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Torque/Force Problem-Help Please?

  1. Jul 16, 2007 #1
    1. The problem statement, all variables and given/known data

    A rotational axis is directed perpendicular to the plane of a square and is located as shown in the drawing. Two forces, 1 and 2, are applied to diagonally opposite corners, and act along the sides of the square, first as shown in part a and then as shown in part b of the drawing. In each case the net torque produced by the forces is zero. The square is one meter on a side, and the magnitude of 2 is 6 times that of 1. Find the distances a and b that locate the axis. Note that a and b are not drawn to scale.
    a = ? m
    b = ? m


    2. Relevant equations

    sum of net torque=Fl

    3. The attempt at a solution

    Ok, so I got this help, but i don't think it's right...anyone else able to check this to see where it's wrong?

  2. jcsd
  3. Jul 16, 2007 #2


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    You are right and the solution is wrong. The theta's in the problem are the angles between the force vector and the vector connecting the axis to the point where the force is applied. It's going to be tough to make them all 90 degrees. On the other hand the question is pretty nasty. There is not a unique solution for a or b! I don't know quite what the person who framed this question was thinking of. I would suggest you just try and find one point by confining your search to the diagonal of the square connecting the two forces. That makes the problem at least manageble and the angles are easy.
  4. Jul 16, 2007 #3

    I agree, I'm finding it impossible to find a and b! I wanted to use the diagonal also, but I only know the large diagonal is the sq.root of 2 and it has a correlation to the sq.root of a squared + b squared....there arent any numbers to use though!
  5. Jul 16, 2007 #4


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    If you are along the diagonal then a=b (at least in the second picture). Compute the distance to the points where the force is applied. sin(theta) for both forces will be the same - just opposite in sign, right?
    Last edited: Jul 17, 2007
  6. Jul 17, 2007 #5


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    From the first setup one gets that

    [tex]b = 6a[/tex]

    From the second setup one gets that

    [tex]\frac{6}{1} = \frac{1 - a}{b}[/tex]

    since the action line of the resultant needs to go through the pivot point in order to produce a zero torque.

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