Atwood Machines — All Variants

What is an Atwood Machine?

An Atwood machine is the simplest pulley problem in physics — two masses hanging from a string over a pulley. It looks toy-like, but every JEE/NEET pulley question is some twist on this setup. We use it to teach Newton’s second law for connected systems, and once we master the standard Atwood, the variants (massive pulley, friction, double pulley, incline-pulley) are just bookkeeping.

The classical Atwood has two masses $m_1$ and $m_2$ on either end of a light, inextensible string passing over a frictionless, massless pulley. If $m_2 > m_1$ , then $m_2$ falls and $m_1$ rises. The acceleration is

$a = \frac{(m_2 - m_1)g}{m_1 + m_2}$

and the tension is

$T = \frac{2m_1 m_2 g}{m_1 + m_2}$

That’s it. Everything below is built on these two formulas plus FBDs.

Key Terms & Definitions

Light string: massless. The tension is the same on both sides of the pulley.

Inextensible string: the two masses share the same speed and acceleration magnitude.

Frictionless, massless pulley: the pulley rotates freely without storing any rotational energy. Tension on either side is equal.

Constraint relation: the equation that links the accelerations of the two masses. For the standard Atwood it is just $a_1 = -a_2$ .

Methods & Concepts

Method 1: Free-body diagrams

Draw the FBD for each mass. Pick a positive direction (say, downward for the heavier mass). Write Newton’s second law for each. Add equations to eliminate $T$ and solve for $a$ . Back-substitute for $T$ .

$a = \frac{(m_2 - m_1)g}{m_1 + m_2}, \quad T = \frac{2m_1 m_2 g}{m_1 + m_2}$

Method 2: System equation (when valid)

If you only need the acceleration, treat the two masses as one system. The net external force along the direction of motion is $(m_2 - m_1)g$ , and the total mass is $m_1 + m_2$ . So $a = (m_2 - m_1)g/(m_1 + m_2)$ directly.

This shortcut works because internal forces (tension) cancel out. It fails for variants where tension differs on the two sides (e.g. massive pulley).

Method 3: Energy/Lagrangian approach

For a massive pulley of moment of inertia $I$ and radius $r$ , energy conservation gives a cleaner derivation. After the heavier mass falls a distance $h$ , the system has KE $\tfrac{1}{2}(m_1 + m_2)v^2 + \tfrac{1}{2}I(v/r)^2$ , equating to PE loss $(m_2 - m_1)gh$ . Differentiate with respect to time to get $a$ .

Solved Examples

Example 1 (CBSE level — Easy)

Two masses $3$ kg and $5$ kg hang from a light string over a frictionless pulley. Find the acceleration and tension. Take $g = 10$ m/s².

$a = (5 - 3)(10)/(3 + 5) = 20/8 = 2.5$ m/s².

$T = 2 \times 3 \times 5 \times 10 / 8 = 37.5$ N.

Example 2 (JEE Main — Medium): Massive pulley

A pulley of mass $M = 2$ kg and radius $r$ (treat as uniform disc, $I = \tfrac{1}{2}Mr^2$ ) carries masses $m_1 = 4$ kg and $m_2 = 6$ kg. Find the acceleration. Take $g = 10$ m/s².

Writing FBDs for each mass and torque equation for the pulley:

$m_2 g - T_2 = m_2 a$ , $T_1 - m_1 g = m_1 a$ , $(T_2 - T_1)r = I\alpha = \tfrac{1}{2}Mr^2 (a/r)$ .

Adding the first two and using the third: $(m_2 - m_1)g = (m_1 + m_2 + M/2)a$ .

$a = (6 - 4)(10)/(4 + 6 + 1) = 20/11 \approx 1.82$ m/s².

The " $M/2$ " term is the pulley’s effective inertia. For a generic pulley, replace $M/2$ by $I/r^2$ .

Example 3 (JEE Advanced — Hard): One mass on an incline

$m_1$ on a smooth incline of angle $\theta$ connected over a pulley at the top to hanging mass $m_2$ . Find $a$ assuming $m_2 g > m_1 g \sin\theta$ .

$m_2 g - T = m_2 a$ (hanging block falls).

$T - m_1 g \sin\theta = m_1 a$ (block on incline moves up).

Adding: $a = (m_2 - m_1\sin\theta)g/(m_1 + m_2)$ .

This is the JEE workhorse — an Atwood with one side replaced by an incline. The factor $\sin\theta$ replaces the $1$ that the simple Atwood has on the lighter side.

Atwood Variants Cheat Sheet

Variant	Effective formula
Standard	$a = (m_2 - m_1)g/(m_1+m_2)$
Massive pulley ( $I, r$ )	$a = (m_2 - m_1)g/(m_1+m_2+I/r^2)$
Incline + hanging	$a = (m_2 - m_1\sin\theta)g/(m_1+m_2)$
With friction $\mu$ on incline	$a = (m_2 - m_1\sin\theta - \mu m_1\cos\theta)g/(m_1+m_2)$
Double Atwood (mass on movable pulley)	requires constraint relation $a_1 + a_2 = 2a_p$

Exam-Specific Tips

For JEE Main, Atwood with massive pulley appears almost every year. Memorise the substitution: replace $(m_1+m_2)$ in the denominator with $(m_1+m_2+I/r^2)$ .

For NEET, the focus is on incline-pulley combos. Always sketch the FBD for the incline block — students drop the $\sin\theta$ factor under exam pressure.

For CBSE Class 11, Atwood is a 3-mark question. Show the FBDs first, write equations clearly, then solve. Step marks are easy if you set up cleanly.

Common Mistakes to Avoid

1. Different tensions on the two sides of a frictionless, massless pulley. They are equal. Two different tensions appear only with a massive pulley.

2. Adding accelerations as scalars without sign convention. Pick a direction first. The two masses have opposite vector accelerations but the same magnitude $a$ .

3. Forgetting $\sin\theta$ on inclines. The component of gravity along the incline is $g\sin\theta$ , not $g$ . This is the single most common error.

4. Using the system shortcut when tensions differ. With a massive pulley, the two-side tensions are unequal. The simple “treat as one mass” trick fails — write three equations (two FBDs + torque equation).

5. Treating the pulley’s mass as if it were attached to the heavier block. It isn’t — the pulley’s inertia adds an effective $I/r^2$ to the system, not to either block individually.

Practice Questions

Q1. Two masses of $2$ kg and $4$ kg are connected by a string over a smooth pulley. Find the acceleration. ( $g=10$ )

$a = (4-2)(10)/6 = 10/3 \approx 3.33$ m/s².

Q2. In Q1, what is the tension?

$T = 2 \times 2 \times 4 \times 10 / 6 = 160/6 \approx 26.67$ N.

Q3. A $5$ kg block on a smooth incline of $30^\circ$ is connected over a pulley to a hanging $4$ kg block. Find $a$ . ( $g=10$ )

$a = (4 - 5\sin 30^\circ)(10)/9 = (4-2.5)(10)/9 = 15/9 \approx 1.67$ m/s². The hanging block falls.

Q4. With a pulley of mass $M = 4$ kg (uniform disc) and masses $3$ kg, $5$ kg, find $a$ .

$a = (5-3)(10)/(5+3+2) = 20/10 = 2$ m/s².

Q5. Two equal masses $m$ on a frictionless pulley. What is $a$ ?

$a = 0$ . The system is in equilibrium for any starting condition.

Q6. In a standard Atwood with $m_1 = 1$ kg and $m_2 = 2$ kg, find the apparent weight of $m_1$ during motion.

Apparent weight = $T = 2(1)(2)(10)/3 = 40/3 \approx 13.33$ N. Real weight is $10$ N — the lighter mass feels heavier while accelerating up.

Q7. For very large $m_2/m_1$ ratio, what is the limiting acceleration?

$a \to g$ as $m_2 \gg m_1$ . Heavy block falls in near free-fall, light block flies up at near- $g$ acceleration.

Q8. If the string in an Atwood machine breaks, what happens?

Both masses go into free fall — each accelerates at $g$ downward. The constraint disappears.

FAQs

Q: Is the tension always less than the weight of the heavier mass?

Yes. If $T \geq m_2 g$ , the heavier mass would not fall. Quick sanity check.

Q: Why does a massless pulley have the same tension on both sides?

Because applying $\tau = I\alpha$ with $I = 0$ gives $\tau = 0$ , so the net torque must vanish. The two tensions are at equal radii but opposite sides, so they must be equal.

Q: How does friction at the pulley axle change the answer?

It introduces a torque opposing motion, making the tensions unequal. The problem becomes algebraically messier but conceptually identical.

Q: Can the two masses ever have the same acceleration vector?

Only if both are zero. A non-zero acceleration is in opposite directions for the two masses (one up, one down).

Q: What if the string is elastic?

Then the two masses don’t share the same acceleration. The setup becomes a coupled-spring problem — out of standard Atwood scope.

Q: When does a pulley problem need both Newton’s laws AND torque equations?

When the pulley itself has appreciable mass. Light pulley → tensions equal, two FBDs suffice. Massive pulley → three equations.