What is the power input of a motor pulling up an elevator?

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magnesium12
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Homework Statement


Once under way at a steady speed, the 1100-kg elevator A rises at the rate of 1 story (2.74 m) per second. Determine the power input Pin into the motor unit M if the combined mechanical and electrical efficiency of the system is e = 0.76.

Here is a link to the diagram for the problem:
https://imgur.com/jf58KT6

Homework Equations


e = Pin/Pout
P = Fv
∑Fy = ma = 0

The Attempt at a Solution


First I tried finding the tension in the rope leading to the bottom two pullies:
∑F = 0 = -W +2T1
T1 = W/2

Then I tried finding T2 and T3:
∑F = 0 = T2 + T3 - T1 = 2T3 -T1

Assuming T2 = T3
0 = 2T3 -T1
T2 = T3 = T1/2 = W/4

So then the tension leading to the motor is W/4 = T2

Calculating the power output of the motor:
P = Tv = (W/4)(2.74m/s) = (1100/4 kg)(9.81m/s/s)(2.74m/s) = 7391.84 J/s = 7.39184 kW

Power input of the motor:
e = Pin/Pout
Pin = e(Pout) = 0.76(7.39184kW) = 5.618 kW (Incorrect answer)

I think I'm going wrong when calculating the tensions in the cables. Can someone please explain how to do this correctly?

Thank you!

 
on Phys.org
magnesium12 said:
finding the tension in the rope leading to the bottom two pullies:
You don't need to worry about the tensions. Just think about the net force on the elevator.
magnesium12 said:
P = Tv = (W/4)(2.74m/s)
That's the velocity of the elevator, not of the cable.
 
The problem may be easier to solve if the diagram is ignored. We have an elevator on which work is being done as it is lifted. We know its mass and how rapidly it is being lifted. We have a mechanism which is doing all of the work lifting the elevator. It has a known efficiency. We have a power source providing all of the energy to allow the mechanism to do its work.

Three black boxes -- elevator, mechanism, power source. Is the information already provided enough to solve the problem without peeking inside the black boxes?