Is Static Equilibrium Really That Simple?

[urk.nono]

Hello! I'm a student from Norway, and I've been struggling with a static equilibrium for a few hours now. The terms I use might not be "right", but I'll try to explain the problem to the best of my ability.

Homework Statement

Find the forces Fa and Fb on this figure:

http://img44.imageshack.us/img44/4505/321k.jpg

Homework Equations

\Sigma F_X = 0
\Sigma F_Y = 0
\Sigma \tau = 0 (sum torque = 0)

The Attempt at a Solution

I solved it by drawing the forces on the figure:
http://img41.imageshack.us/img41/7513/321graph.jpg

After much fiddling with the "relevant equations" I came across the following:

20kN *3,5m - Fb * sin(30) * 5,5m = 0
=> Fb = 37,21kN

OR

20kN * 3,5m - Fb * cos(70) * 5,5m = 0
=> Fb = 37,21kN

Fa * cos(11,75) * 2m - 20m * 3,5m = 0
=> Fa = 35,74kN

I know from my measurements and by checking the solution in the book, it's pretty damn close. By which formula, or by what method I got no clue whatsoever.

I'd appreciate if someone could help me out by pointing me in the right direction

Thanks in advance

Urk

 

[kuruman]

Hi urk.nono, welcome to PF. I am not sure what the second drawing you posted shows. I am used to force vectors represented as arrows, not line segments. Anyway, if you write the x and y components of the force at A as [FSUB]x[/SUB] (to the right) and F_y (up), can you find three equations saying

ΣF_i,x = 0
ΣF_i,y = 0
Σtorques = 0
?

Note: Your work will be much easier if you calculate torques about point A.