Finding the shape of a hanging rope

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The discussion revolves around finding the optimal shape of a hanging rope, suggesting the use of integrals to minimize the area while considering the slope and material length. The integral approach involves calculating the length and potential energy of the rope, emphasizing the need for constraints related to the string length to achieve a minimum. Participants recommend researching the catenary curve and the principle of least action, noting that the latter is more accurately described as the principle of stationary action. Clarifications are made regarding the necessity of minimizing potential energy rather than applying the principle of least action in a dynamic context. The conversation highlights the importance of imposing additional constraints to effectively solve the problem.
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Homework Statement
given two poles at a distance between them and a rope that it's length is bigger than the distance between them, describe the shape of the rope.
Relevant Equations
no relevent equations
i started to think to maybe do an integral to find the minimum area, and then I thought that the area itself is not sufficient because there is more material depending on the slope. so I thought to do an integral depending on the length instead of x.
##dh^{2}=dx^{2}+dy^{2}##

##\int{}f(x)dh= \int{}f(x)\sqrt{dx^{2}+dy^{2}}##
##dy=\frac{dy}{dx}dx=dx f'(x)##
##\int{}f(x)\sqrt{dx^{2}+f'(x)^{2}dx^{2}}=\int{}f(x)\sqrt{f'(x)^{2}+1} dx^{2}##
how can I get a minimum to solvw this question?
 
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You can edit your post to fix latex issues rather than constantly making an entirely new post.
 
Look up the catenary and the principle of least action.
 
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You only have an integral describing the potential energy but without restrictions on the string length. In order to find an actual minimum of the potential energy, you will furthermore need to impose additional constraints for the string length. There are several ways in which you can do this.

docnet said:
Look up the catenary and the principle of least action.
Nothing is moving here so the principle of stationary* action is not really needed. Only minimizing the potential energy.

* The ”principle of least action” is a misnomer. It is more accurate to use ”stationary” as the solution may be a maximum or saddle point as well.
 
Thread 'Chain falling out of a horizontal tube onto a table'
My attempt: Initial total M.E = PE of hanging part + PE of part of chain in the tube. I've considered the table as to be at zero of PE. PE of hanging part = ##\frac{1}{2} \frac{m}{l}gh^{2}##. PE of part in the tube = ##\frac{m}{l}(l - h)gh##. Final ME = ##\frac{1}{2}\frac{m}{l}gh^{2}## + ##\frac{1}{2}\frac{m}{l}hv^{2}##. Since Initial ME = Final ME. Therefore, ##\frac{1}{2}\frac{m}{l}hv^{2}## = ##\frac{m}{l}(l-h)gh##. Solving this gives: ## v = \sqrt{2g(l-h)}##. But the answer in the book...

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