Question

A ladder having a uniform density and a mass *m* rests
against a frictionless vertical wall, making angle 50.0° with the
horizontal. The lower end rests on a flat surface where the
coefficient of static friction is µ_{s} = 0.420. 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

Computing the torque about ladder's foot, we get ( remember, clockwise torque - negative, counter clockwise torque - positive). Torue = force*distance

Fw*sin 50*L = 2m*g*cos 50*x + m*g*cos 50*L/2

where x is the distance of cleaner from foot of the ladder.

solving for Fw , get

Fw = 2m*g*cos 50*x + m*g*cos 50*L/2 / sin 50 *L

This is equal to F_{vertical} = (2m + m )g*
u_{s}

2m*g*cos 50*x + m*g*cos 50*L/2 / sin 50 *L = 3mg*0.420

we can cancel mg as it is present on both sides, we get

2*cos50*x + cos50*L/2 / sin 50*L = 1.26

take cos 50 common from numerator,

cos 50 ( 2x + L/2) / L*sin 50 = 1.26

0.839099 (2x + L/2) / L =1.26

2x + L/2 / L = 1.501609

2x/L + 1/2 = 1.501609

2x/L = 1.001609

x/L = 0.5008

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