Question:
A person of mass 50 kg stands on a weighing scale on a lift. If the lift is descending with a downward acceleration of 9 m/s2 what would be the reading of the weighing scale?
Solution:
The apparent weight on the weighing scale decreases when the lift descends with an acceleration a.
Let W’ be the apparent weight
Therefore,
W’ = R = (mg – ma) = m(g – a)
Therefore, apparent weight, W’ = 50(10-9) = 50 N
Reading of the weighing scale = R/g = 50/10 = 5 kg.