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Suppose I have a single rope attached to a fixed point via a load cell, which gives a number (in kN) based on the load it's experiencing. I take a weight (x) and attach it to the rope at a fixed point and raise it up to a given height and then drop it. The rope has elastic properties and has some elongation.

Now, suppose I repeat the process, but this time I use two ropes hanging in parallel (actually, it's the same rope with a mid-line knot which prevents the tension/elongation in one half being transferred to the other half).

What I need to understand is whether the number on the load cell would be different between the two scenarios, and if so, why?

Ropes diagram

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Is this a question about rock climbing? If so, then the answer is never to use twin ropes, because they're a total pain, and their theoretical advantages never show up in actual climbing. (And note that (1) twin ropes have different physical specs, and (2) the twin ropes will usually not share the load at all equally.)

But anyway, as a pure physics question, the answer to this is not one number. The tension is a complicated function of time in both cases. The masses will move as in simple harmonic motion when the rope is under tension, but this can be interspersed with periods of free fall when the rope is slack.

You could change the question to ask about the maximum tension in the rope. Then conservation of energy tells us (for small damping) that $T\propto \sqrt{k}$, where $T$ is the maximum total tension and $k$ is the equivalent spring constant of the ropes. (To see this, observe that the work done by the rope is fixed, and equals $(1/2)kx^2$. The maximum tension is $kx$, which is therefore proportional to $\sqrt{k}$.) Using two ropes in parallel doubles $k$, and therefore increases the maximum total tension $T$ by $\sqrt{2}$.

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  • $\begingroup$ Can you please elaborate on how the stored potential energy (work) and the maximum tension leads us to $\sqrt{k}$? I don’t see how this comes about. What I could draw from those two equations was that $T \propto \sqrt{\frac1k}$ which is obviously false $\endgroup$ – Shinaolord Oct 14 '19 at 3:25
  • $\begingroup$ +1, though IMHO there are big advantages to twin ropes for rock climbing in some situations, such as when climbing in a party of three. $\endgroup$ – Ben51 Oct 14 '19 at 16:59
  • $\begingroup$ @Shinaolord: The reasoning is what I gave in the parenthetical statement beginning with "To see this..." If you're getting a different result, I would have to see what your reasoning was. $\endgroup$ – user4552 Oct 14 '19 at 20:49
  • $\begingroup$ The rope breaks above a certain potential energy $U_{max}\leq\frac{kx^2}2$ which can be rearranged to what i said above. I don't see how we can incorporate the maximum tension, $kx$. That's the essence of what I don't get, I suppose. Are you setting up something along the lines of $U_{max}= \frac{T_{max} x}{2}$ ? $\endgroup$ – Shinaolord Oct 14 '19 at 22:00
  • $\begingroup$ Would the following work? $U_{max} \leq \frac{kx^2}2 \implies x\propto \sqrt{\frac1k} \implies \frac{T_{max}}k \propto \sqrt{\frac1k} \Longleftrightarrow T_{max} \propto \sqrt{k}$ ? $\endgroup$ – Shinaolord Oct 14 '19 at 22:08
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Allowing for a tidier knot than you have drawn, so the knot itself has no effect on load (i.e. the knot doesn't have any 'give' in it), then things will be as follows:

  • While the weights are falling freely the load cells in both systems will record zero load (for simplicity I am assuming a weightless rope).

  • When the masses reach the bottom of the rope/ropes, the doubled rope load cell will display twice the force than the single rope load cell during the period over which the masses are being decelerated to a halt. The double force, in the folded rope system, will be displayed for half the time as the force in the single rope system. (The above assumes that the masses were dropped from the same initial high above the respective stopping heights of the two systems and that the free length of rope/double-rope is the same in both systems. If you vary the height of the drop and/or the suspension height then things will obviously be different).

  • After the masses have decelerated and are hanging freely, the two load cells will both register identical loads Mg.

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  • $\begingroup$ Thank you!! This is along the lines of what I was thinking. Are you able to elaborate on the following please, and if you have mathematical proof as well, that would be excellent "When the masses reach the bottom of the rope/ropes, the doubled rope load cell will display twice the force than the single rope load cell during the period over which the masses are being decelerated to a halt." I need to know why this occurs $\endgroup$ – A.Benson Oct 13 '19 at 22:15
  • $\begingroup$ During deceleration ropes can be considered as springs (with very high spring constants). For each spring/rope the force as it comes into tension is F=-Kx (for some K), Two springs acting in parallel will produce twice the force as a function of extension, as each can be considered independently, so F= -Kx + -Kx = -2Kx. But note that if you do this with a real rope then each half of the doubled rope will have half the length - in that case the spring constant for each shortened half will double - then the force would go as F=-2Kx for each half, so total would then be 4x as large! $\endgroup$ – Penguino Oct 13 '19 at 23:15
  • $\begingroup$ Thanks. Just to confirm, K = the spring constant and X = the mass of the load?? $\endgroup$ – A.Benson Oct 14 '19 at 0:00
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    $\begingroup$ No, X is the amount the rope is stretched. $\endgroup$ – Ben51 Oct 14 '19 at 0:22
  • $\begingroup$ When the masses reach the bottom of the rope/ropes, the doubled rope load cell will display twice the force than the single rope load cell during the period over which the masses are being decelerated to a halt. No, this is wrong, as shown in my answer. $\endgroup$ – user4552 Oct 14 '19 at 0:56

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