Can the tension in a hanging cable be determined with known length and location?

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    Cable Catenary
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Discussion Overview

The discussion revolves around determining the tension in a hanging cable modeled by the equation y=acosh(x-h)/a+k, given the known length and location of its endpoints. The context includes theoretical exploration and potential applications in computer simulations where the weight of the cable is not considered significant.

Discussion Character

  • Exploratory
  • Technical explanation
  • Conceptual clarification

Main Points Raised

  • One participant seeks to determine the parameter 'a' in the cable's equation to complete the model, noting that the weight of the cable is not a factor in their simulation.
  • Another participant provides a formula for the length of the cable, L = 2a sinh(d/2a), suggesting that 'a' can be numerically approximated using known values of d (distance between endpoints) and L.
  • A participant questions whether the provided equation is applicable for endpoints of unequal height, indicating uncertainty about the suitability of the catenary model used.
  • A later reply introduces a scenario where the force acting on the cable is not gravity but drag, proposing a more complex situation involving non-uniform force distribution and asking if this can be solved.

Areas of Agreement / Disagreement

Participants express differing views on the applicability of the catenary equation for cables with endpoints at unequal heights, and there is no consensus on how to approach the problem of non-uniform forces acting on the cable.

Contextual Notes

There are limitations regarding the assumptions made about the endpoints' heights and the nature of the forces acting on the cable, which remain unresolved in the discussion.

joel_f
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hi~
i need to determine the tension a in the equation of the hanging cable y=acosh(x-h)/a+k with a known length and known location of end points. i figured out how to determine the horizontal translation h but i need a in order to do it and to have the complete equation.

this is for a computer simulation where the weight of the cable is not important. this is why i condensed tension in the formula down to a. if the weight is a requirement then it can be set, in which case i would need to find the horizontal force acting on the cable.

is it possible to find the tension with the given information?
 
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The length of the cable L is given by y=2a\mbox{sinh}\left(\frac{d}{2a}\right), where d is the distance between the end points of the cable (I assumed they were at equal heights based on your equation). You can get a numerical approximation of a from this equation after substituting in the known values of d and L.
 
would this work for end points of unequal height? am i using the wrong equation for the catenary of this type?
 
Hello benorin,

I hope you will still get this. I very much liked your extensive answer of the problem.

What if you would replace the force acting on the cable from being gravity to drag. In other words what if the force is not equally distributed. Like a fish on a fishingrod swimming around the fisherman, or a cable with increasing density towards one end.

Can this be solved?

Regards,
Seuren
 

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