Dynamics – Apparent Weight in an Accelerating Lift A lift (elevator) moves downward with acceleration 9.8 m/s^2. The normal reaction/pressure exerted by a man on the lift floor becomes zero. Is this statement true?

Introduction / Context: Apparent weight in elevators is a standard dynamics concept. The normal reaction (N) from the floor is perceived as 'weight' on a scale. When the elevator accelerates, N differs from the true weight mg.

Given Data / Assumptions:

• Lift acceleration a = 9.8 m/s^2 downward (approximately g).
• Man of mass m standing on the floor; vertical motion only.
• Neglect air resistance; consider ideal contact.

Concept / Approach: For downward acceleration a, the equation of motion is: N + (−mg) = m(−a). Rearranged, N = m(g − a). Apparent weight equals N. If a = g, then N = 0, meaning weightlessness (free fall).

Step-by-Step Solution: Write vertical force balance taking upward positive. Sum of forces: N − mg = −m a. Thus N = m(g − a). Given a = g ⇒ N = m(g − g) = 0.

Verification / Alternative check: This is the classic 'free-fall' condition: all bodies inside accelerate equally, so the contact force vanishes.

Why Other Options Are Wrong: 'False': Incorrect because the derived formula gives N = 0 at a = g. 'Only if the man jumps' or 'mass is zero': Unnecessary; the result follows from dynamics. 'Air resistance' is not part of the contact-force calculation here.

Common Pitfalls: Using N = mg ± ma without sign care; always derive N = m(g − a) for downward acceleration.

Final Answer: True