Statics: finding forces in frame

[SoylentBlue]

Homework Statement

When the piston CD is lowered, ACB and CBE are horizontal, and they ask for the force in piston CD.

Homework Equations

The Attempt at a Solution

So first we look at the entire frame ACDBE, and we sum moments at A:
-(14)(2500) + 24E(suby) = 0; E(suby) = 1458
Then A(suby) is 1041 since they both total 2500
A(subx) = 0 since no horiz forces; E(subx) = 0 since it is a roller.
So far so good.
Now we dismember and end up with 2 legs: CBE, and ACB. We don't know the force at B, and we are looking for the value of C, but we have 2 equations and 2 unknowns, so this should be solvable.
On the bottom leg, we sum moments about A:
4C -14(2500)-18B=0 (assume B is downward)
On the top leg, we sum moments about E:
-6B + 10(2500) - 20C=0 (assume B is opposite and therefore up)
We multiply the top equation by 5 so the C values cancel; top equation becomes
20C-175,000-90B=0
We're left with B= -1562
Which yields a value of 103 for C. The book's answer is 1339 for C.
Point me in the right direction please. Is this an error in logic, math, or something else? Thank you in advance :^)

[/B]

 

[haruspex]

On the bottom leg, we sum moments about A:
4C -14(2500)-18B=0 (assume B is downward)

How is member ACB 'aware' of the 2500lb weight?

 

[SoylentBlue]

Thank you thank you thank you...it was such a D'Oh mistake, but I just couldn't see that I was counting the 2500LB force twice! When I took the 2500LB force away from the bottom arm, the problem almost solved itself, and I got the same answer as the book gives.

 

[haruspex]

Thank you thank you thank you...it was such a D'Oh mistake, but I just couldn't see that I was counting the 2500LB force twice! When I took the 2500LB force away from the bottom arm, the problem almost solved itself, and I got the same answer as the book gives.

Glad to hear it.