I hope I'm asking this in the right place. I have the following problem:
and am to calculate A,B & C so that the system is in static equilibrium. I calculated A without too many problems.
My problem is with B & C
What I have done so far:
I realised that my placing my moment at the point ABF I can eliminate B as an unknown and solve for C.
I calculated the resultant for the diagonal part ($R_1=q_0\cdot l$) and for the vertical part ($R_2=\frac{q_0\cdot 2l}{2}=q_0\cdot l$).
I calculated the distances of $R_1$ and $R_2$ from my moment $M^{ABF}$ using the Pythagorean theory and set them both negative as they are turning my system in a mathematically negative direction.
The force C I added to my moment as $C\cdot 2l$.
My moment calculation is at this point:
$$\Sigma M^{ABF}=2l\cdot C-(\sqrt{(2l)^2+(2l)^2}+\frac{l}{2})\cdot R_1-\sqrt{(2l)^2+(\frac{2}{3}2l})^2\cdot R_2$$
My problem: I don't know what to do with the moment $M$ which is at a distance of $l$ from $M^{ABF}$. I have tried adding as it is, adding it and multiplying by $l$, but both times the answer comes out wrong (I have a numerical answer).
I'm obviously doing something wrong but I don't know what.
In case it helps to have values:
$F=4q_0l$, $M=2q_0l^2$, $\alpha=45°$
$q_0=3kN/m$, $l=2m$