Is this thought experiment posible?

AI Thread Summary
The discussion revolves around a thought experiment involving a cylinder submerged in water and a floating object inside it. Participants debate the feasibility of extracting energy from this setup, with some arguing it contradicts the laws of physics and gravity. The concept of using the float's potential energy when dropped is explored, but it is concluded that any energy gained is minimal compared to the energy required to set up the system. Ultimately, the consensus is that the experiment does not yield "free energy" and serves more as a physics exercise to illustrate energy conservation principles. The conversation highlights the challenges of understanding energy dynamics in such hypothetical scenarios.
lerts
Messages
5
Reaction score
0
Take a 10 m long cilinder 2 m wide, cover the top and put it on water in such way that the level inside the cilinder is 10 m above sea level

Take a 1 square m float,and put it inside the cilinder from sea level it will float 10 m up

Now cover the bottom of the cilinder and then uncover the top, take the float and let it drop

Now cover the top of the cilinder and uncover the bottom and repeat the proces
 
Physics news on Phys.org
It doesn't make a whole lot of sense to me - how do you make the first part happen? If you take a cylinder that is covered at the top and submerge it vertically in water, it will remain empty. Are you trying to draw a vacuum in the cylinder?

This sounds like an attempt at free energy, but it doesn't do anything. Water doesn't just spontaneously move up and down like that.
 
Last edited:
Lol,man..

Hey this is Sk8terBoy,well to beging with ur entire question is defieng the basic laws of gravity lol,and the concept of force x gravitationall pull. So basicly just figure out this equationj and your set man : F x G =2f (1g+2g)-4g= A:smile:
 
It is a half well thought experiment to test a perpetuum mobile. And what is funny with this kind of thing is to show that it doesn't work, without using the energy conservation. It is a good physics exercise.

Here it is: When the float is up in the cylinder, it will take the place of its volume out of the water. Then, when you retire the float, the level of the water will be lower corresponding to the amount of the volume of the water. You can take a float which fits the section of the cylinder, it is better for the energy. The best of energy you can take from the cylinder is the equivalent of the cylinder full then empty. Then, no more "free" energy. The only thing to do now is to optimize your system (you are losing energy allowing the float to go freely to the surface of the water inside the cylinder, you could use this energy for something, as to mount some water. And to prove that, even optimized, there is no "free" energy to resell. Of course you must not forget the initial "investment" of energy to heave the cylinder out of the water.

P.S.: I invented a better one decades ago (really hard to prove).
 
Last edited:
It is a half well thought experiment to test a perpetuum mobile.

Lpfr, I don't follow your reasoning. If the float weighs m and is h metres above the sea, then we can reclaim mgh ( less some losses) by dropping it. Raising it is more or less free, so why can't some energy be extracted by this machine ? It is not trying to be a perpetuum mobile.

This sounds like an attempt at free energy, but it doesn't do anything. Water doesn't just spontaneously move up and down like that.
Russ, you haven't grasped the situation either. The water in the column does not move, it is always held up by atmospheric pressure.
 
Last edited:
Mentz114 said:
Lpfr, I don't follow your reasoning. If the float weighs m and is h metres above the sea, then we can reclaim mgh ( less some losses) by dropping it. Raising it is more or less free, so why can't some energy be extracted by this machine ? It is not trying to be a perpetuum mobile.
Yes, you can reclaim mgh, but it is just a bit of the energy that you invested to put the pipe over the water: Mgh/2, if M is the mass of water in the pipe. You put a lot of energy and you reclaim a small part.

Imagine that your float has a density just a bit less than water. To take out your float from the pipe is the same thing as if you take out the same amount of water from the pipe. How much you can reclaim before the pipe is empty? Lo! Just Mgh/2.
 
OK, I realize there's something dodgy about my reasoning, because I now can't see where the energy comes from.

But you only have to raise the column of water once. After that you can do as many drops as you want.

[edit] Lpfr, obviously you're right. I just need to redo the accounting. I think pushing the float under the tube would probably use up any spare energy. Sigh.
 
Last edited:
Mentz114: Maybe you forget that each time that you take the float or the water from the cylinder, the level of the water in the cylinder diminishes?
 
Last edited:
Yep. I also lost all my wits for a while. I think I'll go and immerse myself in some cold water and see if it wakes me up ! Thanks ...
 

Similar threads

Back
Top