# Homework Help: Force Couples - Computation of Forces

1. May 30, 2010

### SandboxSgt

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. May 30, 2010

### tiny-tim

Welcome to PF!

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.

3. May 30, 2010

### 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.

4. May 30, 2010

### SandboxSgt

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.

5. May 30, 2010

### tiny-tim

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?

6. May 30, 2010

### SandboxSgt

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 ?

7. May 30, 2010

### tiny-tim

Hi SandboxSgt!

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.
I'm confused … what's B, and which point are you taking moments about?

8. May 30, 2010

ERrr E not B! #$#@ I'm so over this question, I saw it in my sleep last night! And I realise I've gotten that wrong... M(a) = (400) + E(4) therefore E = 100N M(e) = (400 x 3) +A(4) therefore A = 300N 9. May 30, 2010 ### tiny-tim eek! you keep changing it! No, one of E(4) and A(4) is wrong … moment isn't distance times force, it's distance cross force. And where did 400 x 3 come from? Goodnight! :zzz: Last edited: May 31, 2010 10. May 30, 2010 ### SandboxSgt Haha Night Mate... Thanks for your help. 11. May 30, 2010 ### SandboxSgt God damnit, its a 100N each! If I could show you the reams of paper I have here, it was the first answer I worked out, but thought.... That can't be right! tiny-tim, thanks for kicking my head into the right direction. 12. May 30, 2010 ### PhanthomJay Now that you have E and A, you must indicate their direction. Now continue to find the forces supported by the pin at C. 13. May 31, 2010 ### SandboxSgt Well after many reams of paper, I have the reactions but am no where near the Forces in C. I know the answer is 224N, but I seem to just push around the moment. Either I'm missing something obvious or I'm just really not getting it. If I have the reactions of A being 100N Up and E being 100N down, whats the next step? What I do is take 100/Cos 45 which is equal to 141.421, move up to C, time 141.421 x 2 as there are two arms, then multiply by sin45 and I get 200 again... I know it's wrong.... Or I take moments, and get 400 at a point 2m below C..... #@$@#$@#$%^!

Last edited: May 31, 2010
14. May 31, 2010

### tiny-tim

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?

15. May 31, 2010

### SandboxSgt

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?

16. May 31, 2010

### tiny-tim

Hi SandboxSgt!

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.

17. May 31, 2010

### SandboxSgt

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* ;)

18. May 31, 2010

### tiny-tim

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.

19. May 31, 2010

### SandboxSgt

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...

20. May 31, 2010

### tiny-tim

uhh? that's what I get

how does it not match up?

21. May 31, 2010

### SandboxSgt

Ok, heres what I did..
for ABC
Sum -> =0
BX = CX
Sum ^ = 0
So Cy = 100

Sum M (B) AC Pos +100-100-Cx = 0
Therefore -CX = 0

For CDE
Sum -> =0
DX = CX
Sum ^ = 0
So Cy = 100

Sum M (D) AC Pos 400-100+100-Cx = 0
Therefore CX = 400

Which is different to what I said above as for the above I didn't use the 400, and I was using the reactions directions not the force in the element

Last edited: May 31, 2010
22. May 31, 2010

### tiny-tim

No, you've got pluses and minuses the wrong way round.

For example …
… that should be Cy = -100

(and it would help if you didn't use the same symbol, Cy, for two opposite forces)

Try again.

23. May 31, 2010

### SandboxSgt

I'm getting sleepy :)

Right
for ABC
Sum ->+ =0
-BX-Cx=0
-Cx=BX

Sum ^+ = 0
-100+Cy=0
Cy=100

Sum(M)(b) AC+
100 + 100 + Cx = 0
Cx = -200

For CDE
Sum ->+ =0
DX+C'x=0
C'x=-Dx

Sum ^+ = 0
100-C'y=0
-C'y=-100

Sum(M)(d) AC+
400 + 100 + 100 - C'x = 0
-C'x = -600

Therefore Cy = 200 and CX = 400

24. May 31, 2010

### SandboxSgt

So I sq 200 and 400
Which gives 160000 and 40000, which I add and sq.root

Which gives me 447.214...

And as that is a single resultant and it's all symetrical, I divide by 2 and get 223.6, which is close enough to 224 that I won't lose any sleep...

BOOOYAH!

Or did I make that last bit up :D

25. May 31, 2010

### tiny-tim

'fraid so

you only divided by 2 because you knew it was twice too much.
I don't understand where either part of that last line comes from.

I thought you'd already found that Cy = 100?

And your equation 400 + 100 + 100 - C'x = 0 (which uses Cy = 100, doesn't it?) should be 400 - 100 - 100 - C'x = 0.