Determine the shape assumed by a heavy cable (chain), fixed…

Determine the shape assumed by a heavy cable (chain), fixed at both ends, hanging under the action of its own weight — the catenary problem.

Step 1: Setup and force balance

Let the cable have uniform weight per unit length $\rho$. Place the origin at the lowest point, $x$ horizontal, $y$ vertical (upward positive). Consider the arc from the lowest point $O$ to a point $P=(x,y)$; it is in equilibrium under the horizontal tension $T_0$ at $O$, the tension $T$ at $P$ (along the tangent, making angle $\varphi$ with the horizontal), and the arc's own weight $\rho s$ acting downward ($s$ = arc length from $O$ to $P$):

\[ T\cos\varphi = T_0, \qquad T\sin\varphi = \rho s. \]

Step 2: The governing differential equation

Dividing the two equilibrium relations and using $\tan\varphi=y'$: $y' = \frac{\rho s}{T_0}.$

Differentiate with respect to $x$, using $ds/dx=\sqrt{1+y'^2}$, and introduce the length parameter $a=T_0/\rho$:

$y'' = \frac{1$.

Step 3: Reduction of order

The equation does not contain $y$ explicitly. Set $p=y'$: $\frac{dp}{\sqrt{1+p^2}} = \frac{dx}{a}.$

Step 4: First integration

The left side integrates to $\sinh^{-1}p$. At the lowest point, $x=0$, the tangent is horizontal, so $p(0)=0$: $\sinh^{-1}p = \frac{x}{a} \ \Rightarrow\ y' = \sinh\frac{x}{a}.$

Step 5: Second integration

Integrating again and choosing the origin so that the lowest point sits at height $y(0)=a$: \[y = a\cosh\frac{x}{a} + C_2, \qquad a = a\cosh0+C_2 \ \Rightarrow\ C_2=0.\]

Step 6: Equation of the catenary

$y = a\cosh\frac{x$, where $a=T_0/\rho$ is the catenary parameter, equal to the height of the lowest point above the directrix.

Step 7: Geometric and physical properties

Arc length from $O$: $s=\int_0^x\cosh(t/a)\,dt = a\sinh(x/a)$, giving the elegant relation $y^2-s^2=a^2$. The tension at $P$ is $T=T_0/\cos\varphi = \rho a\cosh(x/a)=\rho y$ — the tension at any point equals the weight of a length $y$ of cable. The radius of curvature is $R=(1+y'^2)^{3/2}/y'' = a\cosh^2(x/a) = y^2/a = (a^2+s^2)/a$.

For a shallow cable ($x/a\ll1$), $y\approx a+\dfrac{x^2}{2a}$: the catenary is well approximated by a parabola, the shape of a cable under a uniformly distributed horizontal load (e.g.\ a suspension-bridge deck).

Step 8: Verification

From $y=a\cosh(x/a)$: $y'=\sinh(x/a)$, $y''=\dfrac1a\cosh(x/a)$. Then: \[\frac1a\sqrt{1+y'^2} = \frac1a\sqrt{1+\sinh^2\frac{x}{a}} = \frac1a\cosh\frac{x}{a} = y''. \ \checkmark\]