Timeline for How does potential energy work out for floating things?
Current License: CC BY-SA 3.0
7 events
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| Mar 27, 2015 at 16:42 | comment | added | BowlOfRed | 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. | |
| Mar 27, 2015 at 16:39 | comment | added | Name | @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. | |
| Mar 27, 2015 at 16:34 | comment | added | BowlOfRed | 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. | |
| Mar 27, 2015 at 16:32 | comment | added | Name | +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. | |
| Mar 26, 2015 at 4:23 | history | edited | BowlOfRed | CC BY-SA 3.0 |
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| Mar 26, 2015 at 3:39 | comment | added | Gerard | 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. | |
| Mar 25, 2015 at 23:11 | history | answered | BowlOfRed | CC BY-SA 3.0 |