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?
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?

When the lift lowers with an acceleration a, the apparent weight on the weighing scale decreases.

W’ denotes the apparent weight.

Therefore,

W’ = R = (mg – ma) = m(g – a)

As a result, W’ = 50(10-9) = 50 N is the apparent weight.

The weighing scale reading is R/g = 50/10 = 5 kg.