Question

A ladder having a uniform density and a mass m rests against a frictionless vertical wall, making angle 48.0° with the horizontal. The lower end rests on a flat surface where the coefficient of static friction is µ_s = 0.370. 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?

Answer 1

Using Force balance in the vertical direction on the ladder

Fy_net = 0 = N - m*g - M*g

N = Normal force due to lower end

m*g = weight of ladder

M*g = weight of window cleaner

Given that M = 2m

N = m*g + M*g = 3*m*g

Now Using force balance in horizontal direction

Fx_net = 0 = Fh - Ff

Fh = Force due to upper end of ladder in horizontal direction

Ff = Friction due to ladder's lower end

Fh = Ff = us*N

Fh = us*3*m*g

Now Using torque balance about the lower end of ladder

Net Torque = 0 = (L/2)*m*g*cos 48 deg + x*M*g*cos 48 deg - L*Fh*sin 48 deg

M = 2m

x = distance of window cleaner from bottom of ladder

2*m*g*x*cos 48 deg = L*Fh*sin 48 deg - 0.5*L*m*g*cos 48 deg

divide whole equation by (2*m*g*L*cos 48 deg)

x/L = Fh*tan 48 deg/(2*m*g) - 0.25

Since Fh = us*3*m*g

x/L = us*3*m*g*tan 48 deg/(2*m*g) - 0.25

x/L = 1.5*us*tan 48 deg - 0.25

given that us = 0.370

x/L = 1.5*0.370*tan 48 deg - 0.25

x/L = 0.366

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