An ideal lever is a light rod of negligible mass pivoted at the point along its length. This point is called the fulcrum. The lever is a system in mechanical equilibrium.

Two forces $\overrightarrow{\mathrm{F}}_{1}$ and $\overrightarrow{\mathrm{F}_{2}}$ parallel to each other and usually perpendicular to the lever, as shown here, act on the lever at distances $d_{1}$ and $d_{2}$ respectively from the fulcrum as shown in fig.

Let $\overrightarrow{\mathrm{R}}$ be the reaction of the support at the fulcrum. $\overrightarrow{\mathrm{R}}$ is directed opposite to the forces $\overrightarrow{\mathrm{F}}_{1}$ and $\overrightarrow{\mathrm{F}_{2}}$

The forces in upward direction are considered positive and the forces in downward direction are considered negative.

For translational equilibrium

$\mathrm{R}-\mathrm{F}_{1}-\mathrm{F}_{2}=0$

$\therefore \mathrm{R}=\mathrm{F}_{1}+\mathrm{F}_{2}$

The lever force $F_{1}$ is some weight to be lifted. It is called the load and its distance from the fulcrum $d_{1}$ is called the load arm.

Force $\mathrm{F}_{2}$ is the effort applied to lift the load, distance $d_{2}$ of the effort from the fulcrum is the effort arm.

For rotational equilibrium, sum of torque about a effort point should be zero and moment of force $\tau=d \times \mathrm{F}\left[\because \theta=90^{\circ}, \therefore \sin 90^{\circ}=1\right]$

$\therefore d_{1} \mathrm{~F}_{1}-d_{2} \mathrm{~F}_{2}=0$

Here, the anticlockwise moment is considered as positive and the clockwise moment is considered as negative. So $d_{2} \mathrm{~F}_{2}$ taken with negative.

$\therefore d_{1} \mathrm{~F}_{1}=d_{2} \mathrm{~F}_{2}$

Means load arm $\times$ load $=$ effort $\operatorname{arm} \times$ effort

The above equation expresses the principle of moments for a lever.