- #1

magnesium12

- 19

- 0

## 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

*P*in 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 = P

_{in}/P

_{out}

P = Fv

∑F

_{y}= 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 +2T

_{1}

T

_{1}= W/2

Then I tried finding T2 and T3:

∑F = 0 = T

_{2}+ T

_{3}- T

_{1}= 2T

_{3}-T

_{1}

Assuming T2 = T3

0 = 2T3 -T1

T

_{2}= T

_{3}= T

_{1}/2 = W/4

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

_{2}

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!