The forces F1 = 62.4 N and F2 = 20.8 N are applied to the ends of the weightless lever, under the action of which

The forces F1 = 62.4 N and F2 = 20.8 N are applied to the ends of the weightless lever, under the action of which it is in equilibrium. Determinant of the length of the lever, if the length of the smaller arm l1 = 12cm.

F1 = 62.4 H.
F2 = 20.8 H.
l1 = 12cm = 0.12 m.
L -?
The length of the lever L is the sum of the lengths of the arms l1 and l2, to which F1 and F2 are applied, respectively.
L = l1 + l2.
Since the lever is in equilibrium, the law of equilibrium of the lever is valid for it.
When the lever is in balance, the forces are inversely proportional to the shoulders to which they are applied.
F1 / F2 = l2 / l1.
l2 = F1 * l1 / F2.
L = l1 + F1 * l1 / F2.
L = 0.12 m + 62.4 H * 0.12 m / 20.8 H = 0.48 m.
Answer: the length of the arm is L = 0.48 m.