Common catenary

At a Glance

• Frequency: 4 sub-parts across 4 of 13 years (2015, 2021, 2022, 2025)
• Priority tier: T3
• Marks (count): 20 (1), 15 (1), 12 (1), 10 (1)
• Average solve time: ~12 min
• Difficulty mix: hard 2, medium 2
• Section: B | Dominant type: computation/proof

Why This Chapter Matters

Catenary questions appear in Section B at 10–20 marks and are among the harder mechanics problems. Every question derives from three standard catenary formulae; mastering these three formulae plus the geometric meaning of the parameter $c$ converts a daunting question into an algebra problem. The key is to identify which formula pairs are needed for the particular question and to use the hyperbolic identity $\cosh^2-\sinh^2=1$ to eliminate the parameter.

Minimum Theory

The catenary. A uniform heavy chain hanging freely under gravity hangs in a catenary. Taking the lowest point at $x=0$, the equation is $y = c\cosh\!\left(\frac{x}{c}\right),$ where $c = H/w$ is the ratio of the horizontal tension $H$ to the weight per unit length $w$.

Standard formulae. With $s$ = arc length from the vertex and $\psi$ = angle of tangent to horizontal:

Key identities. $\cosh^2\theta-\sinh^2\theta=1$, so $s^2+c^2 = c^2\cosh^2(x/c) = y^2$: the “Pythagorean” form $y^2 = c^2+s^2$ (tension = $T_0\sec\psi = wy$, a useful shortcut).

Question Archetypes

catenary (4 question(s); 2015, 2021, 2022, 2025)

Recognition Cues

• A chain or cable hangs between two points; given some of: length $l$, sag $d$, half-span $a$, parameter $c$.
• “Show that …” or “find the tension at …” or “show $HM$ of tensions…” — all require the three standard formulae above.
• The phrase “catenary parameter $c$” signals that the answer will be expressed in terms of $c$.

Solution Template

1. Set up coordinates: place the vertex at $x=0$; supports at $x=\pm a$ (for symmetric catenary).
2. Write the two formulae that relate the given quantities to the unknowns.
3. Eliminate the intermediate variable (usually $a/c$) using $\cosh^2-\sinh^2=1$.
4. For tensions: use $T=T_0\sec\psi$ (where $T_0 = wc$ is the horizontal tension) and the intrinsic relation $s=c\tan\psi$.

Worked Example 1

2025 Paper 1, 2025-P1-Q5d (10 marks)

Two points $A$, $B$ on the same horizontal line, $2a$ apart. Equal strings $AO$, $BO$ hang from $A$, $B$ and are tied at $O$. Arc length of each string $= l$, depth of $O$ below $AB$ = $d$. Show: $l^2-d^2=2c^2[\cosh(a/c)-1]$.

Setup. By symmetry $O$ is the vertex; $A$, $B$ are the supports at $x=\pm a$.

Arc length: $l = c\sinh(a/c)$, so $l^2 = c^2\sinh^2(a/c) = c^2(\cosh^2(a/c)-1)$.

Sag: $d = c(\cosh(a/c)-1)$, so $d^2 = c^2(\cosh(a/c)-1)^2$.

Subtract: $l^2-d^2 = c^2(\cosh^2(a/c)-1) - c^2(\cosh(a/c)-1)^2 = c^2(\cosh(a/c)-1)\bigl[(\cosh+1)-(\cosh-1)\bigr] = 2c^2(\cosh(a/c)-1).\;\blacksquare$

$\boxed{l^2-d^2 = 2c^2\!\left[\cosh\!\left(\frac{a}{c}\right)-1\right].}$

Worked Example 2

2021 Paper 1, 2021-P1-Q6a (20 marks)

At points $A$, $B$, $C$ on a catenary the inclinations are in AP with common difference $\beta$. Weights of portions $AB$, $BC$ are $\omega_1$, $\omega_2$. Prove: (i) $\text{HM}(T_1,T_2,T_3) = 3T_2/(1+2\cos\beta)$. (ii) $T_1/T_3 = \omega_1/\omega_2$.

Part (i). Tension: $T_i = T_0\sec\psi_i$. Let $\psi_2$ be the inclination at $B$; $\psi_1=\psi_2-\beta$, $\psi_3=\psi_2+\beta$.

Harmonic mean: $\text{HM} = 3/(1/T_1+1/T_2+1/T_3) = 3T_0/(\cos\psi_1+\cos\psi_2+\cos\psi_3)$.

Using sum-to-product: $\cos\psi_1+\cos\psi_3 = 2\cos\psi_2\cos\beta$.

So $\cos\psi_1+\cos\psi_2+\cos\psi_3 = \cos\psi_2(1+2\cos\beta)$.

$\text{HM} = \frac{3T_0}{\cos\psi_2(1+2\cos\beta)} = \frac{3T_0\sec\psi_2}{1+2\cos\beta} = \boxed{\frac{3T_2}{1+2\cos\beta}.}\;\checkmark$

Part (ii). The vertical tension changes by the weight of the segment: $T_0(\tan\psi_2-\tan\psi_1)=\omega_1$ and $T_0(\tan\psi_3-\tan\psi_2)=\omega_2$.

Using $\tan\psi_2-\tan\psi_1 = \sin(\psi_2-\psi_1)/(\cos\psi_2\cos\psi_1) = \sin\beta/(\cos\psi_2\cos\psi_1)$, and similarly for $\omega_2$:

$\frac{\omega_1}{\omega_2} = \frac{\sin\beta/(\cos\psi_2\cos\psi_1)}{\sin\beta/(\cos\psi_2\cos\psi_3)} = \frac{\cos\psi_3}{\cos\psi_1} = \frac{T_0/T_3}{T_0/T_1} = \boxed{\frac{T_1}{T_3}.}\;\blacksquare$

Common Traps

• Mixing up $l$, $d$, and $c$. The sag $d$ is measured from the vertex to the support level (not from the support to any other reference). The arc length $l$ runs from the vertex to the support along the curve. The half-span $a$ is the horizontal distance.
• $\cosh^2-\sinh^2=1$ is the elimination tool. Almost every catenary proof involves writing $l^2=c^2\sinh^2(\cdot)$ and $d^2=c^2(\cosh(\cdot)-1)^2$ and then taking the difference, using this identity.
• Tension = $T_0\sec\psi$, not $T_0/\psi$. The horizontal tension $T_0$ is constant along the catenary; the total tension increases with inclination via $\sec$.
• Weight of a segment = change in vertical tension. $\omega = T_0(\tan\psi_{\text{right}}-\tan\psi_{\text{left}})$ — this is the load-bearing physics that connects weights to inclinations.

Marks-Aware Writing

For 10-mark identity proofs: state the two catenary formulae used, substitute, and show the algebraic steps explicitly. The key step (factoring $(\cosh-1)$ and using $\cosh+1-\cosh+1=2$) must be written out.

For the 20-mark tensions-in-AP question: each of the two parts requires a separate derivation; the sum-to-product formula ($\cos\psi_1+\cos\psi_3=2\cos\psi_2\cos\beta$) and the weight–tangent relation must both be stated explicitly.

For the 12-mark cable span formula (2022): use the small-sag (large-$c$) expansion of $\cosh$ and $\sinh$ to first order in $1/c$; state the expansion explicitly.

Practice Set

• 2015-P1-Q8a (12 m) — — Hint: chain over circular pulley; 2/3 contact means support points are $120°$ from the bottom; tangent angle at leaving point is $60°$; use arc $= c\tan\psi$.
• 2022-P1-Q6a (20 m) — — Hint: expand $\cosh(d/c)\approx1+d^2/(2c^2)$ and $\sinh(d/c)\approx d/c+d^3/(6c^3)$ for large $c$.