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Calculating tension in a taut cable

  1. Feb 14, 2009 #1
    1. The problem statement, all variables and given/known data

    A steel cable is suspended between 2 fixed points that are 10m apart. A force of 12N pulls down at a point 6 m from the end of the cable (ie: the other part of the cable will be 4m). Find the tension in each section of the cable

    2. Relevant equations



    3. The attempt at a solution
    For a typical "hanging weight" question like this, I would use trigonometry to find the angle at which the cable was hanging, then add the 2 tension vectors to give the resultant force of 12N. This force triangle could be solved using the sine law. However, in this case, the cable is horizontal, so there are no angles to calculate, and I don't know how to make the force triangle.
     
  2. jcsd
  3. Feb 14, 2009 #2

    AEM

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    Without some "sag" in the cable the problem makes no sense. A purely horizontal cable would have a tension only in the x, or horizontal, direction and no component in the y direction to support the weight. This is, of course, what you have surmised already.

    So, the next question is: Is there something left out in the wording of the problem?
     
  4. Feb 14, 2009 #3
    Thanks for the quick response :) So, what you are saying is, that this kind of problem can't be solved unless you have a "sag" of some kind in the cable. There may have been an error in the question then. I'm just curious to know what the equal and opposite force would be in a situation like this? The weight is pushing down on the cable. What is the force pushing up?
     
  5. Feb 14, 2009 #4

    Doc Al

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    Staff: Mentor

    The cable will exert an upward force equal to the weight. (The cable tension will have a vertical component, which is why there must be a sag.)
     
  6. Feb 14, 2009 #5
    I did not see anything in the problem statement saying that there was no sag; it simply was not given information. So assume that there is an unknown sag at the point of attachment, and write the equilibrium equations for the point of attachment. Remember that tension itself is constant throughout the cable, even as it passes through the point of attachment. I imagine you may wind up with some messy equations that may require a numerical solution, but there is nothing to prevent that with Maple, Mathematica, or Matlab.
     
  7. Feb 14, 2009 #6
    AnneP, you jumped to a conclusion when you said that the cable was horizontal and that there was no force triangle. Actually, there are two force triangles to deal with, one to the left of the load point, and the other to the right of the load point.
     
  8. Feb 15, 2009 #7

    Doc Al

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    Staff: Mentor

    (1) Without being given the amount of sag, you cannot solve for the tension in the cable segments.
    (2) Don't assume that the tension is constant throughout the cable--in fact it cannot be.
     
  9. Feb 15, 2009 #8
    Thanks for all your thoughts. I am a visual person, so I'm having some trouble getting a grasp of where I am going wrong here without a picture. I've attached some diagrams I
    drew for this problem. Thanks for taking the time to help me out!
     

    Attached Files:

  10. Feb 15, 2009 #9

    Doc Al

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    I don't think you are going wrong at all--the problem statement is missing key information.

    What textbook is this problem from? Have you given the complete statement of the problem, word for word?
     
  11. Feb 15, 2009 #10
    The problem was actually from a worksheet prepared by a teacher, and I stated it word for word. The solution given by the teacher was 11N and 13N for the respective forces on each end of the cable.
     
  12. Feb 15, 2009 #11

    Doc Al

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    Staff: Mentor

    Did the teacher give the solution or just the answers? (If he gave his solution, please post it.)

    Are you expected to know the properties of this particular steel cable? (Its Young's modulus and thickness, for example. I assume not.)
     
  13. Feb 15, 2009 #12
    No solution was given. Just the answers. I don't think the teacher was assuming any particular properties for the cable, so I'm thinking there was probably an error in the question. This was actually a math question, not a physics question, but I thought perhaps there might be some physics principles or properties I didn't know that would enable me to solve the question. From your answers, I am assuming that the question must have been missing information, and that to solve a question like this (grade 12 level math), I would need one additional piece of information. Does this sound likely/reasonable to you?
     
  14. Feb 15, 2009 #13
    AnneP, this is way too much problem for a 12th grade math problem, so rest easy on that point.

    I had previously said that cable tension is constant but Doc Al has pointed out that this cannot be. This is an interesting point. In many circumstances, such as when a load is transferred from one point to another over a cable, the load is applied to the cable through a pulley. Assuming that the load is moving slowly, so that the laws of statics apply (no acceleration), then the cable tension is the same on both sides of the pulley (as I suggested). The subtly in this situation is, however, that the downward load must have a side wise component; otherwise, the load will simply roll to the lowest point on the cable and stop there. To be in equilibrium at some place other than the lowest point, the downward force must actually be along the angle bisector between the two sides of the cable as they leave the pulley, thus creating a line of symmetry about the downward force vector. In short then, if the load is purely vertical, as it appears to be in this problem, then I was in error and Doc Al is correct; the tensions cannot be the same on the two sides of the load.

    I hesitate to say, however, that the problem cannot be solved. One of the critical points is to understand that the connection point moves both downward and laterally. Let the long side be to the left and the short side be to the right of the load. Then denote the displacement of the load point as S downward and D to the right. The following additional notations are used:

    TL = Load to the left
    TR = Load to the right
    LLo = original length on the left side = 6 m
    LRo = original length on the right side = 4 m
    W = applied load = 12 N
    EL = elongation on the left
    ER = elongation on the right
    AE = product of cross sectional area with Young's modulus for the wire

    The following equations are available:

    Horiz Force Sum = - TL * (LLo+D)/sqrt((LLo+D)^2+S^2) + TR * (LRo-D)/sqrt((LRo-D)^2+S^2) = 0
    Vert Force Sum = TL * S/sqrt((LLo+D)^2+S^2) + TR * S/sqrt((LRo-D)^2+S^2) = 0
    EL = TL*LLo/AE = sqrt((LLo+D)^2+S^2) - LLo
    ER = TR*LRo/AE = sqrt((LRo-D)^2+S^2) - LRo

    It is necessary to specify AE because this describes the stiffness of the wire. A very high AE value will mean a very stiff wire and a very small deflection (small S, smaller D), whereas a lower value for AE will give larger values for both S and D.

    Just as a matter of interest, I have run out solutions for this system of equations for the given information and using AE = 650309 N. The resulting tensions are only slightly different from each other:
    TL = 223.99 N
    TR = 224.05 N
     
  15. Feb 16, 2009 #14

    Doc Al

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    Staff: Mentor

    Very reasonable. Simply explain to your teacher how you would solve the problem and ask for the missing information.
     
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