Difficult Mechanics 2 Problem

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    A uniform Lamina is in the shape of an Equilateral triangle ABC, and rests in equilibrium inside a vertical hoop with B and C in contact with the hoop, and A at the center of the circle. Contact is smooth at C but rough at B, where the coefficient of friction is U (mu). The lamina is in limiting equilibrium with B lower than C, and the axis of symmetry through A makes an angle theta with the vertical. By first taking moments about A show that \tan\theta=U*\sqrt3/(\sqrt3-U)
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    (Original post by Electric-man7)
    A uniform Lamina is in the shape of an Equilateral triangle ABC, and rest in equilibrium inside a vertical hoop with B and C in contact with the hoop, and A at the center of the circle. Contact is smooth at C but rough at B, where the coefficient of friction is U (mu). The lamina is in limiting equilibrium with B lower than C, and the axis of symmetry through A makes an angle theta with the vertical. By first taking moments about A show that tan(theta)=U*sqrt(3)/sqrt(3)-U
    What have you tried??
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    (Original post by RDKGames)
    What have you tried??
    I have tried to take the moments about A, and also the moments about C, but when I take the moments about C I have to include not only the the moment of the normal force at B, but also the moment of the force of friction. I have also tried resolving all the forces ( contact forces at B, C, and the weight of the lamina) in the directions perpendicular and parallel to BC, but I can't seem to continue to from here. Any Ideas?
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    (Original post by Electric-man7)
    A uniform Lamina is in the shape of an Equilateral triangle ABC, and rests in equilibrium inside a vertical hoop with B and C in contact with the hoop, and A at the center of the circle. Contact is smooth at C but rough at B, where the coefficient of friction is U (mu). The lamina is in limiting equilibrium with B lower than C, and the axis of symmetry through A makes an angle theta with the vertical. By first taking moments about A show that \tan\theta=U*\sqrt3/(\sqrt3-U)
    Having spent several hours trying to do it via the suggested method and getting it wrong, I resorted to triangle of forces (combined the forces at B into one to reduce the count to three).

    And get \displaystyle \tan \theta = \frac{\sqrt{3}\mu}{\sqrt{3}-2\mu}

    Which although different to the desired result in your first post, does agree with a thread on stackexchange. And the working was a hell of a lot simpler.
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    (Original post by ghostwalker)
    Having spent several hours trying to do it via the suggested method and getting it wrong, I resorted to triangle of forces (combined the forces at B into one to reduce the count to three).

    And get \displaystyle \tan \theta = \frac{\sqrt{3}\mu}{\sqrt{3}-2\mu}

    Which although different to the desired result in your first post, does agree with a thread on stackexchange. And the working was a hell of a lot simpler.
    Thanks a lot, I have eventually managed to use the suggestion, to solve the problem but I can't imagine how this problem was ever on an OCR exam.
 
 
 
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