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Homework Help: Inverse cable problem

  1. Feb 8, 2008 #1
    cable problem

    I'm doing some self study on the subject of cables and I am curious about the reversed process of working out the linear mass density when given the shape of the cable (instead of the usual case of determining the shape of the cable given the density). The text I am currently reading explains an example but I would like to know how it works for other examples.

    1. The problem statement, all variables and given/known data

    Say we have a flexible chain hanging between fixed ends separated by a distance [tex] r_0 [/tex]. The weight of a small element [tex]ds[/tex] located at a position along the chain defined by the angle [tex]\theta[/tex] is [tex]\lambda[/tex][tex]gds[/tex] where [tex]\lambda[/tex] is the linear mass density of the cable and [tex]g[/tex] the acceleration due to gravity.

    I don't have a scanner so I can't scan the diagram of forces, but considering the forces on the element, the equilibrium equations give:

    (1) tangential component: [tex]dT = \lambda g ds sin \theta[/tex]
    (2) normal component: [tex]Td \theta = \lambda g ds cos \theta[/tex]

    dividing (1) by (2) we get

    [tex] \frac {dT}{T} = tan \theta d\theta[/tex]

    i.e. a separable O.D.E thus integrating both sides we get:

    [tex]T = \frac {T_0}{cos \theta}[/tex]

    where [tex]T_0[/tex] equaling the cable tension when [tex]\theta = 0[/tex]

    Now suppose I want to find the linear density for the shape of a parabola [tex]y = x^2[/tex]. I have taken the following approach: (please correct me if I am wrong)

    The attempt at a solution

    From (2), rearranging we get [tex]T \frac{d\theta}{ds} =\lambda g cos \theta[/tex]

    Curvature [tex]\kappa[/tex] is given by [tex]\frac{d \theta}{ds}[/tex] or in the form [tex]\kappa = \frac{y''}{(1+y'^2)^{3/2}}[/tex]

    So we now have

    [tex]T \kappa = \lambda g cos \theta[/tex] (*)

    differentiating our equation for the parabola [tex]y=x^2[/tex], [tex]y'=2x[/tex], [tex]y''=2[/tex] and substituting these into our [tex]\kappa[/tex] equation and rearranging (*) for [tex]\lambda[/tex]:

    [tex]\lambda = T(\frac{\frac{2}{1+(2x)^2)^{3/2}}}{g cos \theta}[/tex]

    Can someone please tell me if I am taking the right approach? Also is there any way to simplify what i have obtained for [tex]\lambda[/tex]?

    Many many thanks in advance!
    Last edited: Feb 8, 2008
  2. jcsd
  3. Feb 9, 2008 #2

    Tom Mattson

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    Staff Emeritus
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    Gold Member

    I'm not exactly sure of where equations (1) and (2) come from, but if you can apply equation (2) to the y=x^2 geometry, then your approach should be fine.
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