1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    User Avatar
    Science Advisor
    Homework Helper

    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

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    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

    User Avatar
    Science Advisor
    Homework Helper

    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

    User Avatar
    Science Advisor
    Homework Helper

    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

    User Avatar
    Science Advisor
    Homework Helper

    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

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    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

    User Avatar
    Science Advisor
    Homework Helper

    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

    User Avatar
    Science Advisor
    Homework Helper

    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

    User Avatar
    Science Advisor
    Homework Helper

    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

    User Avatar
    Science Advisor
    Homework Helper

    uhh? that's what I get :smile:

    how does it not match up? :confused:
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook