Question

A 63 kg window cleaner uses a 7.7 kg ladder that is 5.6 m long. He places one end on the ground 3.2 m from a wall, rests the upper and against a cracked window, and climbs the ladder. He is 1.2 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? em dem (a) Number Units IN lem Units TN em (b) Number (c) Number lem Units To (degrees) em Click if you would like to show Work for this question: Qren Show Works

Answer 1

A)

Balancing the moments about lowest point of ladder,

(7.7 x 9.8 x (3.2/2)) + (63 x 9.8 x (1.2/5.6) x 3.2) = F sqrt(5.6^2 - 3.2^2)

Force, F = 118.39 N

B)

Horizontal component, H = 118.39 N

Vertical component, V = (63 + 7.7) x 9.8 = 692.86 N

Net force, F = sqrt(H^2 + V^2) = 702.9 N

C)

Theta = arctan (V/H) = 80.3°

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