Question

A 73 kg window cleaner uses a 12 kg ladder that is 5.6 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 2.6 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.

(a) When the window is on the verge of breaking, what is the magnitude of the force on the window from the ladder?

(b) When the window is on the verge of breaking, what is the magnitude of the force on the ladder from the ground?

(c) What is the angle of this force on the ladder? (above horizontal)

Answer 1

a)

equating torque about the lowermost point of the ladder,

12*9.8*(2.9/2) + 73*9.8*(2.6/5.6)*2.9= N*sqrt(5.6^2-2.9^2)

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

So, N = 236.6N

So, answer in 2 significant figures = 237N

b)

equating forces along the horizontal direction,

the horizontal component of friction,Fx = N = 237N

and the vertical component of friction,Fy = (12+73)*9.8 = 833 N

So, the net frictional force,F = sqrt(Fx^2+Fy^2) = sqrt(237^2+833^2) = 866.05N

So, answer in 2 significant figures = 866N

c)

so, the angle the force makes = atan(833/237) = 74.11degrees

So, answer in 2 significant figures = 74.1 degrees