The weight with a fluid: center of mass or force at bottom?

1. Dec 24, 2014

Cri85

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

It is a question I asked to myself because I need to reply to another one. I imagined this case:

http://imageshack.com/a/img912/7730/R964Vq.png [Broken]

The step2 needs the energy E1, the step3 needs nothing because the gravity is 0 (I imagine the device between the Earth and the Moon), the air container is fixed, at step4 I recover E1, at step5 I recover an energy E2. At final I have -E1+E1+E2 = E2.

2. Relevant equations

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3. The attempt at a solution

With a solid I'm agree to put the force of weight in the center of mass and if I do that the energy is conserved. But why I need to do that with a fluid ? The force from the weight is at the bottom of the container not in the center of mass. If I use the true force in the bottom of the container, I need to add an upward force ?

Last edited by a moderator: May 7, 2017
2. Dec 24, 2014

Bystander

Wrong. You haven't lowered all the water to the original starting height.

3. Dec 24, 2014

Cri85

Sorry, I changed my message because it's not Fd it's an integration of F.

And what ? the force of pressure is at the bottom of the container, no ? d is the same at step2 and at step4.

4. Dec 24, 2014

Bystander

You have it correct, "NO."

5. Dec 24, 2014

Cri85

6. Dec 24, 2014

Bystander

What is the increment of weight, dw, for each increment of volume in your container? It's not zero at the top, is it?

7. Dec 24, 2014

Cri85

I don't understand the question, could you reformulate it please ?

8. Dec 24, 2014

Bystander

Take nine duplicates of your "box." Arrange them on the table with the original. Place 1/10 of the water in each box. What's the mass of the water? What's the force the water exerts on the table? In terms of the original force.

9. Dec 24, 2014

Cri85

The same.

10. Dec 24, 2014

Bystander

Now, how much work are you going to have to do to put all the water back into the original box?

11. Dec 24, 2014

Cri85

It depends how boxes are arranged. When the box move up or move down (in my cycle), nothing moves inside the container.

12. Dec 24, 2014

Bystander

Last chance: blow it, and this thread gets locked, and you get labelled as a perpetual motion/over unity nut.
How much work do you have to do to lift the water spread among ten boxes back into the original box?

13. Dec 24, 2014

Cri85

Weigth*h/2 but I'm not sure, with h the height of the container

Last edited: Dec 24, 2014
14. Dec 25, 2014

Cri85

Someone could explain to me why the force F is apply in the center of mass and not at the wall ? Why the work at step 4 is not the same than in the step 2 ?

15. Dec 25, 2014

Bystander

All right! Do it your silly way, BOTH directions. Don't use COM lifting, and shift to bottom of mass when lowering. Pick ONE OR THE OTHER.

16. Dec 25, 2014

Cri85

I understood ! if I take a lot of small parts, it's logical, thanks. I thought the force was the same for an altitude for one container with air or not. But it's not the same weight. So I can reply to my another question: if the weight is the same with a fixed bubble inside ?: the answer is no, the weight is lower. Now, with a bubble that is moving up, in one part I would say it's lower and in the other part when the bubble accelerates to the top the water accelerates to the bottom so the weight must be a little higher I don't know the sum. But it's ok for this question, thanks for your patience. I asked myself if I move down the container to the center of the Earth and place the container with air where g=0.

Last edited: Dec 25, 2014