A 70 kg window cleaner uses a 14 kg ladder that is 5.0 m long. He places one end

Question

A 70 kg window cleaner uses a 14 kg ladder that is 5.0 m long. He places one end on the ground 2.3 m from a wall, rests the upper end against a cracked window, and climbs the ladder. He is 3.0 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 is the magnitude of the force on the window from the ladder? When the window is on the verge of breaking, what is the magnitude of the force on the ladder from the ground? What is the angle of this force on the ladder? (above the horizontal)

Explanation / Answer

m = 70 kg

M = 14 kg

costheta = 2.3/5 = 62.6 degree

torque = 0

5*Fg*sin(180 - theta) - 3*mg*sin(180 - (90 -theta)) - 2.3 Mgsin(180 - (90-theta))

Fg = 3mgsin(152) + 2.3Mgsin(152)/ 5sin(117.39)

Fg = 251 N

part b )

force in x direction

Fh = Fg = 251 N

in y direction

Fv = mg + Mg

Fv = 823.2 N

F = sqrt(Fx^2 + Fy^2)

F = 860.062 N

part c )

angle = tan^-1(Fy/Fx) = 73.04

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