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Moments and trusses

  1. Dec 15, 2013 #1
    Hi, i'm new on this website and it seems great :p So, I'm struggling with something really simple... trusses analysis. I'm struggling with moments etc , I've just drawn an example linked in this post , could you try to find the reaction forces (roller pin etc) with some explanation ? Thanks
     

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  3. Dec 15, 2013 #2

    tiny-tim

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    Hi Paladashe! Welcome to PF! :wink:

    (please use the homework forum in future)

    If I'm understanding the question correctly, no truss analysis is involved …

    you have a rigid body with one applied force (plus its own weight), and you have to find the two reaction forces, so just use linear components of force, and moments …

    show us what you get. :smile:
     
  4. Dec 15, 2013 #3
    Thanks Tiny-tim for your response. In fact, I now for vertical external forces and horizontal ones so I divided the 100N vector in two vectors (one with a value of 50N orientated downwards and one with a value of 50•(square root of 3). I found that each pin has a vertical vector orientated upwards with a value of 25 N and for the fixed one, also an horizontal vector with a value of 50•(square roots of 3 orientated to the left). Am I right? I'm sure that I've made some mistake....what value did you find?
     
  5. Dec 15, 2013 #4

    tiny-tim

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    Hi Paladashe! :smile:
    Correct. :smile:
    Nooo …

    you can't assume they're equal, can you?

    all you know is that they add to 50N, so you need one more equation, which is … ? :wink:
     
  6. Dec 15, 2013 #5
    The moment equation, right?
    .... EMa=-50•3•12,1244 + 50(square root of 3)•7 + 4•12,1244•Rb ?
     
  7. Dec 15, 2013 #6
    Where a is the pin fixed and b the roller pin
     
  8. Dec 15, 2013 #7

    tiny-tim

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    Yes. :smile:

    (either about A or about B)
    (where 12.1244 = 14*√3/2)

    Yes …

    or you could avoid splitting the 100 N by doing 100 times the lever arm (the perpendicular distance from the line of the 100 N to the point A). :wink:
     
  9. Dec 15, 2013 #8
    Yes but dont know how to do this because its not perpendicular to the line of action:)
     
  10. Dec 15, 2013 #9
    Yes, but by doing my equation , I found that Rb=50N so R(vertical) is equal to 0??
     
  11. Dec 15, 2013 #10

    tiny-tim

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    use trig :wink:
     
  12. Dec 15, 2013 #11
    yes but it's not working for me :p could you please try to explain why my equation is wrong ??? Please !! I'm getting a headache by trying and trying :p
     
  13. Dec 15, 2013 #12

    tiny-tim

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    I don't get 50 :confused:
     
  14. Dec 15, 2013 #13
  15. Dec 15, 2013 #14

    tiny-tim

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    i mean, using your equation (which is a little confusing :redface:) i don't get 50
     
  16. Dec 15, 2013 #15
    Can you give me your result ? Is it 37,5?
     
  17. Dec 15, 2013 #16

    tiny-tim

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    from your equation (am i misreading it?) i got 25 :confused:
     
  18. Dec 15, 2013 #17
    No as you said before, we can't assume that both pin have the same vertical forces (25-25). Or, if Rby is equal to 25 and that EFy=0 <=> -50+25+Ray=0 :) so Ray is equal to 0 which is impossible
     
  19. Dec 16, 2013 #18

    tiny-tim

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    (just got up :zzz:)
    yes, we can't assume it …

    but that doesn't mean that we can't prove it (if it's true) :wink:
    not following you …

    doesn't -50+25+Ray=0 mean Ray = 25 ? :confused:
     
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