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Moment problem (statics)

  1. Sep 27, 2008 #1
    Diagram is attached.

    P = forces
    M = free-moving moment
    C = looks like a pin connection
    L = distance between A and C, and C and B

    With the system in equilibrium.

    Question: Write the moment equations about point A, B and C and prove that they are all the same.

    so basically I end up with:

    (Counter-clockwise moment is positive.)

    Summation of moments about point A: P(sin(90-theta))(2L) - M - Cx(sin(90-theta))(L) + Cy(sin(theta))(L) = 0

    point B: P(sin(theta))(2L) + Cx(sin(90-theta))(L) - Cy(sin(theta))(L) - M = 0

    point C: P(sin(90-theta))(L) - M + P(sin(theta))(L) = 0

    which means that

    P(cos(theta))(2L) - Cx(cos(theta))(L) + Cy(sin(theta))(L) = P(sin(theta))(2L) + Cx(cos(theta))(L) - Cy(sin(theta))(L) = P(cos(theta))(L) + P(sin(theta))(L) = M.

    I don't see how these can be equal. Any mistakes seen here in the approach to solving this problem? Maybe pin supports Cx, Cy don't create moments?

    Thanks for the help!
     

    Attached Files:

  2. jcsd
  3. Sep 27, 2008 #2

    Redbelly98

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    Please post homework questions in the homework section of PF.
     
  4. Sep 28, 2008 #3

    PhanthomJay

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    Your attempt is excellent. Cx and Cy (at the support point C) do create moments about B or A. A and B are not pin suppports, they are just the ends of the member that rotates about C under the applied forces and couple, until equilibrium is reached. Note that by applying the equilibrium equations sum of Fx = 0 and sum of Fy = 0, you can solve for Cx and Cy in terms of P. That will greatly simplify the equations in a manner that will clarify their equality.
     
  5. Sep 28, 2008 #4
    Sorry for posting in the wrong sub-forum :(

    Thanks for the help!

    My issue is: when I do summation of forces in x and y, I get

    P = (Cy(cos(theta)) + Cx(sin(theta)))/(cos(theta)+sin(theta)) from sum of Fx and

    P = -(Cx(cos(theta))+Cy(sin(theta)))/(sin(theta)-cos(theta)).

    Which I'm not sure if they're equal?

    Either way, I plugged the 1st P into
    P(cos(theta))(2L) - Cx(cos(theta))(L) + Cy(sin(theta))(L) and
    P(sin(theta))(2L) + Cx(cos(theta))(L) - Cy(sin(theta))(L) and
    P(cos(theta))(L) + P(sin(theta))(L)

    (all equal to M, this is the last line of my initial post)

    When I plugged P into those, I got a huge mess that doesn't clarify the equality. Any tips?

    Thanks
     
  6. Sep 28, 2008 #5

    PhanthomJay

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    I am not sure from where you are getting your sum of Fx and Fy equations. Cx and the upper P are the only 2 forces in the x direction, and Cy and the lower P are the only 2 forces in the y direction. So in the x direction, you have Cx -P =0, and in the y direction, you have Cy -P = 0. Solve for Cx and Cy.
     
  7. Sep 28, 2008 #6
    OH! Yeah My sum of Fx and Fy came from looking at my altered diagram... which had the member lying horizontally, so I stupidly assumed that the x-coordinate system was horizontal and that the y-coordinate system was perpendicular to that, simply from looking at my diagram which has been rotated :P

    Yes, with that P = Cx = Cy equation, it can be easily shown that sum of Ma = sum of Mb = sum of Mc. Thanks for the help.
     
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