# Rod in Hemispherical Bowl

1. Jun 23, 2015

### Tracyxyzd

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

A uniform rod AB of length 3R weight W rests inside a hemispherical bowl with radius of R. Determine the angle corresponding to equilibrium.

2. Relevant equations

3. The attempt at a solution

Moment about A: -1.5R*mg cos(theta)+B*AB=0
Sum Fx: Acos(theta)-mg sin(theta)=0
Sum Fy: Asin(theta)-mg cos(theta)+B=0
sin(theta)/R=Sin(180-2theta)/AB

I have 4 equations and 4 unknowns: A, B, AB, theta

Does that seem right?

#### Attached Files:

• ###### bowl.jpg
File size:
18.2 KB
Views:
160
Last edited: Jun 23, 2015
2. Jun 23, 2015

### haruspex

Yes, that all looks correct.
You could make it a bit simpler by taking moments about B instead of A, and leaving out sum Fy. That eliminates force B.
Also, the last equation can be simplified a lot. Just consider how to find AB/2 from R and theta.

3. Jun 24, 2015

### Tracyxyzd

If I take the moment about B, how would I get the perpendicular distance between the y component of the weight and B?
AB/2=Rcos(theta)

4. Jun 24, 2015

### haruspex

Yes, that gives you AB. Subtract half the rod length from that to get the distance from B to the midpoint of the rod, then multiply by cos(theta) to get the horizontal distance.