ARCHETYPE: The Atwood Machine & Variants

Every traction lift you have ridden is an Atwood machine with a motor at the pulley. The car and a counterweight hang over a driven sheave; the counterweight is sized to the car plus roughly half its rated load, so the machine is nearly balanced and the motor supplies only the difference — exactly the $(m_2 - m_1)g$ term. That is why a lift serving forty floors runs on a motor not much bigger than a car engine. Funiculars go one better with two counterbalanced cars on one cable — Atwood machines you can sit inside. And wherever engineers need a small, controllable net force from large opposing ones — testing rigs, counterweighted drawbridges — they are reusing Atwood's insight that a nearly balanced system responds gently.

Geometry first, because in an archetype the geometry is the physics. A light string passes over a pulley on a rigid support: $m_1$ (lighter) on the left, $m_2$ (heavier) on the right. Release from rest: $m_2$ descends, $m_1$ rises, and — the crucial reading — they speed up together. The string cannot stretch, so every centimetre $m_2$ drops is a centimetre $m_1$ climbs. One clock, one speed, one acceleration magnitude for the whole machine.

Two idealizations make the machine solvable with algebra alone, and both earn explicit statements in a board answer:

The string is massless and inextensible. A massless segment with unequal pulls on its ends would suffer infinite acceleration — impossible — so the tension is one number, $T$, all along the string. Inextensible means fixed length: if $m_2$ goes down by $x$, $m_1$ comes up by exactly $x$, so their acceleration magnitudes match at every instant.

The pulley is massless and frictionless. A real pulley needs a torque to spin up, and torque comes from a difference in side tensions. An ideal pulley needs no torque, so it merely redirects the string: same $T$ both sides. (A pulley with mass opens the rotation chapter — the tensions then split.) Look at the free-body diagrams in the figure. Each mass feels exactly two forces: weight down, the same tension $T$ up. For $m_1$ tension wins — it accelerates upward; for $m_2$ weight wins — it accelerates downward. The whole exam-grade analysis is those two sentences turned into equations.

Every Atwood question begins the same way. Choose a positive direction along the string — the direction the system actually moves: down for $m_2$, up for $m_1$. This single choice guarantees both equations use the same $a$ with the same sign, which is where most sign errors die. Newton's second law for the lighter mass, taking its upward motion as positive:

where: $T$ — the single string tension; $m_1$ — the lighter mass; $a$ — the common acceleration magnitude. Read it physically: $m_1$ accelerates upward only because $T$ exceeds its weight. If your final $T$ comes out below $m_1 g$, something is wrong. The same law for the heavier mass, taking its downward motion as positive:

where: $m_2$ — the heavier mass. Weight beats tension here, and the same $T$ and $a$ appear as in (1) — the ideal string and the constraint doing their jobs. Add (1) and (2): the tension, an internal force, cancels, leaving

where: $(m_2 - m_1)g$ — the unbalanced weight driving the system; $(m_1 + m_2)$ — the total inertia being accelerated. This "driving difference over total inertia" reading is the exam superpower: it generalizes to inclines, tables, and lifts without re-deriving anything. Note $a < g$ always — Atwood's whole point. Substitute $a$ back into (1) to get the tension:

where: $T$ is exactly the harmonic mean of the two weights $m_1 g$ and $m_2 g$ — always between them, and always at or below their arithmetic average. For $3$ and $5\ \mathrm{kg}$ at $g = 9.8$: weights $29.4$ and $49$ N, average $39.2$ N, but $T = 36.75$ N. Examiners love offering the average as a distractor.

The variants examiners actually pose

The lift variant. Mount the machine inside a lift accelerating upward at $a_0$. In the lift's frame gravity feels stronger — every $g$ becomes $g + a_0$ ($g - a_0$ for downward acceleration):

where: $a_{\text{rel}}$ — acceleration relative to the lift. The substitution $g \to g_{\text{eff}} = g + a_0$ is the entire trick. A machine in free fall ($a_0 = -g$) has $g_{\text{eff}} = 0$: nothing moves relative to the lift and the tension vanishes.

The double-pulley (double Atwood) variant. Replace $m_2$ by a second, movable light pulley carrying its own string and two masses. One idea unlocks it: a massless pulley must feel zero net force, so the upper-string tension is exactly twice the lower-string tension. Full double-Atwood solutions are JEE-Advanced territory; at board level you need only the $2T$ insight — also exactly why the axle of the classic machine feels $2T$, a favourite one-mark question.

1 · Board level — the standard machine (CBSE/ISC pattern, 3 marks)

Blocks of $3.0\ \mathrm{kg}$ and $5.0\ \mathrm{kg}$ hang from a light inextensible string over a smooth light pulley. The system is released from rest. Find the acceleration and the tension. Take $g = 9.8\ \mathrm{m/s^2}$.

Step 1. State idealizations (this earns method marks): light string + smooth light pulley $\Rightarrow$ same $T$ both sides; inextensible $\Rightarrow$ same $a$.

Step 2. Newton's second law, each block along its own motion. For the $5$ kg block (down positive): $49 - T = 5a$. For the $3$ kg block (up positive): $T - 29.4 = 3a$.

Step 3. Add: $19.6 = 8a \Rightarrow a = \boxed{2.45\ \mathrm{m/s^2}}$.

Step 4. Substitute: $T = 3(9.8 + 2.45) = \boxed{36.75\ \mathrm{N}}$ (≈ $36.8$ N to three significant figures). Sanity: $T$ lies between the weights ($29.4$ and $49$ N) ✓. With $g = 10$ (the usual JEE/NEET convention) the same work gives $a = 2.5\ \mathrm{m/s^2}$, $T = 37.5\ \mathrm{N}$ — use whichever $g$ the paper states.

2 · Board level — half-Atwood on a smooth table (ISC/A-Level connected-particles pattern)

A block of mass $4.0\ \mathrm{kg}$ rests on a smooth horizontal table, connected by a light string over a light pulley at the table's edge to a hanging mass of $2.0\ \mathrm{kg}$. Find the acceleration, the tension, and the speed after the hanging mass falls $0.90\ \mathrm{m}$ from rest. ($g = 9.8\ \mathrm{m/s^2}$.)

Step 1. Table block, horizontal: $T = 4a$. Hanging mass, down positive: $2(9.8) - T = 2a$.

Step 2. Add: $19.6 = 6a \Rightarrow a = \boxed{3.27\ \mathrm{m/s^2}}$.

Step 3. Tension: $T = 4(3.27) = \boxed{13.1\ \mathrm{N}}$. Cross-check from the other block: $T = 2(9.8 - 3.27) = 13.1$ N ✓.

Step 4. Kinematics ride-along: $v^2 = 2as = 2(3.27)(0.90) = 5.88 \Rightarrow v = \boxed{2.42\ \mathrm{m/s}}$. Notice what the table did: it removed $M$'s weight from the driving side (the normal force carries it) but kept its mass on the inertia side.

3 · Competitive level — JEE-Main pattern (~2.5 minutes)

Masses of $4.8$ and $5.2\ \mathrm{kg}$ hang from an ideal Atwood machine, released from rest. Take $g = 10\ \mathrm{m/s^2}$. (a) How long does the heavier mass take to descend $1.0\ \mathrm{m}$? (b) What force does the pulley exert on its axle clamp during the motion?

Step 1. Acceleration by the system shortcut — no FBDs needed for part (a): $a = \dfrac{(5.2-4.8)(10)}{10} = 0.40\ \mathrm{m/s^2}$ ($4$ N of unbalanced weight over $10$ kg total).

Step 2. Time from rest through $s = 1.0$ m: $t = \sqrt{2s/a} = \sqrt{2(1.0)/0.40} = \sqrt{5} = \boxed{2.2\ \mathrm{s}}$.

Step 3. Tension for part (b): $T = \dfrac{2(4.8)(5.2)(10)}{10} = 49.92\ \mathrm{N}$.

Step 4. The axle supports two string segments, each pulling down with $T$: $F_{\text{axle}} = 2T = \boxed{99.8\ \mathrm{N}}$ — less than the $100$ N total weight. The shortfall, $0.16$ N $= (m_2 - m_1)a$: the centre of mass accelerates downward, so the support carries less than the full weight. Timing note: nearly-equal masses are the examiner's hint that $a$ is small and clean — trust equation (3), don't re-derive. Budget ~2.5 min; the only place to slow down is Step 4's $2T$ insight.

4 · Competitive level — NEET/JEE-Advanced stretch: the machine in a lift (~4 minutes)

An Atwood machine with $m_1 = 3.0$ and $m_2 = 5.0\ \mathrm{kg}$ hangs in a lift accelerating upward at $a_0 = 2.0\ \mathrm{m/s^2}$. The string bears at most $40\ \mathrm{N}$; $g = 10\ \mathrm{m/s^2}$. (a) Find the blocks' acceleration relative to the lift and the tension. (b) Does the string survive?

Step 1. In the lift's frame, effective gravity is $g_{\text{eff}} = g + a_0 = 12\ \mathrm{m/s^2}$. Every stationary-machine formula now works with $g \to 12$.

Step 2. Relative acceleration, equation (5): $a_{\text{rel}} = \dfrac{(5-3)(12)}{8} = \dfrac{24}{8} = \boxed{3.0\ \mathrm{m/s^2}}$.

Step 3. Tension: $T = \dfrac{2(3)(5)(12)}{8} = \dfrac{360}{8} = \boxed{45\ \mathrm{N}}$.

Step 4. The stationary machine (at $g=10$) would give $T = 37.5\ \mathrm{N} < 40$ N — safe. But in the accelerating lift $T = 45\ \mathrm{N} > 40\ \mathrm{N}$: the string snaps. The upward acceleration raises the tension by the ratio $g_{\text{eff}}/g = 1.2$, as if gravity had strengthened by 20%. Timing note: the entire problem is the substitution $g \to g + a_0$ — spot it and this is 90 seconds of arithmetic; miss it and you drown in pseudo-force bookkeeping. Lift accelerating downward: use $g - a_0$; free fall: tension dies entirely.

Mixed sign conventions — the number-one mark killer. Writing $T - m_1g = m_1a$ and then $m_2g - T = -m_2a$ (or $+T$ in both) contradicts the constraint. Fix it structurally: pick one positive sense along the string and write both equations in that sense. Don't know which way it moves? Guess — a negative $a$ just means the machine runs the other way; magnitudes stay right.

Forgetting the constraint. Two symbols $a_1$, $a_2$ with no relation leaves two equations, three unknowns. The inextensible string is physics content, not decoration: $|a_1| = |a_2|$, and saying so earns a mark on board papers.

"$a = \frac{(m_2-m_1)g}{m_2}$" and cousins. Dividing the driving force by one mass is the classic hurried-MCQ slip. The whole mass $(m_1 + m_2)$ goes in the denominator. Five-second check: your formula must give $a=0$ when $m_1 = m_2$.

Tension distractors. $T \ne m_2 g$ (the heavy block would not fall), $T \ne m_1 g$ (the light block would not rise), and $T \ne$ the average of the weights — it is the harmonic mean, always below the average. Options offering both $36.75$ N and $39.2$ N for the 3-and-5 machine are fishing with the arithmetic mean.

Pulley axle force. The clamp feels $2T$, not $(m_1 + m_2)g$; they agree only for a balanced machine. For any accelerating Atwood, $2T < (m_1+m_2)g$ — the difference accelerates the centre of mass downward.

Which $g$? CBSE/ISC expect $g = 9.8\ \mathrm{m/s^2}$ unless stated; JEE-Main/Advanced and NEET almost always declare $g = 10$ in the question — use what is stated. AP Physics 1 permits $10$ for estimation; IB uses $9.8$; CIE A-Level uses $9.81$. Never mix values mid-solution: $a = 2.45$ (at $9.8$) next to $T = 37.5$ (at $10$) is a self-contradiction examiners notice.

Calculator habit. Store $a$ unrounded (e.g. $19.6/6 = 3.266\ldots$) before computing $T = Ma$; rounding early shifts the tension enough to miss a "three significant figures" answer box.

Texts. NCERT Physics Class 11, Ch. 4 (Laws of Motion) · HC Verma, Concepts of Physics Vol. 1, Ch. 5 · Morin, Introduction to Classical Mechanics, Ch. 3 (the Atwood family).

Provenance. All results are standard. A verifier should recompute: (3) and (4) by adding the FBD equations; $a = 2.45$, $T = 36.75$ N (3 & 5 kg, $g=9.8$; at $g=10$: $2.5$, $37.5$); half-Atwood $a = 3.27$, $T = 13.1$ N (4 & 2 kg); Example 3's $t = \sqrt{5} = 2.2$ s and $2T = 99.8$ N; Example 4's $a_{\text{rel}} = 3.0$, $T = 45$ N at $g_{\text{eff}} = 12$.

Prerequisites. Newton's Second Law · Free-Body Diagrams · Tension & Ideal Strings. Leads to. Connected Blocks & Constraints · Lifts & Apparent Weight · Pulley With Mass (rotation).