A 73 kg window cleaner uses a 17 kg ladder that is 5.1 m long. Heplaces one end

Question

A 73 kg window cleaner uses a 17 kg ladder that is 5.1 m long. Heplaces one end on the ground 3.0 m from a wall, rests the upper endagainst a cracked window, and climbs the ladder. He is 2.8 m upalong the ladder when the window breaks. Neglect friction betweenthe ladder and window and assume that the base of the ladder doesnot slip. (a) When the window is on the verge ofbreaking, what is the magnitude of the force on the window from theladder?
N

(b) When the window is on the verge of breaking, what is themagnitude of the force on the ladder from the ground?
N

(c) What is the angle of this force on the ladder?
° (above the horizontal) (a) When the window is on the verge ofbreaking, what is the magnitude of the force on the window from theladder?
N

(b) When the window is on the verge of breaking, what is themagnitude of the force on the ladder from the ground?
N

(c) What is the angle of this force on the ladder?
° (above the horizontal)

Explanation / Answer

m g L / 2 * cos 54 + M g * 2.8 cos 54    =clockwise torque about base of ladder (m = ladder, M =painter) 9.8 * (17 * 2.55 + 73 * 2.8) * cos 54 = 1427 Nm N L sin 54 = counterclockwise torque   where N isnormal force of window on ladder N = 1427 / (5.1 * sin 54) = 346 N     alsoequals horizontal force of ground on ladder Fvertical = (M + m) g = 90 * 9.8 = 882N      vertical force of ground onladder   (the window is frictionless) Total force of ground on ladder = (3462 +8822) = 947 N tan = 882 / 346     = 68.6deg    angle of force on ladder due to ground

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