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Force couple supported by line

  1. Oct 18, 2012 #1

    I know how to calculate the reaction to a force couple supported by 2 points. But what if it is supported by a line. See example attached. How do I figure the force distribution along the constant support?


    Attached Files:

  2. jcsd
  3. Oct 18, 2012 #2
    Good evening teleswamp and welcome to Physics Forums.

    Actually neither of your analyses are correct so let us start with the simplest first one and get that correct.

    The block as drawn is not in equilibrium so you cannot apply the equations of equilinrium to it.

    Firstly does you block have no weight?

    Secondly what of the horizontal components of the reactions at the two support points?
    Last edited: Oct 18, 2012
  4. Oct 18, 2012 #3
    Yes, for simplicity assume the block is essentially weightless. Assume it is pin connected on either end. The horizontal reaction would be 10 lbs. And the vertical reaction +10 lbs at one support and -10 lbs at the other.
  5. Oct 18, 2012 #4
    That's very slick and quick, but unfortunately not correct.

    If you are going to get the right answers you need to do the job properly.
  6. Oct 18, 2012 #5
    I guess you are questioning my supports? so pin connection on one end and roller on the other, Pin support vertical reaction is -10 lbs, horz reaction is -10 lbs, roller support is +10 lbs.
  7. Oct 18, 2012 #6
    Sorry, should be -20 vertical at pin, -10 horz at pin and +20 vertical at roller.
  8. Oct 18, 2012 #7
    Now that you understand why you cannot have two pinned joints - you then have three equations and four unknowns so you cannot solve them. Although in this case you can solve the vertical reactions only by using a moment condition as well as the vertical equilibrium.

    Using the roller removes one of the unknowns so you can then solve for the other one.

    Now to your second diagram.

    In order to solve this you have to make some assumptions.

    The assumption I would make would be that there is an even upward pressure.

    This upward pressure may be replaced by a single force acting through the centroid of the line of pressure, equal in magnitude to the integral of pressure times area (or line length in this case).

    Does this help?
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