9. Mechanics — Statics

Properties of common catenary

The common catenary is the curve assumed by a perfectly flexible, inextensible uniform heavy string hanging freely under its own weight between two fixed points.

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

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Definition and statement

Fundamental Equation and Intrinsic Properties of the Common Catenary

The intrinsic equation of a common catenary is given by $s = c \tan \psi$, and its Cartesian equation with respect to vertex-origin axes is $y = c \cosh\left(\frac{x}{c}\right)$, where $c$ is the parameter representing the span-to-tension ratio at the lowest point.

📐 Geometric Interpretation

The curve is symmetric about the $y$-axis passing through its lowest point (vertex), where the tension is minimum and purely horizontal. The parameter $c$ represents the length of the string whose weight equals the tension at the vertex.

⚙️ Algebraic & Structural Meaning

The relation between arc length $s$, intrinsic angle $\psi$, and Cartesian coordinates $(x,y)$ involves hyperbolic functions, establishing deep links between exponential growth and equilibrium under uniform gravity.

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 ($\rho$ or $w$ is constant)	`Entire arc of the string`	Strictly Mandatory	If linear density varies with arc length, the differential equation governing the equilibrium changes from a simple hyperbolic form to a non-uniform integral equation.
Perfect flexibility and inextensibility	`Entire string domain`	Strictly Mandatory	If rigidity or elasticity is introduced, bending moments or Hooke's law extensions occur, invalidating standard catenary tension formulas.
Uniform gravitational field acting vertically downward	`Spatial domain`	Essential for equilibrium	Non-uniform gravity alters the vertical force component per unit arc length, resulting in distorted curves.

Useful results and shortcuts

Use these identities and results when their conditions apply:

Tension at any point

$$T = wy = w(c + s) = wc \sec \psi$$

Rule: The tension at any point is directly proportional to its height $y$ above the directrix, or to $c + s$ measured from the vertex.

Cartesian coordinates and Hyperbolic identity

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

Rule: Instantly relate the ordinate $y$, parameter $c$, and arc length $s$ without integrating.

Application: Use this result in the stated type of problem.

Relation between span ($2l$), dip ($d$), and parameter $c$

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

Rule: Use the parabolic approximation $d \approx \frac{l^2}{2c}$ when the sag is small compared to the span.

Radius of curvature

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

Rule: The radius of curvature at any point equals the normal component intercepted between the curve and the directrix.

Directrix distance from vertex

$$y_{\text{vertex}} = c$$

Rule: The height of the lowest point above the directrix is always equal to the parameter $c$.

Application: Quick conceptual check in multiple-choice questions.

Angle of friction or slope at support

$$\tan \psi = \frac{s}{c}$$

Rule: The slope at any point is given by the ratio of the arc length from the vertex to the parameter $c$.

Application: Standard elimination tool for slope-related options.

Common errors and counterexamples

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

⚠️ COMMON ERROR #01

Confusing Directrix with Axis of Symmetry

Common Pitfall: Candidates take $y=0$ as the directrix instead of $y=c$ when measuring coordinates from the vertex.

Counterexample: Standard form $y = c \cosh(x/c)$ has directrix at $y = 0$, so vertex is at $(0, c)$.

Why it fails: Leads to incorrect height measurements and wrong tension values ($T = wy$ vs $T = w(y-c)$).

🛡️ Exam Defense: Always remember that the directrix lies at a distance $c$ below the lowest point (vertex).

⚠️ COMMON ERROR #02

Misapplying Parabolic Approximation

Common Pitfall: Using $d = \frac{l^2}{2c}$ when the dip $d$ is large or comparable to span $l$.

Counterexample: When $d$ is large, the exact relation $d = c(\cosh(l/c) - 1)$ must be used.

Why it fails: Approximation breaks down for steep catenaries, yielding significant calculation errors.

🛡️ Exam Defense: Check if sag is small ($d \ll l$) before applying parabolic formulas.

Comparison

Concept / Curve	Equations (Cartesian / Intrinsic)	Tension Formula	Geometric Meaning	Exam Identification Clue
Common Catenary	$y = c \cosh(x/c)$, $s = c \tan \psi$	$T = wy = w(c+s)$	Hanging heavy string under gravity	Keywords: uniform weight, flexible string, gravity
Parabola	$y^2 = 4ax$ or $y = \frac{x^2}{4a}$	$T$ varies with uniform load along span	String loaded uniformly horizontally	Keywords: suspension bridge, uniform horizontal load
Uniformly Stretched String	Hooke's law included	$T = T_0 + \lambda \frac{\Delta s}{s}$	Elastic string under self-weight	Keywords: elastic, modulus of elasticity $\lambda$
Catenary of Equal Strength	Constant stress cross-section	$T / A = \text{constant}$	String designed not to break under self-weight	Keywords: equal strength, variable cross-section

Solved examples

Each problem includes a full solution and a shorter method:

CANONICAL PROBLEM 01 Dual-Method Breakdown

A uniform heavy string of length $2l$ hangs from two fixed points at the same level. If the sag is $d$, find the parameter $c$ of the catenary.

(A) $\frac{l^2 - d^2}{2d}$(B) $\frac{l^2 + d^2}{2d}$(C) $\frac{2l^2}{d}$(D) $\frac{l^2}{d}$

📚 Method A: Formal Textbook Derivation

Step 1: Half span is $l$, total height at support is $c + d$. Step 2: Use $(c+d)^2 = c^2 + l^2$. Step 3: Expand $c^2 + 2cd + d^2 = c^2 + l^2 \implies 2cd = l^2 - d^2$ or $c = \frac{l^2 - d^2}{2d}$ (Wait, let's verify exact relation: support is at $(l, c+d)$. Equation is $c+d = c \cosh(l/c)$. Alternatively, using $s^2 + c^2 = y^2$, with $s=l$ and $y=c+d$, we get $l^2 + c^2 = (c+d)^2 = c^2 + 2cd + d^2 \implies 2cd = l^2 - d^2 \implies c = \frac{l^2 - d^2}{2d}$.

⚡ Method B: 25-Second Exam Shortcut

Apply $s^2 = y^2 - c^2 \implies l^2 = (c+d)^2 - c^2 = 2cd + d^2$, hence $c = \frac{l^2-d^2}{2d}$.

💡 Tip: Memorize this standard derivation as it frequently appears directly in UP PGT.

CANONICAL PROBLEM 02 Dual-Method Breakdown

If $T_1$ and $T_2$ are the tensions at the lowest point and at the support of a common catenary respectively, and $w$ is the weight per unit length, what is the weight of the portion of the string between the lowest point and the support?

(A) $\sqrt{T_2^2 - T_1^2}$(B) $T_2 - T_1$(C) $\sqrt{T_1^2 + T_2^2}$(D) $\frac{T_1 + T_2}{2}$

📚 Method A: Formal Textbook Derivation

Step 1: Tension at lowest point $T_1 = wc$. Step 2: Tension at support $T_2 = w(c+s) = wc + ws = T_1 + W$, where $W = ws$ is the weight of the half string. Thus $W = T_2 - T_1$ for vertical component, but total tension relation is $T_2^2 = T_1^2 + (ws)^2 \implies ws = \sqrt{T_2^2 - T_1^2}$.

⚡ Method B: 25-Second Exam Shortcut

Recall force triangle at support: horizontal tension $T_1 = wc$, vertical tension $ws$, resultant tension $T_2$. By Pythagoras theorem, $(ws)^2 = T_2^2 - T_1^2$.

💡 Tip: Distinguish carefully between vertical tension component ($T_2 \sin \psi = ws = W$) and total tension magnitude ($T_2 = \sqrt{T_1^2 + W^2}$). Option B is a common trap for vertical component when magnitude is asked.

Practice questions

Choose an answer to check it and see the solution:

The intrinsic equation of a common catenary is given by $s = c \tan \psi$. What represents the parameter $c$?

A Length of arc from vertex

B Ratio of horizontal tension to weight per unit length

C Radius of curvature at support

D Maximum sag of the string

Derivation & Analysis:

At the vertex, $\psi = 0$, and $T_0 = wc$, so $c = \frac{T_0}{w}$, representing the length of string whose weight equals tension at the vertex.

In a common catenary, the tension at any point $(x, y)$ is given by:

A $wy$

B $wc \cosh(x/c)$

C $w \sqrt{x^2 + y^2}$

D $wc \tan \psi$

Derivation & Analysis:

The fundamental property of a common catenary is that the tension $T$ at any point is equal to the weight of the portion of the string hanging from the directrix to that point, which is $wy$.

The Cartesian equation of a common catenary referred to its directrix as the $x$-axis is:

A $y = c \sinh(x/c)$

B $y = c \cosh(x/c)$

C $y = c \tanh(x/c)$

D $y = c \ln(\sec(x/c))$

Derivation & Analysis:

When the directrix is chosen as the $x$-axis and the $y$-axis passes through the vertex, the equation is $y = c \cosh(x/c)$.

If $\rho$ is the radius of curvature at any point $(x, y)$ on a common catenary, then $\rho$ in terms of $y$ and $c$ is:

A $\frac{y}{c}$

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

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

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

Derivation & Analysis:

We know $\rho = c \sec^2 \psi$. Since $\sec \psi = \frac{y}{c}$, we have $\rho = c \left(\frac{y}{c}\right)^2 = \frac{y^2}{c}$.

The relation between the arc length $s$ measured from the vertex and the Cartesian coordinate $y$ is:

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

B $y = c + s$

C $y^2 = s^2 - c^2$

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

Derivation & Analysis:

From $y = c \cosh(x/c)$ and $s = c \sinh(x/c)$, using $\cosh^2\theta - \sinh^2\theta = 1$, we get $\left(\frac{y}{c}\right)^2 - \left(\frac{s}{c}\right)^2 = 1 \implies y^2 = c^2 + s^2$.

If the span of a catenary is $2l$ and dip is $d$, the length of the endless chain or total string length $2s$ can be approximated for small $d$ by:

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

B $2l + \frac{d^2}{l}$

C $l + \frac{d^2}{2l}$

D $2l$

Derivation & Analysis:

For small dip $d$, the length of half the string is $s \approx l + \frac{d^2}{2l}$ (or using standard expansion $s = l(1 + \frac{8}{3}\frac{d^2}{l^2})$ for total length).

At the vertex of a common catenary, the tangent is:

A Vertical

B Horizontal

C Inclined at $45^\circ$

D Undefined

Derivation & Analysis:

By definition, the vertex is the lowest point of the catenary where the tangent is horizontal (i.e., $\psi = 0$). basin of symmetry.

The normal at any point $(x, y)$ of a common catenary of parameter $c$ cuts the directrix at an angle equal to:

A $\psi$

B $90^\circ - \psi$

C $180^\circ - \psi$

D $2\psi$

Derivation & Analysis:

The angle between the normal and the vertical is $\psi$, which equals the angle the tangent makes with the horizontal.

If the tension at the lowest point of a catenary is equal to the weight of length $c_0$ of the string, then $c_0$ is equal to:

A $c$

B $2c$

C $c/2$

D $0$

Derivation & Analysis:

The parameter $c$ is defined precisely as the length of string whose weight equals the tension at the lowest point.

Q10

What is the shape formed by a uniform heavy cable hanging under its own weight between two supports?

A Parabola

B Catenary

C Ellipse

D Cycloid

Derivation & Analysis:

A cable hanging under its own weight forms a catenary. If loaded uniformly along the horizontal span, it forms a parabola.

Common questions

A common catenary is formed when a string supports only its own weight (load distributed uniformly along the arc length). A parabola is formed when the load is distributed uniformly along the horizontal span (such as in a suspension bridge deck).

It depends on the choice of origin. When the origin is taken at the directrix, $y$ is the height from the directrix, and $T = wy$. When the origin is at the vertex, the height above the vertex is $y'$, and $T = w(c + y')$. Confusing the coordinate axes leads to sign and formula errors.

Master the three core relations: $s = c \tan \psi$, $y = c \cosh(x/c)$, and $y^2 = c^2 + s^2$. Using these identities avoids lengthy integration during the exam.