A 75 kg window cleaner uses a 10 kg ladder that is 5.0 m long. He places one end on the
ground 2.5 m from a a wall, rests the upper end against a cracked window and climbs the ladder. He is 3m 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.
net torque = r1F1 + r2F2 + r3F3
The Attempt at a Solution
i have spent literally two hours on this one question, drawing triangles and diagrams over and over. i think my main problem is that i have no idea what to use for r. is it horizontal or vertical or along the angle of the ladder? every way i try, it doesn't come out to the correct answer. i also lack an understanding of what forces act in this situation.
F1 = weight of man on ladder
F2 = force of... window on ladder?
F3 = force of ladder on window
r1 = ?
r2 = ?
r3 = ?
net torque = (75*9.8)r1 + (10*9.8)r2 + F3r3 = 0