# Static Equilibrium Two Pulleys and Weights With Ring

I have a problem.

Lets say there were two pulleys put at equal height.
You have a metal ring and attach three wires to it.
At the end of each wire, have three weights of mass w.
One weight hangs down, one weight goes over one pulley and the other over the other pulley.
The system goes to equilibrium.

Okay, simple problem:
Look at one weight over the pulley, sum of vertical forces is equal, therefore:
Fy = 0 = Tension - Weight
Tension = weight.

Same for other side.

Now looking at the ring in the middle:
Fx = 0 = Wcos(a) - Wcosb
0 = W[cos(a)-cos(b)]
cos a = cos b
a = b.

So the angles are the same. And you do more math and they turn out to be 30 degrees each.
So if you can picture sort of the wires are 120 degrees apart perfectly with the ring exactly between the pulleys.

Now consider this:
The right pulley is relatively moved higher by h amount. The weight moves to equilibrium (moves left toward the pulley that is lower).

Where is the ring on the x scale? Only other information is you can use a variable like t to represent the horizontal distance between the two pulleys. I guess you have to give ratio or something.

Okay, no problem, continue like we did before.
Look at one weight.
The sum of the vertical force is 0.
Fy = 0 = T - w
T = w

Similar argument for the other pulley.

Now look at the middle ring.

Fx = 0 = W sin(a) - W sin(b)
implies a = b. Uh oh.

I have no idea how to proceed.

I would appreciate a hint (hopefully I described the situation well).

## Answers and Replies

CWatters
Science Advisor
Homework Helper
Gold Member
Nothing wrong with a = b.

Do a drawing of the geometry being careful not to confuse it with a force diagram.

No... this is a force diagram and the angles are clearly different. At least according to this scale diagram
[EDIT] Oh my god your right... Thank you!

CWatters
Science Advisor
Homework Helper
Gold Member
Looks like you have the answer.

The angles can still be the same (eg a=b) but with the ring in a different position. It then becomes a trigonometry problem to find the offset in the x axis.