Why $\frac{W}{2P}$ and not $\frac{2P}{W}$?

Question

I am trying to solve this:

I'm a little bit confused, by symmetry, we can draw something like:

And $\sin \theta = \frac{O}{H}=\frac{P}{W/2}=\frac{2P}{W}$, no? Perhaps there is some equivalence but it's not clear to me.

Answer 1

The vertical component of the force exerted by one of the weights of weight $P$ is $P\sin\theta$, so the two weights between them exert a force of $2P\sin\theta$. So at rest, $W = 2P\sin\theta$, or $\sin\theta = \frac{W}{2P}$. (Note that the horizontal components of the two forces balance, so that $W$ hangs straight down from the middle.)