Correctly identify the apparent weight of a lift which is moving upwards, with acceleration?
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → The apparent weight of the lift shall be more than its actual weight
The apparent weight is the push of the floor on a body in the lift, the reading a weighing machine on the floor would show. When the lift accelerates upwards, the floor must push harder than the weight to give the body that acceleration.
Forces on a body of mass \(m\): its weight \(mg\) downwards and the reaction \(R\) of the floor upwards
The net upward force produces the acceleration \(a\): \(R-mg=ma\)
So \(R=m(g+a)\), more than the actual weight \(mg\)
Why the other options are incorrect
Option (2) The apparent weight of the lift shall be less than its actual weight — Incorrect. This happens when the lift accelerates downwards: \(R=m(g-a)\).
Option (3) The apparent weight of the lift shall be equal to actual weight — Incorrect. This happens when the lift is at rest or moves with uniform velocity (\(a=0\)).
Option (4) The apparent weight of the lift shall be zero — Incorrect. This happens only in free fall (\(a=g\)), when the body feels weightless.