9. Mechanics — Statics

Common catenary

The common catenary is the curve assumed by a perfectly flexible, inextensible, uniform heavy string suspended freely between two fixed points under gravity.

For: UP PGT Mathematics UP TGT Mathematics DSSSB PGT / TGT KVS / NVS PGT LT Grade

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Definition and statement

Intrinsic and Cartesian Equations of the Common Catenary

A uniform heavy flexible string of weight per unit length $w$ hangs freely under gravity between two fixed points. The intrinsic equation relating arc length $s$ (measured from the lowest point $C$) to the tangent angle $\psi$ is given by $s = c \tan \psi$, and the Cartesian equation referred to the lowest point as origin with the axis of $y$ vertically downwards is $y = c \cosh\left(\frac{x}{c}\right)$, where $c = \frac{T_0}{w}$ is the parameter of the catenary representing the span length at the lowest point.

📐 Geometric Interpretation

The curve is symmetric about the vertical axis passing through its lowest point (vertex). The tangent at the lowest point is horizontal ($\psi = 0, s = 0$). As we move along the curve, the slope $\tan \psi$ increases proportionally with the arc length $s$ scaled by the parameter $c$. The ordinate $y$ at any point equals the tension at that point divided by the weight per unit length ($y = T/w$).

⚙️ Algebraic & Structural Meaning

The governing differential equation $\frac{dy}{dx} = \tan \psi = \frac{s}{c}$ integrates via hyperbolic functions. Using the fundamental arc length relation $ds^2 = dx^2 + dy^2$, substitution leads directly to exponential and hyperbolic cosine solutions, connecting calculus, differential equations, and statics.

Conditions and hypotheses

Check each condition and what changes when it is omitted:

Hypothesis / Condition	Domain / Interval	Necessity	What Fails If Condition Dropped
Uniform linear density ($w = \text{constant}$)	`Entire arc of the string`	Strictly Mandatory	If the string is non-uniform ($w$ varies with $s$), the differential equation $\frac{d}{dx}\left(T \frac{dy}{dx}\right) = w$ becomes non-linear with variable coefficients, and standard hyperbolic relations fail.
Perfect flexibility and inextensibility	`Material property across domain`	Strictly Mandatory	If the string possesses bending stiffness (like a thin metallic rod), internal bending moments arise, transforming the catenary problem into an elastica problem involving elliptic integrals.
Equilibrium under gravity alone	`Force system vector field`	Essential for static equilibrium	Presence of lateral wind forces or electromagnetic loads alters the vertical force balance, changing the curve from a catenary to a general funicular curve or parabola.

Useful results and shortcuts

Use these identities and results when their conditions apply:

Tension-Ordinate Relation

$$T = wy$$

Rule: The tension $T$ at any point $(x, y)$ on a catenary is directly proportional to its depth $y$ below the directrix (or height above the lowest point when origin is at vertex, depending on coordinate system; with directrix at $y=0$, $T = wy$).

Application: Directly asked in UP PGT to find tension at supports given the span or length.

Cartesian Coordinates & Arc Length Relation

$$s^2 = y^2 - c^2 \quad \text{or} \quad y = \sqrt{c^2 + s^2}$$

Rule: Instantly compute the ordinate $y$ when arc length $s$ and parameter $c$ are known, skipping hyperbolic integrations.

Application: Frequently used in TGT/PGT numerical problems involving length of string and sag.

Cartesian to Intrinsic Parameter Conversion

$$y = c \sec \psi, \quad x = c \ln(\sec \psi + \tan \psi) = c \psi_{\text{gd}}^{-1}$$

Rule: Relate Cartesian coordinates directly to the angle of the tangent $\psi$ without computing derivatives.

Application: Used in DSSSB PGT problems involving slope and coordinates at supports.

Span and Sag Approximation for Shallow Catenaries

$$s \approx l + \frac{8d^2}{3l}$$

Rule: For a shallow catenary with span $2l$ and central sag $d$, the total length $s$ can be approximated rapidly without solving transcendental equations.

Application: Saves 2 minutes in numerical calculation problems in UP PGT.

Radius of Curvature Property

$$\rho = \frac{c}{\cos^2 \psi} = \frac{y^2}{c}$$

Rule: The radius of curvature $\rho$ at any point is equal to the length of the normal intercepted between the curve and the directrix, and is also equal to $c \sec^2 \psi$.

Application: Combined questions on differential geometry and statics in KVS PGT.

Horizontal Tension Invariance

$$T_0 = H = w c = \text{constant}$$

Rule: The horizontal component of tension is constant everywhere along the catenary and equals $wc$.

Application: Instant elimination tool for horizontal force balance questions.

Common errors and counterexamples

These examples show how a conclusion can fail when a condition is missing:

⚠️ COMMON ERROR #01

Catenary vs. Parabola Confusion

Common Pitfall: Assuming a freely hanging uniform cable forms a parabola. While a cable under a *uniformly distributed horizontal load* (like a suspension bridge cable) forms a parabola, a cable under its *own weight* (distributed along its arc length) forms a catenary.

Counterexample: y = c \cosh(x/c) \neq a x^2 + bx + c

Why it fails: Using parabolic formulas ($y = \frac{w x^2}{2H}$) for a self-weight hanging chain introduces systematic errors in length and tension calculations.

🛡️ Exam Defense: Check the loading: self-weight along arc $\implies$ Catenary (Hyperbolic); vertical load per unit horizontal span $\implies$ Parabola (Quadratic).

⚠️ COMMON ERROR #02

Origin Placement Confusion

Common Pitfall: Mixing up formulas when the origin is chosen at the lowest point versus when the origin is chosen on the directrix.

Counterexample: y = c \cosh(x/c) \text{ (vertex origin)} \quad \text{vs.} \quad y = c \cosh(x/c) \text{ with directrix at } y=0 \implies y = c

Why it fails: Using $s = c \sinh(x/c)$ when coordinates are measured from the directrix leads to completely wrong numerical answers.

🛡️ Exam Defense: Always verify coordinate axes specified in the problem statement before applying $s = c \tan \psi$ or $y^2 = c^2 + s^2$.

Comparison

Concept / Curve	Governing Equation	Tension Relationship	Geometric Meaning	Exam Identification Clue
Common Catenary	$y = c \cosh(x/c)$	$T = wy$ (from directrix)	Hangs under its own uniform weight	Keywords: 'uniform string', 'intrinsic equation', 'cosh'
Parabola (Suspension Bridge)	$y = \frac{w x^2}{2H}$	$T = \sqrt{H^2 + W^2}$	Loaded uniformly horizontally	Keywords: 'weight uniformly distributed horizontally', 'bridge'
Catenary of Equal Strength	$y = c \ln(\sec(x/c))$	Stress is uniform across cross-section	Profile adjusted for uniform tensile stress	Keywords: 'equal strength', 'varying cross-section'
Uniformly Stretched String (Elastic String)	$T = \lambda \frac{s - s_0}{s_0}$	Hooke's law combined with equilibrium	Elastic elongation under weight	Keywords: 'elastic string', 'natural length', 'modulus of elasticity $\lambda$'

Solved examples

Each problem includes a full solution and a shorter method:

CANONICAL PROBLEM 01 Dual-Method Breakdown

A uniform chain of length $l$ is suspended from two points at the same level. If the sag $d$ is small compared to the span $2a$, find the parameter $c$ approximately.

(A) $c \approx \frac{a^2}{2d}$(B) $c \approx \frac{2a^2}{d}$(C) $c \approx \frac{a^2}{d}$(D) $c \approx \frac{d^2}{2a}$

📚 Method A: Formal Textbook Derivation

Step 1: Use relation $\frac{l}{2} = c \sinh\left(\frac{a}{c}\right)$. Step 2: Use approximation $\sinh x \approx x + \frac{x^3}{6}$. Step 3: Relate $d = c \left[ \cosh\left(\frac{a}{c}\right) - 1 \right] \approx \frac{a^2}{2c}$. Step 4: Solve for $c \approx \frac{a^2}{2d}$.

⚡ Method B: 25-Second Exam Shortcut

Recall standard parabolic approximation for shallow catenary: sag $d = \frac{wa^2}{2T_0} = \frac{a^2}{2c}$, hence $c = \frac{a^2}{2d}$.

💡 Tip: Memorize the shallow catenary approximations $d \approx \frac{a^2}{2c}$ and $l - 2a \approx \frac{a^3}{3c^2}$ for rapid answering.

CANONICAL PROBLEM 02 Dual-Method Breakdown

If $s$ is the arc length of a common catenary from the lowest point and $\psi$ is the angle of the tangent with the horizon, then the radius of curvature $\rho$ at $(s, \psi)$ is:

(A) $c \sec \psi$(B) $c \sec^2 \psi$(C) $c \tan \psi$(D) $c \cosh \psi$

📚 Method A: Formal Textbook Derivation

Step 1: Intrinsic equation is $s = c \tan \psi$. Step 2: Radius of curvature is $\rho = \frac{ds}{d\psi}$. Step 3: Differentiating $s = c \tan \psi$ with respect to $\psi$ gives $\rho = c \sec^2 \psi$.

⚡ Method B: 25-Second Exam Shortcut

Since $s = c \tan \psi$, differentiating $\tan$ gives $\sec^2$, instantly yielding $\rho = c \sec^2 \psi$.

💡 Tip: Direct formula recall test. High frequency in UP TGT/PGT.

Practice questions

Choose an answer to check it and see the solution:

The intrinsic equation of a common catenary is given by:

A $s = c \sinh \psi$

B $s = c \tan \psi$

C $s = c \cosh \psi$

D $s = c \sec \psi$

Derivation & Analysis:

By definition, the intrinsic equation of the common catenary with arc length $s$ measured from the vertex is $s = c \tan \psi$, where $\psi$ is the angle of inclination of the tangent.

In a common catenary, if $y$ is the height of any point above the directrix and $c$ is the parameter, then tension $T$ at that point is:

A $T = wc$

B $T = wy$

C $T = ws$

D $T = w \sqrt{y^2 - c^2}$

Derivation & Analysis:

The tension at any point on a common catenary is given by $T = wy$, where $w$ is the weight per unit length and $y$ is the vertical distance from the directrix.

The Cartesian equation of a common catenary with origin at the vertex and $y$-axis downwards is:

A $y = c \sinh\left(\frac{x}{c}\right)$

B $y = c \cosh\left(\frac{x}{c}\right) - c$

C $y = c \cosh\left(\frac{x}{c}\right)$

D $y = c \ln\left(\sec\frac{x}{c}\right)$

Derivation & Analysis:

When the origin is chosen at the vertex (lowest point), $y = c(\cosh(x/c) - 1)$. If the origin is chosen at a distance $c$ below the vertex (i.e., at the directrix), then $y = c \cosh(x/c)$.

If $\psi$ is the angle of the tangent with the horizontal at any point of a catenary, then the Cartesian coordinate $y$ (from directrix) in terms of $\psi$ is:

A $y = c \tan \psi$

B $y = c \sin \psi$

C $y = c \sec \psi$

D $y = c \cos \psi$

Derivation & Analysis:

From equilibrium conditions, $T \cos \psi = T_0 = wc$ and $T = wy$, which gives $wy \cos \psi = wc \implies y \cos \psi = c \implies y = c \sec \psi$.

The horizontal tension at any point of a common catenary is:

A Variable along the string

B Equal to $wc$

C Equal to $wy$

D Equal to $ws$

Derivation & Analysis:

The horizontal component of tension $H = T \cos \psi = wc$ is constant throughout the length of the common catenary.

The relationship between arc length $s$, ordinate $y$ (from directrix), and parameter $c$ is:

A $s^2 = y^2 + c^2$

B $y^2 = s^2 + c^2$

C $s = y + c$

D $y = \sqrt{s^2 - c^2}$

Derivation & Analysis:

Since $y = c \cosh(x/c)$ and $s = c \sinh(x/c)$, using the hyperbolic identity $\cosh^2\theta - \sinh^2\theta = 1$, we get $y^2 - s^2 = c^2$, hence $y^2 = s^2 + c^2$.

The radius of curvature $\rho$ at any point of a common catenary in terms of ordinate $y$ and parameter $c$ is:

A $\rho = \frac{y}{c}$

B $\rho = \frac{y^2}{c}$

C $\rho = \frac{c^2}{y}$

D $\rho = \sqrt{y^2 - c^2}$

Derivation & Analysis:

Radius of curvature is $\rho = \frac{ds}{dy} \cdot \text{something}$ or via formula $\rho = \frac{y^2}{c}$ since $\rho = c \sec^2 \psi$ and $y = c \sec \psi \implies \rho = c\left(\frac{y}{c}\right)^2 = \frac{y^2}{c}$.

If a heavy uniform string of length $2l$ is suspended from two points at the same level and the tension at the supports is double the tension at the lowest point, the angle $\alpha$ which the string makes with the horizon at the supports is:

A $\frac{\pi}{6}$

B $\frac{\pi}{4}$

C $\frac{\pi}{3}$

D $\frac{\pi}{2}$

Derivation & Analysis:

Tension at supports is $T = wy$, tension at vertex is $T_0 = wc$. Given $T = 2 T_0 \implies wy = 2wc \implies y = 2c$. Since $y = c \sec \alpha$, we have $\sec \alpha = 2 \implies \alpha = \frac{\pi}{3}$.

The Cartesian coordinates $x$ and $y$ (measured from the vertex) of a common catenary are connected by the differential equation:

A $\frac{dy}{dx} = \frac{x}{c}$

B $\frac{dy}{dx} = \sinh\left(\frac{x}{c}\right)$

C $\frac{dy}{dx} = \cosh\left(\frac{x}{c}\right)$

D $\frac{dy}{dx} = \frac{s}{c}$

Derivation & Analysis:

When origin is at the vertex, $y = c(\cosh(x/c) - 1)$, differentiating gives $\frac{dy}{dx} = \sinh(x/c)$.

Q10

What curve is formed by a uniform heavy string hanging freely under gravity?

A Parabola

B Catenary

C Ellipse

D Brachistochrone

Derivation & Analysis:

By definition, the curve formed by a uniform heavy flexible string hanging under gravity is the common catenary.

Common questions

The confusion arises because for small spans and small sags (shallow catenary), the hyperbolic cosine curve $y = c \cosh(x/c)$ can be expanded via Taylor series as $y \approx c + \frac{x^2}{2c}$, which is a parabola. However, mathematically and mechanically, they are distinct: catenary is governed by self-weight distributed along arc length, whereas a parabola is governed by a vertical load distributed uniformly along the horizontal span.

Memorize the triumvirate of fundamental relations: (1) $s = c \tan \psi$, (2) $y^2 = c^2 + s^2$, and (3) $T = wy$. Knowing these three allows you to solve over 80% of standard catenary problems without deriving hyperbolic identities from scratch.

Use standard approximations for small sags ($d \approx \frac{a^2}{2c}$ and $s \approx 2a + \frac{8d^2}{3a}$) and memorize standard values of hyperbolic functions or convert them to exponential forms ($e^x$ and $e^{-x}$) for quick polynomial-like algebraic manipulation.

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