A 75 kg window cleaner uses a 10 kg ladder that is 5 m long. He places one end o

Question

A 75 kg window cleaner uses a 10 kg ladder that is 5 m long. He places one end on the ground 2.5 m from a wall, rests the upper end against a cracked window, and climbs the ladder. He is 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 the magnitude of the force on the window from the ladder, the magnitude of the force on the ladder from the ground, and the angle (relative to the horizontal) of that force on the ladder?

Explanation / Answer

costheta = 2.5/5

theta = 60 degrees

In equilibrium net torque about the base = 0

75*9.8*3*costheta + 10*9.8*5/2*costheta = N2*L*sintheta

(75*9.8*3*cos60) + (10*9.8*5/2*cos60) = N2*5*sin60

N2 = 283 N

++

(b)

N1 = (m1 + m2)*g

N1 = 85*9.8 = 833 N

(c)

90 degrees

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