Explain the construction and working of an ideal lever and also explain the principle of moments of force.

(N/A) An ideal lever is a light rod of negligible mass pivoted at a point along its length. This point is called the fulcrum. The lever is a system in mechanical equilibrium.
Two forces $\overrightarrow{F}_{1}$ and $\overrightarrow{F}_{2}$ parallel to each other and perpendicular to the lever act on the lever at distances $d_{1}$ and $d_{2}$ respectively from the fulcrum,as shown in the figure.
Let $\overrightarrow{R}$ be the reaction of the support at the fulcrum. $\overrightarrow{R}$ is directed opposite to the forces $\overrightarrow{F}_{1}$ and $\overrightarrow{F}_{2}$.
The forces in the upward direction are considered positive and the forces in the downward direction are considered negative.
For translational equilibrium:
$R - F_{1} - F_{2} = 0$
$\therefore R = F_{1} + F_{2}$
The lever force $F_{1}$ is the weight to be lifted. It is called the load,and its distance from the fulcrum $d_{1}$ is called the load arm.
Force $F_{2}$ is the effort applied to lift the load,and the distance $d_{2}$ of the effort from the fulcrum is the effort arm.
For rotational equilibrium,the sum of torques about the fulcrum should be zero. The moment of force is $\tau = d \times F$ (since $\theta = 90^{\circ}$,$\sin 90^{\circ} = 1$).
Taking anticlockwise moments as positive and clockwise moments as negative:
$d_{1} F_{1} - d_{2} F_{2} = 0$
$\therefore d_{1} F_{1} = d_{2} F_{2}$
This means: $\text{Load arm} \times \text{Load} = \text{Effort arm} \times \text{Effort}$.
This equation expresses the principle of moments for a lever.