4) A 30 ft ladder weighing 80 lbs rests against a slippery (frictionless) wall.

Question

4) A 30 ft ladder weighing 80 lbs rests against a slippery (frictionless) wall. The foot of the ladder is 20 feet from the wall. The coefficient of static friction between the ground and the bottom of the ladder is 0.5, and the coefficient of sliding friction is 0.4. A person weighing 160 lbs climbs up the ladder. (a) Calculate the horizontal contact force Fx of the wall on the ladder, as well as the vertical contact force Fy and the frictional force f on the ladder by the ground required to hold the person when they are half way up. (b) How far up the ladder can the person climb before the ladder will be in danger of sliding out?

Explanation / Answer

a] By balancing torque about lower end, Fx * h = mg*10 + Mg*10

Fx * sqrt(30^2-20^2) = 80*10+160*10 =2400

Fx = 2400/sqrt(30^2-20^2) = 107.3 lb answer

Fy = mg+Mg = 80+160 = 240 lb

friction f = Fx = 107.3 lb answer

b] maximum friction f = uN = 0.5*240 = 120

Fx = f=120 lb

balancing torque about bottom of ladder, 120*sqrt(30^2-20^2) = 80*10 + 160*(x/30)*20

2683-800 = 320x/3

x = [2683-800]*3/320 = 17.65 ft answer

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