1. Nov 10, 2009

### Want to learn

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

A 80 kg ladder is 3.00 m in length is placed against the wall ant an unknown angle. The center of gravity of the ladder is 1.2 m from the base of the ladder. The coefficient of friction between the base and the ladder is 0.400. No friction between the wall and the ladder. What is the minimum angle the ladder makes so it would not slip and fall?

2. Relevant equations

2 conditions of equilibrium

Force equations

3. The attempt at a solution

This is what I tried. I figured that the normal force is $$N = mg$$ which equals in this problem then 784N.Then I find out the Force of friction $$F_f = \mu*N$$ in this case it's (0.400)*(784) = 313.6 Then I created this. Tclockwise = T counterclockwise

313.6 x 3sin(x) = 784 x 1.2cos(x)

I end up getting tan (1) = 0.017... which doesn't make any sense :(.

What am I doing wrong.

Thanks

2. Nov 10, 2009

### semc

Hello~ Maybe you want to sum up the torque created by gravity and friction? Since its static equilibrium so summation torque should be 0. Your forces are correct but the moment arm is wrong so the answer is wrong.

3. Nov 10, 2009

### Want to learn

How would I find the correct lever arm??

Could you elaborate on the adding up part also?

Sorry for asking such silly questions, but I just don't seem to understand.

4. Nov 10, 2009

### semc

Alright maybe you can google moment arm but the way i do it is shortest distance from direction of force applied to the torque.
Adding up? Just do like you normally do for translational equilibrium only this time its F.d

Nope no silly questions. Silly questions are questions you don't ask

5. Nov 10, 2009