts a force F on one end of a rod at a distance a from a fulcrum was lifted by asin(ϕ) height. The work performed must be
W=Fasin(ϕ)

On the other end of a rod, on distance b from the fulcrum, force f pushed the lever down by bsin(ϕ). The work performed equals to
W=fbsin(ϕ)

Both works must be the same, so
Fasin(ϕ)=fbsin(ϕ)
or
Ff=ba

From the last formula we derive that the advantage of using a lever depends on a ratio between lever ends' distance from fulcrum. The more the ratio is - the more advantage we have and more weight we can lift.