Physics
posted by Chong on .
A 15 kg uniform ladder leans againt a wall in the atrium of a large hotel. The ladder is 8m long, and it makes an angle as 60 degree with the floor. The coefficient of static friction between the floor and the ladder is u+ 0.45. Assume the wall is frictionless.
a) How far along the ladder can a 60kg painter climb before the ladder starts to slip?
b) Therefore, at where height above the ground can the same painter climb before the ladder starts to slip ?

I am going to say g = 10 m/s^2
base of ladder from wall = 4 m
top of ladder at 4 sqrt 3 = 6.93 m
Take moments about top point of ladder
total weight = 10 (15+60) = 750 N
Friction force = .45*750 = 338 N
Clockwise:
750 N * 4 m = 3000
Counterclockwise:
x is horizontal distance painter to wall
338*6.93 + 150*2 + 600(x)
so
600 x = 3000  2342  300 = 358
x = .596 m
along ladder to top = 2x = 1.19 m
so along ladder from bottom = 8 1.19 = 6.8 m from floor along ladder to painter
height of painter = (6.8)/2 sqrt 3 = 5.9 m