Question

(b) Discuss the apparant weight of a person in a lift/elevator when-
(i) lift is stationary (ii) left is moving up wards
(iii) lift is moving downwards.​

Answer 1

Answer:

When lift accelerates up, pseudo force acts downwards, hence it increases apparent weight. W 1>mgWhen lift is stationary, W 2=mgWhen lift falls freely, it accelerates with g downwards, causing an upwards pseudo force in the frame of the lift equal to mg. Hence total force is 0. So weight is 0. W 3=0

When lift accelerates down, pseudo force acts upwards, hence it decreases apparent weight WW4 <mg, but also the acceleration is less than g,thereforeW 4=m(g−a)>0

Hence, W 3<W 4 <W 2<W 1

Answer 2

ANSWER

Apparent weight of a person in lift will be m(g+a) while moving upwards whereas it will be m(g−a) while moving downwards.

Hence, as per the given information,

m(g+a)=2m(g−a)

g+a=2g−2a

Hence, a=

3

g