Answer in units of degrees

Question

A 17.7 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 0.22 . The angle between the horizontal and the ladder is θ . The person wants to climb up the ladder a distance of 0.91 m along the ladder from the ladder’s foot. What is the minimum angle θmin (between the horizontal and the ladder) so that the person can reach a distance of 0.91 m without having the ladder slip? The acceleration of gravity is 9.8 m/s 2 . Answer in units of ◦

Answer 1

The friction force cannot exceed FMG , where F=0.22, M=17.7 kg and G=9.8 (gravity). At the moment the ladder is about to slip, the friction force reaches that value. The wall pushes on the ladder with the same FMG force. Let calc the torque applied to the ladder at that moment. It is MGH cos(θ) pushing "down" (H=0.91m) and FMGL sin(θ) supporting (where L is the length of the ladder which we do need to know!). So tan(θmin) = (H/L) / F. When θ goes below θmin, support becomes not enough. The M and G are not part of the equation.