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Free Body Diagram of 3-Bar Linkage. Splitting the force?

  1. Mar 3, 2017 #1
    1. The problem statement, all variables and given/known data
    Hi, please see the attached file. Im trying to find the force as illustrated in the diagram.

    2. Relevant equations
    Picture1.png


    3. The attempt at a solution
    I've found the force for the 4deg beam and therefore the horizontal beam however I am unsure how to take into account the third angled beam.

    Thank you in advanced for any help. It is much appreciated.
     
  2. jcsd
  3. Mar 3, 2017 #2

    Nidum

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    As drawn that mechanism is just floating in space with all the links loose relative to one another .

    Any applied forces will distort the mechanism and accelerate the whole thing away from the starting location .

    You need to have some constraints somewhere .
     
    Last edited: Mar 3, 2017
  4. Mar 3, 2017 #3
    Hi Nidum, does this explain it better? The long horizontal beam represents an actuator. The mechanism is to tip a container.

    Picture2.png
     
  5. Mar 3, 2017 #4

    Nidum

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    That's better . We don't want our problems propelled into outer space .

    The mechanism won't actually function with the 4 deg angle . Linkages like that typically need 25 deg + to work .

    Do you want to alter the problem to be a bit more realistic ?
     
    Last edited: Mar 3, 2017
  6. Mar 3, 2017 #5
    Alright, handy to know! :)

    So if i change the 4 deg angle to 25, I get: 9/(sin(25)) = 21.3KN on the upper angled beam.
     
  7. Mar 3, 2017 #6

    Nidum

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    Let's see a proper free body diagram for the linkage with all the forces acting shown clearly ..
     
  8. Mar 3, 2017 #7
    Pic3.png

    This ok?
     
  9. Mar 3, 2017 #8

    Nidum

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    There are three missing forces .....?
     
  10. Mar 3, 2017 #9
    Pic4.png
    This better? Not sure what other forces I need.
     
  11. Mar 3, 2017 #10

    Nidum

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    Left hand side of diagram ? Two horizontal forces + one vertical force ?
     
  12. Mar 3, 2017 #11
    The horizontal component of F1? at either end of the beam? Then the vertical at the end of the first angled beam?
     
  13. Mar 3, 2017 #12

    Nidum

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    Reaction forces from container fixing pin and ground fixing pin . Four in total but you are given one so three to find .

    The mechanism and force system is symmetric about c/l so not much work to do .
     
    Last edited: Mar 3, 2017
  14. Mar 3, 2017 #13
    Something like this?



    PIC5.png
     
  15. Mar 3, 2017 #14

    Nidum

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  16. Mar 3, 2017 #15
    Ah that makes more sense. How would I go about finding the X and Y components at the lower ground fixing pin? Would I find the force going along the lower angled beam (hypotenuse) and go from there?

    Also, once I've found the X and Y components, how do I go about finding the force required "F ??".

    Thanks
     
  17. Mar 4, 2017 #16

    Nidum

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    For the complete mechanism what can you say about the sum of externally applied forces in the X direction and in the Y direction ?
     
  18. Mar 4, 2017 #17
    So the force required "F ??" Is taken from the the sum of the X components and the sum of the Y components for the container and linkage fixing pin?
     
  19. Mar 4, 2017 #18

    Nidum

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    Not quite - sum in X direction and sum in Y direction are separate calculations .
     
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