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I have hemp rope which is hanging horizontally between two winches 140 m far away from each other.

Is it physically possible that the rope won't rip when I turn the winches so the rope will be even (or hanging only 20 cm in the middle)?

I am looking for a formula which considers the weight of the rope.


My attempt so far is this with a formula which doesn’t consider the weight of the rope, only additional weight in the middle of the rope.

enter image description here

F(o) = Tension in Newton
F(g) = weight in the middle of the rope
F(z) = tensile force for two ends
F(s) = initial tension

l = 140 m (Length of the distance between winches)
h = 0,2 m (maximal sag)
m = 0 kg (additional weight onto the rope)
F(s) = 6000 N 

My problem is that I always get 18.000 N (1835.489178 kg) since F(z) is always 0. enter image description here

Because with this formula my hemp rope won’t rip since a hemp rope with a diameter of 4,6 cm (which weighs 1,5 kg/meter) can hold 1870 kg. So theoretically the rope can be tied horizontally. But the rope itself weighs 210 kg and if I assume only 1,5 kg in the middle of the rope I get F(o) = 25.725 N (= 2620 kg) - my rope rips.

How can I consider the weight of the rope in the formula? Either the total weight (as 210 kg) or the weight of 1 m rope (as 1,5 kg).

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  • $\begingroup$ Between two wenches? $\endgroup$ Feb 4, 2017 at 1:39
  • $\begingroup$ winches --- sorry $\endgroup$ Feb 4, 2017 at 1:40
  • $\begingroup$ en.wikipedia.org/wiki/Catenary $\endgroup$
    – user126422
    Feb 4, 2017 at 4:43
  • $\begingroup$ @AlbertAspect thank you for the link. But I am not a physicist at all; which formula do I have to use? I have no clue from the wiki-page. thank you. $\endgroup$ Feb 4, 2017 at 8:42

1 Answer 1

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There is no "formula" to find the tension on the rope. You need to be able to solve an equation using numerical methods. You want to find $a$ from this equation:

$h=a(cosh(L/2a)-1)$

where $h$ is the vertical distance between the minimum height at the center of the rope and the edge of the rope, and $L$ is the length of the rope.

Once you find $a$ you can calculate the tension as $T=aw$, where $w$ is the weight of the rope per unit length.

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  • $\begingroup$ Is gravity for the tension irrelevant? I found a formula F=wg(h+a) (w= weight of the rope per meter; g=gravity; h=sag; a as above; de.wikipedia.org/wiki/Kettenlinie_(Mathematik)#Beispiel) The unit would be Newton, right? $\endgroup$ Feb 5, 2017 at 9:04
  • $\begingroup$ I cannot follow german, but the result looks fine to me on an overview, and seems essentially the same. First, in both solutions gravity appears in the same place (the g in my solution appears because your w is mass density and mine is weight density (a factor g of difference). Also, I made the approximation that h<<L, which should be good in your case, to simplify the solution, but I did not realize that this approximation in the end seems not to be necessary. $\endgroup$
    – user126422
    Feb 5, 2017 at 14:06

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