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Impossible to solve? [appears simple but isn't]

  1. Jun 29, 2011 #1

    Femme_physics

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    1. The problem statement, all variables and given/known data

    http://img835.imageshack.us/img835/7509/appearsimple.jpg [Broken]

    Long story short: A is fixed to the ground. Point C is a pin joint. W = 200 [N]
    The rest is what you see is what you get.

    3. The attempt at a solution

    I get too many unknowns no matter what I do.

    If I look at only external forces (Which is what we were told to do with every "frame" problem such as this, this is what happens:

    Sigma Ma = 1.5W - 1.5By +Ma = 0

    (2 unknowns!)

    If I do sigma Mb, I get 3 unknowns!

    I can't build enough equations with similar unknowns to solve systems of equations.


    Even when I look at each member individually I don't have enough to solve for anything.
     
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Jun 29, 2011 #2

    SteamKing

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    You've only shown one equation for static equilibrium. This might be that rare case where you have two equations in two unknowns (shudder). You'll get better guidance if you show all your work.
     
  4. Jun 29, 2011 #3

    Femme_physics

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    Well, here is my full dead end:


    Sigma Ma = 1.5W - 1.5By +Ma = 0
    Sigma Fy = Ay + By - W = 0
    Sigma Fx = Ax + Bx = 0

    Member AE:

    Sigma Mc: -2Ax +2T +ma = 0
    Sigma Fx: Ax + Cx - T
    Sigma Fy: Ay - Cy

    Member BD:

    Sigma Fy: -W +By +Cy = 0
    Sigma Fx: T -Bx - Cx = 0
    Sigma Mc: -1.5By +2Bx +1.5W -2W = 0


    Dead ends everywhere
     
  5. Jun 29, 2011 #4

    I like Serena

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    Hi Fp! :smile:

    What kind of a joint is C?
    Could it perhaps give you an extra equation?

    And do you know of a relationship between T and W?
     
  6. Jun 29, 2011 #5

    Femme_physics

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    Well T is in fact W. I just for clarity wrote it differently but since it provoked you to ask a question maybe it wasn't so clear. But the tension is equal to the weight.

    Joint C is a regular joint. Allows rotation but does not allow linear movement at the X and Y planes (2 resistance directions). Same as in all our frames exercises. Joints, typical, good ol' joints.

    Didn't I write all possible equations? Well, I can do the formula for sum of all moments on all the joints in each member and in external forces but I doubt that would give me any more insight.
     
  7. Jun 29, 2011 #6

    I like Serena

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    C doesn't look like a regular joint.
    It looks like it can slide freely along its length.

    Edit: With C as a regular joint you cannot solve the system.
    This is called a statically undetermined system.
     
    Last edited: Jun 29, 2011
  8. Jun 29, 2011 #7

    DaveC426913

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    What exactly is being asked to be solved here?

    1] The diagram, taken at face value, depicts a rigid structure. Nothing happens.

    So, presumably, C can slide vertically, otherwise there's nothing to solve.


    2] Assuming C can slide, presumably the issue is whether the weight drops or rises.

    Is that what's supposed to be solved?
     
  9. Jun 29, 2011 #8

    SteamKing

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    Writing your moment equation about point A is an unexpected choice. However, writing a moment equation about point B would, perhaps, allow you to solve for the reaction at point C.
     
  10. Jun 29, 2011 #9

    SteamKing

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    To amplify my previous suggestions, instead of looking at the entire mechanism, draw a Free body diagram for the arm BD. By putting BD in equilibrium, it would then be possible to draw a Free body diagram for the arm AE and write its equations of equilibrium.
     
  11. Jun 29, 2011 #10

    SammyS

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    I agree with SteamKing.

    Assuming the pulley is frictionless and that the pin at C slides without friction in the slot, you should be able to find the force that the pin at C exerts on member BD. In doing this, I would suggest that you find the moment, MB, about point B (torque about point B).
     
  12. Jun 30, 2011 #11

    Femme_physics

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    Thanks, I solved it :smile:

    All the frames exercises I solved so far had a regular good ol' joint that wasn't allowed sliding. Now we have a joint that's allowed sliding, so it became confusing because I was taken aback.

    Also, we were told always to start by external forces, but here we start with internal forces, which confused me even more! But anyway, thanks to these replies and my classmate I finally was able to pin that problem down. :approve: Excuse the pun :wink:

    Thanks a bunch! :smile:


    For reference I'm attaching my friend's solution since I'm at work and have no scanner, but we got the same results
     

    Attached Files:

  13. Jun 30, 2011 #12

    SammyS

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    Looks good to me !!!
     
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