Statics Question -- Bar and roller resting on a circular surface

[dksland]

Homework Statement

[/B]

Homework Equations

The Attempt at a Solution

[/B]
Honestly, I don't even know how to begin this problem. I've drawn myself some free body diagrams, but I'm uncertain of all of them. The weight of the bar is just (0,-mg) at .9r from the pivot point. But I don't know how to calculate the weight that's perpendicular to the pivot point because there's no angles so that doesn't help me. Is there an x component to the normal force? I feel like there should be, but I don't know how to calculate it because there is no x component to the weight. This was the first of 9 problems... I did the other 8 no problem. This one though... this one blows my mind for some reason.

 

[mfb]

The problem statement is hard to read.

You can calculate the angle based on the given distances and some trigonometry.

 

[haruspex]

Is there an x component to the normal force?

The normal force, by definition, is at right angles to the surface, so yes, there is an x component.
You have three forces: the normal force at A, the weight, and the joint reaction at O.
You have three unknowns: the magnitudes of the normal force, the horizontal component of the joint reaction, and its vertical component. Create symbols for them.
You have three equations: horizontal force balance, vertical force balance, and torque balance. Try to write those equations.

 

[mfb]

You have three unknowns: the magnitudes of the normal force, the horizontal component of the joint reaction, and its vertical component. Create symbols for them.
You have three equations: horizontal force balance, vertical force balance, and torque balance. Try to write those equations.

You will still need the angle as parameter to set up an equation that relates torque and normal force (because the direction of the normal force is not known in advance).

 

[haruspex]

You will still need the angle as parameter to set up an equation that relates torque and normal force (because the direction of the normal force is not known in advance).

I'm not sure what you are saying there. The direction of the normal is just a matter of geometry, no?

 

[mfb]

Yes, that's what I said in post 2 already, and it has to be calculated to solve the problem.