Force Couples - Computation of Forces

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




3. The attempt at a solution

No idea where to begin!

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.

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.

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.

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?

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 ?

tiny-tim

Hi SandboxSgt!

I'm about to go to bed!
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?

SandboxSgt

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

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!

SandboxSgt

Haha Night Mate...

Thanks for your help.

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.

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.

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..... #@$@#$@#$%^!

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?

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?

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.

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* ;)