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I realize this isn't possible, but I can't see why not, especially if you change the model a little bit so that the balls simply travel through a tube of water on the way up, rather than exactly this model.

Please be clear and detailed. I've heard explanations like "the balls wouldn't move" but that doesn't do it for me - I really do not see why the balls on the right would not be pulled/pushed up, and the rest of the chain wouldn't continue on.

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    $\begingroup$ A good question for learning physics! As an occasional physics tutor, it keeps me sharp to explain why some crackpot scheme won't work, or lead student to an insight on that. BTW, nice illustration, but the greedy guy belongs on the personal finance SE site, no? $\endgroup$
    – DarenW
    Nov 17, 2010 at 4:37
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    $\begingroup$ 1. Build this machine. 2. ???? 3. PROFIT!!! $\endgroup$
    – Vortico
    Dec 21, 2010 at 21:43
  • $\begingroup$ There was a similar question posted here: physics.stackexchange.com/questions/244880/… It was posted afterwards, but it deals with a real company that claims to build such machines for real customers. Fascinating! $\endgroup$
    – Ilja
    Apr 16, 2016 at 8:33
  • $\begingroup$ The funniest thing to me about this diagram is the (labeled) seal on the right, preventing water in the column from "escaping" to the ocean, which is pictured as having the same water level. $\endgroup$
    – Gregor Thomas
    Jun 15, 2017 at 16:32

3 Answers 3

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The balls are entering the water well below the surface. The pressure there is much higher than at the surface. The work needed to push the balls into the water at this depth cancels the work gained when they float back up.

We can ignore the gravitational force on the balls since gravity pulls down as much as up as you traverse the loop.

Mathematically, if the balls enter the water at depth $d$, the pressure is $g \rho d$ with $g$ gravitational acceleration and $\rho$ the density of water.

The work done to submerge the balls is then the pressure times their volume, or $W_{ball} = g \rho V d$.

The force upwards on the ball is the weight of the water they displace which is $g \rho V$, and the work the water does on the balls is this force times the distance up that they travel, or $W_{water} = g \rho V d$.

The work the ball does on the water is the same as the work the water does on the ball. No free energy.

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    $\begingroup$ +1 It took me some lost sleep to realize that the force needed to push the ball into the water was equal to the bouyancy of a volume of air equal to a cylinder the size of the ball and going all the way from the bottom to the top. So if the bouyancy comes from a volume of air less than such a cylinder, such as a string of balls, it can't even start. $\endgroup$
    – Mike Dunlavey
    Nov 17, 2011 at 13:57
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    $\begingroup$ ... or to put it another way, it will actually run the wrong way, as water runs in to fill the empty container! $\endgroup$
    – Mike Dunlavey
    Nov 17, 2011 at 14:00
  • $\begingroup$ I solved this problem using a cuboid. I actually changed the design of the 'project' $\endgroup$
    – raindrop
    Dec 21, 2012 at 5:11
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The energy needed to submerge a ball is equal to the energy gain from other ball to emerge from the water on the other side, so any waste on friction drives the process impossible.

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The energy needed to push a sphere into the water at the dotted line is $$E=Fs=\int PA ds=P_{water}V_{sphere}$$

I leave this incomplete because Mark Eichenlaub already answered the question. Note that since it required some energy to put the balls in the configuration shown in the questioner's diagram, essentially you'd just get that energy back by letting the water flow into the empty space and cutting the air balls so that they can float to the top. If left in its current state it should remain stationary due to force equilibrium. (no perpetual motion).

enter image description here

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