Apparent weight = the normal reaction force R exerted on a person by the surface of contact (scale, floor), which is what the weighing machine reads. Actual weight = mg (always constant, depends only on mass and g). Derivation: apply F_net = ma to the person. Take upward as positive. R − mg = ma (net force = ma). Therefore: R = m(g + a) when net acceleration is upward; R = m(g − a) when net acceleration is downward.

• The weighing machine reads NORMAL FORCE (R), not actual weight (mg). This is key to all lift problems. When a question says 'what does the weighing machine show?', it asks for R, not mg.
• Net acceleration direction determines which formula: upward (any case) → R = m(g+a); downward (any case) → R = m(g−a). The direction of velocity is irrelevant. A lift moving upward but slowing down has downward acceleration → R = m(g−a).
• NEET: 'A person weighs 60 kgf at rest. The lift accelerates up at 2 m/s². Apparent weight?' → R = m(g+a) = 60×(10+2) = 720 N = 72 kgf. The increase in apparent weight = ma = 60×2 = 120 N.

Scenario → Net acceleration direction → Formula for R: (1) At rest: a=0, R=mg. (2) Uniform velocity (up or down): a=0, R=mg. (3) Accelerating up: a↑, R=m(g+a). (4) Decelerating while going down: a↑ (lift slowing down → net force upward), R=m(g+a). (5) Accelerating down: a↓, R=m(g−a). (6) Decelerating while going up: a↓ (lift slowing down → net force downward), R=m(g−a). (7) Free fall (a=g, down): R=m(g−g)=0. Heaviest in cases 3 and 4. Lightest in case 7. Weightless in case 7.

• Critical deceleration cases: 'Lift decelerates going UP' = lift velocity is upward but decreasing = acceleration is DOWNWARD (opposing motion) = R = m(g−a) [person feels lighter, even though moving up]. 'Lift decelerates going DOWN' = velocity is downward but decreasing = acceleration is UPWARD = R = m(g+a) [person feels heavier, even though moving down].
• When can apparent weight exceed real weight? Any time the lift has an upward net acceleration — whether lifting off, or braking to a stop while going down. Maximum apparent weight = m(g + a_max) where a_max is the maximum upward acceleration.
• NEET assertion-reason: 'A person feels weightless in a freely falling lift.' A: TRUE (R=0). 'This means the person has no mass.' R: FALSE. Apparent weight is zero but ACTUAL weight (mg) and mass remain unchanged. Weightlessness = absence of normal force, not absence of mass or gravity.

Free fall = lift accelerating downward at exactly a = g. Applying R − mg = m(−g): R = 0. The normal reaction is zero — the person does not press on the floor, and the floor does not push back. This is 'apparent weightlessness.' The actual weight mg still acts (gravity hasn't changed) — but there is no contact force. A weighing machine in a freely falling lift reads zero. The same condition occurs in a satellite in orbit (continuous free fall).

• Can apparent weight be zero without free fall? Only if a = g downward (i.e., free fall). For any smaller downward acceleration (a < g): R = m(g−a) > 0 (person still presses on the floor). For a > g (lift falling faster than free fall): R = m(g−a) < 0 — this means the person would be pressed against the ceiling. Physical interpretation: the person leaves the floor.
• Weightlessness in orbiting satellites: astronauts in the International Space Station are in continuous free fall (orbiting = constantly falling toward Earth while moving sideways fast enough to avoid hitting it). Their apparent weight = 0 (R = 0), though true weight (gravitational force mg) still acts — it provides the centripetal force for orbital motion.
• NEET: 'A coin is placed in a lift. In which scenario does the coin leave the floor?' → When lift's downward acceleration > g (a > g). In normal free fall (a = g) the coin is on the verge of leaving. For a > g: the lift floor moves away faster than the coin falls → coin floats above the floor (or hits the ceiling).