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Applied Torque On A Beam

  1. Dec 7, 2012 #1
    1. The problem statement, all variables and given/known data
    Calculate the net torque (magnitude and direction) on the beam in the figure below about the following axes.

    (a) an axis through O, perpendicular to the page

    (b) an axis through C, perpendicular to the page


    2. Relevant equations



    3. The attempt at a solution

    I was able to solve part (b), but was unsuccessful with part (a). For part (a), I took note of the fact that the 30 N force applied at a 45 deg. angle would produce zero torque of the beam, because it is being applied at the axis of rotation. This means we have only two forces acting on the beam, that are actually producing a torque.

    So, [itex]\Sigma\tau=25N\sin(30^{\circ})\times2.0m-10N\sin(20^{\circ})\times4.0m[/itex] However, this yields the incorrect value. What did I do wrong?
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     

    Attached Files:

  2. jcsd
  3. Dec 7, 2012 #2

    tiny-tim

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    Hi Bashyboy! :smile:
    Draw the complete triangle, and you'll see that it isn't 30° :wink:
     
  4. Dec 7, 2012 #3
    Are you trying to say that I am suppose to find the angle with respect to the horizontal? If so, why do I have to use that particular angle?
     
  5. Dec 7, 2012 #4

    tiny-tim

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    because the line from O is horizontal :smile:

    the line from O to the point of application of the right-hand force makes an angle of 20° with the force

    what angle does the line from O to the point of application of the left-hand force make with that force? :wink:
     
  6. Dec 8, 2012 #5
    A 45 deg. angle?
     
  7. Dec 8, 2012 #6

    tiny-tim

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    oh, i forgot there were 3 forces :redface:

    i should have said, what angle does the line from O to the point of application of the middle force make with that force?
     
  8. Dec 8, 2012 #7
    A 60 deg angle?
     
  9. Dec 8, 2012 #8

    haruspex

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    Yes.
     
  10. Dec 8, 2012 #9
    Okay, my question, then, is why do we have to take the angle with respect to the horizon?
     
  11. Dec 8, 2012 #10

    haruspex

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    It isn't with respect to the horizon. A torque about a point is produced by a force acting through a line at some offset from the point, i.e. not through the point. The magnitude of the torque is given by the magnitude of the force multiplied by the distance from the point to the line of action, i.e. as measured at right angles to the line of action. If we trace back that middle force through its line of action it will miss O by distance OC sin(60o).
    That's a very nonvectorial way of looking at things. The vectorial way is to say that the force is a vector, a vector from O to point in the line of action is OC (any such vector will do), so the torque is the cross-product of the two vectors.
     
  12. Dec 9, 2012 #11

    tiny-tim

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    yup … torque = displacement x force

    so you always use the displacement, ie the distance from your origin (of moments) to the point of application of the force,

    and the cross product explains why it's always the sin of the angle between them :smile:
     
  13. Dec 9, 2012 #12

    PhanthomJay

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    There are multiple ways to calculate torques of a force about a point on an axis, and you have been given a couple already. There are others, like resolving the forces into their components and using the component perpendicular to the position vector to calculate the torque, since the component parallel produces none. You seemed to have done part b corectly, so what is the problem with part a? You may have sined when you should have cosined, maybe...recheck your work and use the method most comfortable and simple for you.
     
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