Weights weighing 14 kg and 26 kg were suspended at the ends of a thin homogeneous rod 2 m long.

Weights weighing 14 kg and 26 kg were suspended at the ends of a thin homogeneous rod 2 m long. At what distance from the middle of the rod should the support be placed so that the rod is in equilibrium.

Weights weighing 14 kg and 26 kg were suspended at the ends of a thin homogeneous rod 2 m long. At what distance from the middle of the rod should the support be placed so that the rod is in equilibrium

m1 = 14 kg.

m2 = 26 kg.

l = 2 m.

r -?

The condition for the balance of the lever is the equality of the moments of forces M1 and M2, which act on the ends of the lever: M1 = M2.

The moment of force M is determined by the formula: M = F * l, where F is the force, l is the lever arm.

The force of gravity acts on the lever: F = m * g.

The equilibrium condition will take the form: m1 * g * l1 = m2 * g * l2.

l2 = l – l1.

m1 * l1 = m2 * (l – l1).

m1 * l1 = m2 * l – m2 * l1.

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

l1 = 26 kg * 2 m / (14 kg + 26 kg) = 1.3 m.

l2 = 2 m – 1.3 m = 0.7 m.

r = l / 2 – l2.

r = 2 m / 2 – 0.7 = 0.3.

Answer: the support should be offset from the center of the rod by 0.3 m towards the larger load.