What is the angle of equilibrium for a rod in a hemispherical bowl?

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The discussion focuses on determining the angle of equilibrium for a uniform rod resting in a hemispherical bowl. The user presents equations related to moments and forces, seeking confirmation on their approach. Feedback suggests simplifying the problem by taking moments about point B instead of A, which would eliminate one variable. Additionally, the equations can be refined to better relate the rod's length and angle to the bowl's radius. The conversation emphasizes the importance of correctly calculating distances and components to find the equilibrium angle.
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Homework Statement



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

Homework Equations

The Attempt at a Solution


[/B]
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?
 

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Tracyxyzd said:
Does that seem right?
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.
 
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)
 
Tracyxyzd said:
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)
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.
 
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