A 80 kg window cleaner uses a 18 kg ladder that is 5.9 m long. He places one end

Question

A 80 kg window cleaner uses a 18 kg ladder that is 5.9 m long. He places one end on the ground 2.9 m from a wall, rests the upper end against a cracked window, and climbs the ladder. He is 2.7 m up along the ladder when the window breaks. Neglect friction between the ladder and window and assume that the base of the ladder did not slip. (a) Find the force exerted on the window by the ladder just before the window breaks (b) Find the magnitude and direction of the force exerted on the ladder by the ground just before the window breaks ° (above the horizontal)

Explanation / Answer

angle made by the ladder with the wall cos theta = 2.9/5.9 = 60.55 deg

Net torque T = 0

T = 5.9 *Fg * sin (180-60.55) -(2.7 * mg * sin (190-90-60.55) -(2.9 * Mg sin (180-90-60.55) = 0

Fg = 263.16 N

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Horizontal force Fx = Fg = 263.16 N

Fy = mg + Mg

Fy 18* 9.8 + 80* 9.8

Fy = 960.4 N

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