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Modeling a hanging chain as a PDE

  1. Sep 14, 2015 #1
    1. The problem statement, all variables and given/known data

    A flexible chain of length [itex]\ell[/itex] hangs from one end at [itex]x=0[/itex] but oscillates horizontally. Let the [itex]x[/itex] axis point downwards and the [itex]u[/itex] axis point to the right. Assume that the force of gravity at each point of the chain equals to the weight of the part of the chain below the point and is directed tangentially along the chain. Assume that oscillations are small. Find the PDE satisfied by the chain.

    2. Relevant equations

    I will liberally use Newton's law [itex]\textbf F=m\textbf a[/itex]. I will use [itex]g[/itex] as the gravitational acceleration.

    3. The attempt at a solution

    The longitudinal component of tension must counteract gravity, hence

    [tex]\frac{T}{\sqrt{1+u_x^2}}=-\int_x^\ell\rho g\,dx'=\rho g(x-\ell)[/tex]

    Observing that oscillations are small, we have [itex]\sqrt{1+u_x^2}\to1[/itex], and so [itex]T=\rho g(x-\ell)[/itex]. The transversal component of tension at any given [itex]x[/itex] is

    [tex]\frac{Tu_x}{\sqrt{1+u_x^2}}=\int\rho u_{tt}\,dx[/tex]
    [tex]\frac{\partial}{\partial x}\frac{\rho g(x-\ell)u_x}{\sqrt{1+u_x^2}}=\frac{\partial}{\partial x}\int\rho u_{tt}\,dx[/tex]

    Once again using the small oscillations to remove the denominator:

    [tex]\rho gu_x+\rho g(x-\ell)u_{xx}=\rho u_{tt}[/tex]

    Which is my PDE. Is this correct?
     
  2. jcsd
  3. Sep 19, 2015 #2
    Thanks for the post! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?
     
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