## Force Couples - Computation of Forces

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

3. The attempt at a solution

No idea where to begin!
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 Blog Entries: 27 Recognitions: Gold Member Homework Help Science Advisor Hi SandboxSgt! Welcome to PF! 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? If so, start by taking moments about a convenient point, to find the weight (in newtons) supported at E.
 Recognitions: Gold Member Homework Help Science Advisor 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.

## Force Couples - Computation of Forces

 Quote by tiny-tim Hi SandboxSgt! Welcome to PF! 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? If so, start by taking moments about a convenient point, to find the weight (in newtons) supported at E.
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...

 Quote by PhanthomJay 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.
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.

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

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 Quote by SandboxSgt Morning Tim!
Hi SandboxSgt!

I'm about to go to bed!
 This is where I get confused, in a force couple isn't it a moment itself?
That's correct … whichever point you take moments about, a couple always goes in as a pure moment, of the same amount.
 So for A i get Sum M(a) = (400) + B(4) therefore A = 100N ?
I'm confused … what's B, and which point are you taking moments about?
 ERrr E not B! #$#@ I'm so over this question, I saw it in my sleep last night! And I realise I've gotten that wrong... M(a) = (400) + E(4) therefore E = 100N M(e) = (400 x 3) +A(4) therefore A = 300N Blog Entries: 27 Recognitions: Gold Member Homework Help Science Advisor  Quote by SandboxSgt M(a) = (400) + E(4) therefore E = 100N M(e) = (400 x 3) +A(4) therefore A = 300N eek! you keep changing it! No, one of E(4) and A(4) is wrong … moment isn't distance times force, it's distance cross force. And where did 400 x 3 come from? Goodnight!  Haha Night Mate... Thanks for your help.  God damnit, its a 100N each! If I could show you the reams of paper I have here, it was the first answer I worked out, but thought.... That can't be right! tiny-tim, thanks for kicking my head into the right direction.  Recognitions: Gold Member Homework Help Science Advisor Now that you have E and A, you must indicate their direction. Now continue to find the forces supported by the pin at C.  Well after many reams of paper, I have the reactions but am no where near the Forces in C. I know the answer is 224N, but I seem to just push around the moment. Either I'm missing something obvious or I'm just really not getting it. If I have the reactions of A being 100N Up and E being 100N down, whats the next step? What I do is take 100/Cos 45 which is equal to 141.421, move up to C, time 141.421 x 2 as there are two arms, then multiply by sin45 and I get 200 again... I know it's wrong.... Or I take moments, and get 400 at a point 2m below C..... #@$@#$@#$%^!
 Blog Entries: 27 Recognitions: Gold Member Homework Help Science Advisor Morning SandboxSgt! 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?
 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?

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