A force of 40H is applied to the left arm of the lever, and a weight of 3 kg is suspended to the right arm at a distance of 20 cm

A force of 40H is applied to the left arm of the lever, and a weight of 3 kg is suspended to the right arm at a distance of 20 cm from the fulcrum. Find the length of the arm if it is in balance. Consider the lever weightless.

Given:

F1 = 40 Newton – force applied to the left arm of the lever;

m = 3 kilograms is the mass of the load applied to the right arm of the lever;

dl = 20 centimeters = 0.2 meters – the distance from the fulcrum to the right arm of the lever;

g = 10 Newton / kilogram – acceleration of gravity.

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

Since, according to the condition of the problem, the lever is in equilibrium, and in addition, the lever itself is weightless, then, according to the rule of the lever, we find:

F * (L – dl) = m * g * dl;

F * L – F * dl = m * g * dl;

F * L = F * dl + m * g * dl;

F * L = dl * (F + m * g);

L = dl * (F + m * g) / F = 0.2 * (40 + 3 * 10) / 40 =

= 0.2 * (40 + 30) / 40 = 0.2 * 70/40 = 14/40 = 0.35 meters.

Answer: The length of the arm is 0.35 meters.