Recent content by SandboxSgt

Argh, I thought I'd put this to rest... If i put 400 in CDE, i don't get -200 or is that purely because my arrows are wrong...

Right, here's my solution... [PLAIN]http://sphotos.ak.fbcdn.net/hphotos-ak-snc3/hs261.snc3/27687_408007628600_532198600_4318550_8123271_n.jpg Everything seems to marry up, I was adding the Cy to the C'y and the Cx to the C'x... Have I stretched the truth with this one? I think I need to...

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

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

Ok, here's 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...

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...

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* ;)

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...

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...

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.

Haha Night Mate... Thanks for your help.

ERrr E not B! #$#@ I'm so over this question, I saw it in my sleep last night! And I realize I've gotten that wrong... M(a) = (400) + E(4) therefore E = 100N M(e) = (400 x 3) +A(4) therefore A = 300N

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 ?

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, I am 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...

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