Why is lever in mechanical equilibrium? Let's consider a lever with a fulcrum (see image below), such that it's load arm and load are $d_1$ and $F_1$, respectively, and it's effort arm and effort are $d_2$ and $F_2$, respectively. Now my book states the principle of moments by saying that the system is in mechanical equilibrium and hence,
$$d_{1}F_{1}=d_{2}F_{2}$$
My question is why the lever is in mechanical equilibrium?

 A: Two conditions for equilibrium

My question is why the lever is in mechanical equilibrium?

There are two conditions for equilibrium:

*

*vector sum of all forces acting on the body is zero; this is first Newton's law of motion for translational motion;

*object must have no tendency to rotate; sum of torques due to all the external forces acting on the object must be zero;


Equilibrium conditions for your example
In your example there are three forces acting on the body: (i) $F_1$ downwards at the far-left end, (ii) $F_2$ downwards at the far-right end, and (iii) $n$ upwards at the fulcrum.
The first condition for equilibrium is
$$F_\text{net} = F_1 + F_2 - n = 0 \tag 1$$
The torque is defined as force times distance (lever) $\vec{\tau} = \vec{r} \times \vec{F}$. The "$\times$" symbol here denotes vector product, which means that only perpendicular component of force $\vec{F}$ on distance $\vec{r}$ gives torque. The second condition of equilibrium must be satisfied for any point. Let's write the net torque with respect to the fulcrum, taking clockwise as positive rotation direction:
$$\tau_\text{net} = (-F_1 \cdot d_1) + (n \cdot 0) + (F_2 \cdot d_2) = -F_1 d_1 + F_2 d_2 = 0 \tag 2$$
You can repeat the above procedure for the far-left end
$$\tau_\text{net} = (F_1 \cdot 0) + (n \cdot d_1) + (-F_2 \cdot (d_1 + d_2)) = 0$$
The first condition for equilibrium from Eq. (1) gives $n = F_1 + F_2$, and with this the above equation gives the same condition for equilibrium as in Eq. (2).

Homework
Try repeating the procedure for second condition for equilibrium for any other point, for example halfway between the fulcrum and the far-right end. You should get the same result as in Eq. (2). You are free to choose your own positive rotation direction, just make sure to use it consistently!

Energy considerations
You can also derive the condition for equilibrium by considering energy (work). For system not to change its translational or rotational velocity, the total net work done on the system must be zero. This follows directly from the work-energy theorem
$$\Delta K = W$$
Work is defined as a scalar product of force and displacement $dW = \vec{F} \cdot d\vec{r}$. In your example, for some change of angle $\Delta \theta$, the displacement the left-end does is $r_1 = d_1 \Delta\theta$ in the counterclockwise direction, and the displacement the right-end does is $r_2 = d_2 \Delta\theta$ in the clockwise direction. Note that $r_1$ and $r_2$ are arc lengths. Since two displacements are in the opposite direction, one of the displacements is positive and the other is negative. The total work is then
$$W = F_1 r_1 - F_2 r_2 = 0$$
which gives $F_1 d_1 = F_2 d_2$.
A: If you are comfortable with conservation of energy, then consider that if $F_2$ causes the lever to rotate through a small angle, then the work input from $F_2$ must be equal to the work done by the lever against $F_1$.  (Consider the arc lengths.)
