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Homework Help: Reaction at a fixed support

  1. Oct 18, 2008 #1
    Hi all;

    An L-shaped bar is constrained by a fixed support at A while the other extreme C is free. Determine the reaction at the point A when the bar is loaded by the depicted forces.


    [tex]AB=3m, BC=2m, CD=1m[/tex]
    [tex]R_{Cx}=6N, R_{Cy}=2N, R_{Cz}=3N[/tex]

    Either I'm doing something wrong or there's more than enough data in the problem. I assume we just have to write the three equations of equilibrium:




    Then we just calculate their resultant in order to find [tex]R_B[/tex]. If so, then why would we need all those distances? :confused:

    Maybe my assumptions are plain wrong, I'm really confused... Help appreciated. Thanks.
  2. jcsd
  3. Oct 18, 2008 #2


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    This is not really my field, but should you be considering torques as well as forces?
  4. Oct 20, 2008 #3
    I'm not sure... That's why I was asking. Are there any other reactions that I didn't consider? Thanks.
  5. Oct 20, 2008 #4


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    As per Redbelly98's comment, you need to consider torques.
    Your translational equilibrium equations will correctly give you the force reactions at the fixed support, but the loadings also produce moment reactions at that support (also called torques or couples). Thus, you must sum all moments = 0 about the support to find the moment reactions at that support. And you have to look separately at moments about each axis, if they exist.
  6. Oct 20, 2008 #5
    When in three dimensions, there are actually six equations of equilibrium (three in 2D). You have moment equations about each axis direction. Note that you can solve mathematically without looking at it about each axis, but it just makes it easier sometimes to break it up into components (like you could break a force into x, y, and z force components).

    From your post, I'm guessing you're not too familiar with moments. A moment is simply a twisting force. Like the other guys said, it's also called torque (a couple is slightly different). If you push a door in a straight line, it's still going to swing about the hinge.. right? But you pushed it in a straight line. This is because your force caused a moment about the door hinge. Now if you were to push the same door a few centimeters away from the hinge, it would take a lot more effort to get it to swing at the same speed. This is because the same force creates less of a moment at this distance. The formula is:

    Moment = Force * Perpendicular distance.

    It's important that you take the perpendicular distance. If you push at right angles to the door (straight on), then the distance would be that along the door between your hand and the hinge.

    Moments are positive when counter-clockwise and negative when clockwise.

    Sorry if that was confusing or long winded.

    As for your problem.. it's simple because each force is is a single plane. It might help you to reset point A to the origin.
    Last edited by a moderator: Oct 20, 2008
  7. Oct 20, 2008 #6


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    Presumably the OP has been taught about torques or moments in class (otherwise such a problem would not have been assigned), so beyond giving basic information like this, I wouldn't proceed giving the solution any further than this. People learn more the more they have to do themself.


  8. Oct 20, 2008 #7

    It just seemed from his post that he hadn't (or wasn't present when they taught it), so thought doing the first part would help give him the idea.
  9. Oct 20, 2008 #8


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    Yeah, it can be a tough call sometimes knowing how much information to give out. I had hoped after post #2 that just hearing the word "torque" would trigger a memory of something that had been covered in class. It's possible moments or torques were not covered by the teacher, but that would make it odd for this question to be assigned in the first place.

    Perhaps some clarification from disclaimer would help us know what to do to help out.
  10. Oct 21, 2008 #9
    Right, at the time I didn't realize that there should be six reactions at a fix support. Thanks.
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