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Moment at different points

  1. Sep 19, 2016 #1
    1. The problem statement, all variables and given/known data
    For the moment about B , Cy(0.8) +Ay(0.8) + Cx(0.6) = 0 , right ?

    Why the author only give ,-Ay(0.8) = 0

    Is it wrong ?

    2. Relevant equations


    3. The attempt at a solution
     

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    Last edited: Sep 19, 2016
  2. jcsd
  3. Sep 19, 2016 #2
    because the moment Cy(0.8) +Ay(0.8) can cause the member to turn about B , right ?
     
  4. Sep 19, 2016 #3

    billy_joule

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    Cy and Cx don't act on the boom so don't appear in the fbd for that member.
     
  5. Sep 19, 2016 #4
    What do you mean by boom?
     
  6. Sep 19, 2016 #5

    billy_joule

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    Read your first attachment.
    The fbd for the boom appears at the bottom of your 3rd attachment.
     
  7. Sep 19, 2016 #6
    OK, saw it.. Why they don't act on the 'boom' ?
     
  8. Sep 20, 2016 #7

    billy_joule

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    Because point C isn't part of the boom.
    would the forces you apply to your keyboard appear in a free body diagram of your neighbours kitchen sink?
    A fbd of the boom should include only the forces acting on the boom. Those are the forces at A and B.
     
  9. Sep 20, 2016 #8
    What is the boom actually??
     
  10. Sep 20, 2016 #9
    Then, why is point C given in the problem?
     
  11. Sep 20, 2016 #10

    billy_joule

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    The rod starts at C and ends at B.
    The boom starts at A and ends at B.
    Call them member BC and member AB if you prefer.
     
  12. Sep 20, 2016 #11

    billy_joule

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    Edit:Double post.
     
  13. Sep 20, 2016 #12
    We need to consider moment at C due to all members in the problem, right? So, we need to consider the forces at C, right?
     
  14. Sep 20, 2016 #13

    billy_joule

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    Your second attachment takes moments about C for the entire structure.

    The question asks you to find the reaction forces, so yeah, we need to consider the forces at C, and find them.
     
  15. Sep 20, 2016 #14
    No, I am asking why for moment about B, ' we don't have to consider force Cx and Cy??
     
  16. Sep 20, 2016 #15

    billy_joule

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    You'll have to be more specific.

    The worked solution takes moments about B for the boom only.

    Cx and Cy don't act on the boom so are not included in the free body diagram for the boom and so are not included in any subsequent equilibrium equations.

    If we were to take moments about B for the rod, or for the entire structure then Cx and Cy would be relevant.
     
  17. Sep 20, 2016 #16
    What is the boom here?
     
  18. Sep 20, 2016 #17

    billy_joule

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    See post #10
     
  19. Sep 20, 2016 #18
    ok , why we have to consider 1 'boom' only during calculation of moment about B ?? for moment about C , we consider 2 boom , right ? That's why the Ax(0.6) and -30(0.8)
     
  20. Sep 20, 2016 #19

    billy_joule

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    It's just the way the author choose to do it.
    It's a quick way to show Ay is zero.
    It could be done in literally infinite different ways, we can take moments about any point in space for any or all static members and the answer is always zero. The trick is to find the most convenient point to make the math quick and easy.
     
    Last edited: Sep 20, 2016
  21. Sep 20, 2016 #20
    So, this is the special case? If not, then my moment about B , Cy(0.8) +Ay(0.8) + Cx(0.6) = 0 would be correct?
     
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