A 70 kg window cleaner uses a 16 kg ladder that is 5.6 m long. He places one end on the ground 2.0 m from a wall, rests the upper end against a cracked window, and climbs the ladder. He is 3.5 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) When the window is on the verge of breaking, what is the angle of that force on the ladder?
tau_net = 0.
tau = r x F
The Attempt at a Solution
I was able to get part (a), which was 193.914N. But I don't get part (b). I drew a forces diagram as shown (attached file). I resolved the force on the ladder from the ground into horizontal and vertical components because otherwise this force will have no torque, and how could I calculate the force then?
[FONT=verdana, helvetica, sans-serif]I count ccw as positive and cw as negative, and the system is not moving so net torque is 0.
So I put the origin where the ladder meets the vertical wall. Then there will only be 1 F_N, and furthermore, only one ccw torque: F_N(x). Since the angle it makes is the complement of theta, I can use cos (theta) instead of sin theta.
The full equation would then look something like this:
tau_net = F_N(x)*5.6*cos(theta) - F_N(y)*5.6*sin(theta) - F_gl*5.6/2*sin(theta) - F_gp*(5.6-3.5)*sin(theta) = 0.
We can calculate theta since we're given the hypotenuse 5.6 and one of the legs 2.
But this gives me two unknowns — F_N(x) and F_N(y). How can I solve this?
Also, for (c), I've no idea which force they're talking about...
Any help would really be appreciated![/FONT]