Do you guys know if there's some kind of weight distribution that fits a problem like this one?

Let there be a string with two different sections with different linear densities, like the following image, where the concentrated loads may be taken as being positioned in the center of each string section. So I can then calculate the equivalent catenary and be able to use the catenary equations.

I need to know if there's a way to find an equivalent linear density that can be applied in the middle of the complete string.

So far I have tried the obvious assumption that there's a linear distribution but the following equation yields a somewhat not credible result, it seems that, because a large part of the system's total mass is concentrated near x = 0 coordinate.

enter image description here

I'm wondering that there may be some kind of exponential behaviour to the linear density distribution, as exemplified by the m(x) curve, but I would like to be more sure of it.

Thanks for your attention,


Problem setting and additional graphic data

The first image depicts the problem, the second one is the linear density distribution along the x axis and the third one is the concentrated weight distribution along the x axis.

  • $\begingroup$ I think there might be some big misconceptions lurking in your question. A catenary is described by $y=a \cosh(x/a)$, where $a=T_0/(\lambda g)$, $T_0$ is the horizontal component of the tension, and $\lambda$ is the linear mass density. If you have two separate sections of rope with different linear mass density, then you have two separate catenaries with different $a$ constants that you have to join together. Could you clarify what you mean by "apply the equivalent linear density", or "concentrated loads may be taken as [...]"? $\endgroup$
    – David
    Aug 10, 2022 at 22:37
  • $\begingroup$ Hello David, I guess I may reformulate the question because the catenary calculation is the second part of the problem, I'm trying to find some solution tha allows the string with two different linear densities to be considered as one equivalent string with an equivalent concentrated load. After this step I'm looking to deal with the catenary as its usually done, as you have said, a catenary equation depends on the linear densisty of the string per its formulation, $\endgroup$ Aug 12, 2022 at 14:15
  • $\begingroup$ so do you want to approximate the true solution as a single catenary curve with a single mass density constant? You've written the true $\mu(x)$ already! It's one constant for one segment of rope, and a different constant for the next segment. $\endgroup$
    – David
    Aug 12, 2022 at 15:49

1 Answer 1


The catenary equation can be derived by considering the minimization of the functionals, \begin{align} f[y]&=\int_0^{l_2}\mu gy\sqrt{1+\left(\frac{\mathrm dy}{\mathrm dx}\right)^2}\,\mathrm dx \tag{pot. e.}\\ g[y]&=\int_0^{l_2}\sqrt{1+\left(\frac{\mathrm dy}{\mathrm dx}\right)^2}\,\mathrm dx\tag{length} \end{align} where $l_2$ is the point we're hanging a chain of length $l_1$.

Next, we can define the functional, $$ S[y]=f[y]+\lambda g[y]=\int_0^{l_2}\mathcal{L}(y,\,y',\,x)\,\mathrm{d}x $$ where $\lambda$ is a parameter that helps us determine the correct chain length and $y'=\mathrm{d}y/\mathrm{d}x$. From this definition, we have that our Lagrangian is, $$\mathcal{L}\triangleq\left(\mu gy+\lambda\right)\sqrt{1+(y')^2}.$$ So by applying the Legendre transform, we get our Hamiltonian, \begin{align} \mathcal{H}&=y'\frac{\partial\mathcal{L}}{\partial y'}-\mathcal{L} \\ &=y'\left(\mu gy+\lambda\right)\cdot\frac{y'}{\sqrt{1+(y')^2}}-\left(\mu gy+\lambda\right)\sqrt{1+(y')^2}. \end{align}

If we then let the Hamiltonian be the constant total energy (density), $\varepsilon$, then we end up with, $$\left(\frac{\mu gy+\lambda}{\varepsilon}\right)^2=1+(y')^2$$ Under the constant mass density case, we can trivially solve this to get that common $\propto\cosh(x)$ function.

In the non-constant density, the ODE is a little different, $$\frac{\mathrm dy}{\mathrm dx}=\left(\left(\frac{g\mu(x)y(x)+\lambda}{\varepsilon}\right)^2-1\right)^{1/2}\tag{1}$$ and may be a bit more complicated to solve, possibly requiring some numerical solution to fit the two boundary conditions ($y(0)=y_0$ and $y(l_2)=y_2$) along with the condition that $g[y]=l_1$ (i.e., length is fixed).

Since you are interested in an effective mass density when given two different-but-uniform ropes, your mass density is, $$ \mu(x)=\begin{cases} \mu_1 & x<x_p \\ \mu_2 & \text{otherwise} \end{cases} $$ where $x_p$ is the joint. This seems to point you towards having two separate linear cases and an additional constraint that $y(x_p^-)=y(x_p^+)$ (i.e., the heights at the joint as measured from above and from below must match). This likely will require some numerical minimization methods to satisfy.

Once you have that solution, it may be possible to add an additional parameter of the equivalent mass density, $\mu_\text{equiv}$, by fitting the constant-density case with the additional condition that $y(x_p)=y_p$ (i.e., the curve must also pass through the joint), though I don't know if this would result in a solution or not.


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