Weights of 0.7 kg and 2.8 are suspended at the ends of the lever in equilibrium

Weights of 0.7 kg and 2.8 are suspended at the ends of the lever in equilibrium. The total length of the arm is 1 m. Determine the distance from the fulcrum to the greater force.

m1 = 0.7 kg.

m2 = 2.8 kg.

g = 9.8 m / s ^ 2.

l = 1 m.

l2 -?

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 lever is acted upon by gravity: 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 = 2.8 kg * 1 m / (0.7 kg + 2.8 kg) = 0.8 m = 80 cm.

l2 = 1 m – 0.8 m = 0.2 m = 20 cm.

Answer: the distance to the greater force is l2 = 20 cm.