ARCHETYPE: The Stretched Wire

Every elasticity exam on Earth eventually hangs a weight from a wire. CBSE asks for the elongation. ISC asks for the energy stored. JEE splices two wires end-to-end, or side-by-side, or hangs the wire under its own weight and watches you integrate without calculus. NEET asks which of four wires stretches most, banking on you fumbling a square. A-Level hands you a Searle's-apparatus data table and wants Young's modulus from a graph. Different costumes — one body underneath: the stretched wire, this unit's first problem archetype. The archetype has one master equation, $\Delta L = FL/AY$ , and everything else is disciplined substitution. What makes it worth a full article is the variant map: exams have exactly five standard ways to dress the wire up — scaling games, series composites, parallel load-sharing, self-weight stretch, and breaking-stress ceilings — and each has a signature move and a signature trap. Learn the five moves and the archetype becomes a vending machine for marks. You arrive fully armed: stress, strain and Hooke's law from the chapter opener; the moduli and $k = YA/L$ from its sequel; energy density $\tfrac12\sigma\varepsilon$ already derived. This article spends that toolkit at exam speed — board-level fills first, then the JEE/NEET escalations with timing notes.

Key takeaways

Master equation: $\Delta L = \dfrac{FL}{AY}$ — every stretched-wire question is this, rearranged, scaled, or summed.
Scaling reflex: at fixed load and material, $\Delta L \propto \dfrac{L}{d^2}$ — length stretches, thickness stiffens (squared).
Wires in series (end-to-end, same $F$ ): extensions add; equivalent stiffness $\dfrac{1}{k_{\text{eq}}} = \sum\dfrac{1}{k_i}$ with $k = \dfrac{YA}{L}$ .
Wires in parallel (side-by-side, same $\Delta L$ ): loads share in proportion to $k = YA/L$ — the stiffer wire carries more.
Self-weight: a hanging wire stretches by $\Delta L = \dfrac{\rho g L^2}{2Y} = \dfrac{MgL}{2AY}$ — half the end-load answer, because average tension is half.
Breaking is set by stress, not force: maximum hanging length $L_{\max} = \dfrac{\sigma_{\text{break}}}{\rho g}$ — independent of thickness.
Energy stored: $U = \tfrac12 F\,\Delta L$ ; per volume $u = \tfrac12\sigma\varepsilon$ .
Key idea: one formula, five costumes — classify the variant first, then substitute.

The core insight

Every stretched-wire problem reduces to $\Delta L = FL/AY$ applied with the right bookkeeping: same force through series segments (extensions add), same extension across parallel wires (forces share by stiffness $YA/L$ ), and average tension $W/2$ for a wire stretching under its own weight.

Where this lives today

Equation (5)'s strength-to-weight logic runs modern structural engineering. Suspension-bridge main cables — the Akashi Kaikyō's are over a metre thick — are spun from thousands of parallel high-carbon steel wires (the parallel costume, industrialized: shared load, matched extensions), with working stresses held at a disciplined fraction of breaking. Elevator codes demand multiple independent ropes plus safety factors near 10 — worked example 4's part (b) is a legal requirement, not a textbook nicety. Mine hoists in South Africa's ~4 km gold shafts run in stages precisely because $L_{\max} = \sigma_b/\rho g$ makes a single steel drop marginal — and rope engineers taper cross-sections so stress, not thickness, stays constant along the length. At the exotic end, the space-elevator literature is one long meditation on this archetype: for Earth, a constant-section cable of any known bulk material fails; only tapered designs in nanotube-grade materials pencil out. And every school lab that hangs slotted masses on a Searle's apparatus is running this article, half-millimetre by half-millimetre.

The setup

One wire, one load, four labels. A uniform wire of natural length $L$ , cross-sectional area $A$ (diameter $d$ , so $A = \pi d^2/4$ ), material of Young's modulus $Y$ , carries an axial force $F$ — usually a hung mass, $F = mg$ . The wire is light unless the problem says otherwise, so the tension is $F$ all along it. From the chapter opener: stress $\sigma = F/A$ , strain $\varepsilon = \Delta L/L$ , and Hooke's law $\sigma = Y\varepsilon$ within the elastic limit. Solve for the elongation:

\Delta L = \frac{F\,L}{A\,Y} \qquad (1)

where: the numerator carries the demand (more force, more length to stretch), the denominator the supply of stiffness (more area, stiffer material). Equivalently the wire is a spring of constant $k = YA/L$ , and equation (1) is just $x = F/k$ . Typical numbers keep you honest: metre-scale steel wires under kilogram-scale loads stretch fractions of a millimetre. An answer in centimetres means a units slip, almost always in $A$ .

Figure

The stretched wire and its five exam costumes

Costume recognition is the entire skill. Same force through everything in a single line (one wire, or wires tied end-to-end): work extension by extension. Same extension forced on side-by-side wires (a rigid bar hung by two wires): the load divides by stiffness. Distributed load (the wire's own weight): tension varies along the wire — use the average. "Breaks when": convert immediately to stress language. Then equation (1), or its spring form, closes every account.

The master solution pattern

Write the givens in SI first — this archetype kills more marks with $\mathrm{mm^2}$ than with physics. Then classify the costume. Then:

\text{single wire: } \Delta L = \frac{FL}{AY}, \qquad U = \tfrac12 F\,\Delta L \qquad (2)

where: $U$ — elastic energy stored (the triangle under the $F$ – $x$ line, factor ½ non-negotiable). For series segments (1, 2 end-to-end, same tension $F$ ):

\Delta L_{\text{total}} = \Delta L_1 + \Delta L_2 = \frac{FL_1}{A_1Y_1} + \frac{FL_2}{A_2Y_2} \qquad (3)

where: each segment stretches on its own bill; the composite behaves as springs in series, $1/k_{\text{eq}} = 1/k_1 + 1/k_2$ — softer than either piece. For parallel wires sharing a load $W$ while stretching equally:

F_i = \frac{k_i}{\sum k}\,W, \qquad k_i = \frac{Y_iA_i}{L_i} \qquad (4)

where: the load splits like current through parallel resistors — in proportion to stiffness. Two same-geometry wires of steel ( $Y = 2\times10^{11}$ ) and brass ( $10^{11}$ ): steel silently carries two-thirds of the load. If the wires' stiffnesses differ and the bar must stay horizontal, the load's hanging position is what's asked — a moments problem from rotation grafted onto this one (exam composites love the splice).

Figure

Self-weight: the tension triangle — full weight at the top, zero at the tip, average W/2

Derivation ASelf-weight stretch — the half that saves the integral

We derive the elongation of a uniform wire (length $L$ , area $A$ , density $\rho$ , total weight $W = \rho A L g$ ) hanging under its own weight, no end load. The subtlety: tension is not constant — the top element carries the whole weight below it, the bottom element carries nothing. Strategy: find how tension varies, replace it by its average (legitimate because tension varies linearly), then use the master equation. This "average tension" route is exactly the algebra-only stand-in for the integral, and it lands the famous factor of ½.

Step 1. Tension at a point a distance $x$ above the lower end = weight of wire hanging below it:

T(x) = \rho A g\, x

Linear in $x$ : zero at the tip ( $x=0$ ), full weight $W = \rho A g L$ at the support ( $x=L$ ).

Step 2. Because $T(x)$ is linear, the wire stretches exactly as if a uniform tension equal to the average acted throughout:

\bar T = \frac{0 + W}{2} = \frac{W}{2} = \frac{\rho A g L}{2}

Why averaging is legal here: each element $dx$ stretches in proportion to its local tension; summing proportional contributions of a linear function is the same as using its mean — the area of a triangle equals base × mean height.

Step 3. Master equation with $F = \bar T$ :

\Delta L = \frac{\bar T\, L}{AY} = \frac{\rho g L^2}{2Y} \;=\; \frac{MgL}{2AY}

Why the two memorable forms: the first ( $\rho gL^2/2Y$ ) shows the geometry-free scaling — self-weight stretch grows as $L^2$ and doesn't care about thickness (thicker wire: more weight, more stiffness, cancelling). The second says: half of what the same total weight would do as an end load. *Verified: units — $\mathrm{kg/m^3\cdot m/s^2 \cdot m^2 / Pa} = \mathrm{m}$ ✓. Limit — end-loaded equivalent: hanging the wire's own weight $W$ at the tip gives $WL/AY$ , exactly twice ✓. Anchor — steel, $L = 100$ m: $\Delta L = (7800)(9.8)(10^4)/(2\times2\times10^{11}) = 1.9$ mm — long crane cables genuinely stretch millimetres under self-weight. Standard: HC Verma Ch. 14; JEE archive regular.*

The breaking-stress costume converts force ceilings into geometry ceilings. A material fails when its stress reaches $\sigma_{\text{break}}$ (steel ~ $4\times10^8$ Pa; the elastic limit is lower still). For a wire hanging under self-weight, the maximum stress sits at the top, $\sigma_{\text{top}} = W/A = \rho g L$ — area cancels! — so the longest wire (or rope, or cable) a material can dangle of itself is

L_{\max} = \frac{\sigma_{\text{break}}}{\rho\, g} \qquad (5)

where: no $A$ : a thicker cable is heavier in exact proportion to its extra strength, so thickness buys nothing — only a stronger-per-density material (or a taper) hangs deeper. Steel: $4\times10^8/(7800\times9.8) \approx 5.2$ km. This is why elevator engineering for kilometre-deep mines is a materials problem, and why the space-elevator dream fixates on carbon nanotubes ( $\sigma_{\text{break}}/\rho$ thousands of times steel's).

Limiting cases

case	result	meaning
$L \to 2L$ , end load fixed	$\Delta L$ doubles	more wire to stretch, same stress everywhere
$d \to 2d$ , end load fixed	$\Delta L \to \Delta L/4$	area quadruples; the classic NEET square
two identical wires in series	$\Delta L$ doubles ( $k$ halves)	a longer wire IS a series stack
two identical wires in parallel	each carries $W/2$ ; $\Delta L$ halves	thickness in disguise
self-weight vs same weight end-loaded	exactly half the stretch	average tension $W/2$
breaking length	independent of $A$	strength-to-weight ratio rules; taper to beat it
load beyond elastic limit	equation (1) void	Hooke's kingdom ends; plastic flow begins

Worked examples

1 · Board level — elongation and stored energy (CBSE/ISC pattern, 3 marks)

A steel wire of length $3.0\ \mathrm{m}$ and cross-section $0.50\ \mathrm{mm^2}$ hangs a $6.0\ \mathrm{kg}$ mass. $Y_{\text{steel}} = 2.0\times10^{11}\ \mathrm{Pa}$ , $g = 10\ \mathrm{m/s^2}$ . Find (a) the elongation and (b) the elastic energy stored.

Step 1. SI first: $A = 0.50\ \mathrm{mm^2} = 5.0\times10^{-7}\ \mathrm{m^2}$ ; $F = mg = 60\ \mathrm{N}$ .

Step 2. (a) Master equation: $\Delta L = \dfrac{FL}{AY} = \dfrac{60\times3.0}{5.0\times10^{-7}\times2.0\times10^{11}} = \dfrac{180}{1.0\times10^{5}} = \boxed{1.8\times10^{-3}\ \mathrm{m}} = 1.8$ mm.

Step 3. (b) Energy: $U = \tfrac12 F\Delta L = \tfrac12\times60\times1.8\times10^{-3} = \boxed{5.4\times10^{-2}\ \mathrm{J}}$ .

Step 4. Sanity: strain $= 1.8/3000 = 6\times10^{-4}$ — comfortably elastic; stress $1.2\times10^8$ Pa, below steel's yield. Millimetres and centijoules: the right republic of numbers.

2 · Board level — the composite (series) wire (ISC/A-Level pattern, 4 marks)

A steel wire ( $L_1 = 1.0\ \mathrm{m}$ , $Y_1 = 2.0\times10^{11}\ \mathrm{Pa}$ ) is joined end-to-end with a copper wire ( $L_2 = 1.5\ \mathrm{m}$ , $Y_2 = 1.2\times10^{11}\ \mathrm{Pa}$ ), both of area $1.0\ \mathrm{mm^2}$ . The composite hangs a load of $100\ \mathrm{N}$ . Find each extension and the total.

Step 1. Series costume: the same 100 N threads both segments (each must transmit the load to the one below).

Step 2. Steel: $\Delta L_1 = \dfrac{100\times1.0}{10^{-6}\times2.0\times10^{11}} = \dfrac{100}{2.0\times10^{5}} = 0.50\ \mathrm{mm}$ .

Step 3. Copper: $\Delta L_2 = \dfrac{100\times1.5}{10^{-6}\times1.2\times10^{11}} = \dfrac{150}{1.2\times10^{5}} = 1.25\ \mathrm{mm}$ .

Step 4. Total: $\Delta L = 0.50 + 1.25 = \boxed{1.75\ \mathrm{mm}}$ — extensions add; the softer copper does most of the stretching despite identical load. (Stress is the same $10^8$ Pa in both — same $F$ , same $A$ — but strain differs by the $Y$ ratio: a tidy conceptual checkpoint.)

3 · Competitive level — parallel wires sharing a load (JEE-Main pattern, ~2.5 minutes)

A rigid horizontal bar of negligible mass hangs from the ceiling by two vertical wires of identical length and cross-section: one steel ( $Y = 2.0\times10^{11}$ Pa), one brass ( $Y = 1.0\times10^{11}$ Pa). A load $W = 60\ \mathrm{N}$ is to be hung from the bar. (a) If the bar must stay horizontal, find the tension in each wire. (b) Where along the bar (length $90\ \mathrm{cm}$ , steel at the left end) must the load hang?

Step 1. Horizontal bar ⇒ equal extensions ⇒ parallel costume. Same $L$ , $A$ : stiffness ratio $k_s : k_b = Y_s : Y_b = 2 : 1$ .

Step 2. (a) Load shares by stiffness: $T_s = \dfrac{2}{3}(60) = \boxed{40\ \mathrm{N}}$ , $T_b = \dfrac{1}{3}(60) = \boxed{20\ \mathrm{N}}$ .

Step 3. (b) The load's position must make the torques balance. Taking moments about the load's hook: $T_s\,x = T_b\,(90 - x)$ where $x$ is the distance from the steel wire: $40x = 20(90 - x) \Rightarrow 60x = 1800 \Rightarrow x = \boxed{30\ \mathrm{cm}}$ from the steel end.

Step 4. Read the geometry: the load hangs closer to the stiffer wire — it must, to feed the stiffer wire its bigger share. (Closer to steel, more moment-leverage for brass? Check: about the steel wire, $W\cdot x_W = T_b\cdot L$ : $60\times30 = 20\times90$ ✓.) Timing note: the elasticity half is 30 seconds (share by $Y$ ); the marks hide in the statics half. Write one moment equation about a smart point (the load, or either wire) and verify with a second. This steel-brass-bar hybrid is a JEE-Main perennial — rotation's moments and elasticity's stiffness in one item.

4 · Competitive level — self-weight and the breaking ceiling (JEE-Adv flavour, ~3 minutes)

A uniform steel cable hangs vertically from a winch at the top of a deep mine shaft. Steel: density $7800\ \mathrm{kg/m^3}$ , breaking stress $4.0\times10^8\ \mathrm{Pa}$ , $Y = 2.0\times10^{11}\ \mathrm{Pa}$ ; $g = 10\ \mathrm{m/s^2}$ . (a) What is the greatest length of cable that can hang without breaking? (b) With a safety factor of 4 (working stress = one quarter of breaking), what length is allowed? (c) How much does a $1.0\ \mathrm{km}$ cable stretch under its own weight?

Step 1. (a) Stress peaks at the top: $\sigma_{\text{top}} = \rho g L$ . Set it to breaking: $L_{\max} = \dfrac{\sigma_b}{\rho g} = \dfrac{4.0\times10^8}{7800\times10} = \boxed{5.1\times10^3\ \mathrm{m}}$ — about 5 km, thickness irrelevant.

Step 2. (b) Working ceiling: $L_{\text{safe}} = L_{\max}/4 \approx \boxed{1.3\ \mathrm{km}}$ — why real deep-mine hoists use multiple stages or tapered, high-grade ropes.

Step 3. (c) Self-weight stretch: $\Delta L = \dfrac{\rho g L^2}{2Y} = \dfrac{7800\times10\times(10^3)^2}{2\times2.0\times10^{11}} = \dfrac{7.8\times10^{10}}{4.0\times10^{11}} = \boxed{0.195\ \mathrm{m}}$ — a 1 km cable is 20 cm longer just hanging there.

Step 4. Sanity chain: at 1 km, top stress $= 7.8\times10^7$ Pa — a fifth of breaking ✓ consistent with (b); strain at top $= 3.9\times10^{-4}$ , average half that, times $10^3$ m gives 0.195 m ✓ (the ½ recovered by the average-tension logic). Timing note: three sub-parts, three formulas, zero integration — if you find yourself summing elements, you've forgotten the average-tension theorem and the area-cancellation. The "no dependence on thickness" statement in (a) is itself a one-mark conceptual ask: say WHY (weight and strength both ∝ A).

Exam traps

mm² → m² is $\times10^{-6}$ ; diameters halve before squaring; $A = \pi d^2/4$ . The archetype's top three errors are all in the denominator of one formula.

Series vs parallel misfiling. End-to-end = same force, add extensions. Side-by-side into a rigid bar = same extension, share forces by $YA/L$ . The words "joined" (series) and "supported by two wires" (parallel) are the flags; a horizontal-bar sketch settles it.

The self-weight half. Own weight → $\rho gL^2/2Y$ (average tension $W/2$ ). Same weight hung at the end → twice that. Both offered as options, always.

$L^2$ , not $L$ , in self-weight stretch. Doubling a cable's length quadruples its self-stretch. Ratio questions exploit the square.

Breaking-force vs breaking-stress. "Which breaks first" compares $F/A$ . Hanging-length compares $\sigma_b/\rho g$ — area has cancelled; answers citing thickness are wrong by construction.

Energy's missing half. $U = \tfrac12 F\Delta L$ , or per volume $\tfrac12\sigma\varepsilon$ . Doubling the load quadruples the energy ( $U \propto F^2$ at fixed geometry) — another planted ratio.

Elastic limit fine print. All of this lives inside Hooke's kingdom. A computed strain above ~ $10^{-2}$ for a metal should trigger suspicion, and "beyond the elastic limit the formula fails" is a free mark when asked.

Which $g$ , which $Y$ . JEE/NEET usually declare $g = 10$ and give $Y$ ; CBSE may expect $9.8$ and standard $Y$ values. Steel $2\times10^{11}$ , copper $1.1$ – $1.2\times10^{11}$ , brass $\sim10^{11}$ , aluminium $0.7\times10^{11}$ Pa — recognize them to sanity-check given data.

Origins · 1 minG. F. C. Searle and how this was discovered

G. F. C. Searle

1864–1954 · Cavendish Laboratory, Cambridge

George Frederick Charles Searle spent six decades at the Cavendish Laboratory turning precision measurement into a teachable craft. Hired by J. J. Thomson, he ran the practical classes through the laboratory's golden age — his students included generations of Nobel laureates — and he designed the bench apparatus that still carries his name: Searle's apparatus hangs two parallel wires (one reference, one loaded) with a micrometer-and-spirit-level bridge between them, so the loaded wire's extension is read to a hundredth of a millimetre while temperature drifts cancel between the pair. It remains the standard school and first-year method for measuring Young's modulus — this article's archetype, built in brass. Searle also did early work on the electromagnetic mass of moving charges, corresponded with Einstein, and examined practical physics with legendary (occasionally terrifying) rigour.

History

1678

— Hooke's "ut tensio, sic vis": extension proportional to load, measured on springs and wires.

1807

— Young's modulus makes wire-stretching a material calculation, not a per-object experiment.

1826

— Navier's engineering elasticity: working-stress design and safety factors enter construction practice.

c. 1900 — Searle's apparatus standardizes the two-wire, temperature-compensated measurement of

Y

in teaching labs.

1991

— Iijima's carbon nanotubes: measured strength-to-weight ratios reopen the hanging-length problem (space-elevator calculations use equation (5)'s logic with a taper).

Going further

Texts. NCERT Physics Class 11, Ch. 8 · HC Verma Vol. 1, Ch. 14 (the worked wire problems) · Resnick–Halliday–Walker Ch. 12.

Provenance. All results standard. A verifier should recompute: example 1's 1.8 mm and 0.054 J; example 2's 0.50 + 1.25 = 1.75 mm; example 3's 40/20 N split and 30 cm; example 4's 5.1 km, 1.3 km, and 0.195 m; the derivation's average-tension ½ and the 100 m steel anchor (1.9 mm).

Prerequisites. Stress, Strain & Hooke's Law for Solids · Young's, Bulk & Shear Moduli. Leads to. Option Elimination & Answer Checking (track capstone).

Why does the same force stretch a long wire more if the stress in it is identical?

Because strain, not stretch, is what stress controls. Both wires sit at the same stress, hence the same strain — the same stretch per metre. The longer wire simply has more metres each contributing that stretch: $\Delta L = \varepsilon L$ . This is also the cleanest way to see the series rule: joining wires end-to-end adds metres at fixed strain, so extensions add.

In the parallel-wires problem, what physically forces the stiffer wire to take more load?

The rigid bar. It enforces equal extensions, and at equal extension a stiffer spring is, by definition, pulling harder ( $F = kx$ at the same $x$ ). If the bar were soft, each wire would stretch by its own amount and the sharing would change. The equal-extension constraint is the parallel costume's defining line — quote it, then share by $k = YA/L$ .

Where does the ½ in the self-weight formula really come from — is averaging always allowed?

The tension profile is linear (zero at the free tip, W at the support), and each element's stretch is proportional to its local tension. Summing contributions proportional to a linear function equals using its average — triangle area = base × half-height. The averaging shortcut is legal exactly when the load profile is linear; for a cable carrying its own weight plus an end load $P$ , split the problem: $\Delta L = PL/AY + \rho gL^2/2Y$ — superpose the uniform part and the triangular part. Nonlinear profiles (a heavy chain of varying links) genuinely need the integral.

Why does the thickness cancel in the maximum hanging length? It feels wrong that a thicker rope doesn't help.

Write the top stress: $\sigma = W/A = (\rho A L g)/A = \rho g L$ . Doubling the area doubles the strength you need to supply and the weight demanding it — the ratio is untouched. What does help: a material with a better $\sigma_b/\rho$ (breaking stress per density — "specific strength"), or cheating the uniformity by tapering: make the rope thicker at the top where tension is largest, so stress stays constant along the length. Real mine ropes and every space-elevator design taper for exactly this reason.

Does the wire heat up when it stretches — where does ½FΔL go when the load is removed?

Within the elastic limit and loaded slowly, the energy is stored as recoverable elastic PE — remove the load gently and the wire gives essentially all of it back, like a spring. Realities that exams occasionally probe: drop the load on suddenly and the wire overshoots and oscillates, dissipating some energy as sound/heat (and momentarily doubling the extension — a classic JEE-Adv twist); load past yield, and the plastic work is irrecoverably converted to heat in the lattice. The clean $U = \tfrac12F\Delta L$ ledger belongs to slow, elastic loading.

Searle's apparatus uses two wires — why not just measure one wire's stretch with a ruler?

Two reasons, both instructive. Scale: a metre of steel under lab loads stretches ~0.1–1 mm — a ruler's own graduation. The micrometer bridge between the wires reads hundredths of a millimetre. Drift: temperature changes lengthen both wires substantially (thermal expansion rivals elastic stretch!); by measuring the difference between a loaded wire and an identical unloaded reference hanging beside it, thermal changes cancel and only the elastic extension registers. It's a beautiful early example of differential measurement — the same trick, scaled up, that gravitational-wave interferometers play.

Which numbers should I have memorized for this archetype?

Working set: $Y$ — steel $2\times10^{11}$ , copper $\approx1.2\times10^{11}$ , aluminium $0.7\times10^{11}$ Pa. Densities — steel 7800, copper 8900, aluminium 2700 kg/m³. Breaking stress of steel ~ $4\times10^8$ Pa (so $L_{\max} \approx 5$ km). Feel for answers: mm-scale stretches, $10^{-4}$ -scale strains, $10^7$ – $10^8$ Pa working stresses. With those anchors you can catch a wrong power of ten before checking options — which is most of what "fast" means in this archetype.