Two balls weighing 1kg and 5kg are held together by a weightless rod. The center of the first ball is 90cm

Two balls weighing 1kg and 5kg are held together by a weightless rod. The center of the first ball is 90cm from the center of the second. At what distance from the center of the lighter ball is the center of gravity of the system?

m1 = 1 kg.

m2 = 5 kg.

L = 90 cm = 0.9 m.

L1 -?

When the lever is in equilibrium, the moments of forces from opposite sides of the lever are equal to each other: M1 = M2.

M1 = F1 * L1 = m1 * g * L1.

M2 = F2 * L2 = m2 * g * L2.

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

m1 * L1 = m2 * L2.

We express the length of the entire lever by the formula: L1 + L2 = L.

L2 = L – L1.

m1 * L1 = m2 * (L – L1).

m1 * L1 = m2 * L – m2 * L1.

m1 * L1 + m2 * L1 = m2 * L.

The distance from the center of the lighter ball to the center of gravity of the system L1 will be determined by the formula: L1 = m2 * L / (m1 + m2).

L1 = 5 kg * 0.9 m / (1 kg + 5 kg) = 0.75 m.

Answer: L1 = 0.75 m.