A lift of mass 200 kg is moving up with an acceleration of 4 m`cdot s^(-2)`. What is the tension in the lift cable? If the lift moves down with the same acceleration, what will be the tension in that case?
[ g = 9.8 m`cdot s^(-2)` ]

Explanation

Let the tension in the cable be T when the lift is moving with an acceleration a. The equation of motion for the lift is,
T - mg = ma
or, T = m(g +a)
Substituting the values for m, g and a in equation (1), for upward motion,
T = 200(9.8 + 4)
= 200 `xx 13.8` = 2760 N
For the lift moving downwards, a = -4 m`cdot s^(-2)`,
T = 200(9.8 -4) = 200 `xx` 5.8 = 1160 N.