A 83 kg window cleaner uses a 8.5 kg ladder that is 4.9 m long. He places one en

Question

A 83 kg window cleaner uses a 8.5 kg ladder that is 4.9 m long. He places one end on the ground 2.0 m from a wall, rests the upper end against a cracked window, and climbs the ladder. He is 2.5 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. 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?

Explanation / Answer

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 2.0m from the wall, then the top of the ladder is (4.9² - 2.0²) m = 4.47 m from the ground.

CW moments: (8.5kg * 9.81m/s² * ½*4.9m) + (83kg * 9.81m/s² * (2.5/4.9)*2.0m) = 1035.14 N·m

CCW moments: Fw * 4.47m

Then the force on the wall Fw = 1035.14 N·m / 4.47m = 231.57 N (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 231.57 N, the ground must also exert a horizontal force of 231.57 N (otherwise the ladder would be translating to the left). Vertically we have

Fg =( 83kg * 9.81m/s²) + (8.5kg * 9.81m/s²) = 897.6 N

so the magnitude of the force on the ladder from the ground is

R = (897.6² + 231.57 ²) N = 927 N

(c) = arctan(897.6/231.57) 75.53º

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