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Calculating torque.

  1. Feb 24, 2013 #1
    1. The problem statement, all variables and given/known data
    A body is located at (-2m, 0, 4m). An force is applied at this body Fx=6N. Determine the torque.


    3. The attempt at a solution
    First of all I apologize for my insane paint skills. Yes, I'm very talented, thank you.
    Torque.png
    Ok, I thought I new how to calculate torque, but this question confused me greatly.

    Now here's the thing, there are two ways of calculating Torque (which are really the same thing, but still, two ways):
    One of those is using the right hand rule to determine the sign, getting the magnitude of R and using the magnitudes and sin of the angle between them to calculate Torque.
    The other is getting the vectors and doing vectorial product, using matrix determinant.

    Sometimes I do in both ways to sort of 'proof-read' my results, and in this case I'm finding conflicting results.

    Let's just get the pointless calculations out of the way:

    [itex]r = 4.47m \\ \theta = 63.43°[/itex]

    Ok, with all that said, let me show you what I did:

    1)
    [itex]\tau = 4.47*6*sin(63.43°)*(-1) = -23.98N.m[/itex]

    Why the (-1)? Well, because of the right-hand rule (and I think the error lies here). The rule states that I should put my fingers along the r dirirection, and curl them towards the force F, therefore, by the drawing we can all see that it would be negative.

    2) (calculating the determinant)
    [itex]\tau = (4*F_x + 2F_z) \\ F_z = 0 \\ \tau = 4*6 = (24 N.m) J[/itex]
    (ignore the difference between 24 N.m and 23.98 N.m, this is just because of the bad precision involved in the calculations)

    So, I believe that the second method is the correct one (since I directly calculate the determinant), but my question then is, why the right-hand rule fails?
     
    Last edited: Feb 24, 2013
  2. jcsd
  3. Feb 24, 2013 #2
    Is this the full question?
    I would assume you would be dealing with an extended body when working out the torque.
     
  4. Feb 24, 2013 #3
    Ops, that was stupid of me, the torque is in relation to the origin.
     
  5. Feb 24, 2013 #4
    OK, so you have the force applied to the point (-2m, 0, 3m) and the body rotates about the origin.
    Then, how do you get r=4.47m?
    Is the point (-2m, 0, 4m) ?
    Edit : OK, I can see this is true from the diagram
     
    Last edited: Feb 24, 2013
  6. Feb 24, 2013 #5
    Just looking at your diagram again - make sure the directions of your axes are correct.
    If the x- and y-axes are in the plane of the page as indicated, then the z-axis should be pointing towards you, ( out of the plane of the page ).

    When you evaluate the determinant in your second method, you should include the unit vectors in the calculation in order to see the direction of the torque.
     
  7. Feb 24, 2013 #6

    Sorry, the diagram is mistaken, I meant to draw x and z axis (as there's no component y in position). I'm going to edit the Torque too, I'm sorry, I made a considerably messy post, but I encountered (I'm going to edit that)
    [itex] (24 N.m)J[/itex]

    So in short, the 2nd method tells me that it is 24 N.m in the positive direction of the y axis (Not represented in the image) while the first tells me that it is 24 N.m in the negative direction of the y axis.
     
  8. Feb 24, 2013 #7
    The first method gives the same direction as the second.
    I think your problem is correctly visualising the orientations of the 2 vectors - this can be difficult in a 3D problem.
     
  9. Feb 24, 2013 #8
    If your vertical axis is the z-axis, then the y-axis goes into the page.
    If you bring the force vector down to the origin and curl the position vector towards the force vector, your thumb should point into the page ( ie in the positive y-direction )
     
  10. Feb 24, 2013 #9
    OHHH I see! I decided on a very bad axis distribution, that confused me completely!

    Thanks a lot, that's clear now !
     
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