A lift of mass 400 kg is hung by a wire. Calculate the tension in the wire when the lift is (a) at rest, (b) moving upward with a constant acceleration of 2.0 `m//s^(2)` and (c) moving downward with an acceleration 2.0 `m//s^(2)`.

Explanation

Lift at rest T = W = mg = 4000 N
(b) Moving up with 2 m/s, a = 0 ,
T = W = mg = 4000 N
( c) Moving up with a = 2m/`s^(2)`
T = mg + ma = 4000 `+ 400 xx 2 = 4800 N`
(d) Moving down with a = -2m/`s^(2)`,
T = mg - ma = 4000 - 400 `xx` 2 = 3200 N