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Find angle for the ring to be in equilibrium

1. Homework Statement
Find the angle Theta so that the system is in equilibrium
Mass of each block: 2 Kg
Mass of ring: 4π Kg
Sin título.png


2. Homework Equations
Static equilibrium:
Rotational equilibrium

3. The Attempt at a Solution
Static equilibrium:
2g + 2g + 4πg = Normal
N = 4g + 4πg
Sin título - copia.png


Taking the torque with respect to the center of mass:

(N*(cosθ + sinθ)*2R*√2)/√2 * 3π + g*R(cosθ-sinθ) = [ 2*g*(cosθ+sinθ)*( (R/√2) + (2R√2)/3π) ] / √2 + 2*g*cosθ *(R + 2R/3π) + 2g*sinθ*(2R/3π)

This doesn't lead me anywhere.
Thanks for your help
 

BvU

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Dag Anton, :welcome:

Center of mass is way too complicated: depends on ##\theta##. Take the support point or the center of the circle
 
Last edited:

haruspex

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Not sure whether you figured this out, but I believe the given mass 4π is for the complete ring, so the 3/4 ring shown has mass 3π.
It will be easier to think about if you add in the missing quarter ring, plus another placed symmetrically on the other side to compensate. Then you only have to deal with that quarter ring balancing the two suspended masses.
What are the x and y coordinates of the mass centre of the quarter ring, relative to the circle's centre?
 

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