Question
A man of mass stands on a weighing machine inside a lift. The lift accelerates upward at , then moves with constant velocity, then decelerates at before stopping. Find the reading of the weighing machine in each phase. Take .
Solution — Step by Step
The weighing machine shows the normal force pressing on it, not the man’s actual weight. So we need in each phase from Newton’s second law on the man.
Taking up as positive: .
So .
The machine reads .
Acceleration is zero, so , reading . The man reads his true weight.
Deceleration means acceleration points downward (opposite to velocity). So .
, reading .
Final answer: Readings are in the three phases.
Why This Works
The weighing machine never measures gravity directly. It measures the contact force between you and itself. In a non-inertial frame (accelerating lift), that contact force adjusts to provide the net force needed for the lift’s acceleration.
A neat sanity check: if the lift were in free fall (), — apparent weightlessness. If it accelerated up at , — you feel twice as heavy. Our problem sits between these limits.
Alternative Method
We can use the lift’s frame and add a pseudo-force on the man. Equilibrium in the lift’s frame gives (up phase) directly, no equations of motion needed.
Common Mistake
Students often confuse “decelerating while moving up” with “accelerating down”. They are the same thing direction-wise — acceleration vector points opposite to velocity. The sign of in must be negative here, giving a lower reading. Many students plug in and get the wrong for phase 3.