Question

A ladder of length 5m has mass of 25 Kg. the ladder is leaned against a frictionless vertical wall at an angle of 10 degrees with the vertical. A repairman with a mass of 82Kg needs to stand on an upper rung of the ladder that is 1 m from the end. What is the minimum coefficient of friction between the ladder and the floor such that the ladder doesn’t slip?

Answer 1

Require that the moment about the upper support point be zero. The vertical force component at the lower end of the ladder is (25 + 82)*g = 1049 J, since the upper support point is frictionless.

The distance of the lower end of the ladder from the wall is
X = 5m * sin 10 = 0.868 m.
The repairman stands in the ladder at a horizontal distance of 1.0 m sin 10 = 0.1736 m from the wall.

Assume the Center of Mass of the ladder is midway between the ends. The CM is then 5 sin 10 = 0.868 m from the wall.
25*0.868*g + 82*0.1736*g = 107*g*Us(5 cos10

g cancels out. Solve for the required static friction coefficient Us

Answer 2

Thank you very much