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

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Homework Help Overview

The problem involves a uniform rod resting inside a hemispherical bowl, with the goal of determining the angle of equilibrium. The context is centered around concepts of static equilibrium and forces acting on the rod within the geometry of the bowl.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster presents equations related to moments and forces acting on the rod, questioning the validity of their approach. Some participants suggest alternative methods, such as taking moments about different points to simplify the problem.

Discussion Status

The discussion is active, with participants providing feedback on the original poster's equations and suggesting simplifications. There is an exploration of different approaches to the problem, but no consensus has been reached on a specific method or solution.

Contextual Notes

Participants are navigating the complexities of the problem, including the relationships between the rod's length, the radius of the bowl, and the angle of equilibrium. There is an emphasis on ensuring all forces and moments are accounted for in the analysis.

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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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