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Rod in Hemispherical Bowl

  1. Jun 23, 2015 #1
    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?
     

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    Last edited: Jun 23, 2015
  2. jcsd
  3. Jun 23, 2015 #2

    haruspex

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    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.
     
  4. Jun 24, 2015 #3
    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)
     
  5. Jun 24, 2015 #4

    haruspex

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    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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