Two boys swing on a board thrown over a log. The length of the board is 5.5 m.

Two boys swing on a board thrown over a log. The length of the board is 5.5 m. Where should the fulcrum of the board be so that the boys are in balance, if the mass of one is 48 kg, and the other is 40 kg?

L = 5.5 m.

g = 9.8 m / s2.

m1 = 48 kg.

m2 = 40 kg.

L1 -?

L2 -?

The condition for the balance of the log is the equality of the moments of forces M that act on the log from different sides of the log: M1 = M2.

The moment of the force M is the product of the force F, which acts, at the smallest distance from the line of action of the force to the axis of equilibrium L: M = F * L.

The forces of the boys’ weights act on the log from different sides, the value of which is determined by the formula: F1 = m1 * g, F2 = m2 * g.

m1 * g * L1 = m2 * g * L2.

m1 * L1 = m2 * L2.

L1 + L2 = L.

L1 = L – L2.

m1 * (L – L2) = m2 * L2.

m1 * L – m1 * L2 = m2 * L2.

m1 * L = m1 * L2 + m2 * L2.

L2 = m1 * L / (m1 + m2).

L2 = 48 kg * 5.5 m / (48 kg + 40 kg) = 3 m.

L1 = 5.5 m – 3 m = 2.5 m.

Answer: the balance point of the log will be at a distance L1 = 2.5 m from a boy with a mass of m1 = 48 kg, and at a distance of L2 = 3 m from a boy with a mass of m2 = 40 kg.