Is the Pulley System in Equilibrium?

[engineer248]

Homework Statement

This is the given diagram. All weights are the same, the gray round things are pulleys, the black dot is simply a connection of the ropes.

Question: The system is in equilibrium. Solve for x and y in terms of b and h.

http://img27.imageshack.us/img27/9724/physicsps1.png"

Homework Equations

Sum(F_x) = 0
Sum(F_y) = 0

The Attempt at a Solution

Since nothing is moving, the horizontal forces must be equal for both strings. Both are under the same tension, therefore the ropes' angles are the same, so that the sum of the sines (the vertical upward forces) equal the drape force of the center weight.

So far my solution. Only in the diagram, the right rope goes up way steeper than the one on the left. Am I missing something? What about (y/x) = (h+y)/(b-x)? These are the tangents of the triangles, sort of. But I always get stuck somewhere, nothing seems to get me anywhere useful.

I'm looking forward to your ideas!

 

[PhanthomJay]

Homework Statement

This is the given diagram. All weights are the same, the gray round things are pulleys, the black dot is simply a connection of the ropes.

Question: The system is in equilibrium. Solve for x and y in terms of b and h.

Homework Equations

Sum(F_x) = 0
Sum(F_y) = 0

The Attempt at a Solution

Since nothing is moving, the horizontal forces must be equal for both strings. Both are under the same tension, therefore the ropes' angles are the same, so that the sum of the sines (the vertical upward forces) equal the drape force of the center weight.

So far my solution. Only in the diagram, the right rope goes up way steeper than the one on the left.

it's a diagram that may not be drawn to scale.

Am I missing something?

Yes, what angle does the rope make with the x-axis at the center weight?

What about (y/x) = (h+y)/(b-x)?

Looks good, based on similar triangles, not tangent triangles.

But I always get stuck somewhere, nothing seems to get me anywhere useful.

I'm looking forward to your ideas!

Simplify your solution by noting the relationship between x and y.

 

[engineer248]

Thanks for your answer, Jay! You at least seem to know where this is going.

So I know that the angles must be equal. The angle formed with the horizontal must be

$$atan\left({y}\over{x}\right)$$ or $$asin\left({y}\over{\sqrt{x^2+y^2}}\right),$$

which is equal to

$$atan\left({h+y}\over{b-x}\right)$$ or $$asin\left({h+y}\over{\sqrt{(b-x)^2+(h+y)^2}}\right),$$

I was hoping to get something out of this, but this is exactly the point where I get stuck. Equating the first and the second equations is a dead end. I just can't seem to get x (or y) isolated on one side of the equation. Can you give me one more hint?

 

[PhanthomJay]

As you note, you have the right equations, but you are going nowhere because you are missing a key point. You noted earlier that the horizontal components of the cord tensions at the central weight tie point must be equal. You also noted that the vertical components of the cord tensions at that point must sum to the weight at that point. And you also noted the cord angles with the horizontal at that point must be equal. Therfore, the vertical cord tensions at that point not only sum to the weight at that point, but also, they too must be equal. So if the horiz and vert components are equal, then the angle must be ________degrees, and y/x = _____?