Tension & Ideal Strings

Every crane lift, every tug-of-war, every pulley problem you will ever solve runs on a force you can never see acting at a single point: tension. Push a block and you can point at the contact. But a rope pulls everywhere along itself at once — pinch it at any cross-section, imagine cutting it there, and the two sides must be pulling each other with equal force, or the rope would fly apart. Tension is not one force; it is an unbroken chain of third-law pairs running the whole length of the string. Exams compress this richness into a beautifully useful fiction: the ideal string — massless, inextensible, perfectly flexible. Accept the fiction and two mighty simplifications follow. The tension becomes a single number for the whole string, even over a pulley; and the string's fixed length chains the motions of whatever hangs at its ends. Those two facts power the Atwood machine, connected-block trains, lift cables, and half the Laws of Motion paper on every board from CBSE to A-Level. But examiners also love the moment the fiction breaks. Give the rope mass, and tension is no longer one number — it grows toward the support of a hanging rope, and tapers along a dragged one. JEE-Main asks the "tension at distance $x$ " question nearly every cycle. This article builds both layers: the ideal string and its two superpowers, then the honest rope and its tension profile — with the sign conventions, traps, and timing habits that convert rope problems into reliable marks.

Key takeaways

Tension is the pulling force a string transmits along itself: cut it anywhere (mentally) and the two sides pull each other with equal magnitude $T$ — a third-law pair at every cross-section.
Strings only pull. A string can never push, and a slack string carries $T = 0$ .
An ideal string (massless, inextensible) carries one tension throughout — even over an ideal (massless, frictionless) pulley, which merely redirects the pull.
Inextensibility is a constraint: the string's fixed length forces connected objects to share one speed and one acceleration magnitude along the string.
A rope with mass has varying tension: hanging, $T$ grows linearly from bottom to top; dragged by a force, $T$ tapers from the pulled end to the free end — each cross-section carries only what is behind it.
Key idea: an ideal string transmits one tension unchanged along its length and enforces one acceleration magnitude on everything it connects — one number $T$ , one number $a$ .

The core insight

A string transmits force by tension — equal-and-opposite pulls at every internal cross-section; in the massless, inextensible limit the tension is a single value along the entire string (even across an ideal pulley) and the fixed length forces all connected bodies to share one acceleration magnitude.

Where this lives today

The world's cables are tension-profile engineering at scale. A deep-mine hoist rope or a skyscraper lift cable is thickest where equation (3) says the load peaks — near the winding drum, where it carries the car plus kilometres of its own steel; designers speak of "rope self-weight" as the limiting term, exactly your $mgy/L$ . Suspension-bridge mains are spun from thousands of parallel wires because the tension at the towers, set by the whole deck hanging below, exceeds what any single castable member could take. Climbing ropes are certified by impact force — the peak tension when a fall is arrested, your $m(g+a)$ with a deliberately stretchy rope keeping $a$ survivable. And the proposed space-elevator tether is the ultimate tension-profile problem: the taper needed so each cross-section carries only what hangs beyond it is exponential, and no manufactured material yet sustains it — a 21st-century problem whose first equation is the one you cut a rope to write.

The physical picture

Hold a book by a string and ask what holds it up. Not your hand — your hand is half a metre away, touching only the top of the string. The honest answer runs down the string link by link. Your hand pulls the top segment up; that segment pulls the next one up; and the bottom segment pulls the book up with a force we call the tension $T$ . By Newton's third law the book pulls back down on the string, the bottom segment pulls down on the one above it, and so on back to your hand. Tension is this chain of paired pulls. The string does not "contain" a force the way a box contains marbles — it transmits one, cross-section by cross-section.

Figure

Cut the string anywhere — a third-law pair of equal pulls appears at every cross-section

Three properties define the exam-grade ideal string, and each buys you something concrete:

Massless. Take any little segment of the string and apply Newton's second law: (pull from above) − (pull from below) = (segment's mass) × (acceleration). If the mass is zero, the two pulls must be equal no matter how the string accelerates — otherwise a zero mass would suffer infinite acceleration. Chain this argument along the string, segment by segment: the tension is one number everywhere. This is the single most-used idealization in mechanics.

Inextensible. The string's length is fixed. If one end moves 10 cm along the string, the other end must too — instantly. So two blocks joined by a taut string share one speed and one acceleration magnitude along the string's line. This is not decoration: it is an equation, the constraint, and stating it earns method marks. Its full exploitation — pulley trains, movable pulleys, block chains — is the business of the Atwood Machine and Connected Bodies archetypes.

Perfectly flexible. The string transmits force only along itself. It cannot push sideways, cannot carry compression, and goes slack (with $T = 0$ ) rather than resist shortening. A rod, by contrast, can push and pull; a spring can do both while stretching — the spring's own story is next in the chapter. Add the ideal pulley — massless, frictionless — and the tension survives the turn. A pulley with mass would need a difference in side tensions to spin up (that torque is a rotation-chapter problem), and friction at the axle would eat some pull; the ideal pulley needs neither, so it changes the direction of the tension without changing its size. One string, one $T$ , even bent through 180°.

The math: one number, then a profile

For the ideal string the mathematics is a single powerful statement:

T_{\text{everywhere}} = T \qquad (1)

where: $T$ — the one tension carried by the whole string, whatever its shape over ideal pulleys. Read it as a license: to find $T$ , cut the string anywhere convenient and apply Newton's second law to either side. For a block of mass $m$ hanging in equilibrium, cutting just above the block gives $T - mg = 0$ , so $T = mg = 49$ N for 5 kg at $g = 9.8$ . If the block accelerates upward at $a$ , the same cut gives:

T = m(g + a) \qquad (2)

where: $g + a$ — effective gravity in the accelerating case; the cable of a lift accelerating upward carries more than the weight, which is why cables are rated far above their static loads. Downward acceleration replaces the sign: $T = m(g - a)$ , reaching zero in free fall — a slack cable and the weightless stomach both report the same physics.

Derivation AWhy a massless string carries one tension throughout

We prove equation (1) — the uniform-tension rule — including its survival across an ideal pulley. The strategy is Newton's second law applied to an arbitrary piece of the string itself, then the observation that "massless" makes the net force on any piece vanish. Watch how the pulley case needs one extra idea: the pulley must exert only forces perpendicular to the string's direction of travel.

Step 1. Isolate a straight segment of string. Forces along the string: tension $T_1$ pulling one way at one cut, $T_2$ pulling the other way at the other cut. Newton's second law along the string:

T_1 - T_2 = m_{\text{seg}}\, a

Step 2. Set $m_{\text{seg}} = 0$ (massless string). Then $T_1 = T_2$ for any acceleration $a$ — the tension cannot change from one point to the next. Why: this is stronger than an equilibrium argument — it holds even while the string whips around accelerating at 100 m/s². Zero mass means zero force needed, means the pulls at the two cuts always match.

Step 3. Chain the argument. Every segment has equal end-tensions, so the tension is one value along the entire straight string.

Step 4. Around an ideal pulley: take a tiny arc of string on the pulley. The pulley (frictionless) can push only perpendicular to the string — a normal force. Along the string's direction the only forces are the two end-tensions of the arc, and the arc is massless, so they are equal. Marching arc by arc around the wheel: the tension enters and leaves the pulley unchanged. Why: friction is exactly a force ALONG the surface — banning it is what protects the tension. A rusty pulley or a rope wrapped on a rough post loses this protection, which is why a sailor can hold a ship with two turns of rope around a bollard. *Verified: limits — a real string with small mass $\lambda$ per length shows a small tension difference $\lambda\,\ell\,a$ across a length $\ell$ , vanishing as $\lambda \to 0$ , consistent with Step 2. And a pulley with mass demands $T_1 \ne T_2$ to angularly accelerate, confirming which idealization each conclusion leans on.*

When the rope has mass — the tension profile

Real ropes weigh something, and then equation (1) dies in an orderly, examinable way. The principle stays Newtonian: cut the rope where you are asked about, and the tension there must account for everything hanging (or accelerating) beyond the cut. For a uniform rope of mass $m$ and length $L$ hanging from a ceiling, a cut at height $y$ above the bottom supports the rope below the cut — mass $m y/L$ :

T(y) = \frac{m g}{L}\, y \qquad (3)

where: $y$ — distance from the bottom end. The profile is linear: zero at the free bottom end (nothing left to hold), $mg$ at the ceiling (the whole rope). Hang a block $M$ from the bottom and every cross-section carries $M$ 's weight too: $T(y) = Mg + (my/L)g$ .

Figure

Tension profiles of a real rope — linear growth when hanging, linear taper when dragged

The dragged rope is the mirror image. Pull a rope of mass $m$ , length $L$ along a smooth floor with force $F$ : the whole rope accelerates at $a = F/m$ . Cut at distance $x$ from the pulled end; the tension there is the force needed to accelerate the remaining mass $m(L-x)/L$ :

T(x) = F\left(1 - \frac{x}{L}\right) \qquad (4)

where: $x$ — distance from the end where $F$ is applied. Full force at the front, zero at the free tail. JEE-Main's favourite disguise is "find the tension at the midpoint": $F/2$ , by symmetry of the formula, independent of the rope's mass — a one-line answer if you know the profile, a five-minute derivation if you don't.

Go deeper · Derivation B · 1 minTension at a point in a dragged rope

Derivation BTension at a point in a dragged rope

We derive the taper formula (4), the standard JEE-Main "rope pulled by a force" result. Strategy: find the common acceleration of the whole rope first, then cut at the point of interest and apply Newton's second law to the far piece — the piece with fewer forces on it. The trap to watch: measuring $x$ from the wrong end silently flips the answer to $Fx/L$ .

Step 1. Whole rope: mass $m$ , one horizontal force $F$ (smooth floor). Common acceleration:

a = \frac{F}{m}

Step 2. Cut the rope at distance $x$ from the pulled end. The far piece has length $L - x$ , hence mass $m' = \dfrac{m(L-x)}{L}$ . Why: choose the far piece because only ONE horizontal force acts on it — the tension at the cut. The near piece works too but carries both $F$ and $T$ , one more term to juggle.

Step 3. Newton's second law for the far piece, with the same $a$ (inextensible rope):

T(x) = m' a = \frac{m(L-x)}{L}\cdot\frac{F}{m} = F\left(1 - \frac{x}{L}\right)

Step 4. Read the limits directly: $T(0) = F$ (the cut at the pulled end carries everything), $T(L) = 0$ (nothing beyond the tail), $T(L/2) = F/2$ (the midpoint drags exactly half the rope). Verified: the rope's mass cancels from the midpoint answer — $F/2$ for any uniform rope — and summing the two pieces' equations, $F - T = (mx/L)a$ plus $T = m'a$ , returns $F = ma$ ✓. With gravity instead of $F$ (rope hanging), the same cut-and-count logic gives the linear growth of equation (3).

Which equation answers which question

question	tool
tension in a light string holding a static load	cut above load: $T = mg$
light string, load accelerating up/down	$T = m(g \pm a)$
one light string over an ideal pulley	same $T$ both sides — cut anywhere
heavy rope hanging, tension at height $y$	$T = (my/L)g$ (+ any load below)
heavy rope dragged by $F$ , tension at $x$	$T = F(1 - x/L)$
two blocks, one string — accelerations	constraint: one $a$ magnitude (see Atwood / Connected Bodies)

Limiting cases

case	result	meaning
string slack	$T = 0$	strings never push; slack = out of the problem
load in free fall	$T = 0$	cable and load fall together; nothing to transmit
rope mass $\to 0$	profile flattens to one $T$	the ideal string recovered
dragged rope, $x = L/2$	$T = F/2$	midpoint hauls exactly half the rope, any mass
pulley with mass	$T_1 \ne T_2$	tension splits to torque the wheel — rotation chapter
rough post, rope wrapped	$T$ drops along the wrap	friction eats tension — why capstans hold ships

Common trap"The tension doubles because both sides of the rope pull"

Tug-of-war intuition breaks here more marks than any rope ever broke. A string pulled with 100 N from each end carries a tension of 100 N — not 200 N. The two pulls are not additive loads; they are the third-law pair that defines a tension of 100 N. Test it with the massless-segment law: any piece of the string feels 100 N leftward and 100 N rightward — net zero, as a massless segment requires — and the force transmitted across any cut is 100 N. A spring balance hooked between two teams each heaving with 100 N reads 100 N. If one team suddenly pulled with only 80 N, the rope would accelerate; while it does, the massless rope still has a single tension at each instant. When an exam option says "200 N," it is fishing for exactly this reflex.

Worked examples

1 · Board level — two blocks hanging in a chain (CBSE/ISC pattern, 3 marks)

Blocks of $3.0$ kg and $2.0$ kg hang from a rigid support, the 3 kg block above, connected by light string 1 to the support and light string 2 between the blocks. Find the tensions in both strings. ( $g = 9.8\ \mathrm{m/s^2}$ .)

Step 1. Cut string 2 (between the blocks): below the cut hangs only the 2 kg block, in equilibrium: $T_2 = 2.0 \times 9.8 = \boxed{19.6\ \mathrm{N}}$ .

Step 2. Cut string 1 (above everything): below this cut hang both blocks, $5.0$ kg total: $T_1 = 5.0 \times 9.8 = \boxed{49\ \mathrm{N}}$ .

Step 3. Cross-check with the 3 kg block's own FBD: up $T_1$ , down $mg + T_2$ (string 2 pulls it down — the third-law partner of holding the 2 kg block): $49 = 29.4 + 19.6$ ✓. The cut-and-count habit — "tension here supports everything beyond here" — solved this without simultaneous equations.

2 · Board level — the lift cable (A-Level/AP connected pattern)

A lift of mass $600$ kg accelerates upward at $1.2\ \mathrm{m/s^2}$ . Find the cable tension. What does it become when the lift moves upward but decelerates at $1.2\ \mathrm{m/s^2}$ ? ( $g = 9.8\ \mathrm{m/s^2}$ .)

Step 1. Accelerating up, equation (2): $T = m(g + a) = 600(9.8 + 1.2) = \boxed{6600\ \mathrm{N}}$ .

Step 2. Moving up but slowing ⇒ acceleration points down: $T = m(g - a) = 600(8.6) = \boxed{5160\ \mathrm{N}}$ .

Step 3. Sanity: static weight is $5880$ N — the answers bracket it, heavier when "gravity effectively strengthens," lighter when the lift eases off. The passenger-side version of this bracketing is the Lifts & Apparent Weight archetype.

3 · Competitive level — tension inside a dragged rope (JEE-Main pattern, ~2 minutes)

A uniform rope of mass $2.0$ kg and length $4.0$ m lies on a smooth floor. A horizontal force of $12$ N is applied at one end. Find (a) the tension at the midpoint, (b) the tension $1.0$ m from the free end.

Step 1. Whole rope: $a = F/m = 12/2.0 = 6.0\ \mathrm{m/s^2}$ .

Step 2. (a) Midpoint by the profile: $T = F/2 = \boxed{6.0\ \mathrm{N}}$ — or honestly: far half has mass $1.0$ kg, so $T = 1.0 \times 6.0 = 6.0$ N ✓.

Step 3. (b) $1.0$ m from the free end means the far piece has mass $2.0 \times (1.0/4.0) = 0.50$ kg: $T = 0.50 \times 6.0 = \boxed{3.0\ \mathrm{N}}$ .

Step 4. Formula check with $x = 3.0$ m from the pulled end: $T = 12(1 - 3/4) = 3.0$ N ✓. Timing note: the entire archetype is "acceleration of the whole, then Newton on the far piece." Under 2 minutes. If you find yourself integrating along the rope, stop — uniform ropes need only ratios of masses.

4 · Competitive level — heavy rope plus hanging load (JEE-Main/Advanced stretch, ~3 minutes)

A uniform rope of mass $1.0$ kg and length $2.0$ m hangs from a ceiling with a $4.0$ kg block tied to its lower end. Take $g = 10\ \mathrm{m/s^2}$ . Find the tension (a) at the bottom of the rope, (b) at the midpoint, (c) at the top. (d) If instead the whole assembly is pulled upward so it accelerates at $2.0\ \mathrm{m/s^2}$ , what is the new midpoint tension?

Step 1. Static case — cut and weigh what hangs below. (a) Bottom: just the block: $T = 4.0 \times 10 = \boxed{40\ \mathrm{N}}$ .

Step 2. (b) Midpoint: block + half the rope ( $0.5$ kg): $T = 4.5 \times 10 = \boxed{45\ \mathrm{N}}$ .

Step 3. (c) Top: block + whole rope: $T = 5.0 \times 10 = \boxed{50\ \mathrm{N}}$ .

Step 4. (d) Accelerating at $a = 2.0\ \mathrm{m/s^2}$ upward: every cut must both support and accelerate what is below it — replace $g$ by $g + a = 12$ : midpoint $T = 4.5 \times 12 = \boxed{54\ \mathrm{N}}$ . Timing note: notice the answer pattern 40/45/50 — linear in the hanging mass, the signature of equation (3). Part (d) is one substitution, $g \to g + a$ , the same effective-gravity move as the lift cable. Three minutes total; drawing three separate FBDs is correct but slow — cut-and-count is the archetype.

Exam traps

Tension drawn as a push. On an FBD the string force always points away from the body along the string — the string pulls the body toward itself. A "tension" arrow pushing a block is an automatic FBD error, and it happens most in pulley diagrams where the string turns a corner.

Doubling the tension at a wall or in tug-of-war. A rope pulled by 100 N at each end carries 100 N (see the pitfall box). Same trap in disguise: a rope tied to a wall and pulled with 100 N — the wall pulls back with 100 N and the tension is 100 N, identical to the two-team case.

Using one $T$ for a heavy rope. The moment a question gives the rope mass, the single-tension license is revoked. Signal words: "uniform rope/chain of mass $m$ ," "tension at the midpoint," "at distance $x$ ." Cut at the asked point and count what lies beyond.

Measuring $x$ from the wrong end. Equation (4) uses distance from the pulled end; the mirror formula $T = Fx/L$ uses distance from the free end. Both appear as MCQ options. Anchor yourself with the limits: the pulled end must carry the full $F$ .

Forgetting a string can go slack. If your algebra returns $T < 0$ , the string has gone slack in reality: set $T = 0$ and re-solve the motion. Rods can push (negative "tension" is fine for a rod); strings cannot.

String-2-pulls-down amnesia. In stacked hanging systems, the string below a block pulls it down (third-law partner of supporting the lower block). Omitting it under-counts the upper tension — the Example 1 cross-check catches this in five seconds.

Which $g$ . CBSE/ISC: $9.8\ \mathrm{m/s^2}$ unless stated; JEE/NEET papers almost always declare $g = 10$ ; CIE A-Level uses $9.81$ . Use the stated value and never mix two in one solution.

Origins · 1 minLeonhard Euler and how this was discovered

Leonhard Euler

1707–1783 · Berlin & St. Petersburg Academies

Leonhard Euler turned ropes and strings from craftsmen's lore into mathematics. In his mechanics he formalized the cut principle this article leans on — isolate any piece of a body and let internal forces appear as external pulls on the cut faces — the very move that makes "tension at a point" meaningful. His analysis of a rope wrapped on a rough post produced the capstan equation, by which each extra turn multiplies the holdable force: the reason two loops around a bollard let one sailor hold a ship against the tide. With Daniel Bernoulli he derived the vibrating string's mechanics, from which the wave speed $\sqrt{T/\mu}$ — tension over mass per length — controls every instrument with a string in it. Euler produced more mathematics than anyone in history, much of it after going blind: "now I will have fewer distractions."

History

1673

— Huygens analyzes the compound pendulum; cords and constraints enter precise mechanics.

1687

— Newton's Principia treats strings and pulleys via the third law — tension as paired pulls.

1691

— Jacob Bernoulli, Huygens and Leibniz solve the hanging-chain (catenary) shape, where tension varies along the curve.

1765

— Euler's capstan equation quantifies rope-on-post friction: tension can drop exponentially along a wrap.

1820

s — Navier and Cauchy generalize the cut principle into stress — tension per unit area — founding continuum mechanics.

Going further

Texts. NCERT Physics Class 11, Ch. 4 (Laws of Motion) · HC Verma, Concepts of Physics Vol. 1, Ch. 5 · Kleppner & Kolenkow Ch. 2 (the massless-string argument done honestly).

Provenance. All results are standard. A verifier should recompute: Example 1's $T_2 = 19.6$ , $T_1 = 49$ N with the third-law cross-check; Example 2's $6600$ / $5160$ N; Example 3's $6.0$ / $3.0$ N via both the mass-ratio and the profile formula; Example 4's $40/45/50$ N ladder and the $g \to 12$ midpoint $54$ N.

Prerequisites. Newton's Third Law & Interaction Pairs · Free-Body Diagrams · Weight & Normal Force. Leads to. The Atwood Machine & Variants · Connected Blocks, Strings & Constraints · Monkey Climbing a Rope. Deep dive (UG). Tension & Ideal Strings (UG treatment, with the heavy-rope calculus).

Is tension a force? Which way does it point?

Tension is the magnitude of the pulling force a string exerts at a cut or an attachment — so "the tension is 40 N" describes a pair of forces, each 40 N, pulling the two sides of any cross-section toward each other. On a specific body's FBD it becomes one force: 40 N directed away from the body, along the string. The string pulls whatever it touches toward itself; it can never push. If you remember only one drawing rule: tension arrows leave the body along the string.

Why exactly does massless mean the same tension everywhere?

Newton's second law on any string segment: $T_1 - T_2 = m_{\text{seg}}a$ . Zero mass forces $T_1 = T_2$ whatever the acceleration — an unequal pair would give a massless segment infinite acceleration. Repeat along the string and one number rules it all. The contrapositive is just as useful: whenever tension must differ (heavy rope, massive pulley, rope on a rough post), some mass or friction is there soaking up the difference.

Does the tension change when the string bends over a pulley?

Over an ideal pulley — massless, frictionless — no: the pulley pushes only perpendicular to the string, so nothing along the string changes and $T$ survives the turn; only direction changes. Over a massive pulley the two sides must differ to provide the torque that spins the wheel ( $T_2 > T_1$ when accelerating) — that is the Pulley With Mass problem of the rotation chapter. Over a rough post (no wheel at all), friction along the contact eats tension continuously — Euler's capstan effect. "Light frictionless pulley" in a question is your license for one $T$ ; anything else is a warning label.

What happens to the tension if the string suddenly breaks or goes slack?

It becomes zero instantly — tension is a constraint force, existing only while the constraint (taut string) exists. The bodies then move under their remaining forces: a lift in a snapped-cable nightmare is in free fall with $T = 0$ ; the two Atwood masses become independent projectiles. In problems, "the string is cut" means: erase $T$ from every FBD and re-solve from that instant with positions and velocities carried over.

How is a string different from a spring or a rod?

Three force personalities. A string pulls only, with whatever tension the constraint demands, and its (ideal) length never changes. A rod both pushes and pulls — it can carry compression — still at fixed length. A spring pulls and pushes but only by changing length: its force is determined by its stretch through Hooke's law $F = kx$ , not by the demands of the constraint. That is the deep difference: string tension is read off the problem (it adjusts), spring force is read off the geometry (it is what the stretch says, no more). The spring's own article is next.

For a hanging rope with a load, why is the tension not just the load's weight everywhere?

Because each cross-section holds everything below it — and "everything" includes rope. At the very bottom, only the load hangs, so $T = Mg$ there. Halfway up, the cut holds the load plus half the rope's weight. At the ceiling, everything. The profile climbs linearly from $Mg$ to $Mg + mg$ . In the ideal-string limit ( $m \to 0$ ) the profile flattens to the single value $Mg$ — the fiction and the truth agree exactly when the rope's weight is negligible next to the load's.

Why do exam ropes always say "light and inextensible"? What breaks if they aren't?

"Light" buys one tension for the whole string (else: profiles, as in equations 3–4). "Inextensible" buys one shared acceleration magnitude (else: the ends move differently while the string stretches, and you need spring physics to track it). Drop both and you're doing waves-on-a-string and elasticity — real physics, but not this chapter's. The exam phrase is a contract: one $T$ , one $a$ . Honour it in your first line and cash it in the algebra.