Simple Statics Force diagram problem using moments. I have a statics problem in which my approach differs from the suggested solution, so I was wondering if anyone can point out my flawed logic.  Below is the question.
Two identical uniform rods $ab$ and $bc$, each of length $2l$ and weight $W$, are smoothly hinged at $b$.  Ends $a$ and $c$ rest on a smooth horizontal floor.  The system is kept in
equilibrium in a vertical plane by a light inextensible string joining $a$ to the midpoint of $bc$. the and between $ab$ and $bc$ is $2\alpha$.  Show the tension of the string is $T = \frac{W}{4} \sqrt{9\tan^2\alpha + 1}$.
So below is the force diagram I have created.

The suggested solution for this problem resolves the $\vec{j}$ components of the system as:
$$ R_a + R_c = 2W $$
without taking into consideration the resolved tension force, $T$, of the light inextensible string.
Also taking moments about the point $a$ of the system abc the solution is:
$R_c \cdot 4l \sin \alpha = l \sin\alpha \cdot W + 3l\sin \alpha \cdot W $$
Once again no mention of the tension.
So I am I wrong to include the tension force for the $\vec{j}$ components of the system and  the moments about $a$.
Any help appreciated.
 A: 
It is true that the tension force $T$ in the string is a single force. But, this tension in the string exerts two equal forces of $T$ on the system at both ends, i.e. at $a$ and at $d$, as shown in the diagram.
If you still have doubts, remove the string and apply two unequal forces, say $T_1$ and $T_2$, but acting along the same straight line at $a$ and $d$. When you resolve the forces acting on the system in $\underline{i}$ and $\underline{j}$ directions assuming that the system is in stable equilibrium, you get the two equation shown below.
$$R_a + R_c = 2W +T_2\sin(\alpha) – T_1\sin(\alpha)$$
$$T_1\cos(\alpha) - T_2\cos(\alpha) = 0 \qquad\rightarrow\qquad T_1 = T_2$$
Since $T_1 = T_2$, the first equation reduces to
$$R_a + R_c = 2W$$
When you are taking moments about a given point, you can ignore all the forces passing through that point, because such forces generate zero moments. Therefore, you have to consider neither $T_1$ nor $T_2$ as they both pass through $a$.
A: Calling the forces indexed by point of application
$$
\cases{
T_a = T(\cos\beta,\sin\beta)\\
F_a = (0,R_a)\\
F_b = (0,R_b)\\
F_c = (H_c,V_c)\\
F_{g_1} = (0,-W)\\
F_{g_2} = (0,-W)\\
a = (0,0)\\
b = 2l(\sin\alpha,0)\\
c = 2l(\sin\alpha,\cos\alpha)\\
g_1 = \frac 12 (a+c)\\
g_2 = \frac 12(b+c)\\
}
$$
with $\tan\beta = \frac 13\cot\alpha$ we have the equilibrium equations
$$
\cases{
F_a+T_a+F_{g_1}+F_b = 0\\
F_{g_1}\times (g_1-a)+F_c\times(c-a) = 0\\
}
$$
and
$$
\cases{
-F_b+F_{g_2}-T_a + F_c = 0\\
(-F_c)\times(b-c)+(F_{g_2}-T_a)\times(g_2-c) = 0
}
$$
Solving the corresponding linear system we got
$$
\cases{
R_a = \frac{W (\cos (\alpha -\beta )+6 \cos (\alpha +\beta ))}{6 \cos (\alpha ) \cos (\beta )-2 \sin (\alpha ) \sin (\beta )} \\
R_b = \frac{W (\sin (\alpha ) \sin (\beta )+5 \cos (\alpha ) \cos (\beta ))}{6 \cos (\alpha ) \cos (\beta )-2 \sin (\alpha ) \sin (\beta )} \\
H_c = \frac{2 W \sin (\alpha ) \cos (\beta )}{\sin (\alpha ) \sin (\beta )-3 \cos (\alpha ) \cos (\beta )} \\
V_c = -\frac{W \cos (\alpha -\beta )}{6 \cos (\alpha ) \cos (\beta )-2 \sin (\alpha ) \sin (\beta )} \\
T = \frac{2 W \sin (\alpha )}{3 \cos (\alpha ) \cos (\beta )-\sin (\alpha ) \sin (\beta )} \\
}
$$
and using the relationship $\tan\beta = \frac 13\cot\alpha$ we have finally
$$
T = \frac{3}{4} W \tan \alpha \sqrt{\frac{\cot^2\alpha}{9}+1} = \frac W4\sqrt{9\tan^2\alpha+1}
$$
