Newton's Laws and tension problem

Question

An elevator of 4850 kg is to be designed so that the maximum acceleration is 0.068g. What are the maximum and minimum forces the motor should exert on the supporting cable?

Answer 1

F_max = = mg + ma = m(g + 0.068g) a = m(g + 0.068g) = 1.068mg = 1.068·9.81m/s²·4850kg = 50814 N

F_min = mg - ma = m(g - 0.068g) = 0.932mg = 0.932·9.81m/s²·4850kg = 44343 N