Question

A 14.5 m uniform ladder weighing 510 N rests against a frictionless wall. The ladder makes a 55.0° angle with the horizontal.

(a) Find the horizontal and vertical forces the ground exerts on the base of the ladder when an 810 N firefighter is 4.10 m from the bottom.
Magnitude of the horizontal force
N Direction

towards the wallaway from the wall    

Magnitude of the vertical force
N Direction

updown    

(b) If the ladder is just on the verge of slipping when the firefighter is 9.20 m up, what is the coefficient of static friction between ladder and ground?

Answer 1

The sum of the horizontal forces = 0 => Fx - Fw = 0 => Fx = Fw where Fw is the horizontal force of the wall on the ladder and Fx is the horizontal force on the foot of the ladder pointing towards the wall.
The sum of the vertical forces = 0 = 510 + 810 - Fy => Fy =1320N
Fx = Fy*µs where Fy is the upward force of the floor on the foot of the ladder and µs is the coefficient of static friction.
When the man is at 4.1 m from the foot of the ladder
Balance the torques about the foot of the ladder
Fw*sin 55*14.5 = 510*cos55*7.25 + 810*cos55*4.1
Fw = (510*cos55*7.25 + 810*cos55*4.1)/(sin55*14.5) = 323.89 N
Fx = 323.89 N <--------------- (a)
Fy = 1320N <--------------- (a)
When the man is 8.9m from the foot of the ladder
Now balance the torques about the foot of the ladder:
Fw*sin55*14.5 = 510*cos55*7.25 + 810*cos55*9.2
Fw = (510*cos55*7.25 + 810*cos55*9.2 )/(sin55*14.5) = 538.411 N = Fx = Fy*µs = 1320*µs
Fx = 538.411N <----------
Fy = 1320N <-----------
Fw/Fy = µs = 538.411/1320 = 0.4079 <----------- (b)