Question

A 80 kg window cleaner uses a 18 kg ladder that is 5.6 m long. He places one end on the ground 2.4 m from a wall, rests the upper end against a cracked window, and climbs the ladder. He is 3.5 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. (a) When the window is on the verge of breaking, what is the magnitude of the force on the window from the ladder? Incorrect: Your answer is incorrect. N (b) When the window is on the verge of breaking, what is the magnitude of the force on the ladder from the ground? N (c) When the window is on the verge of breaking, what is the angle of that force on the ladder? Incorrect: Your answer is incorrect. ° (above the horizontal)

Answer 1

I'll assume the ladder is leaning to the right. Sum the moments about the base of the ladder -- that way the friction force and the normal force at the base of the ladder can be ignored. If the ladder's base is 2.4m from the wall, then the top of the ladder is √(5.6² - 2.4²) m = 5.059 m from the ground.
CW moments: 18kg * 9.8m/s² * ½*5.6m + 80kg * 9.8m/s² * (3.5/5.6)*2.4m = 1669.92 N·m
CCW moments: Fw * 5.059m
Then the force on the wall Fw = 1669.92N·m / 5.059m = 330.08N ← (a)

(b) To get the force on the ladder from the ground, sum the vertical and horizontal forces. Since the wall exerts a horizontal force on the ladder of 330 N, the ground must also exert a horizontal force of 330 N (otherwise the ladder would be translating to the left). Vertically we have
Fg = 80kg * 9.8m/s² + 18kg * 9.8m/s² = 960.4 N
so the magnitude of the force on the ladder from the ground is
R = √(960.4² + 330²) N = 1015.6 N

(c) Θ = arctan(960.4/330) ≈ 71º