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* = 2*m* 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_{w}sin49L = 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_{y}xµ_{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

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