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

• Cri85

## Homework Statement

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.

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## 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 ?

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at step4 I recover Fd,
Wrong. You haven't lowered all the water to the original starting height.

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

Wrong. You haven't lowered all the water to the original starting height.
And what ? the force of pressure is at the bottom of the container, no ? d is the same at step2 and at step4.

And what ? the force of pressure is at the bottom of the container, no ?
You have it correct, "NO."

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

Cri85
I don't understand the question, could you reformulate it please ?

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.

The same.

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

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

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?

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

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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 ?

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.

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.

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