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I am trying to solve following physics exercise but I cannot find my error.

A rope with a cross section $A$, density $\rho$ and a length $l$ is hanging (staticly). I want to calculate the total length the rope extends under its own weight, under the assumption that the density/total weight and cross section stays the same when stretched, and we know hooke's law:

$$\frac{\Delta l}{l} = \frac{F}{AE}$$

So in physicists manner I tried to use following approach: We use a coordinate $x$, with $x=0$ at the bottom end, $x=l$ at the top end of the rope. We consider a infinitesimal length of rope of length $dx$ at the position $x$. The force pulling on this piece is $F = \rho V g = \rho Axg$, so the elongation of this piece of rope is given by

$$\Delta l = \frac{F}{AE} dx = \frac{\rho A xg}{AE} dx = \frac{\rho xg}{E} dx$$

When summing over all those $dx$ we should get an integral for the total elongation $\Delta L$:

$$\Delta L = \int_0^l \Delta l = \int_0^l \frac{\rho x g}{E} dx = \frac{\rho g}{2E} l^2$$

Is this correct so far? The manipulation of those infinitesimal pieces $dx$ etc. always seems very dubious to me, but thats how our physics professors teach it. Is there any ressource that sheds any light on how to work with those?

EDIT: I think the units are not consistent, lets look first at:

$$\left[\frac{\rho}{E}\right] = \frac{kg/m^3}{Pa} = \frac{s^2}{m^2}$$

This would imply $[\Delta L] = s^2$ which doesn't make any sense at all.

flawr
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  • aren't you missing a factor of 1/2 when integrating? – EHH Nov 05 '15 at 16:20
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    If you're bothered by the infinitesimals you could check out this book – Ryan Vitale Nov 05 '15 at 16:27
  • Using infinitesimals can be confusing at first, like a lot of new concepts it's important that you really make sure you try and understand them. This understanding will be gained through both the study of notes (like the ones suggested above) but also a key thing will be practicing using them in lots of examples. It's also worth going and asking your professor about aspects of the course you'd like clarifying. – EHH Nov 05 '15 at 16:35
  • Remember that you can make sense of 'infinitesimals' with a rigorous definition of the Riemann sum. – Simon S Nov 05 '15 at 16:49
  • @EHH Thank you for those hints, I do think a lot of my confusion is because I am studying maths but now took my first physics lecture. And I quite often find myself thinking about how things like those could be mathematically be justified or else why would anyone do this? – flawr Nov 05 '15 at 16:55
  • There is certainly a very different approach in the two subjects, and both are worthwhile and often instruct the other. The physics approach (not universally but certainly commonly) is often to look at a situation from a logical and practical point of view, given that an object is a bit like lots of small versions of that object stuck together we apply our equations to one of those small objects and add/integrate the results to get the result for the whole. This is justified because it works, but is not rigorous, however it can be made rigorous and has indeed been made so in various ways... – EHH Nov 05 '15 at 17:04
  • ... however often the rigorous solution is less intuitive and can sometimes be less 'useful' in understanding a problem, although it is very important because really it is only at that point where we can be safe in the knowledge there is not some error in logic. This interplay has been going on for a long time now, initially involving people doing something that works using dubious mathematics, then when the maths is sorted out properly it often leads to new and beautiful theory. – EHH Nov 05 '15 at 17:08
  • So back to my question, one thing I forgot to add first was that I think the units of my result are incorrect, so there must be an error somewhere, and I suspect the error is in how I used those $dx$ etc. I just updated the question. – flawr Nov 05 '15 at 17:08
  • Is your force term where the error is? $F=ma$, $\rho V$ is mass but you need to multiply by acceleration due to gravity to get the force. – EHH Nov 05 '15 at 17:16
  • Oh thanks, that is it! I am almost disappointed now that the answer was so simple=) – flawr Nov 05 '15 at 17:26
  • Glad to help, and you should feel pleased that you did all the tricky bits fine! – EHH Nov 05 '15 at 17:40

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