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Math Help - physics problem

  1. #1
    Member Chokfull's Avatar
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    Unhappy physics problem



    OK so you have the diagram here, with only a bit of the left side cut off, but it's just the wall. the coefficient of static friction between B and the table is .25, B weighs 711 N. find the maximum weight of block A, such that the system is stationary.

    I got this:
    the rope attatched to B exerts a horizontal force on B, so for B to just barely not move ther force must be 177.75 N, and this is the tension in the rope. Therefore, that is the force exerted on the knot by the wall, so the vertical force must be 177.75*\sin 30^o, and this is the weight of the block. The book says im wrong, but i dont see where i erred.
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  2. #2
    MHF Contributor
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    Hi



    Equilibrium of B
    T_B - F = 0
    - m_B g + N = 0

    Just at the rupture of equilibrium F = fN

    Therefore T_B = F = fN = f m_B g

    Equilibrium of the knot
    T_B - T \cos \alpha = 0

    T_A - T \sin \alpha = 0

    Therefore T_A = T \sin \alpha = T_B \tan \alpha = f m_B g \tan \alpha

    Equilibrium of A
    - m_A g + T_A = 0

    m_A = \frac{T_A}{g} = f m_B \tan \alpha
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  3. #3
    Member Chokfull's Avatar
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    Talking

    okay, thanks. My main problem was that I haven't worked with more complex rope systems, and i wasn't sure how the force on block B worked. I was thinking that maybe, because gravity was pulling block A down, the force on block B may have been partially downwards.
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