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Calculate the force on the other support

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  1. Nov 28, 2014 #1
    1. The problem statement, all variables and given/known data
    A uniform thin rod of weight 100 N is supported horizontally by two bricks at its
    ends. At t=0 one of these bricks is kicked out quickly. Calculate the force on the
    other support immediately thereafter.

    2. Relevant equations
    I'm not sure if I am on the right track with this equation since I don't have the length and from what I know, I need the length for torque problems: Mg(L/2)sin(theta)

    3. The attempt at a solution
    I thought since each brick has a force of 50N that if one brick is locked the force on the supporting brick is -50N since it's a downward force. Please help me with the equation and how to go about it. Thanks
     
  2. jcsd
  3. Nov 28, 2014 #2

    haruspex

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    Follow that through and see what happens. Maybe the length will not feature in the answer.
    (From dimensional analysis, you can predict that it won't. Given an acceleration (g) a mass and a length as input, and a force to be the result of multiplying and dividing these and raising them to powers, the only combination that works is acceleration * mass * constant.)
     
  4. Nov 30, 2014 #3
    So what would my constant be? My weight would be 50N since it's for one brick, and my acceleration is 9.81m/s. I'm confused about the constant.
     
  5. Nov 30, 2014 #4

    haruspex

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    You need to determine that by working through the usual equations.
    Create unknowns for mass, length, angular acceleration and write out the standard equations for linear acceleration versus force and angular acceleration versus torque.
     
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