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UPSC 2025 Maths Optional Paper 1 Q5d — Step-by-Step Solution

10 marks · Section B

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

Question

Given that $A$ and $B$ are two points in the same horizontal line distant $2a$ apart. $AO$ and $BO$ are two equal heavy strings tied together at $O$ and carrying their weight at $O$. If $l$ is length of each string and $d$ is depth of $O$ below $AB$, then show that the parameter $c$ of this catenary, in which the strings hang, is given by $l^2 - d^2 = 2c^2\left[\cosh\left(\dfrac{a}{c}\right) - 1\right]$.

Technique

Use the common catenary $y = c\cosh(x/c)$ with parameter $c=H/w$ (horizontal tension over weight per unit length), apply the standard arc-length and sag relations, then eliminate using the identity $\cosh^2-\sinh^2=1$.

Solution

A uniform heavy string hangs in a catenary $y = c\cosh\!\left(\dfrac{x}{c}\right)$, with the lowest point at $x=0$, $y=c$, and parameter $c$.

By symmetry the two equal strings $AO$, $BO$ form a single symmetric catenary whose lowest point is $O$ (the junction carrying the load lies at the vertex). Take the origin of the catenary at its vertex, so the lowest point is at $x=0$, height $c$.

The supports $A,B$ are symmetric about the vertical through $O$ at horizontal distance $a$ on each side, i.e. at $x=\pm a$.

Standard catenary relations (vertex at $x=0$):

1. Arc length from the vertex ($x=0$) to the support ($x=a$):

Since each string runs from $O$ (the vertex) to a support, the length of each string is

2. Vertical rise (sag) from the vertex to the support level:

Since $AB$ is the support level and $O$ is the vertex, the depth of $O$ below $AB$ is

Eliminate $a/c$. Write $C \equiv \cosh\!\left(\dfrac{a}{c}\right)$. From (1) and the identity $\sinh^2 = \cosh^2 - 1$,

From (2),

Subtract:

Answer