A uniform ladder `5.0 m` long rests against a frictionless, vertical wall with its lower end `3.0m` to from the wall. The ladder weighs `160 N`. The coefficient of static friction between the foot of the ladder and the ground is `0.40`. A man weighing `740 N` climbs slowly up the ladder.
What is the actual frictional force when the man has climbed `1.0 m` along the ladder?

Options

A. `3m`
B. `5m`
C. `2.7m`
D. `1.25m`

Answer

A

Explanation

Since the wall is frictionless, the only vertical forces are the weights of the man and the ladder, and the normal force.
For the vertical forces to balance., `N_(1)=w_(1)+w_(m)=160N+740N=900N` and the maximum frictional forces is `mu_(s)N_(2)=(0.40)(900N)=360N`. Note the that ladder makes contact with the wall at as height of `4.0 m` above the ground. balancing torques about the point of contact with the ground.

`(4.0m)_(2)=(1.5m)(160N)+(1.0m)(3/5)(740N)=684N.m` so `N_(2)=174.0N.` This horizontakl force about must be balanced by the frictional force. Setting the frictional fore, and hence `n_(1)` equal to the maximum of `360 N` and solving for the distance `x` along the ladder `(0.4m)(360N)=(1.50m)(160N)+x(3/5)(740N)` so `x=2.70m`