# Floating and sinking

1. Apr 11, 2006

### touqra

Can an object sinks in a liquid, but not to the bottom of the liquid, just to the middle? Can Archimedes principle predicts whether the object sinks partially or sinks totally to the bottom?

2. Apr 11, 2006

### Cyrus

Well, depending on the relative density of your body to its fluid, it can sink like a stone if the density is greater, it can float depending on how much less the density is compared with that of the fluid, or it can have exactly the same density, in which case it will have Neutral Buoyancy. In that case, it wont 'sink' to the middle. You would have to physically move it to the middle, but once you did, it would stay there and not move up nor down. Neutral Buoyancy is what submarines use to stay underwater without ascending or desending.

3. Apr 11, 2006

### HallsofIvy

Staff Emeritus
No, not sink to the middle as you describe it. If the specific gravity (density) of the object is less than that of water, it will float at the top of the water with just a portion below the surface. If the specific gravity of the object is greater than that of water, the object will sink to the bottom. If an object has density exactly that of water, then it will float anywhere in the water- you could place in the middle and it will float there, but there would be no reason for it to sink to the middle and then stay there.

Of course, temperature can change the density of water. If you have a layer of water significantly colder than that just above it, you could have an object with density greater than that of warmer water, less than that of the colder water. Then the object would sink through the warmer water to "sit" upon the layer of colder water.

4. Apr 11, 2006

### DaveC426913

Another factor to consider: objects are not perfectly rigid - they will compress as they are subjected to pressure, meaning their density will increase, meaning they will sink more. The opposite is true - as they rise, they will become more bouyant.

This means that equilibrium is unstable and is difficult to achieve without active compensation. Divers are well aware of this. They sink faster the deeper they go, as their bodies and equipment are compressed.

5. Apr 11, 2006

### Staff: Mentor

...so theoretically, though, if you had an object that was rigid and a fluid that was somewhat compressible, you could end up with a "float to the middle" situation. Blimps don't really need it (since they are powered), but they could maintain a stable altitude on that principle alone, since they are constant volume.

6. May 1, 2006

### jasc15

air. a helium balloon will only reach a certain altitude, when the air density is equal to that of helium. as far as liquids, since they are highly incompressible, this phenomenon will likely not be observed anywhere.

Last edited: May 1, 2006
7. May 1, 2006

### Staff: Mentor

http://www.weathershop.com/lcm_galileo_thermometers.htm [Broken]

Last edited by a moderator: May 2, 2017
8. May 1, 2006

### Clausius2

That does not contradict what HIvy said. The thermometer works such that it contains a liquid (dangerous by the way) which density is sensible to room temperature changes. A chain of balls of different densities are submerged in the tube. When room temperature changes, the lowest of the floating balls in the upper part of the cylinder indicates the correct temperature. Why is that?. Assume $$\rho=\rho_o$$ is the liquid density and $$\rho_d$$ the ball density. A perturbation in density $$\rho=\rho_o+\epsilon$$ (i.e. decreasing in temperature) will make some of the lowest balls in the lower part (which before had a greater density but now smaller) to be lifted. Those balls which $$\rho_d>>\rho$$ will remain at the bottom. Those balls which $$\rho_d<<\rho$$ will remain at the top. That ball which $$\rho_d\sim \rho$$ but $$\rho_d<\rho_o$$ will be lifted and will "apparently" stop by some level until the final density state of the liquid $$\rho$$ is achieved. You should visualize this as an unsteady process.

And I said "apparently" because of the next. The equation of motion of a particle submerged in a viscous fluid is

$$M\ddot z=C(\dot z)\dot z^2+(\rho-\rho_d)gV$$ where z is the upwards vertical coordinate (Inertia=Viscous+Buoyancy).

If $$\rho\sim \rho_d$$, since the terminal velocity is proportional to that density difference it is very small. So that the ball is not really stopped at that z coordinate.

The last equation of motion comes to demonstrate HIvy words, because there's no solution for a ball at rest and a disparity of densities at the same level.

Equilibrium levels may be reached in stratified flows, as it occurs in the atmosphere with clouds and pollutants.

Last edited by a moderator: May 2, 2017
9. May 21, 2006

### disregardthat

a unopened box (metal with 33 cl) with beer floats in swimming pool water! :P

10. May 21, 2006

### disregardthat

sorry, i mean that it doesnt float, or sink. it stays where you put it