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CW/CCW Torques

  1. Feb 10, 2009 #1
    1. The problem statement, all variables and given/known data
    Find all cw and ccw torques. I don't know how to post the picture, but I was absent from my physics class today for AMC so I missed out. First problem: A horizontal massless rod of length 6 meters is pivoted about its center. There is a 2kg point mass on each end of the rod.


    2. Relevant equations
    torque = Fd


    3. The attempt at a solution
    So I looked at the textbook, but they only provide examples with cylinders. So I just need one example for this and I think I can pretty much do the rest. Do I calculate the Force for each one? or do I use the moment of Inertia?
     
  2. jcsd
  3. Feb 10, 2009 #2

    LowlyPion

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    Welcome to PF.

    You mean it's a see-saw.

    If it is in equilibrium then the sum of the Torques are 0.

    The F*2m on one side is equal to the F*2m on the other.
     
  4. Feb 10, 2009 #3
    oh. thanks. lol that's that only example where they're both equal. so uh.

    1. The problem statement, all variables and given/known data
    Find all cw and ccw torques. Find the net torque. Find the angular acceleration. A differential horizontal rod has a length of 4 meters and a mass of 5kg. It is pivoted about its center.


    2. Relevant equations
    torque = Fd
    torque = (I)(alpha)


    3. The attempt at a solution
    I found Fg = -49 N
    torque = Fxd = -98 N/m
    angular acceleration = torque / moment of inertia = -98 / (20/3) = -14.7 m/s^2?
    is this right?
     
  5. Feb 10, 2009 #4

    LowlyPion

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    Does the rod have mass?
     
  6. Feb 11, 2009 #5
    yeah 5kg
     
  7. Feb 11, 2009 #6

    LowlyPion

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    If it is pivoted about the center where is the net force?

    Is there a weight on one end as well?
     
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