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Homework Help: Force Couples - Computation of Forces

  1. May 30, 2010 #1
    Hey folks, I'm rather stuck. I have a complete mental blank on this question. Having the force couple first has confused me I think

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

    Compute the forece supported by the pin at C for the frame subjected to the 400-N.m couple

    2. Relevant equations

    [PLAIN]http://sphotos.ak.fbcdn.net/hphotos-ak-snc3/hs549.snc3/30037_407472593600_532198600_4297857_5544101_n.jpg [Broken]


    3. The attempt at a solution

    No idea where to begin!
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. May 30, 2010 #2

    tiny-tim

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    Welcome to PF!

    Hi SandboxSgt! Welcome to PF! :smile:

    I'm not sure I understand what's happening at A and E.

    Is E not resting on anything, but supporting an unknown weight?

    And is A rotating frictionlessly about an axle resting on a fixed surface? :confused:

    If so, start by taking moments about a convenient point, to find the weight (in newtons) supported at E. :smile:
     
  4. May 30, 2010 #3

    PhanthomJay

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    Start by calculating the force reactions at A and E using the basic equilibrium equations. Then use free body diagrams of each member and show your attempt at a solution.
     
  5. May 30, 2010 #4
    G'day tiny-tim,

    A acts as a roller, so has only a vertical reactive force, E is a pin therefore both a vertical and horizonal force...


    PhanthonJay, im just not sure how to start that, if I have a Force Couple of 400N.m shown where it is, how do I translate that into something I can start to find the reactions at A and E.
     
  6. May 30, 2010 #5

    tiny-tim

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    ok, then do A first instead of E …

    start by taking moments about a convenient point, to find the reaction at A

    what do you get? :smile:
     
  7. May 30, 2010 #6
    Morning Tim!

    This is where I get confused, in a force couple isn't it a moment itself? So for A i get Sum M(a) = (400) + B(4) therefore A = 100N ?
     
  8. May 30, 2010 #7

    tiny-tim

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    Hi SandboxSgt! :smile:

    I'm about to go to bed! :zzz:
    That's correct … whichever point you take moments about, a couple always goes in as a pure moment, of the same amount. :smile:
    I'm confused … what's B, and which point are you taking moments about? :confused:
     
  9. May 30, 2010 #8
    ERrr E not B! #$#@ I'm so over this question, I saw it in my sleep last night!

    And I realise I've gotten that wrong...

    M(a) = (400) + E(4) therefore E = 100N
    M(e) = (400 x 3) +A(4) therefore A = 300N
     
  10. May 30, 2010 #9

    tiny-tim

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    eek! you keep changing it! :redface:

    No, one of E(4) and A(4) is wrong … moment isn't distance times force, it's distance cross force. :wink:

    And where did 400 x 3 come from? :confused:


    Goodnight! :zzz:
     
    Last edited: May 31, 2010
  11. May 30, 2010 #10
    Haha Night Mate...

    Thanks for your help.
     
  12. May 30, 2010 #11
    God damnit, its a 100N each!

    If I could show you the reams of paper I have here, it was the first answer I worked out, but thought.... That can't be right!

    tiny-tim, thanks for kicking my head into the right direction.
     
  13. May 30, 2010 #12

    PhanthomJay

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    Now that you have E and A, you must indicate their direction. Now continue to find the forces supported by the pin at C.
     
  14. May 31, 2010 #13
    Well after many reams of paper, I have the reactions but am no where near the Forces in C. I know the answer is 224N, but I seem to just push around the moment. Either I'm missing something obvious or I'm just really not getting it.

    If I have the reactions of A being 100N Up and E being 100N down, whats the next step?

    What I do is take 100/Cos 45 which is equal to 141.421, move up to C, time 141.421 x 2 as there are two arms, then multiply by sin45 and I get 200 again... I know it's wrong....

    Or I take moments, and get 400 at a point 2m below C..... #@$@#$@#$%^!
     
    Last edited: May 31, 2010
  15. May 31, 2010 #14

    tiny-tim

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    Morning SandboxSgt! :wink:

    You need to use moments for each rod separately

    Let the reaction force on rod CDE (from rod ABC) be x right and y up …

    what do you get? :smile:
     
  16. May 31, 2010 #15
    I'm not sure I understand...

    My reactions are at point A and E, if I take moments about B for ABC I get C = 0N
    If I take moments about D I get C = 0N

    I'm missing this bit in the thought process. I believe there are no horizontal forces on any of the points as there is no externally applied horizontal force?
     
  17. May 31, 2010 #16

    tiny-tim

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    Hi SandboxSgt! :smile:

    I assure you that I've done the problem myself (and got the correct answer, 224N), and there is a horizontal component.

    Remember, if the reaction force on rod CDE (from rod ABC) is x right and y up, then the reaction force on rod ABC (from rod DEF) is x left and y down …

    the internal forces always add to zero. :wink:
     
  18. May 31, 2010 #17
    Thats the issue, I'm not sure how to get the reaction of one rod on the other...
    Once I've worked out the reactions at A and E, does that mean the 400 is no longer touched or... I don't know...have spent close to 10 hours on this *cries* ;)
     
  19. May 31, 2010 #18

    tiny-tim

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    Just do what I said … split the reaction into x and y components, add the moments for ABC, to get 0, and add the moments for CDE, to get -400.

    That will give you two equations for two unknowns (x and y).

    Show us what you get. :smile:
     
  20. May 31, 2010 #19
    Ok, so If I'm thinking right, I get the Y component of C to = 100 and the X component to = 200...
    Where looks to be not right because it doesn't match up with what you just said...
     
  21. May 31, 2010 #20

    tiny-tim

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    uhh? that's what I get :smile:

    how does it not match up? :confused:
     
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