How/where to take moments to solve a static equilibrium equation

A 200×250 mm panel of mass 20 kg is supported by hinges along edge AB. Cable CDE is attached to the panel at C, passes over a small pulley at D, and supports a cylinder of mass m. Neglect the effect of friction.

For those who cannot see the picture. The axis are oriented as follows, x+ is pointing out and to the right, y+ is pointing up and z+ is out and to the left. $$A=[0,0,.25]$$ $$B=[0,0,0]$$ $$C=[.2sin(\Theta),-.2cos(\theta),.125]$$ $$D=[.2,.1,0]$$

I have $$\sum F_x=0=A_x+B_x+\frac {mg}{\lVert CD\rVert}(.2-.2sin(\theta))$$

$$\sum F_y =0=A_y+B_y-192.2+\frac {mg}{\lVert CD\rVert}(.1+.2cos(\theta))$$

$$\sum F_z=0=A_z+B_z+\frac {mg}{\lVert CD\rVert}(-.125)$$

I am having trouble deciding where I want to make my moment in order to create a system I can solve for m in terms of $$\theta$$.

• You can take moments about anywhere you like, but it's usually a good plan to choose points where some of the force vectors have zero moment, and therefore don't appear in your equations. – alephzero Oct 2 '16 at 3:11

Since you are only interested in the mass $m$, all you have to do is to set the moment about the hinge AB to zero for equilibrium.
This means you have to consider only the two forces $M g$ (panel weight) and $m g$, since only they contribute to the relevant moment.
For reference, my calculations give me the result for $m$ as $$\frac{5 M \sin (\theta )}{\sin (\theta )+2 \cos (\theta )}$$
The plot shows that $m$ increases with $M$ and $\theta$ as expected.
An interesting confirmation is that the current configuration will not work for $\theta \gtrsim 2$.