Atwood MachineJEE Main

Concept

Two masses hang from a light string over a frictionless pulley. The heavier side falls, the lighter rises — both share the same acceleration magnitude because the string is inextensible. The string's tension is the same throughout and lies between the two weights.

Key formula

Derivation

Newton's second law on each mass (taking m₂ down as positive):

$$m_2 g - T = m_2 a, \qquad T - m_1 g = m_1 a$$

Adding eliminates $T$ to give $a$; substituting back gives $T$. Note $T$ equals the harmonic-mean weight $2m_1m_2g/(m_1+m_2)$ — always between $m_1g$ and $m_2g$.

Scenarios to explore

• Atwood Machine — Two hanging masses over a pulley — acceleration & tension.

Real-world applications

• Elevator counterweights use the same constraint physics.
• Lab measurement of g using slow, controllable accelerations.
• Any rope-over-pulley system: wells, cranes, gym cable machines.

JEE exam tips

• If the pulley has mass (moment of inertia), tensions on the two sides differ — a favourite JEE Advanced twist.
• System trick: a = (net driving force)/(total mass) = (m₂−m₁)g/(m₁+m₂).
• Equal masses ⇒ a = 0 but T = mg ≠ 0.

Common mistakes

• Assuming tension equals the weight of either mass — it is between them.
• Giving the two masses different acceleration magnitudes.
• Forgetting the pulley must be massless & frictionless for T to be uniform.

Exam traps to avoid

• Reading scale attached to pulley reads 2T, not (m₁+m₂)g — less, when accelerating.
• As m₂ → ∞, a → g (not beyond) and T → 2m₁g.