Question

A fully loaded, slow-moving freight elevator has a cab with a total mass of 1200 kg, which is required to travel upward 35 m in 3.5 min, starting and ending at rest. The elevator's counterweight has a mass of only 940 kg, so the elevator motor must help pull the cab upward. What average power is required of the force the motor exerts on the cab via the cable

Answer 1

425.1 W

Step-by-step explanation:

We are given;

Counter mass of elevator; m_c = 940 kg

Cab mass of elevator; m_d = 1200 kg

Distance from rest upwards; d = 35 m

Time to cover distance; t = 3.5 min

Now, this elevator will have 3 forces acting on it namely;

Force due to the counter weight of the elevator; F_c

Force due to the cab weight on the elevator; F_d

Force exerted by the motor; F_m

Now, from Newton's 2nd law of motion,

The force exerted by the motor on the elevator can be given by the relationship;

F_m = F_d - F_c

Now,

F_d = m_d × g

F_d = 1200 × 9.81

F_d = 11772 N

F_c = m_c × g

F_c = 940 × 9.81

F_c = 9221.4 N

Thus;

F_m = 11772 - 9221.4

F_m = 2550.6 N

Now, the average power required of the force the motor exerts on the cab via the cable is given by;

P_m = F_m × v

Where v is the velocity of the elevator.

The velocity is calculated from;

v = distance/time

v = 35/3.5

v = 10 m/min

Converting to m/s gives;

v = 10/60 m/s = 1/6 m/s

Thus;

P_m = 2550.6 × 1/6

P_m = 425.1 W