A stationary elevator and its contents have a combined mass of 3000Kg. The elevator is suspended by a single cable. (Assume three significant digits.) Draw a free-body diagram of the elevator and calculate the values of all the forces that are acting on it when at rest. If the elevator is ascending at a speed of 3.0 m/s, what are the values of the forces acting at this point? If the elevator is descending at 3.0 m/s2, what are the values of all the forces acting at this point? For a i did---F_g=mg = 3000*9.8=29400N (Down) F_g=〖-F〗_T F_T=29400 N (Up) For c i did---Net Force downwards=mg-T downward acceleration= Net force/m= (mg-T)/m a = g-T/m T= (g-a)m = (9.8-3)3000= 20400N(Up) Net Force acting upwards= mg= 3000*9.8= 29400N(Down) Is this right? How to do part b?what equation should i use?