Question

A worker of mass 85 kg has been doing repairs on a building, working from a ladder of mass 35 kg, which has its center of gravity at its midpoint. Wrapping up his workday, the worker starts to move the ladder when he notices that he left his 15-kg toolbox attached at a point 0.75 m from the top. In a hurry, he re-positions the ladder carelessly, with its top resting at 8.0 m above the ground and its base 4.5 m from the wall.   Climbing back up, the worker notices with alarm that the ladder begins to slip when he is three-quarters of the way to the top. Later, when he tells you this story, you offer that you can analyze the static friction force that kept his ladder in place up to the point of slipping. Under the assumption that there is no friction between the ladder and the wall, what is the coefficient of static friction between the ladder and the ground?

Include a sketch and a free-body diagram. Also provide some evaluation, verifying units and assessing whether the answer is reasonable.

Answer 1

Here is the scenario.

Mg is weight of the person (acts as 3L/4) and

mg is weight of the ladder ( acts at L/2)

f is friction force, F is normal force from wall

Taking torque about point where ladder touches the gorund

F * L * cos = Mg * 3L/4 * sin = mg * L/2 * sin

angle is at the top ( with the ladder and wall)

we can cancel 'L' as it appears on both sides, so

F * cos = Mg * 3/4 * sin = mg /2 * sin

F = Mg * 3/4 * tan   + mg/2 * tan

where

tan = 4.5 / 8

as the ladder is not moving

F = f

Mg * 3/4 * tan   + mg/2 * tan = u * (Mg + mg)

4.5 / 8 ( 0.75Mg + 0.5mg) = u * (Mg + mg)

put M = 85 kg and m = 35 kg

4.5 / 8 ( 0.75 * 85 * 9.8 + 0.5 * 35 * 9.8 ) = u * 1176

u = 0.38