A person of mass m is on the floor of a lift.The lift is moving down with an acceleration ‘a’. Then :
a) the net force is acting in downward direction and is equal to mg
b) the force mg must be greater than reaction force
c) the man appears to be lighter than his true weight by a factor (a/g)
As observed from the ground , net force = (mg-N)=ma
As observed from the lift , net force = N+ma -mg=0 ⇒ N = mg-ma < mg
Apparent weight = N < mg