Question

17) To get up on the roof, a person (mass 70.0 kg) places a 6.00-m aluminum ladder (mass 10.0 kg) against the house on a concrete pad with the base of the ladder 2.00 m from the house. The ladder rests against a plastic rain gutter, which we can assume to be frictionless. The center of mass of the ladder is 2 m from the bottom. The person is standing 3 m from the bottom. What are the magnitudes of the forces on the ladder at the top and bottom?

Answer 1

l = 6m,   d = 2m,   x = 3m, point G = center of mass pf ladder, OG = 2m = (l/3), M = mass of the person = 70 kg, m = mass of the ladder = 10 kg

The angle subtended by the ladder with the ground satisfies

Let be the normal reaction at the wall, let be the normal reaction at the ground, and let be the frictional force exerted by the ground on the ladder, as shown in the diagram. Consider the torque acting on the ladder about the point where it meets the ground (Point O). Only three forces contribute to this torque: the weight, of the ladder, which acts half-way along the ladder; the weight, , of the person, which acts a distance along the ladder; and the reaction, , at the wall, which acts at the top of the ladder. The lever arms associated with these three forces are , , and , respectively. Note that the reaction force acts to twist the ladder in the opposite sense to the two weights. Hence, setting the net torque to zero, we obtain,

which yields,

The condition that zero net vertical force acts on the ladder yields

Hence,