# What factors affect the mechanical advantage of a lever?

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
$\frac{F}{f} = \frac{b}{a}$

#### 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 $\phi$ 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 $a \cdot \sin \left(\phi\right)$ height. The work performed must be
$W = F \cdot a \cdot \sin \left(\phi\right)$

On the other end of a rod, on distance $b$ from the fulcrum, force $f$ pushed the lever down by $b \cdot \sin \left(\phi\right)$. The work performed equals to
$W = f \cdot b \cdot \sin \left(\phi\right)$

Both works must be the same, so
$F \cdot a \cdot \sin \left(\phi\right) = f \cdot b \cdot \sin \left(\phi\right)$
or
$\frac{F}{f} = \frac{b}{a}$

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.