Question

Romeo takes a uniform 11.5-m ladder and leans it against the smooth wall of the Capulet residence. The ladder's mass is 24.0kg, and the bottom rests on the ground 2.80m from the wall. When Romeo, whose mass is 70 kg, gets 87.0 percent of the way to the top, the ladder begins to slip. What is the coefficient of static friction between the ground and the ladder?

Answer 1

For the ladder to keep from slipping, sum of all forces has to be zero

Friction force = f = Fw

Normal force = Fn = (M + m)g

here two gravitational will tend to twist the ladder

torque from ladder weight = 5.75mg costheta

Similarly Romeo creates a torque = 10.005 Mg costheta

Finally, to keep the ladder from twisting, the wall also provides a torque

Tw = Fw * 11.5 * sin theta

Fw = f = mu*Fn = mu * (M + m ) g

11.5 * mu * (M + m ) *g sintheta = 5.75mg costheta + 10.005 Mg costheta

mu = (5.75m + 10.005 M ) cot theta/( 11.5 ( M+m) )

theta = cos^-1(2.80/11.5) = 75.91 degree

mu = 0.195