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Torques in equilibrium w/ angle

  1. Jan 14, 2013 #1
    on a meter stick the pivot point is placed at 1/4 its length; predict the force needed to balance the meter stick by pulling upward on the end of it with a string making an angle of 30 degrees w/ respect to the vertical.

    Me trying to solve it:
    sigma T=0
    T1+T2=0
    T1 = 0 <--this is the pivot point, i think it equals 0 b/c we what to reach equilibrium
    T2 = rFsin(theta)
    0+rFsin(theta)
    F=-0/(rsin(theta)) ???
     
  2. jcsd
  3. Jan 14, 2013 #2

    haruspex

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    It's the '=0' in 'T1+T2=0' that says we're to reach equilibrium. T1 is the torque from the gravitational force. Where is the centre of mass of the stick? How far from the pivot point? What torque does it exert?
    As for the string, it's 30 degrees to the vertical. Be careful about sine versus cosine. (As a check, I always think to myself, what if the specified angle were zero? Would the force be zero or max?)
     
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