This article is translated from a Chinese article on my Zhihu account. The original article was posted at 2019-12-18 11:20 +0800.

Review middle school contents!

Suppose a uniform heavy elastic rope has mass $m$, original length $L_0$, and stiffness $k$. Find the mass distribution and length of it when hung vertically.

When the elastic rope has its original length, divide it into $n$ equal segments, where $n$ is large so that when the elastic rope is stretched, the mass distribution within every segment is uniform.

When the rope is hung vertically, the tension experienced by every segment is


Therefore, the length of every segment is

\[\Delta l_j=\frac{F_j}{nk}+\frac{L_0}n=\frac{mg\left(n-j\right)}{n^2k}+\frac{L_0}n.\]

Therefore, the position of the end of every segment is

\[x_j=\sum_{s=1}^j\Delta l_s=\frac{2nL_0kj+2nmgj-mgj-mgj^2}{2n^2k}.\]

The equation can be solved to get


The (linear) mass density of every segment is

\[\rho_j=\frac{m}{n\Delta l_j}=\frac{m}{n\left(\frac{mg\left(n-j\right)}{n^2k}+\frac{L_0}n\right)}.\]

Express $j$ here with $x_j$, and we have


The length of the rope is


Take the limit $n\to\infty$, and we have

\[L=\frac{mg}{2k}+L_0,\] \[\rho\!\left(x\right)=\pm\frac{mk}{\sqrt{L_0^2k^2+m^2g^2+2mgk\left(L_0-x\right)}}.\]

The plus-minus sign here should take plus sign.

To test, calculate $\int_0^L\rho\!\left(x\right)\mathrm dx=m$.

The conclusion can be written in a more beautiful way: