# Homework Help: Finding Power of an Elevator

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1. Feb 13, 2016

### CGI

1. The problem statement, all variables and given/known data

2. Relevant equations
I think Power = Work/ Time = (Force x Distance)/Time or Force x Speed.

3. The attempt at a solution
F = ma, so if velocity is constant, a = 0. So that would mean that the force is zero, so would that make the motor have zero power? That doesn't really sound right. If someone could help me out, I would really appreciate it. Thanks in advance!

2. Feb 13, 2016

### Rx7man

F = ma, but you're forgetting about gravity working more on one side than the other.

Gravitational acceleration works on both the elevator and counterweight.. if they're the same mass, it cancels out, so it won't accelerate.. If they're a different mass, the motor will have to do (or absorb) work.

3. Feb 13, 2016

### CGI

Alright. I got for the counter weight, that the sum of forces in the y direction with up as positive, is T - W = 0 or T = W for the FBD of the counter weight.
For the Elevator, I have that 3T - W = 0 or that 3T = W for the FBD at the elevator.

Would these be correct?

4. Feb 13, 2016

### Rx7man

I'm a little lost.. What is T and W here, and why 3T?

If it's not accelerating, the motor has to do work on the difference of the two weights... so moving the elevator 5000 lb elevator up is moving the 2200 lb counterweight down, The net weight you're lifting will be 3300 lbs @ 15ft/s

5. Feb 13, 2016

### CGI

Oh wow, I'm sorry. T in my FBD stands for tension in the cables and W stands for weight. In the elevator, I thought that it would be 3T - W = 0 because there were three cables lifting that elevator up. As for my counterweight, I thought it would T - W = 0 since there is only one cable so the tension would have to be equal to the weight.

But okay, if the net force of gravity on the system is 3300lbs, could I just multiply 3300 by 15 and have the Power generated by the motor?

6. Feb 13, 2016

### Rx7man

Forget about the tension in the cables.. simplify it... if you look closely at the diagram, you'll notice it's a 1:1 movement ratio for the elevator : counterweight

The tension on the cable to the motor is half the weight of the elevator minus counterweight, but the tension in the cable is irrelevant to work.. the motor needs to move twice as far at half the load which means the work stays the same.

Yes, the motor is making 3300lbs * 15ft/s of work :)

7. Feb 13, 2016

### CGI

Oh okay that makes things so simple! So when it says that there is acceleration upward, how would that change the power from the motor?

8. Feb 13, 2016

### Rx7man

The motor will have to do more work.. and it'll need to apply more force to accelerate both masses.

Just as an exercise (and it'll give you a hint), how fast would the elevator accelerate if the cable to the motor broke?

9. Feb 13, 2016

### CGI

It would just fall so wouldn't it just be the acceleration due to gravity?

10. Feb 13, 2016

### Rx7man

Nope... there's gravity acting on both weights, so the net force will be the difference of gravitational forces, but both weights need to accelerate at the same speed (one up, one down evidently)

11. Feb 13, 2016

### Rx7man

Well, yeah, OK, the acceleration will be due to gravity, but the acceleration will not be 32ft/s^2

12. Feb 13, 2016

### haruspex

3300?

13. Feb 13, 2016

### CGI

Oh that's right, the net weight would be 2800lbs.

And I don't think I'm understanding the way the acceleration would affect the elevator. I am so sorry :(

14. Feb 13, 2016

### haruspex

You cannot simply add an extra power term to handle the acceleration. The power required for an acceleration depends on the current speed.
Easiest is to use power = force x velocity. What force would be required from the motor to achieve the desired acceleration?

Last edited: Feb 13, 2016
15. Feb 14, 2016

### CGI

If I used F = ma with (a) as 3, would the force needed just be 3 x 5000?

16. Feb 14, 2016

### haruspex

In the equation F=ma, it is important to remember exactly what F represents.

17. Feb 14, 2016

### CGI

F in this case is just the applied amount of force on the elevator, no?

18. Feb 14, 2016

### haruspex

19. Feb 14, 2016

### CGI

Okay so F stands for the vector sum of forces on an object. So if I used F = ma, I would have to add up all the forces acting upon the elevator, which would not just be the force of gravity. Is that headed somewhere in the right direction?

20. Feb 14, 2016

### haruspex

Yes. F=ma will tell you the net force, and from that you can deduce the force provided by the motor.

21. Feb 14, 2016

### Rx7man

the total power needed will be the power needed to lift it at it's current rate, plus additional power to accelerate it... Don't forget the effects of the counterweight either.

22. Feb 14, 2016

### haruspex

As I wrote in post #14, that doesn't really help. The power required to accelerate depends on current speed, not just on the desired acceleration rate. Finding the force that would give the required acceleration is as simple a route as any.

23. Feb 14, 2016

### Rx7man

You are right, but sometimes separating things helps me visualize how to solve the problem (it works for me)