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Equation for gravitational torque

  1. Nov 20, 2011 #1
    1. The problem statement, all variables and given/known data
    A thin rod (uniform density and thickness)
    with mass M and length L attached to floor at a fixed location by a frictionless hinge.

    while balanced vertically the gravitational torque acting on the rod is
    a.) zero
    b.) 1/2 MgL
    c.)1/3 MgL
    d.)1/4 MgL
    e.)1/6 MgL

    2. Relevant equations

    I know that torque is F x D(or L)
    F=Ma
    and a is g in this case,
    so T= MgL

    3. The attempt at a solution

    as stated above I got to T=MgL
    but I don't understand where the fractions are coming from? can someone please explain this?
     
  2. jcsd
  3. Nov 20, 2011 #2
    You forgot something when computing the cross product, there's a trig function in there somewhere ;)
     
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