Ladder Problem: Solving for the Minimum Angle

[Want to learn]

Ladder Problem :(

Homework Statement

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?

Homework Equations

2 conditions of equilibrium

Force equations

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

 

[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.

 

[Want to learn]

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.

I'm guessing your are talking about this part 784 x 1.2cos(x) ??
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.

 

[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

 

[ideasrule]

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)

Around which point are you taking the torque? After you define your point, make sure that ALL the forces not acting on that point are taken into consideration. Here, you didn't include the normal force, but the torque due to the normal force can't be 0 if the torque due to friction is non-zero.