Apparent weight in an accelerating lift (elevator): A man’s apparent weight is less than his real weight when the lift is moving downward under which condition?

Introduction / Context:Apparent weight is the normal reaction N exerted by the lift floor on the person. In a non-inertial frame undergoing vertical acceleration, N differs from the true weight W = m * g, and this is often tested in elevator physics questions.

Given Data / Assumptions:

• Mass m of the person; gravitational acceleration g.
• Lift accelerates vertically with magnitude a.
• Positive upward direction convention.

Concept / Approach:

Equation of motion: ΣF = m * a. For downward acceleration (a negative with upward positive), we have N − m * g = m * a → N = m * (g + a). If the acceleration is downward, a = −|a|, hence N = m * (g − |a|) < m * g. Thus apparent weight decreases. For upward acceleration, N = m * (g + |a|) > m * g. At uniform speed (a = 0), N = m * g (no change).

Step-by-Step Solution:

Verification / Alternative check:

Free-body diagram of the person confirms direction of forces and the resulting normal reaction magnitude under different acceleration signs.

Why Other Options Are Wrong:

(a) Uniform speed implies a = 0, so no change. (c) Momentum alone does not determine N; acceleration does. (d) and (e) correspond to upward acceleration, which increases apparent weight.

Common Pitfalls:

Confusing velocity with acceleration; taking downward retardation as “less weight” when it actually means upward acceleration and greater apparent weight.

Final Answer:

With downward acceleration