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Cable question (seems off )

  1. Feb 7, 2007 #1

    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.
  2. jcsd
  3. Feb 7, 2007 #2
    No, they cant be equal. Why? sum forces in any direction and do some thinking.

    Post thoughts, and I will help you.
  4. Feb 7, 2007 #3
    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?
  5. Feb 7, 2007 #4
    Why should they?
  6. Feb 7, 2007 #5
    Well if you consider the eye bolt as the object of analysis, would you not have something like the following picture?


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

    Or is it more like this?

    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.
  7. Feb 7, 2007 #6
    What class is this for?
  8. Feb 7, 2007 #7
    Structural Mechanics I, taken after Statics and during Dynamics.
  9. Feb 7, 2007 #8
    What is the first thing they teach you in statics?
  10. Feb 7, 2007 #9
    Sum of the forces = 0, sum of the moments = 0.
  11. Feb 7, 2007 #10
    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.
  12. Feb 7, 2007 #11
    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.
  13. Feb 7, 2007 #12
    Aha, the tension is equal!

    Sorry for leading you the wrong way! -actually it makes sense now.
  14. Feb 7, 2007 #13
    Can you explain why I'd be getting x-component values that differ so much then?
  15. Feb 7, 2007 #14
    Because the x-component values are *not* the same. It is a component of what?
  16. Feb 7, 2007 #15
    It's a component of the overall tension.
  17. Feb 7, 2007 #16
    So does that mean the components are equal?
  18. Feb 7, 2007 #17
    Well I assume they should be equal in order to maintain equilibrium...
  19. Feb 7, 2007 #18
    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:

    Remember, tension is a vector.
  20. Feb 7, 2007 #19
    I should note that the angles on each side of the weight are not equal. And yes, I know tension is a vector. :-p
  21. Feb 7, 2007 #20
    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, dont balance). It is the hook that completes the balance of the system.
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