At the end of a lever of length L, weights of m1 and m2 are suspended. By neglecting the mass

At the end of a lever of length L, weights of m1 and m2 are suspended. By neglecting the mass of the lever, find the shoulders of strength.

Given:

L is the total length of the lever;

m1 is the mass of the first cargo;

m2 is the mass of the second cargo;

g is the acceleration of gravity.

It is required to determine l1 and l2 – the shoulders of the forces.

By the condition of the problem, we neglect the mass of the lever itself. Then, given that the lever is in equilibrium, according to the rule of the lever, we have the following relationship:

F1 * l1 = F2 * l2;

m1 * g * l1 = m2 * g * l2;

m1 * l1 = m2 * l2.

Taking into account that l2 = L – l1, we get:

m1 * l1 = m2 * (L – l1);

m1 * l1 = m2 * L – m2 * l1;

m1 * l1 + m2 * l1 = m2 * L;

l1 * (m1 + m2) = m2 * L;

l1 = m2 * L / (m1 + m2).

l2 = L – l1 = L – m2 * L / (m1 + m2) = (m1 * L + m2 * L – m2 * L) / (m1 + m2) =

= m1 * L / (m1 + m2).

Answer: the leverage is l1 = m2 * L / (m1 + m2) and l2 = m1 * L / (m1 + m2).