# Cable question (seems off )

stinlin
http://img185.imageshack.us/img185/8830/cablesyv1.gif [Broken]

You're looking at the design for my recent lab we did. Now, you would think, given the geometry of the loadings (i.e. post heights, eye bolt height, weight heights, distances from a specified origin) that the x components of the tension in the middle cables (on the left and right side of the eye bolt) would be equal, no?

BUT! After using my data and calculating the x components of these tensions, I'm getting that one set is about twice the other set of x values. Is there a physical reason for this, or is there some unknown error? I've checked and rechecked my calculations, but NOTHING is making sense.

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Cyrus
No, they can't be equal. Why? sum forces in any direction and do some thinking.

stinlin
Well I thought that they had to be equal because if you were to analyze the forces on the eye bolt, oh...But that could be considered a moment connection for this case (i.e. a connection that can resist forces in x,y,z and moments in x,y,z) so that maybe this connection adds to the forces in the x direction, thus making the tension values not equal?

Problem is, I'm still not seeing why the x components wouldn't cancel each other out. If the report on this data wasn't due tomorrow afternoon, I'd go through and try sectioning off portions of the cable and calculation tensions there.

Can you shed some more light on this by chance?

Cyrus
Problem is, I'm still not seeing why the x components wouldn't cancel each other out.

Why should they?

stinlin
Well if you consider the eye bolt as the object of analysis, would you not have something like the following picture?

http://img506.imageshack.us/img506/8306/cables2jn4.gif [Broken]

Although that almost wouldn't make sense since there's no y-component...

Or is it more like this?
http://img506.imageshack.us/img506/7025/cables3dl1.gif [Broken]

In the first test we performed, there was no eye bolt, so the cable went directly to the other weight. I may be making the wrong assumption that the tensions must equal out. I'm kind of analyzing this as a system like the hanging chandelier example we're all so familiar with (the fixture hangs from n cables, find the tensions in each, and all the x-components there equal zero).

Also, the system is in static equilibrium, so I figured that analysis at any point on the cable system should come out to being in static equilibrium.

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Cyrus
What class is this for?

stinlin
Structural Mechanics I, taken after Statics and during Dynamics.

Cyrus
What is the first thing they teach you in statics?

stinlin
Sum of the forces = 0, sum of the moments = 0.

Cyrus
Hmmm, can you show me how you got the tension values?

Actually, all I care about is what are the angles you got on each segment.

stinlin
Alright - for the left side going to the bolt, I found the angle to be 65.59* in run 1 and 48.74* in run 2. For the right cable, I found the angle leading to the bolt to be 41.67* and 17.281* for runs 1 and 2 respectively.

If it helps, the weights in run 1 were 10 lbs and 8 lbs and in run 2, 3 lbs and 5 lbs.

Cyrus
Aha, the tension is equal!

Sorry for leading you the wrong way! -actually it makes sense now.

stinlin
Can you explain why I'd be getting x-component values that differ so much then?

Cyrus
Because the x-component values are *not* the same. It is a component of what?

stinlin
It's a component of the overall tension.

Cyrus
So does that mean the components are equal?

stinlin
Well I assume they should be equal in order to maintain equilibrium...

Cyrus
When you have a single continuous cable, the tension is the same all throughout the cable. The reason I asked you for the angle is because the component of tension in the y direction must balance each mass.

For the 10lb mass at an angle of 65.59 this means: 10=2T*sin(65.59)

Solving for T gives the value at T=5.49lbs.

Now, do a force balance at the 8lbs mass. 8=2*5.49*sin(theta)

Solving for theta yields: 46.76 degrees, which is very close to what you got:
41.67.

Remember, tension is a vector.

stinlin
I should note that the angles on each side of the weight are not equal. And yes, I know tension is a vector.

Cyrus
Well I assume they should be equal in order to maintain equilibrium...

This is the error in your thinking. You should have looked more carefully at that eyehook. When you added the 10lb weight, you would have noticed that the rope around the eyehook would have slowly swung in the direction of the 10lb mass. In other words, it would not be directly in line with the middle of the eye hook anymore. This would allow the eye hook to provide a reaction force in the x-direction to compensate for the imbalance in the x-direction due to only the ropes themselves (which as you said, don't balance). It is the hook that completes the balance of the system.

Cyrus
I should note that the angles on each side of the weight are not equal. And yes, I know tension is a vector.

Oh...what were all the angles then?

Cyrus
No, this is wrong. The angles on each side of the weight have to be equal. Sorry, its the law of statics. Do a force balance on any weight and section off the rope. If they are not equal, the mass will move in the direction of the imbalance.

stinlin
I see you're argument, but we also had a situation like this:

http://img239.imageshack.us/img239/7162/cablesxwr3.gif [Broken]

Those aren't the exact weights, but that's essentially what the cable loading looked like. The angles are DEFINITELY different in this case. I should remind you that this is one cable that goes from the left side, through the eye bolt, to the right side. The middle bolt is the highest point, followed by the right bolt, and then the left.

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Cyrus
What were the angles on the 10lb weight in that case?

stinlin
It's only a 5 lb weight, and the angles are 17.281 and 31.492 (left angle, right angle)

stinlin
And I don't want to sound conceited, but our first test that we ran had an error percent of about 2%, so my thoughts are that we definitely made the right measurements and calculations. I could be TOTALLY off though...

stinlin
Oh yes, and my group member just told me he got values that were about 2 times in size also.

Cyrus
Can you draw it out with all the angles labeled. I don't know where they go.