College Physics Statics and Torque question

Question

To get up on the roof, a person (mass 70kg) places a 6m alumin ladder (mass 10kg) against the house on a concrete pad with the base of the ladder 2m 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 2m from the bottom. The person is standing 3m from the bottom. what are the magnitutes of the forces on the ladder at the top and bottom?

Answer 1

Let d = 2m: distance from the base of the ladder to the house

L = 6m: length of the ladder

L_cm = 2m: center of mass of the ladder from the bottom

M = 70kg: mass of person

m = 10kg: mass of ladder

x = 3m: distance from the person to the bottom of the ladder

The angle θ subtended by the ladder with the ground is calculated using

θ = cos^-1(d/L) = 70.5°

Let N₂ be the normal reaction at the wall, and N₁ be the normal reaction at the ground, and let f be the frictional force exerted by the ground on the ladder. Consider the torque acting on the ladder about the

point where it meets the ground. Only three forces contribute to this torque:

the weight, mg, of the ladder, which acts 2m from the bottom the ladder; the weight, Mg, of the person, which acts a distance x along the ladder; and the reaction, N₂, at the wall, which acts at the top of the ladder. Note that the reaction force acts to twist the ladder in the opposite sense to the two weights.

thus, setting the net torque to zero, you get

mgL_cmcosθ + Mgxcosθ - N₂Lsinθ = 0

==> N₂ = (mgL_cmcosθ + Mgxcosθ)/Lsinθ ==.... plug in the numbers and the results gives you the force at the top of the ladder

Now using the condition that zero net vertical force acts on the ladder, you get

N₁ -mg - Mg = 0 ==> N₁ = (m+M)g. plugging in the values of m, M and g = 9.8m/s² gives you the magnitude of the force at the bottom of the ladder.