A body hangs from a spring balance supported from the roof of an elevator.
(a) If the elevator has an upward acceleration of `2.45 ms^(-2)` and the balance reads 50N, what is the true weight of the body?
(b) Under what circumstances will the balance read 30N?
(c) What will be the balance reading if the elevator, cable breaks?

Explanation

When the lift moves upward with an acceleration, then the apparent weigth is
`W'=W+ma or W'=W+W/g a =W(1+(a)/(g))`
Here `W'=50N, a=2.45 ms^(-2), g=9.8 ms^(-2)`
`50=(1+(2.45)/(9.8))=W xx 5/4 implies W=4/5 xx 50 = 40 N`
(b) Again if a is upward acceleration, then
`W'=W(1+a/g) gives 30=40(1+a/g)`
`implies (a)/(g)=30/40 -1 =-1/4 implies a=-g/4`
i.e. lift has downward with acceleration `g//4`.
(c ) when the elevator cable breaks, the lift falls freely with acceleration t.
Therefore, apparent weight `W'=W(1-g/g)=0N`.