Apparent Weight in a LiftFoundation

What the scale reads when the elevator accelerates — interactive Physics simulation for IIT-JEE.

Concept

A weighing scale does not measure gravity — it measures the normal force it exerts on you. When the lift accelerates upward the floor must push harder ( $N > mg$ , you feel heavy); accelerating downward it pushes less ( $N < mg$ ). In free fall $N = 0$ : weightlessness.

Key formula

N = m(g + a) \quad (a > 0 \text{ up}), \qquad N = m(g - a) \quad (a \text{ down}), \qquad N = 0 \text{ in free fall}

Derivation

Newton's second law on the person (up positive): $N - mg = ma$ , so $N = m(g+a)$ .

Moving up or down at constant velocity gives $a=0$ → $N = mg$ : the scale is honest only in unaccelerated lifts. Direction of velocity is irrelevant; only acceleration matters.

Scenarios to explore

Apparent Weight in a Lift — Why the scale lies when the elevator accelerates.

Real-world applications

Astronaut 'weightlessness' in orbit — perpetual free fall, not zero gravity.
g-force ratings in aircraft pull-ups and roller coasters.
Drop towers create ~4 s of microgravity for experiments.

JEE exam tips

Cable snap ⇒ a = −g ⇒ N = 0, and a pendulum inside stops oscillating (g_eff = 0).
Effective gravity g_eff = g + a (lift frame) governs pendulum periods inside: T = 2π√(L/g_eff).
Scale reading in kgf = N/g — divide by real g, not g_eff.

Common mistakes

Thinking a lift moving DOWN always makes you lighter — a lift decelerating while moving down makes you HEAVIER (a points up).
Confusing velocity with acceleration.
Saying gravity 'switches off' in orbit — g is ~90% of surface value at ISS altitude.

Exam traps to avoid

Maximum scale reading is at the START of an upward journey (accelerating up), not mid-journey.
If a > g downward (rocket thrust down), the person leaves the floor — N cannot go negative.