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Torques and Rotational Equilibrium

  1. Mar 18, 2009 #1
    1. The problem statement, all variables and given/known data
    The mass m1 is 0.55 kg and it is located at x1 = 30 cm. The pivot point is represented by the solid triangle located at x = 45 cm. The mass of the meter stick (mms = 0.40 kg) is located at its geometric center, xms = 50 cm. The mass m2 is 0.35 kg and it is located at x2 = 80 cm. Calculate the net torque (in N⋅m with the proper sign) due to these three weights. Use g = 9.8 m/s2.


    2. Relevant equations
    T=r*Fsin(theta)


    3. The attempt at a solution
    Where do I even begin?
     
  2. jcsd
  3. Mar 18, 2009 #2

    LowlyPion

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    This might be helpful:
    http://en.wikipedia.org/wiki/Torque

    That might help you generally determine the sign of your torque.

    For the rest of it depending on which side the weights are located from the point of interest, then you just add the moments together (or subtract as appropriate). For the center of mass of the ruler, just use its distance to the pivot as your moment arm.
     
  4. Mar 18, 2009 #3
    I'm still really confused.
     
  5. Mar 18, 2009 #4

    LowlyPion

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    You have a see saw don't you? It's not like I can see your picture.

    Torque is force acting over the distance of the moment arm.

    So you have weights. m1 for instance is .55 kg*9.8 and it's (.45m -.3m) away from the pivot. That means the torque of m1 is .55*(.15)*9.8 The sign will be from r X F, the vector cross product, which since r is to the left and F is down looks to me by the right hand rule to be up and out of the page. Counter clockwise being + here.

    Figure the moments for the remaining weights and add them up.
     
  6. Mar 18, 2009 #5
    Oh, I see. Thanks
     
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