Suppose a non compressible container. Is it’s buoyant force the same at 10 feet below the water surface as it is at 100 feet? If it is different. Is it a very small difference? I suspect it is the same. Water being only a tiny bit compressible.
It’s a very small difference, because, as you said, water is quite resistant to compression.
The object whose buoyancy you’re evaluating may well be more compressible than water. As the object gets deeper, it grows smaller under the increasing pressure - and thus more dense. If the object’s density changes more rapidly than water’s (and since water compresses very little, that’s not uncommon), then bouyancy will change with depth. robby mentioned in another thread recently that this is a noticeable effect on a submarine. “The deeper you go, the heavier you get.”
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Good point - it’s not just the density of the water that matters.
“Suppose a non compressible container” ← I thought this was the question. Why talk about things like submarines when the question specifically excluded them? 
Thanks for confirming my thought. Another question for this instance.
Suppose you have a round port at the bottom of a 100 foot tower filled with water. You want to push a barrel through this port. ( It has a perfect no friction seal around the barrel ). The area of the port is say 1 square foot. So you can calculate the pressure upon that 1 square foot. As you push the barrel through, more of the surface of the barrel is exposed to pressure inside the water tower. But is the pressure to push against, still only that of the 1 square foot port area? I suspect that the pressure on the sides of the barrel have no effect on the force needed to push it. Just that 1 square foot of port area pressure.
Right, the pressure on the sides of the barrel will (almost) cancel out. I put “almost” there, because the pressure on the bottom surface will be greater than that on the top surface, due to its greater depth, so there will be some upward force as well (though that won’t make it any harder to push the barrel in). And this upward force is exactly what buoyancy is.
Thanks again for confirmation.
I seldom ask or try to answer questions here. But I do enjoy just reading a lot of them. That also gives me a good idea of who knows what, for good answers when I do ask.
I took “container” to mean the container holding the water, rather than the object within the water whose buoyancy was being tested.
If that’s not correct, perhaps the OP could clarify.
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I took the “container” to be the buoyant object being immersed in the water. In the OP the water seems to be in its natural state, with an air/water interface at the top and the Earth’s gravity well beneath.
As to Gus’ point, the whole point of the quibble is that a non-compressible container is like a spherical cow: it can’t actually exist in nature. In any physics question it’s OK to explicitly say we’re ignoring this or that factor of reality for simplicity. But this can go very far wrong when the questioner decides to ignore what’s actually the meat of the matter. Doubly so if it’s done implicitly or inadvertently.
In this case, as brad_d said in post #3, it’s the differential compressibility of the water and the test object immersed in the water that will cause any buoyancy change plus or minus with increasing depth. So we can get either result falsely depending on whether the simplifying assumption is that water is incompressible or that the test object is incompressible.
From a practical point of view were I tasked with developing an non-compressible container I would make one that was hollow and fill it with the most non-compressible fluid I could come up with. Maybe water, and that makes the question trivially easy, since the density of the water inside the container would match that of the water outside the container.
Yeah. The ideal non-compressible container for this experiment is a ziplock bag full of water with a pinhole in it.
How much air must be added into a SEALED 55 gallon barrel to cause it to float on end at the depth of 5 feet or less in fresh water? Back in the days of 55 gallon barrels supporting most private docks, these barrels getting away and just floating with the currents were a major problem for outboard engines.
I specified that the container be non compressible to get at the most basic answer. So there might be fewer diversions in the answers, such as it would be crushed, compressed and so forth. Which would lead to great variation in the actual buoyancy at depths. I knew I would likely be further questioned as to the details of the container unless I precluded such things about it. I did not specify that it is not filled with lead.
The situation I am asking the question about, does not require such precision of details to get the useful answer.
That does not mean I don’t find the usual meandering discourse that happens in these things to be interesting and often informative. Especially when everyone goes about it nicely.
I did investigate myself, before asking. But wanted some further back up.
Thanks again.
Google says an empty steel 55 gal drum weighs about 40 lbs. They also say 1.2 gallons of water weighs 40 lbs for round numbers. For round numbers air at sea level pressure weighs nothing compared to water or steel.
Putting that together a 55 gallon steel drum that’s got 1.2 gallons of air in it and the rest is full of the local water will have neutral buoyancy. Note that a “55 gallon” steel drum has more internal volume than that. It’s meant to hold 55 gallons of product plus some airspace on top. Typically about 2.5 gallons worth of air space.
So the usual scenario goes like this: is a dock is built using completely air-filled drums to support the load. Over time water will slowly leak in and push air out. Given enough time with no maintenance the barrels will fill with water and the dock will sink.
If a barrel gets loose before all but 1.2 gallons of water leak in it’ll float partly above the water. If it gets loose after there’s less than 1.2 gallons of air left it’ll sink to the bottom. Somewhere in around that 1.2 gallon figure the barrel will be near enough to neutral buoyancy to float tangent to or just beneath the surface.
I believe that in practice very few barrels will spend much time truly beneath the surface and not on the bottom. Once they get slightly below the surface they slightly compress which leads to two things: decreased buoyancy and increased rate of water infiltration. That’s not a long-term stable spot to sit.
What I *can *believe is that lots of barrels can float low enough above the rough surface as to be unseen until hit. After the boat runs over it it pops up to the surface astern and the person says “Wow, I hit a submerged barrel.” IMO nope, at least not usually. The person hit an awash barrel he didn’t see out ahead.
Thanks for that explanation.
I wanted to bring up water temperatures that are significantly different on some days at just a few feet of depth change, time with warm on top at various levels and sometimes the cold will be at the top and only a few feet thick. Some strange conditions on a Corps Of Engineers hydro power generating lake. But that is getting a bit picky and doing the “Yeah Butt…” thing which is not really cool.
I wanted to inject that at times in the past with many summers living at the lake all summer, things that only happen once in a while get noticed when you have several large families with 6-8 kids who are real water babies.
I was very lucky to be in that situation as I managed to live through it and the experiences have held me in good stead of the course of the years.
Once again, thanks for the logical & understandable explanation. You are very good at doing that.
Something went wrong with those numbers somewhere. 1.2 gallons of water is only about ten pounds.
@Chronos: Good catch.
Yeah, for <reasons:stupid> I screwed that up by a factor of exactly 4 net of my rounding. The right answer isn’t ~1.2 gallons, but rather 4 x ~1.2 gallons = ~4.8 gallons. Or very roughly a barrel that’s filled with 8% air and 92% water will be real close to neutral buoyancy.
@Gus: Glad to help. 'Twas fun. Net of my screw-up. :smack:
You’re right that a lake that’s real warm on the surface and a bunch colder a few feet down will tend to trap some stuff a few feet beneath the surface. Stuff that’s just heavy enough to sink in the less-dense warm water, but just light enough to float on the more-dense cold water. The difference is pretty subtle: 10 degrees C or 18 degrees F of temperature difference around typical lake temps gives about 5 parts per thousand change in water density.
Given that for very round numbers a barrel full of water weights about 400 lbs, even a hefty ~20 degree F difference between the warm and cool layer will only supply 2# of differential floating. IOW, if it’s floating nicely tangent right below the surface and more than 2# ~= 1qt more water leaks in, it’ll free-fall right through a 20F colder layer. But if only a 1# ~= 1 pint leaks in, now it’ll happily sit someplace down in the middle of that warm-ish layer. At least until the air inside cools and shrinks so the pressure reduction inside plus the increased water pressure outside due to the depth leaks the other pint into the barrel more quickly. Then down she goes & no stopping until the bottom.
See Thermocline - Wikipedia for more.
My comment about stuff not floating fully submerged for long also needs some context. If you set a sealed barrel full of air loose on the surface of the lake it might take (total WAG) 2 years to leak enough water to sink fully. During that two years it’s out there as a hazard to navigation it might well be floating near but below the surface for a couple weeks. Compared to two years, two weeks is a small fraction of the time. But may still be plenty long enough for boaters to run into it once or twice on a heavily traveled section of lake.
Depending on what grows in the lake water your barrel might also be accreting mussels or weeds or something. Weeds & algae weigh very close to what water does; some even float. So not much net impact from those. Conversely, mussel shells are heavy.