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UPSC 2022 Maths Optional Paper 1 Q6a — Step-by-Step Solution

20 marks · Section B

Common catenary · Dynamics & Statics · asked 4× in 13 yrs · Read the full method →

Question

A cable of weight $w$ per unit length and length $2l$ hangs from two points $P$ and $Q$ in the same horizontal line. Show that the span of the cable is $2l(1-2h^2/(3l^2))$, where $h$ is the sag in the middle of the tightly stretched position.

Technique

Catenary equations $y=c(\cosh(x/c)-1)$ and $s=c\sinh(x/c)$; small-sag (large-$c$) expansion to second order in $d/c$; eliminate $c$ to relate span and arc length via sag.

Solution

Setup. Cable forms a catenary $y=c(\cosh(x/c)-1)$ measured from the lowest point (vertex at origin, axis along $y$). The total arc length is $2l$ between the suspension points, and sag $h$ at $x=0$ vertically below the support level.

Let span = $2d$, so supports are at $(\pm d, h)$.

Step 1 — Catenary equations

Standard catenary identities (with vertex at origin):

• Arc length from vertex to $x$: $s=c\sinh(x/c)$.
• Height above vertex: $y=c(\cosh(x/c)-1)$.

At the support $x=d$, $y=h$ (the support height above the vertex, which equals the sag from the support level down to the lowest point):

Half-arc length to support:

Step 2 — Small sag assumption (“tightly stretched”)

“Tightly stretched” means the cable is nearly horizontal, so $h\ll l$. This corresponds to large $c$ (large radius of curvature) and small $d/c$.

Expand $\cosh$ and $\sinh$ for small $d/c$:

• $\cosh u\approx 1+u^2/2+u^4/24$.
• $\sinh u\approx u+u^3/6$.

So $h=c[\cosh(d/c)-1]\approx c\cdot\dfrac{(d/c)^2}{2}\!\left[1+\dfrac{(d/c)^2}{12}\right]\approx\dfrac{d^2}{2c}+\dfrac{d^4}{24c^3}$.

To leading order: $h\approx\dfrac{d^2}{2c}$, so $c\approx\dfrac{d^2}{2h}$.

And $l=c\sinh(d/c)\approx c\!\left(\dfrac{d}{c}+\dfrac{1}{6}\!\left(\dfrac{d}{c}\right)^3\right)=d+\dfrac{d^3}{6c^2}$.

Step 3 — Express span $2d$ in terms of $l$ and $h$

From $h\approx d^2/(2c)$: $c\approx d^2/(2h)$, so $c^2\approx d^4/(4h^2)$ and $d^3/c^2\approx d^3\cdot 4h^2/d^4=4h^2/d$.

$l\approx d+\dfrac{d^3}{6c^2}\approx d+\dfrac{1}{6}\cdot\dfrac{4h^2}{d}=d+\dfrac{2h^2}{3d}$.

Solve for $d$ in terms of $l,h$:

So span $2d\approx 2l-\dfrac{4h^2}{3l}=2l\!\left(1-\dfrac{2h^2}{3l^2}\right)$.

Answer