Question

A ladder having a uniform density and a mass m rests against a frictionless vertical wall, making angle 49.0° with the horizontal. The lower end rests on a flat surface where the coefficient of static friction is µ_s = 0.410. A window cleaner with mass M = 2m attempts to climb the ladder. Given a ladder length of L, what fraction of the length of the ladder (x / L) will the worker have reached when the ladder begins to slip?
x / L =

Answer 1

The horizontal forces of the wall against the top of the ladder F_w

and the force of friction at the foot of the ladder F_x.

The horizontal forces must add to zero so F_w = F_x
The vertical forces must sum to zero, F_y is the vertical force of the floor on the foot of the ladder.
2mg is the weight of the window cleaner and mg is the weight of the ladder.

Therefore F_y =2mg+mg= 3mg.
F_wsin49L = F_w = mgxcos49(2mgxcos49x(x) + mgxcos49x(L/2) , where x is the distance up the ladder the cleaner is from the foot of the ladder.
by simplyfying the above equation we get
(2x+L/2)/Lxsin49 = F_x = F_yxµ_s =3mgx0.41
mg is on both sides and can be eliminated and cos49/sin49 = cot49 = 1/tan49
(2x+L/2)/L x cot49 = 3x0.41
2x/L + 1/2 = tan49 x 1.23
2x/L = (1.414) – (1/2)
x/L = 0.914/2 = 0.457