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Simplification of force and couple system

  1. Sep 19, 2015 #1
    1. The problem statement, all variables and given/known data
    In the figure 4-35 c , the bar is rotating at point b , but the book gave the taking moment at point b .. This is confusing.... When we are taking moment at point b , the bar at particular point wouldn't move, am I right?

    2. Relevant equations


    3. The attempt at a solution
     

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  3. Sep 19, 2015 #2

    SteamKing

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    You're confused because you've jumped over a couple of steps.

    Start with figure 4-35(a). Someone is holding a bar which has a vertical load applied at the opposite end. The point of the sequence of figures 4-35(a)-(c) is to show how the moment created by F at the opposite end of the bar can be replaced by the same force F located to the end being held plus a couple whose magnitude is equal to the force F multiplied by the distance d.
     
  4. Sep 19, 2015 #3
    the diagram 4-35c is confusing , it shows the point B is moving ... It should show point A is moving , right ? Since B is the pivot ...
    What does -FB means ? recative force of Force B ? Or we apply the same force at point B in opposite direction ?
     
  5. Sep 19, 2015 #4

    SteamKing

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    You're still confused. Nothing is moving. The straight arrows merely indicate the direction of the force vectors. The curly arrows are there to indicate that a couple is present and the direction (CW or CCW) of the moment.

    The force -F which is shown at B indicates the reaction to the force F applied at the free end of the rod. It is the applied force F and its corresponding reaction -F which are responsible for creating the couple M, shown in figure 4-35(c). The thing to keep in mind here is all three diagrams represent a static situation.

    A suggestion: don't just look at the diagrams and expect to fully comprehend what is being described. There is a discussion in the text just above these figures which would answer most of your question here if you would read it while looking at the diagrams.
     
  6. Sep 19, 2015 #5
    Ok, the f at point a cancel off -f at point b , causing the moment to occur at point b ...but the text gives taking moment about point b , doesn't it means the pivot is situated at point b ?
     
  7. Sep 19, 2015 #6

    SteamKing

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    Yes. The point is, the couple can be positioned anywhere along the length of the rod.
     
  8. Sep 19, 2015 #7

    haruspex

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    It depends what you mean by a pivot. If you are thinking of it as the stick undergoing an actual rotation as a result of these applied forces, with the pivot being the point that remains stationary, you have no idea where the pivot will be. It depends on the mass distribution along the stick.
    If you simply mean the point that you choose to take moments about for analytical purposes, then as SteamKing posted, it doesn't matter which point you pick.
    What the text is saying is that if you apply two forces F and -F along lines of action a distance d apart then no matter where you put your axis the net moment will be Fd. That is the nature of a couple - it does not act around any particular axis.
     
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