Seeing a ship canal lift in TV spawned the following thought:

"Oh my, this sure is a heavy ship and it can be lifted by simply adding water. What a marvellous machinery. After all, I could lift that ship with my teacup and a bit of water in it"

Which in turn spawned the following thought experiment:

Given some basin of know area, I can pour my teacup filled with water into it. The water level will be raised by a certain amount, depending on the volume and area. Seems legit so far.

The water level will always rise by that amount, no matter what is floating in the basin. I have my rubber duck floating in my basin and it is raised by the teacup amount after pouring. Now let's add a big lead duck (or a ship if you want, but ducks are that much cooler, aren't they?) that floats in my basin. I still raise all that floating stuff by the teacup amount, even if I added the ship.

And this is where things started to feel a bit dubious.

How is it possible that I "invest" a (constant) teacup worth of potential energy, yet "gain" the potential energy of rising an arbitrary big mass that is floating in my basin by the constant amount caused by the water from the tea cup?

That right there doesn't sound dubious any more, that sounds plain wrong to me. Because wouldn't that allow the teacup ship lifter gain more potential energy out of the process than he needs to re-rise the water into his teacup?

Where did my thought experiment took a turn for the worse?


I try to describe my problem in more detail: A big ship needs more space, so the basin gets bigger, right?

I'm afraid that's not the case and pretty much the core of my problem: The geometric properties seem to be independent enough to allow a "gain" in potential energy.

Here's how: Let's say the basin area is constant. This means the water level rises a constant amount (depending on how much water is in the teacup) Now how does the constant basin support "arbitrary large masses"? By varying its depth. In order to make a bigger mass float, it has to displace more volume (of water). As the area is fixed, the ship/boat/duck can extend its volume in the third dimension (down) to displace more water, thus supporting more mass to float, yet being raised to the same height as the basin area stays constant.

  • $\begingroup$ Your teacup actually has a tremendous amount of potential energy, just not in a particularly useful form. $\endgroup$ – Dave Coffman Mar 25 '15 at 20:33
  • $\begingroup$ Not sure if this is trivial or not, but you're doing work on the water to add it to the basin $\endgroup$ – Sean Mar 25 '15 at 20:39
  • $\begingroup$ @DaveCoffman I'm not sure I understand you, could you please elaborate or answer? $\endgroup$ – Name Mar 25 '15 at 21:15
  • $\begingroup$ @Sean what exactly does that imply? I guess doing work on the water could heat it up, but I don't know how this explains a heavier mass being lifted by a smaller one, both masses being pretty much independent. $\endgroup$ – Name Mar 25 '15 at 21:21
  • $\begingroup$ Here's one thing you're missing: When you dump your teacup full of water into the basin it doesn't just raise the level of the ship or the toy duck: It also raises the level of the entire surface of water in the basin. $\endgroup$ – Solomon Slow Mar 25 '15 at 22:01

How is it possible that I "invest" a (constant) teacup worth of potential energy, yet "gain" the potential energy of rising an arbitrary big mass that is floating in my basin by the constant amount caused by the water from the tea cup?

The mass cannot be "arbitrarily big". Since it is floating, it has a net density that is less than that of the water. Therefore the mass of the ship is identical to the mass of water that is "missing" from the lake/pond.

If your teacup takes an empty pond from height $y$ to $y'$, then it would take one with a ship floating in it to the same height. The mass of an empty pond at height $y$ and one with a shipping barge in it are identical.

If the barge weighed more than the empty pond, it would not be floating.

To address the specific concern of raising a ship that is longer/deeper in a pond, let's imagine that we have two cylinders in a pond. Each has identical densities and cross section, but one is twice the length (and twice the mass) of the other. As both are floating, the larger one extends deeper into the pool. In fact, the lowest point will be twice as deep as the other (independent of the actual density).

For a moment, imagine that the water does not flow, but maintains its location as we directly lift the cylinders (like it's jello). We raise both cylinders by $1cm$. Since one is twice the mass, it takes twice the total energy to raise it that amount.

But before we let the water move around, we fill in the void underneath the cylinders with the water from our teacup. As we are at the shore, dropping the water down to the void created by moving them will release energy. And in fact dropping the water down to the void under the more massive cylinder will release twice the energy. So raising a taller ship is not a problem, because water drops down farther to displace it

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  • $\begingroup$ I think the OP has explained this in his/her edit. We can make the pond as deep as we want, while maintaining the cross-sectional area. In this manner, we will still raise the pond by the same $y$ when we add the teacup of water. However, in this situation, the shipping barge can be arbitrarily massive since we can make the pond deep enough to have enough water for the barge to displace to float. $\endgroup$ – Gerard Mar 26 '15 at 3:39
  • $\begingroup$ +1 for the jello theory (I can see a whole new field of physics evolving) I guess I have to crunch some numbers to see if this works out. $\endgroup$ – Name Mar 27 '15 at 16:32
  • $\begingroup$ I still think it's easier to think of a pool at a particular height, whether empty or with something floating on it, as always having the same mass. Then to raise that mass by a certain amount will always take the same energy. $\endgroup$ – BowlOfRed Mar 27 '15 at 16:34
  • $\begingroup$ @BowlOfRed you mean as if the whole thing was frozen, then lift it to a certain height and add the teacup water underneath? Maybe that makes the calculation easier indeed. $\endgroup$ – Name Mar 27 '15 at 16:39
  • $\begingroup$ Exactly. It's easy for your eye to see the ship being lifted. But an equivalent way to look at it is that the entire body of water is being lifted. As the boat weighs exactly as much as the water it displaces, then there is no difference in energy for lifting the boat or lifting the water that would be there if the boat were missing. $\endgroup$ – BowlOfRed Mar 27 '15 at 16:42

Think about a body floating in a cylinder of water. The body has a diameter that's only a little less than the cylinder. If it's a small body in a small cylinder, like your rubber duck, your teacup full of water lifts it up a lot. If it's a big body in a big cylinder, like your ship, your teacup full of water lifts it up a little. There's nothing dubious about it. Like everything in physics, it's simple once you see how simple it is.

  • $\begingroup$ This is a good thought, that also came to my mind. The problem is that the cylinder doesn't have to become bigger (in area) to support a bigger mass that floats on the liquid in the cylinder. I updated the question to explain that. $\endgroup$ – Name Mar 25 '15 at 21:05

Well, I guess see floating things are yet to reach the ground level, because ofcourse an object submerged at the bottom most part of the seabed would have less potential energy than the object which is floating right? Implying that floating objects do have potential energy

  • $\begingroup$ Try to put some thought and effort into your answers. Short comments should be left in the comments section. $\endgroup$ – hft Mar 26 '15 at 6:12

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