Tension and Ideal Strings

The ideal string never retired; it just changed clothes. Every tower crane lift begins with equation (3) — hoist cables are certified for the dynamic load $m(g+a)$, not the static weight, which is why crane controllers ramp acceleration gently and why a snatched load can snap a cable rated far above the cargo's weight. Elevator engineering still runs on Atwood's architecture: a counterweight (sized at car weight plus roughly 40–50% of rated load) hangs on the other side of the sheave, so the motor drives only the imbalance, a direct industrial dividend of equation (5). And the capstan exponential is the quiet hero of climbing and sailing — a belayer's modest grip, multiplied by $e^{\mu\theta}$ through a friction device, arrests a falling climber generating kilonewtons. The research frontier speaks fluent tension too. Cable-driven parallel robots — the camera flying over a stadium on four thin wires — solve string statics hundreds of times a second, with the $T \geq 0$ constraint as the central design headache: wires can only pull, so every manoeuvre must keep all cables in positive tension. Tendon-driven surgical robots route forces around corners the way an ideal string rounds a pulley. At the exotic end, proposals for space elevators live and die on the heavy-rope result: a uniform cable to geostationary orbit would need its tension to grow along tens of thousands of kilometres of its own weight, which is why every serious design tapers the cable exponentially and why the search continues for materials with the required strength-to-density. The bead-chain picture from the top of this article scales, literally, to the sky.

Zoom into any taut rope and you find a tug-of-war repeated a trillion times. The rope is a chain of material elements, each bonded to its neighbours by intermolecular forces that resist stretching. Pull the rope's ends apart and every bond stretches by a whisker; every element then pulls its left neighbour rightward and its right neighbour leftward, with the two pulls forming a Newton's-third-law pair across every cross-section. Tension is the magnitude of that pair. It is why the force you apply at one end of a clothesline reappears, undiminished, at the far end — the rope is a bucket brigade for force.

Two idealizations turn this microscopic mess into a single clean number. Massless means the string's own inertia is negligible next to the objects it connects. Apply Newton's second law to any little segment: net force equals mass times acceleration, and if the mass is zero the net force must be zero too — for any finite acceleration. So the pull on the segment's left end must exactly match the pull on its right end, link by link, all the way along the cord: the tension is uniform. Inextensible means the length never changes. That sounds like a statement about material stiffness, but its real job is geometric: a fixed total length is an equation relating the positions of everything the string touches, and differentiating it twice ties all their accelerations together. An ideal pulley — massless, frictionless — adds one more gift. The wheel touches the rope only through forces perpendicular to it (normal forces from the groove), so it can bend the rope around a corner without gripping it along its length; and having no inertia, it needs no net torque to spin up. Both sides of the wheel therefore carry the same tension: the pulley redirects the force without taxing it. That is the entire magic of pulleys — they let one tension act in two different directions. Keep one prohibition in view from the start: a string can pull but never push. Tension is carried by stretched bonds; slacken the string and the bonds simply relax — there is no way for a floppy cord to transmit compression. In the algebra this appears as the rule $T \geq 0$: a calculation that returns negative tension is the mathematics politely telling you the string has gone slack and your taut-string diagram is wrong.

Everything a free-body diagram needs to know about a string is its pull on the body it touches. The force is directed along the string, away from the body, toward the rest of the cord — a string tied to a crate on your left pulls the crate leftward, no exceptions. Writing that as a vector:

where: $\vec{T}$ — the force the string exerts on the attached body; $T$ — the tension magnitude at that end; $\hat{u}$ — the unit vector pointing from the attachment point along the string. The inequality is physics, not pedantry: it encodes "strings don't push," and checking it after solving is how you catch a slack string. A rigid rod, by contrast, drops the inequality — rods happily carry compression. Why is $T$ a single number for an ideal string? Because Newton's second law, applied to a small piece of string, leaves it no choice. Take a segment of length $\mathrm{d}s$ and mass $\mathrm{d}m$ somewhere along a straight cord, pulled by tension $T_{\text{right}}$ at its right face and $T_{\text{left}}$ at its left face:

where: $a$ — the segment's acceleration along the string. This innocuous line is the master equation of every string argument in this article: feed it $\mathrm{d}m \to 0$ and it forces uniform tension; feed it $\mathrm{d}m = \lambda\,\mathrm{d}s$ with real linear density $\lambda$ and it generates the tension profile of a heavy rope. Both cases are worked out below.

With uniformity established, the simplest applications read off instantly. A mass hanging at rest from a vertical string: the only forces on it are weight $mg$ down and tension $T$ up, and equilibrium demands they cancel. Put the same mass in an elevator accelerating upward at $a$ and the tension must both support the weight and supply the acceleration:

where: upward acceleration makes $T > mg$ (the cable works harder), downward acceleration makes $T < mg$, and free fall ($a = -g$) gives $T = 0$ — a plummeting elevator's cable, if intact, goes slack rather than pulling its car down. Note what tension is not: it is not always $mg$. It is whatever value Newton's second law requires, recomputed for each situation. The second string rule is geometry. If a cord of fixed total length $L$ is split by a pulley into segments $\ell_1$ and $\ell_2$, then

where: the overdots are time derivatives. In words: one side lengthens exactly as fast as the other shortens, so the two connected bodies share a single speed and a single acceleration magnitude. This is a constraint equation — it carries no forces, only geometry — and it is what lets a two-body pulley problem be solved with one acceleration unknown instead of two. In systems of several pulleys the same bookkeeping (total string length constant) yields relations like $a_1 = 2a_2$; the constraint, not the tension, is where those famous factors of 2 come from.

Which equation answers which question

In 1784 the Reverend George Atwood wanted to slow gravity down so that pendulum-clock timing could resolve it: two nearly equal masses on a string over a pulley fall with acceleration $(m_2 - m_1)g/(m_1 + m_2)$, as small as you please. The machine remains the perfect first outing for the two string rules, because it uses each exactly once — uniform tension couples the forces, inextensibility couples the motions.

Both results in one display, since they anchor every pulley problem to come:

where: $m_2$ is the heavier mass. Read the tension physically: it is always strictly between $m_1 g$ and $m_2 g$ when the system moves — the string supports the light mass more than gravity demands and the heavy mass less. And notice the machine's original purpose in the formula: with $m_1 = 990\ \mathrm{g}$ and $m_2 = 1010\ \mathrm{g}$, the acceleration is $g/100$ — Atwood's "slowed-down gravity," measurable with a pendulum clock and a metre stick in 1784.

Tie a stone to a string and whirl it: now the tension's job is to bend the velocity, supplying the centripetal force $mv^2/r$ toward the centre. For the conical pendulum — a bob circling on a string that sweeps a cone of half-angle $\theta$ from the vertical — the tension does two jobs at once, and resolving it along the vertical and toward the centre gives the pair

where: $r = \ell\sin\theta$ is the radius of the horizontal circle for string length $\ell$. Divide the two lines and the tension drops out, $\tan\theta = v^2/rg$ — the string finds whatever angle lets its tension simultaneously hold the bob up and turn it. From the first line, $T = mg/\cos\theta > mg$: a whirling string always pulls harder than the plain weight, and diverges as the cone flattens toward horizontal — the reason a string can never whirl a mass in a perfectly horizontal circle with nothing else supporting it. Swing the circle into a vertical plane and the geometry adds a twist that the $T \geq 0$ rule turns into physics. At the top of the loop both the weight and the tension point toward the centre, so $T_{\text{top}} = m v_{\text{top}}^2/r - mg$. The string can help gravity but never oppose it there; if $v_{\text{top}}^2 < gr$ the formula would need $T < 0$, which a string cannot supply — it goes slack and the stone falls out of the circle. The minimum speed to keep a taut string at the top,

where: at exactly this speed the string carries zero tension and gravity alone provides the centripetal force. For a bucket of water on a $0.9\ \mathrm{m}$ rope that threshold is $\sqrt{9.81 \times 0.9} = 3.0\ \mathrm{m/s}$ — swing faster than one revolution per couple of seconds and the water presses into the bucket instead of raining on you.

Real ropes are matter, and equation (2) with $\mathrm{d}m = \lambda\,\mathrm{d}s \neq 0$ tells you exactly what changes: the tension is no longer uniform, but varies at just the rate needed to accelerate — or, hanging in equilibrium, to support — the rope's own flesh.

where: $y$ — height of the cut above the block; the profile runs linearly from $mg$ at the bottom to $(m + M)g$ at the ceiling hook. The rule of thumb this encodes — tension at a point equals the weight of everything hanging below it — is worth more than the formula. It is why a climbing rope is most stressed at the anchor, why a deep-mine hoist cable must taper (its top must carry kilometres of its own steel), and why the "massless" approximation is excellent for a 60-metre, 3.9-kilogram climbing rope holding a 70-kilogram climber: the tension varies by only 38 N along its length against a 687 N load — a five-percent correction.

Friction at a curved surface breaks the other half of the idealization, and does it exponentially. A rope wrapped by angle $\theta$ around a rough post can sustain tensions differing by up to the Euler–Eytelwein (capstan) factor, $T_2/T_1 = e^{\mu\theta}$: with $\mu = 0.3$, three full turns ($\theta = 6\pi$) give $e^{5.65} \approx 286$ — a deckhand's 50 N grip on the slack side can hold more than a tonne of ship on the taut side. The ideal pulley is precisely the $\mu \to 0$, zero-inertia limit in which that ratio collapses to 1.

1 · The elevator cable (confidence builder)

An 850 kg elevator car accelerates upward at $1.2\ \mathrm{m/s^2}$. Find the cable tension, compare it with the tension at constant speed, and find the tension while the car brakes on the way down at the same rate. Use $g = 9.81\ \mathrm{m/s^2}$.

Step 1. Free-body diagram of the car: tension $T$ up, weight $mg = 850 \times 9.81 = 8339\ \mathrm{N}$ down. Newton's second law along the vertical, up positive: $T - mg = ma$.

Step 2. Accelerating upward, $a = +1.2\ \mathrm{m/s^2}$: $T = m(g + a) = 850 \times (9.81 + 1.2) = \boxed{9.36\ \mathrm{kN}}$ — about 12% more than the weight.

Step 3. Cruising at constant speed (any speed): $a = 0$, so $T = mg = \boxed{8.34\ \mathrm{kN}}$. The cable cannot tell 0 m/s from 5 m/s — only acceleration changes tension.

Step 4. Braking while moving down means the acceleration points up, again $a = +1.2\ \mathrm{m/s^2}$: the tension is the same $\boxed{9.36\ \mathrm{kN}}$ as in Step 2. The heavy-load moments for a hoist cable are speeding up upward and slowing down downward — both are upward accelerations. (Only if the car accelerated downward at $1.2\ \mathrm{m/s^2}$ would the cable relax to $850 \times 8.61 = 7.32\ \mathrm{kN}$.)

2 · An Atwood machine, numbers all the way through (exam level)

Masses $m_1 = 3.0\ \mathrm{kg}$ and $m_2 = 5.0\ \mathrm{kg}$ hang from an ideal string over an ideal pulley, released from rest. Find the acceleration, the tension, the speed of $m_2$ after it falls 1.5 m, and the time that fall takes.

Step 1. Acceleration from equation (5): $a = \dfrac{(m_2 - m_1)g}{m_1 + m_2} = \dfrac{2.0 \times 9.81}{8.0} = \boxed{2.45\ \mathrm{m/s^2}}$ — a quarter of $g$, because only 2 kg of weight imbalance must drag 8 kg of total inertia.

Step 2. Tension from equation (5): $T = \dfrac{2 m_1 m_2 g}{m_1 + m_2} = \dfrac{2 \times 3.0 \times 5.0 \times 9.81}{8.0} = \boxed{36.8\ \mathrm{N}}$. Sanity check the window: $m_1 g = 29.4\ \mathrm{N} < 36.8\ \mathrm{N} < m_2 g = 49.1\ \mathrm{N}$ ✓ — the string over-supports the light mass and under-supports the heavy one.

Step 3. Cross-check with $m_1$'s own equation: $T = m_1(g + a) = 3.0 \times (9.81 + 2.45) = 36.8\ \mathrm{N}$ ✓.

Step 4. Kinematics of the fall (constant acceleration from rest): $v = \sqrt{2 a\,d} = \sqrt{2 \times 2.45 \times 1.5} = \boxed{2.71\ \mathrm{m/s}}$, and $t = v/a = 2.71/2.45 = \boxed{1.11\ \mathrm{s}}$. By the constraint (4), $m_1$ is rising at exactly this same 2.71 m/s — one string, one speed.

3 · The bucket at the top of the loop (exam level)

A 1.2 kg bucket of water is whirled in a vertical circle of radius 0.9 m. Find the minimum speed at the top for the rope to stay taut, and the rope tension at the bottom if the bucket passes it at 5.0 m/s.

Step 1. At the top, weight and tension both point down toward the centre: $T + mg = m v_{\text{top}}^2 / r$. The rope's contribution obeys $T \geq 0$, so the slowest taut passage has $T = 0$.

Step 2. Minimum speed from equation (7): $v_{\text{top}}^{\min} = \sqrt{g r} = \sqrt{9.81 \times 0.9} = \boxed{2.97\ \mathrm{m/s}}$. Slower than this and no rope in the world keeps the water in — the string would need to push the bucket outward, which it cannot.

Step 3. At the bottom, tension points up (toward the centre) and weight down: $T - mg = m v^2 / r$, so $T = m\left(g + \dfrac{v^2}{r}\right) = 1.2 \times \left(9.81 + \dfrac{25.0}{0.9}\right) = 1.2 \times 37.6 = \boxed{45.1\ \mathrm{N}}$.

Step 4. Read the asymmetry: 45.1 N at the bottom is nearly four times the bucket's 11.8 N weight, while at the top the rope may carry nothing at all. Whirled systems always stress their string hardest at the lowest point — where the tension must both hold the weight and bend the fastest part of the motion.

Texts. Morin, Introduction to Classical Mechanics, ch. 3 (Using F = ma) — Atwood machines and their many-pulley mutations · Kleppner & Kolenkow, An Introduction to Mechanics, ch. 2–3 — constraint forces, the whirling block, the capstan · Taylor, Classical Mechanics, ch. 1 — Newton's laws with strings and distributed loads · Feynman, Lectures on Physics, vol. I ch. 12 — what a force law is and what idealizations buy.

Provenance. All results are standard textbook mechanics. A verifier should reproduce: (1) uniform tension from $T_{\text{right}} - T_{\text{left}} = (\mathrm{d}m)a$ in the $\mathrm{d}m \to 0$ limit, and $T_1 = T_2$ across a massless wheel from $\tau = I\alpha$ (Morin ch. 3); (2) the Atwood pair $a = (m_2 - m_1)g/(m_1+m_2)$, $T = 2m_1m_2g/(m_1+m_2)$ by adding the per-body equations; (3) the heavy-rope profile $T(y) = (m + My/L)g$ by the cut principle (Taylor ch. 1); (4) the capstan law $T_2/T_1 = e^{\mu\theta}$ by integrating $\mathrm{d}T = \mu T\,\mathrm{d}\theta$. Worked numbers on this page ($T = 9.36$ and $8.34\ \mathrm{kN}$ for the 850 kg elevator; $a = 2.45\ \mathrm{m/s^2}$, $T = 36.8\ \mathrm{N}$, $v = 2.71\ \mathrm{m/s}$, $t = 1.11\ \mathrm{s}$ for the 3/5 kg Atwood; $v_{\min} = 2.97\ \mathrm{m/s}$ and $T = 45.1\ \mathrm{N}$ for the 0.9 m vertical circle; capstan factor $e^{0.3 \times 6\pi} \approx 286$) are illustrative arithmetic recomputed from the boxed formulas with $g = 9.81\ \mathrm{m/s^2}$.

Builds on: Newton's Second Law · Newton's Third Law · Vectors, Coordinate Systems and Kinematics · Gravitational Force Near Earth's Surface. Leads to: Normal Force · Circular Motion and Centripetal Acceleration · Newton's First Law and Inertial Frames.