At the ends of the weightless lever a force of 40 and 240 N acts on the shoulder of a lesser

At the ends of the weightless lever a force of 40 and 240 N acts on the shoulder of a lesser force of 6 cm. determine the length of the arm.

Given:

F1 = 40 Newton – force acting on the larger arm of the lever;

F2 = 240 Newton – force acting on the smaller lever arm;

dl2 = 6 centimeters = 0.06 meters – smaller lever arm.

It is required to determine l (meter) – the length of the lever.

Since the problem statement is not specified, we assume that the lever is in equilibrium.

Then, according to the lever rule, we find the length of the larger lever arm:

F1 * dl1 = F2 * dl2;

dl1 = F2 * dl2 / F1 = 240 * 0.06 / 40 = 6 * 0.06 = 0.36 meters.

Then the length of the lever will be equal to:

l = dl1 + dl2 = 0.36 + 0.06 = 0.42 meters (42 centimeters).

Answer: The length of the arm is 42 centimeters.