Note he said "stack together closely" Displaced does not mean "pushed out of the cup" Displaced means still in the cup or pool.
If the heavier upper cup appears to be floating in the lower cup it is actually sitting on a film of water help between the cups by the surface tension of the water and is NOT floating at all.
This demonstration fails as soon as the top cup is a slightly different shape to the lower container and the top one WILL sink to the bottom if its total weight is more than the weight of the lower liquid or if the the displaced liquid can spread away as in a lake.
Be it gas (BALOONs), steel ships, or plastic cups, Archimedes Principle still holds true.
That would have qualified me as an expert lathe operator - "operate the longitudinal feed with your left hand and the cross feed with your right hand while applying cutting fluid with your other hand." Gerry :-)} London, Canada
Don't need to. Buoyancy has to do with pressure. The so-called "Archimedes' Principle" has an implicit assumption that the liquid is laterally unconstrained. In that case the upward pressure on the object is a function of its depth of submersion, and therefore of the displacement volume.
If an object is immersed in a vessel only slightly larger, then the area of the object is much larger than the area of the surrounding liquid in the limited volume between the vessel and the tank. As the object sinks in the vessel, the displaced liquid goes into the surrounding annular area, and since this area is smaller than the area of the object, the liquid will rise in the outer vessel considerably more than the object sinks. The result is that the object can have a submerged depth (distance from surface to bottom) sufficent to float it while displacing considerably less than its weight of liquid.
Note that this does not depend upon surface tension at all.
In the particular case of a 25 cm dia cylindrical vessel immersed in a 26 cm dia tank, 1 liter of liquid in the 25 cm dia vessel would float when it has displaced only 0.082 liter of similar liquid in the tank. This is because of the ratio of areas of volume in which displacement can occur (region between vessel and tank) to area of the displacing object -- which is 0.082!
If the liquids are water, the water in the outer tank rises by 2.037 cm. This depth makes the pressure on the bottom of the vessel .02 N/cm^2, which when multiplied by the bottom area of the vessel (490.8 cm^2) is 9.807 N -- which is the weight (not mass) of 1 liter (1 kilogram) of water. Therefore, it floats.
It will sink until the outside water is at the same height as the inside water when the displacement is equal to the inside water Whatever the shape of the vessels.
The error in the plastic cup experiment is that the inner plastic cup is a very small part of the total weight and is almost the samd density as the water.
Try this experiment with Glass tumblers where the glass is somewhat denser than the water and of significant volume itself. The inside one will sink past the depth of the inside water and until it has displaced its total weight (mass)(water + glass) or it has touched the bottom.
Comments like that aren't helpful, even if they're accurate. (Yours is not.)
It will indeed. The depth of 1 liter in a 25 cm dia vessel is 2.037 cm. Fancy that! The point is that the floated vessel need not displace its weight of liquid to reach that depth in this situation. When you launch a boat in a lake, the lake level does in fact go up, though far too little to observe. The boat actually doesn't need to quite displace it's own weight of water to float, but the difference is too small for Archimedes to notice -- or perhaps care about.
The vessel need not displace its own weight of liquid in order to raise the liquid to that level, because the area in the space between vessel and tank is considerably smaller than the area of the vessel. Displacement volume is this smaller area * depth while buoyant force is the area of the vessel * depth. Floatation happens when pressure due to depth (distance from bottom of floating object to liquid surface) balances the weight of the floated object. QED.
In the case of non-cylindrical shapes, buoyant force is proportional to the integral of D(A) dA where D(A) is the depth of each differential submerged area parallel to the surface.
Red herring. Thus far we have negelected the weight of the vessel by either treating it as having the same specific gravity as that of the liquid or as though its material is infinitesimally thin, to keep things simple and not obscure the underlying princples with detail clutter. Having the vessel be of heavier material doesn't invalidate the principles, just makes the math a bit messier.
Shall we make the vessel of depleted uranium so John can be right in at least one particular case?
I have tried to show the physics of this situation. If you prefer to argue with belief and opinion, please feel free to believe (and assert) whatever you like.
Experiments are most enlightening if one understands what the experiment is telling him, and why it turns out as it does.
Buoyancy is a consequence of the hydrostatic pressure of a fluid acting on a body. It has nothing to do with the volume of the fluid that is supporting the body beyond the requirement that there be enough fluid for the body to reach equilibrium, either by floating or resting completely submerged on the bottom of a container.
Displacement is *not* the volume of fluid raised by a floating or submerged object, it's the volume of the object below the surface of the fluid.
An object that weighs exactly the same as the volume of water it displaces neither floats nor sinks. If placed on the surface of the water it will remain there, but if dragged down below the surface and released, it will remain where it was released.
One which weighs less than the volume of water it displaces floats; one which weighs more sinks.
Depends on what you regard as "significant". If "enough to result in buoyancy" is "significant" then that's exactly what I contend, and (short of typing a bunch of equations in ASCII) have proven. Those who won't follow the proof would also ignore the equations. I do have the equations in MathCAD.
Yes. And if the liquid level is raised significantly when the body is immersed deeply enough to float, then it doesn't take much displaced fluid volume to float it.
Only in an unconstrained pool of constant surface height.
Only in an unconstrained pool of constant surface height.
This is not a valid experiment for the situation posed, because allowing it to overflow maintains a constant surface height as in an unconstrained pool.
You'd have me do an experiment rather than follow my reasoning, even though Ig has already done an experiment and posted the results?
OK, I'll do that. T'will be interesting to see how you refute it.
(5 minutes later...)
I put 150 cc of water in a polyethylene jar that weighs 26.0 grams empty. I then put the jar in a slightly larger empty beaker, and injected water from a calibrated syringe into the beaker until the jar (with 150 cc of water in it) just floated. The amount of water necessary to float the jar (with load of 150 cc of water, 176 gram total) was 27 cc. Note: water weighs about 1 gram per cc.
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