What is the minimum angle θmin (between the horizontal and the ladder) so that the person can reach a distance of 1.1 m without having the ladder slip?

Question

A 20.3 kg person climbs up a uniform ladder with negligible mass. The upper end of the ladder rests on a frictionless wall. The bottom of the ladder rests on a floor with a rough surface where the coefficient of static friction is 0.23 . The angle between the horizontal and the ladder is θ . The person wants to climb up the ladder a distance of 1.1 m along the ladder from the ladder’s foot. The acceleration of gravity is 9.8 m/s^2 .

Answer 1

Sam, we meet again!

This is a problem in static  equilibrium, which means that the total force and the total torque on the ladder are zero. There are four forces on the ladder: the normal force from the wall, the weight of the person, the normal force from the floor, and the frictional force from the floor. At the critical angle θmin at which the ladder slips, the frictional force from the floor reaches its maximum value of mu_s *N. 

So we have 4 equations: F_x=0, F_y=0, tau=0, and (friction from floor)=mu_s* (normal force from floor)

Our 4 unknowns are: normal force from wall, Normal force from floor, Frictional force from floor, and θmin.

Solve!

Cheers,

Sam