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The first thing we do, since this is a static situation, is to choose our system, choose the point about which we want to sum the torques, and do it. Our system can be the pipe and the marquee. The point about which we sum our torques can be the point where the pipe connects to the wall – this way, we don't need to know the reaction forces on the wall when we do some torque sum.

Following through with it, we have $$\sum\tau_{z}=\tau_{T}-\tau_{g}=0,$$ where $\tau_{T}=hT\sin\theta$ is the torque generated by the tension force and $\tau_{g}=\frac{L}{2}\mathcal{M}g$ is the torque generated by the weights. Now, using the fact that $\sin\theta=\frac{H}{\sqrt{h^{2}+H^{2}}}$ and a little bit of algebra, we can find: $$h=\frac{L\mathcal{M}gH}{\sqrt{4\sigma^{2}\pi^{2}r^{4}H^{2}-L^{2}\mathcal{M}^{2}g^{2}}}.$$

Using our maximum stress of $\sigma_{\rm max}\approx 0.44\times 10^{9}\,{\rm Pa}$ (therefore the maximum tension) we can find the minimum distance $h_{\rm min}\approx 0.47\,{\rm m}$. Any smaller $h$ will result in the cord snapping. This tension carries all the way through the cord to the hook on the wall. Thus, the reaction force on the wall-cord interface is equal and opposite to the tension.