A man of mass "m" is standing on the floor of the lift. Find his apparent weight when:
(i) elevator is rest
(ii) elevator is accelerating upwards
(iii) elevator is accelerating downwards.

Explanation

(i) When the elevator is at rest: The acceleration of the man is zero. Hence the net force acting on the man is zero. When Newton's secons law is applied with respect to inertial frame on the man,
`vec(F)_(G)+vec(N)=0`
`-mghat(j)+Nhat(j)=0`
By comparing the components, we can write
`N-mg=0(or)N=mg`
Since weight, W = N, the apparent weight of the man is equal to his actual weight.
(ii) When the elevator is accelerating upwards: If an elevator is moving with upward acceleration then with respect to inertial frame when Newton's second law is applied on the man,
`vec(F)_(G)+vec(N)=mvec(a)`
The above equation can be written in terms of unit vector in the vertical direction,
`-mghat(j)+Nhat(j)=mahat(j)`
By comparing components,
`N=m(g+a)`
Therefore, apparent weight of the man is greater than his actual weight.

(iii) When the elevator is accelerating downwards: If the elevator is moving with downward acceleration `(vec(a)=-ahat(j))`, by applying Newton's second law on the man, we can write
`vec(F)_(G)+vec(N)=mvec(a)`
In terms of unit vector in the vertical direction, we get the above equation as
`-mghat(j)+Nhat(j)=mahat(j)`
By comparing the components,
`N=m(g-a)`
`therefore` Apparent weight `[W=N=m(g-a)]` of the man is lesser than his actual weight.