Question

A 67 kg window cleaner uses a 11 kg ladder that is 5.9 m long. He places one end on the ground 2.7 m from a wall, rests the upper end against a cracked window, and climbs the ladder. He is 3.3 m up along the ladder when the window breaks. Neglect friction between the ladder and window and assume that the base of the ladder does not slip. When the window is on the verge of breaking, what are (a) the magnitude of the force on the window from the ladder, (b) the magnitude of the force on the ladder from the ground, and (c) the angle (relative to the horizontal) of that force on the ladder?

Answer 1

(a)what is the magnitude of the force on the window from the ladder?

For balancing torques, let's make the point where the ladder touches the ground be the reference point.
Also, let's make the angle the ladder makes with the ground = θ.

(torque due to weight of ladder) + (torque due to weight of window cleaner) = (torque due to window pushing on top of ladder)

(11 kg)(9.81 m/s^2)(5.9 m / 2)sin(90° - θ) + (67 kg)(9.81 m/s^2)(3.3 m)sin(90° - θ) = F(5.9 m)sinθ
(11 kg)(9.81 m/s^2)(5.9 m / 2)cosθ + (67 kg)(9.81 m/s^2)(3.3 m)cosθ = F(5.9 m)sinθ
(11 kg)(9.81 m/s^2)(5.9 m / 2)(2.7 m / 5.9 m) + (67 kg)(9.81 m/s^2)(3.3 m)(2.7 m / 5.9 m) = F(5.9 m)(sqrt[(5.9 m)^2 - (2.7 m)^2] / 5.9 m)

F = 216.68 N

(b) what is the magnitude of the force on the ladder from the ground?

The horizontal component due to friction will be opposite in direction, but have the same magnitude of the force on the window: 216.68 N.

The vertical component of force will equal the weight of the ladder and window cleaner:
(11 + 67 kg)(9.81 m/s^2) = 765.18 N

total force = sqrt[(216.68)^2 + (765.18)^2]
total force = 795.27 N

(c)What is the angle of this force on the ladder?
tanφ = 765.18 / 216.68
φ = 74.19° (measured upward from ground)