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Magnitude of torque due to gravity

  1. Oct 28, 2009 #1
    1. The problem statement, all variables and given/known data
    A person carries a long pole of mass 9 kg and length 4.4 m. Find the magnitude of the torque on the pole due to gravity.



    2. Relevant equationssin(60) and cos(60), Torque=Force_gravity*r*sin(theta)



    3. The attempt at a solution
     
  2. jcsd
  3. Oct 28, 2009 #2

    tiny-tim

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    Hi dorian_stokes! :smile:
    erm :redface:

    where is he holding the pole? and at what angle to the horizontal? :confused:
     
  4. Oct 28, 2009 #3
    he's holding it at the end and the pole is at a 60 degree angle. Sorry for not explaining it, I don't have a picture.
     
  5. Oct 29, 2009 #4

    tiny-tim

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    (just got up :zzz: …)

    ok, then …
    … torque = force times perpendicular-distance-to-the-line-of-application-of-the-force :wink:

    (and cos = adj/hyp, sin = opp/hyp)

    … what do you get? :smile:
     
  6. Jun 20, 2011 #5
    wouldn't it follow the formula torque=-mgrsin(x)? because we're given the angle that the pole makes with the ground, which makes the force that is perpendicular to the center of mass of the pole, opposite to the angle that we're given... and because the distribution of the pole's mass is uniformly distributed, our r is actually r/2. right?
     
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