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A couple is two forces acting in parallel

  1. Apr 10, 2008 #1
    [​IMG]
    A couple is two forces acting in parallel and opposite directions.
    F1=20N
    F2=22N
    F3-18N
    Couple=15N
    I have to determine the magnitude and direction of a single force that has the same effect on the beam as F1+F2+F3+ the couple. To do this would I find the sum of all the x and y components of these forces, then find the resultant of the x and y and make it negative?
    Also, im having trouble understanding the diagram. Which ways is the couple exerting force on the beam?
     
  2. jcsd
  3. Apr 11, 2008 #2

    tiny-tim

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    Hi Ry122! :smile:

    Yes … except why make it negative?

    You're being asked to find one force which will replace everything else … it should give the same result!

    (This isn't like finding a reaction force, which is always opposite.)

    Once you've found the magnitude and direction of the force, F say, you must then find the point of application (the place where the force is to be applied to the beam).

    So you need to find the point where the moment of F will be the same as the torque of 15N clockwise at the right-hand end of the beam. :smile:
     
  4. Apr 11, 2008 #3
    a what distance is the couple being applied to the beam?
     
  5. Apr 11, 2008 #4

    tiny-tim

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    According to the digram, the 15N is being applied to the very right-hand end of the beam.

    Actually, looking at the diagram more carefully, I'm rather confused: all the forces seem to be the wrong way round.

    The normal force is acting towards the support which we see underneath the beam, instead of upwards; and the other two forces - which presumably come from ropes - are also acting down instead of up.

    I wonder whether the 15N is the wrong way round as well? :confused:

    Is that diagram from the book, or is it your own diagram?
     
  6. Apr 11, 2008 #5
    it's from the book. how can you tell it is being applied to the end? all i can see is the word couple but no arrows indicating where it is being applied.
     
  7. Apr 11, 2008 #6

    tiny-tim

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    There's a light grey clockwise curved arrow just to the right of the beam.

    An arrow for torque can't "point" anywhere, since it's circular … so they draw it round something, in this case the end of the beam, to show where it is. :smile:
     
  8. Apr 11, 2008 #7
    If it helps, the beam is supported at A and B
     
  9. Apr 11, 2008 #8
    Isnt it only the force perpendicular to the beam that would effect the magnitude of the torque? Not the y component as well like I said in my first post.
     
  10. Apr 12, 2008 #9

    tiny-tim

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    Hi Ry122! :smile:

    No … any two or more forces which don't go through the same point (three or more if they're all in the same plane, of course, since two will always meet) will produce a torque.
    I didn't understand what you meant by that, and I still don't.

    To find the resultant torque, add the torques of all three given forces to the given couple, and find where on the beam F must be applied to give the same torque. :smile:

    erm … can you show what you've done to find F, so we can see how you're doing so far? :smile:
     
  11. Apr 12, 2008 #10
    Sorry i meant to say that only the y component would have an impact on torque. This is because there is no vertical length given for an x component to act on.With momentum problems in my text book you have to resolve each of the acting forces into y and x components then multiply the y component by the horizontal distance from the force to the fulcrum. And For the x component of the force you multiply it by the vertical distance of the acting force from the fulcrum. In this problem there is no vertical length given to multiply the x component by.
     
  12. Apr 12, 2008 #11

    tiny-tim

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    Ah … I see what you mean now. It was "Not the y component as well" which confused me!

    Yes, that's right … and you seem to have all the ingredients now, so … get mixing! :smile:
     
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