### Author Topic: Buoyancy  (Read 3174 times)

0 Members and 1 Guest are viewing this topic.

#### leafy

• CoT Emeritus
• LV10 31337 u53r (Next: 2000)
• Posts: 1554
• Rating: +475/-97
• Seizon senryakuuuu!
##### Buoyancy
« on: December 14, 2012, 02:22:59 am »
Hey guys, this might be a pretty trivial problem, but I've been puzzling over it for a while. It goes something like this - let's say you have a perfectly rectangular block of something light that you place perfectly flat at the bottom of an equally flat container, then fill it up with water. Since there's no water under the block, there shouldn't be anything to push back up on it, so it should just stay there.

It seems pretty counterintuitive, and I can't really imagine it happening. But I've also been thinking that the normal force at the bottom of the container should negate that downward fluid force, and make it rise again. But what if you have a container exactly equal to the dimensions of the block except for the height, you put the block at the bottom, and you fill the top with water such that only one side is touching the water. Are these two situations equivalent?

Of course, there could be other reasons why the block would stay there (negative pressure, hydrogen bonds, etc.), but negating all that, what should happen?
In-progress: Graviter (...)

#### Builderboy

• Physics Guru
• CoT Emeritus
• LV13 Extreme Addict (Next: 9001)
• Posts: 5673
• Rating: +613/-9
• Would you kindly?
##### Re: Buoyancy
« Reply #1 on: December 14, 2012, 03:42:55 am »
It doesn't sound that counter-intuitive to me, it sounds like a suction cup!  Since there is no way for the water to get underneath the block, there are exactly 2 forces acting on the block.  The force of the water (which goes straight down because water is only pushing on the top and sides), and the normal force.  The normal force in this case would be equal and opposite to the force of the water, meaning the block would stay in it's place.  The block would only rise if an external force was applied to it that was equal or greater than the force of the water.  All of this sounds a lot like the behavior of a suction-cup to me

#### AngelFish

• Is this my custom title?
• LV12 Extreme Poster (Next: 5000)
• Posts: 3242
• Rating: +270/-27
• I'm a Fishbot
##### Re: Buoyancy
« Reply #2 on: December 14, 2012, 03:44:11 am »
You can actually test this for yourself in real life. What happens with a perfectly rigid block is that there's a differential pressure. The top of the block has the weight of the water on it, the bottom of the block has only the static pressure of the container's bottom (which is precisely equal in magnitude to the force exerted downward) , and the sides have evenly distributed pressure all around. There's no net upward force! The block will stay on the bottom unless water somehow gets underneath to exert a buoyant force. Remember that archimedes' principle only applies to an object suspended in fluid.
∂²Ψ    -(2m(V(x)-E)Ψ
---  = -------------
∂x²        ℏ²Ψ

#### leafy

• CoT Emeritus
• LV10 31337 u53r (Next: 2000)
• Posts: 1554
• Rating: +475/-97
• Seizon senryakuuuu!
##### Re: Buoyancy
« Reply #3 on: December 14, 2012, 10:19:39 am »
Thanks for the great explanation, guys! It seems a bit less counterintuitive to me now
In-progress: Graviter (...)