The pulley system for an elevator is shown in the figure on the left. This mechanical system consists of 5 pulleys and 2 electric motors. The maximum allowable weight for the elevator is 5 kip.

a. What is the minimum required force for each motor?

b. If one of the electro-motors is removed and instead the cable is directly attached to the wall, what will be the minimum required force for the remaining single motor?



a. As with all statics problem, a free-body diagram will assist in solving the problem. In this example, all forces acting on the elevator cabin is first analyzed. The 5000 lb weight is divided evenly between the cables due to symmetry. Consequently the force of each cable will be

     P = 5000 / 2 = 2,500 lb


  Next, assume the cable force in each motor is F. The diagram on the left shows the pulley system with the external forces and with the elevator forces.


The free-body diagram for pulley number 2 is shown on the left. Note that the free-body diagram for the pulley number 4 would be similar. Summing the forces in the vertical direction gives,

     ΣFy = 0
     0 = 2 T2 - 2,500 lb
     T2 = 1,250 lb

Since the cable is continuous,

     T2 = F = 1,250 lb


b. If one of the electro-motors is removed, then the cable is directly attached to the wall. The free-body diagram and the solution procedure would be the same. In this case the reaction force at the wall is equal to 1,250 lb. Therefore removing one of the motors will not affect the final answer. Also, by removing one of the electro-motors, both cost and energy will be saved for the consumer and manufacturer.