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Components of Torque

  1. Oct 18, 2010 #1
    1. The problem statement, all variables and given/known data
    The force F = (15x - 22y)N is applied at r = (4.0x + 7.0y) meters. What torque (about the origin) does this force produce?

    2. Relevant equations
    Torque = r * F * sin theta
    r = magnitude of radius, F = magnitude of F, theta = smallest angle between the directions of r and F

    3. The attempt at a solution
    I was wondering what "about the origin" was referring to. Theoretically, shouldn't I be able to find the magnitudes of F and r with (ax2 + ay2)1/2 and the angles they make with the positive x axis with (tan theta = ay/ax) and then figure out the smallest angle between them from that and calculate the torque?

    Is it just saying the origin is the rotational axis so I don't have to worry about trying to translate the equations to some other rotational axis?
     
  2. jcsd
  3. Oct 18, 2010 #2

    collinsmark

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    Homework Helper
    Gold Member

    Yes, that is a completely valid way to solve the problem. Find magnitudes of F and r, and the angle between them, then plug those into your formula. Then use the right hand rule to determine the direction of the resulting torque.

    Or, alternately (which is what I would do because I think it's a little easier), to use the definition of torque, which is

    [tex] \vec \tau = \vec r \times \vec F [/tex]

    and recognize that the definition of the cross product is,

    [tex]
    \vec a \times \vec b =
    \left|
    \begin{array}{ccc}
    \hat x & \hat y & \hat z \\
    a_x & a_y & a_z \\
    b_x & b_y & b_z
    \end{array}
    \right|
    [/tex]

    where the right side of the equation is the determinate of the matrix.

    Either way should give you the correct answer.
    Essentially, yes. That's a good way to think of it. The length vector r has two points. One point at the origin and the other at (4.0, 7.0, 0.0). [And the force is applied at point (4.0, 7.0, 0.0).]
     
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