Consider the setup shown in the diagram below in which we have a uniform rod of mass $10\mathrm{kg}$ attached to the wall at $\mathrm{D}$ by a light inextensible string and pivoted at $\mathrm{A}$. A mass of $5\mathrm{kg}$ is attached at $\mathrm{B}$.
Note that we also have $\mathrm{AB}=2, \mathrm{AC}=0.5$ and $\mathrm{CD}=1$.
I want to find the magnitude of the reaction force at $\mathrm{A}$.
First I've taken moments about $\mathrm{A}$ as follows:
\begin{align*} & \frac{1}{2}T=5g\cdot2\cos\theta + 10g\cdot\cos\theta \\ \implies & T=40g\cos\theta\end{align*} where $\theta$ is the angle between $\mathrm{AB}$ and the horizontal. I've also found that $\theta=\frac{\pi}{2}-\arctan{2}$.
To $3$ significant figures, this gives me $T=438$. (which I'm told is correct)
Now in order to find the reaction at $A$, I think I need to use $F=ma$ in the horizontal direction. That is, $$T\cos\left(\frac{\pi}{2}-\theta\right)=N$$
However, apparently this is not the correct value of $N$. Where have I made my mistake?