1. Oct 30, 2009

### pdrjuarez

1. The problem statement, all variables and given/known data

A uniform 12.0 m long ladder weighing 125 N rests against a smooth vertical wall. The bottom of the ladder makes an angle of 67 degrees with the floor. A bucket of paint with a mass of 14.0 kg rests on a rung, 7 m from the bottom end of the ladder. What is the frictional force exerted on the bottom of the ladder?

2. Relevant equations
Torque=F distance

3. The attempt at a solution
Ok so I took both the bucket and the ladder itself and calculated their force perpendicular to the ladder using the bottom of the ladder as the center point, and from that got the clockwise torque. Now that I have the total clockwise torque, the counterclockwise torque has to be the same, right? So the Torque would have to equal the frictional force times the distance (T=Ff * d) But what exactly would the distance be? wouldnt it be 0 because it's at the bottom of the ladder, which is the center point? Help please I don't know exactly what to do next

2. Oct 31, 2009

### ehild

The torque of the friction is 0, but you have to include the torque of the normal force from the wall, at the top of the ladder, and you get it from the equilibrium of forces: it's magnitude is equal to the force of friction.

ehild

3. Oct 31, 2009

### pdrjuarez

ahhh, that makes so much sense now
thank you