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A question about torque

  1. Feb 27, 2010 #1
    1. The problem statement, all variables and given/known data

    I'm working through a problem:

    "A particle of mass m is attached to a rod of negligible mass and length L. The rod is attached to the ground at point O. If the rod makes an angle [tex]\Theta[/tex] with the horizontal as it falls, calculate the torque about point O.

    Here's a rough sketch of the diagram:


    My book says that the equation for torque is [tex]\tau=Frsin\theta[/tex] where [tex]\theta[/tex] is the angle between the force and position vector.

    The answer for this is [tex]Lmg cos\theta[/tex], but I don't understand why it's cosine. The book answer says to solve it in one of two ways: by multiplying the perpendicular component of the force vector to the lever arm, or by multiplying the perpendicular component of the lever arm to the force (indicating these two methods are equivilent).

    How do you decide whether to use cosine or sine?

  2. jcsd
  3. Feb 27, 2010 #2


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    Homework Helper

    You basically do it like they said in your book

    If you do it like this, the component of the force perpendicular to the lever arm is mgcosθ. Do you understand how the force is split in to this components and a sine component parallel to the arm?

    The length of the lever arm perpendicular to the downward force mg is Lcosθ. This one should be easier to see.
  4. Feb 27, 2010 #3
    Thanks for the help.

    I do understand how the force is split with cosine and sine. My problem is figuring out which angle to use. Is it similar to assigning force vectors in an inclined plane diagram?

  5. Feb 27, 2010 #4


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    This equation sums it up pretty nicely. Where's the force vector (hint: the force is gravity)? Where's the position vector (hint: it's the vector from the pivot to the object)? Which angle is the angle between them?
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