Two masses m₁ > m₂ hang over a fixed frictionless pulley by a light inextensible string. m₁ descends with acceleration a; m₂ ascends with the same acceleration a (string inextensible = same |acceleration| but opposite directions). FBD of m₁ (taking downward as positive): m₁g − T = m₁a ... (1). FBD of m₂ (taking upward as positive): T − m₂g = m₂a ... (2). Add (1) and (2): (m₁−m₂)g = (m₁+m₂)a → a = (m₁−m₂)g/(m₁+m₂). From (2): T = m₂(a+g) = m₂×[(m₁−m₂)g/(m₁+m₂) + g] = m₂g×[(m₁−m₂+m₁+m₂)/(m₁+m₂)] = 2m₁m₂g/(m₁+m₂).

• Verify T < m₁g: T = 2m₁m₂g/(m₁+m₂). Is T < m₁g? → 2m₂/(m₁+m₂) < 1 → 2m₂ < m₁+m₂ → m₂ < m₁ ✓ (true since m₁ > m₂). Physical meaning: m₁ is accelerating downward, so the net force on it is downward (m₁g > T). The string tension is less than m₁g. Verify T > m₂g: 2m₁/(m₁+m₂) > 1 → 2m₁ > m₁+m₂ → m₁ > m₂ ✓. Physical meaning: m₂ is accelerating upward, so T > m₂g. The string pulls m₂ up harder than gravity pulls it down.
• Special cases: (a) m₁ = m₂ = m → a = 0, T = mg. System in equilibrium; string tension equals weight of each mass. Addition of any small mass to m₁ starts the system into motion. (b) m₂ = 0 → a = g, T = 0. m₁ is in free fall (no opposing mass); string is slack. (c) m₁ >> m₂ → a ≈ g, T ≈ 2m₂g. m₁ barely affected by light m₂; m₂ experiences 2mg upward tension (nearly doubles its weight).
• NEET quick computation: if m₁ = 3 kg, m₂ = 1 kg, g = 10 m/s²: a = (3−1)×10/(3+1) = 20/4 = 5 m/s². T = 2×3×1×10/(3+1) = 60/4 = 15 N. Check: T < m₁g = 30 N ✓, T > m₂g = 10 N ✓. Acceleration check: a = 5 = (m₁−m₂)g/(m₁+m₂) = 2×10/4 ✓.

Apparent weight is the force a body exerts on its support (string, scale). In an Atwood machine, the support is the string, so apparent weight = T for both masses. Pressure on pulley axle: the pulley is held up by two string segments; the total force on the pulley axle = 2T (both tension forces pull the pulley downward). This is less than (m₁+m₂)g — the system is lighter than if both masses just hung statically, because the masses are accelerating. Alternatively: pulley force = 2T = 4m₁m₂g/(m₁+m₂). NOTE: 4m₁m₂/(m₁+m₂) ≤ m₁+m₂ with equality when m₁ = m₂ (harmonic ≤ arithmetic mean). So pulley force ≤ (m₁+m₂)g.

• Direction of acceleration: always the heavier mass descends, lighter mass ascends. At t = 0 (system released from rest): m₁ starts moving downward. If at any point during motion m₁ reaches the floor: the string goes slack, m₂ then undergoes free fall upward for a moment, then falls back down. The pulley system has a phase change. In NEET: usually asked only for the initial acceleration phase, not what happens after m₁ hits the floor.
• For the Atwood machine with a pulley of moment of inertia I and radius R (rotational effects): the effective equation of motion becomes a = (m₁−m₂)g / (m₁+m₂+I/R²). This is NOT part of standard NEET, which treats pulleys as massless and frictionless. JEE Advanced includes massive pulleys.
• Ratio problems: NEET may ask 'm₁/m₂ ratio for acceleration to be g/3' → a = g/3 → (m₁−m₂)/(m₁+m₂) = 1/3 → 3m₁−3m₂ = m₁+m₂ → 2m₁ = 4m₂ → m₁/m₂ = 2. General formula: a/g = (m₁−m₂)/(m₁+m₂) = (r−1)/(r+1) where r = m₁/m₂. Knowing this ratio is useful for NEET questions that give a/g and ask for m₁/m₂.

Incline Atwood (m₁ on smooth incline θ, m₂ hanging): Driving force = m₂g − m₁g sinθ (if m₂ causes downward pull and m₁ is pulled up slope). a = (m₂ − m₁ sinθ)g/(m₁+m₂) (if m₂g > m₁g sinθ). Equilibrium condition: m₂ = m₁ sinθ. T = m₁m₂g(1+sinθ)/(m₁+m₂). Rough incline variant: include friction. Drive force = m₂g − m₁g sinθ − μm₁g cosθ (if m₁ is being pulled up rough slope). Gross check: if drive force < 0, m₂ insufficient to pull m₁ up incline — direction reverses.

• Multiple pulley Atwood (movable pulley — one mass hangs from movable pulley, two masses hang from strings): constraint equation changes. For a movable pulley (one side fixed), if m hangs from the movable pulley and m₁, m₂ are on the two strings: constraint: acceleration of movable pulley = (a₁+a₂)/2 (average of the two string accelerations). This is a JEE-level topic but occasionally appears in NEET as a conceptual question. For NEET: the simple Atwood (fixed pulley) is the primary focus.
• Atwood machine on accelerating platform: if the pulley accelerates upward at A, the effective gravity becomes g+A. Formulas: a_rel = (m₁−m₂)(g+A)/(m₁+m₂). T = 2m₁m₂(g+A)/(m₁+m₂). If platform accelerates downward: replace g+A with g−A. At A = −g (free fall of platform): a = 0, T = 0. This is the weightlessness scenario — all forces vanish.
• Atwood machine with three masses on each string: system has string from ceiling, m₁ attached at bottom, then separate string to m₂, which connects to m₃ hanging lower — this is a series of Atwood or a vertical string chain. This is a combined multi-body Newton's Law problem. For NEET: usually at most two or three masses in one system. Apply systematic FBD for each mass.