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Tricky Torque Problem

  1. Nov 15, 2012 #1
    1. The problem statement, all variables and given/known data
    65zm9i.jpg

    The writing was just my reasoning, which I don't think is right.
    Assume the beam is massless.

    2. Relevant equations
    T = Fd sin θ
    F = mg
    net force must be zero.
    net torque must be zero.

    3. The attempt at a solution
    d * 1/2F + 980 = d * (sin 37) T
    1/2 F + 980 = (sin 37) T
    I don't know what to do from here.
    I can mathematically solve the problem to an extent, but conceptually I don't understand how to do the problem. Can someone explain to me how to do the problem from scratch and the reasoning for steps (like what forces are acting on the beam)?
     
  2. jcsd
  3. Nov 15, 2012 #2

    Doc Al

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    Staff: Mentor

    What's F? I thought the beam was massless.

    To solve this sort of problem, apply the conditions for equilibrium. The net torque = 0 is one of them. What are the others?
     
  4. Nov 15, 2012 #3
    ah, so there is no F, meaning it would just be 980 = sin 37 T, where T is the tension.
    net force = zero, forces left = forces right, forces up = forces down, clockwise torques = counterclockwise torques, x and y components of force may separately be set to 0. And T cos 37 would give the force exerted by the wall, I assume?
     
  5. Nov 15, 2012 #4

    Doc Al

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    Staff: Mentor

    Right.
    Exactly.
     
  6. Nov 15, 2012 #5
    And compression would thus be opposite to the force exerted by the wall, for a net force of zero?
     
  7. Nov 16, 2012 #6

    Doc Al

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    Staff: Mentor

    Yes, the net force is zero. The force exerted by the wall is the compression.
     
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