A load weighing 10 kg is suspended on the left shoulder of the weightless lever.

A load weighing 10 kg is suspended on the left shoulder of the weightless lever. In order to balance the lever, on its right shoulder, which is 2 times longer than the left one, it is necessary to suspend a weight

M1 = 10 kilograms is the mass of the load suspended from the left arm of the lever;

g = 10 Newton / kilogram (approximate value);

x2 = 2 * x1 – the right arm of the lever is twice as long as its left arm.

It is required to determine M2 (kilogram) – what weight should be suspended from the right shoulder in order to balance the lever.

By the condition of the problem, the lever has no own weight. Then, by the rule of the lever:

M1 * g * x1 = M2 * g * x2;

M1 * x1 = M2 * x2;

M1 * x1 = M2 * (2 * x1);

M1 * x1 = 2 * M2 * x1;

M1 = 2 * M2;

M2 = M1 / ​​2 = 10/2 = 5 kilograms.

Answer: in order to balance the lever, you need to hang a weight equal to 5 kilograms on the right shoulder