Final position of weight hanging from two opposed pulleys on ropes Let's say I have the situation pictured below. The red and green ropes are tied to immobile supports. The ends of the red and green ropes are attached to frictionless rings. The blue rope is tethered infinitely far below and passes through the two rings. A weight is attached to the other end of the rope. The rings are not attached to each other.

Now, when the weight is dropped, what is the final orientation?
My gut instinct thought this would happen.

However, I tested this out and this happens instead.

This leaves me with two questions:

*

*Why is this the final state?


*How would you go about predicting this final orientation?
(So I can predict other situations with different rope lengths and immobilization points)
 A: The final orientation comes with free body diagram.
Draw a free body diagram and you will see that the two horizontal component of red and green ropes cancelling each other ,which leaves you with the two vertical component which is balanced when the weight of the body hanged on left hand side and the weight of the blue rope on the right hand side.

System comes in final orientation when
Tsin(theta)=Mg.
I have taken mass of the body M.
Remember,the blue rope also exerts downward force because of it's weight.
A: A system will want to lower its potential energy to reach a stable equilibrium.
In this case, we can just suppose that all the mass is focused at the weight, whose potential energy is directly proportional to how high it is above the ground. The closer you can get it to the ground, the lower will its potential energy be, the more stable will the system be.
So, how close can you get the weight to the ground? With your current setup, it's when the ropes form $45^\circ$ angles with the horizontal, such that the weight is as low as possible.
This configuration results in the lowest gravitational energy, and thus the most stable configuration.
Your prediction, however, is not wrong -- the above reasoning holds because we assumed a very ideal system. Notice how you write "serious tension in the rope." In the simulation you ran, the tension in the ropes was free to change. Thus, when the object got lowered, the tension actually decreased, a lot. There is nothing holding the ropes at such a tension.
In practice, you could attach the hoops together by a rope of some initial length $L$ (so I'm modifying your system here a bit) such that there is a constraint to keep the two other ropes at high tension, then, upon dropping a weight, the outcome will look more like your prediction. You can model strings as springs (with very large spring constants) that can store a lot of elastic potential energy by being stretched very slightly. In that case, if you want to lower the system's potential energy, there is a catch: by lowering the gravitational energy (by lowering the mass), you're stretching the string connecting the two hoops, which thereby increases the system's elastic potential energy.
The general approach to these types of problems is to write out the system's potential energy, then minimize it, accounting for your constraints.
These problems are called constrained variation problem, which can be quite complex and require calculus. Here is a very famous example (involving some Lagrange multipliers) which is the catenary problem, though note there are also many simpler problems (like the one you pointed out today that don't require advanced calculus).
A: The answers here have inspired me to write the answer I was looking for. Thanks for the help though, I needed the inspiration.
Theorem 1: If the rings are not touching and are not slack, they will always bisect the angle the blue rope makes.
Proof: Rotate the picture such that the bisection angle is vertical. The horizontal forces on the ring can be easily seen to balance each other. Unless the ring/runner is vertical (bisecting) there will be an unbalanced horizontal force.

Theorem 2: If the rings aren't slack, and setting the angles of two adjacent rings to the bisection angle causes them to cross. The rings will lock together very near the position where they are both fully extended.
Proof: Drawing the force diagram shows that the rings will be pushed together due to the not-at-the-bisection-force. Once touching, their not-at-the-bisection-forces must balance. Subtle shifts to the location allow the forces to balance.


Theorem 3: Rings will go slack if the rope passes closer than the length of the runner on its final path.
Proof: Since runners provide purely tensile strength, they will not act on the rope if it is closer to their anchor point than their fully extended length.
This set of proofs get pretty close to fully describing the situation. I think there needs to be a Theorem 4 also related to when rings go slack (Theorem 3 assumes you have the final solution). However, I don't think that's needed till you have at least 3 rings.
Code:
Anyways. I coded up a simple jupyter notebook based on Theorems 1 and 2. It just iteratively sets the runner angles closer to the bisection angle and usually converges. I left out the slack theorem to make the code cleaner. At the moment, if a runner goes slack, the system just oscillates.
All interesting situations with 2 rings:

Finally, here's the algorithm mostly converging on a much more complicated problem.

Code availability:
Here's the jupyter notebook in 3 formats. The first one is interactive and might die. The second two should be perma-links.
https://hub.gke2.mybinder.org/user/bcov77-runner_angles-8bqm46n0/lab/tree/runner_angles2.ipynb
https://nbviewer.jupyter.org/github/bcov77/runner_angles/blob/master/runner_angles2.ipynb
https://github.com/bcov77/runner_angles/blob/master/runner_angles2.ipynb
A: Because the rings are frictionless, the tension is the same all along the blue rope.
At equilibrium, the horizontal leg must be horizontal by symmetry—you can flip the setup horizontally and in terms the forces, you have exactly the same thing, so the rings must be at the same height.
Given these facts—that the tension is equal in each of the blue legs and that they are purely vertical/horizontal, you can see that in order to provide a balancing force, the red/green ropes can only be oriented at forty-five  degrees. Anything else and they’d pull harder horizontally than vertically or vice versa, and you’d end up with a net force on the rings (so they’d move).
A: It's simple – in your prediction diagram, there is nothing supporting the weight.  Do a free body diagram of one of the rings to prove this to yourself.
Therefore the weight falls until the red and green ropes have a vertical component that can take the weight.
See if you can tell why the tilt of the red and green ropes must be 45°.
Is this an assignment for school?
