5

Suspending a cable produces a hyperbolic cosine shape, but what happens if we orientate the problem in such a way where the rope begins at $(0,y)$ and ends at $(x,0)$ and we specify a rope of length $m$ such that the rope can hug the boundaries if there is enough slack. Intuitively we know the maximum possible length of the rope is $y+x$ (completely hugs boundary) and the minimum is $\sqrt{x^2+y^2}$ (taught). So what is the resulting shape for a rope length in between these boundaries? I have never seen this problem in any text . I did some reading and I suspect we have to use Hamiltonian mechanics and a multiplier but what is the curve.

Daniel Fischer
  • 206,697
mymathrp
  • 51

1 Answers1

1

Wherever the cable hangs in the air its shape is an arc of a catenoid $$\gamma:\quad y=y_0+p\left(\cosh{x-x_0\over p}-1\right),\qquad p>0.$$ Here $(x_0,y_0)$ denotes the apex of the catenoid. The length $l$ of such an arc beginning at $(x_1,y_1)$ and ending at $(x_2,y_2)$ computes to $$l=p\>\left|\sinh{x_2\over p}-\sinh{x_1\over p}\right|\ .$$

When the cable is suspended freely between $A=(0,a)$ and $B=(b,0)$ then the position of the apex will depend on the given total length $\ell$, and there is a characteristic length $\ell_0$ for which the apex is exactly at $B$ (unfortunately $\ell_0$ can only be computed numerically from the given data $a$ and $b$).

When $\ell<\ell_0$ then the apex is to the right of $B$, and the slope of the cable arriving at $B$ is still negative. Introducing the $x$-axis as "floor" will therefore have no effect on the cable.

When $\ell>\ell_0$ then the apex is (below and) to the left of $B$; therefore introducing the $x$-axis as "floor" creates a new situation. I suspect that we shall see the following: The shape of the cable is an arc of a catenoid $$\gamma_c:\quad y=p\left(\cosh{x-c\over p}-1\right),\qquad 0<c<b,$$ beginning at $A$ and ending at the apex $(c,0)$ of $\gamma_c\ $, and then proceeds along the $x$-axis to $B$. In order to find $c$ we have to solve the system $$p\left(\cosh{c\over p}-1\right)=a\ ,\qquad p\ \sinh{c\over p}+(b-c)=\ell$$ for $p$ and $c$.

Christian Blatter
  • 226,825