A 75 kg window cleaner uses 10 kg ladder that is 5.0 m long. He places one end o

Question

A 75 kg window cleaner uses 10 kg ladder that is 5.0 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 3.0 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 are (a) the magnitude of the force on the winodw 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?

The answers, in 2 significant figures, are: (a) 280N, (b) 880 N, (c) 71 degrees

What is the procedure in obtaining these answers?

Explanation / Answer

a)

equating torque about the lowermost point of the ladder,

10*9.8*(2.5/2) + 75*9.8*(3/5)*2.5 = N*sqrt(5^2-2.5^2)

where , N = normal force that the glass exerts on the ladder,

So, N = 282.9 N <--------------answer(part (a))

So, answer in 2 significant figures = 280 N

b)

equating forces along the horizontal direction,

the horizontal component of friction,Fx = N = 282.9 N

and the vertical component of friction,Fy = (10+75)*9.8 = 833 N

So, the net frictional force,F = sqrt(Fx^2+Fy^2) = sqrt(282.9^2+833^2) = 879.7 N <---------answer

So, answer in 2 significant figures = 880 N

c)

so, the angle the force makes = atan(833/282.9) = 71.2 degrees <-------------answer

So, answer in 2 significant figures = 71 degrees

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