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From: Edward Green on 2 Apr 2006 12:02
Bell Foster wrote:

> > As the jar is pushed further down the water pressure increases.
> > This compresses the air in the jar, reducing its volume and thus
> > reducing the volume of water displaced. This reduces the
> > buoyancy force on the jar.
>
> How does what you describe reduce the buoyance in the jar though? For
> that to happen, the volume of the jar would have to decrease, and it
> stays the same...

Perhaps you are thinking of the force on the jar itself, vs. on the jar
and trapped air, considered as a unit? The jar itself, as you note,
doesn't change volume.

Ok, let's follow this through:

If the jar contained no air and were open to the surrounding water, the
net upward force exerted by the medium would be equal to the weight of
water displaced by the (actual structural volume of) the jar. If the
jar contains an air bubble however, notice the following: in the water,
there exists a vertical pressure gradient, determined by the density of
the water. Inside the air bubble, however, the pressure gradient is
less, owing to the lesser density of the air. Hence, while at the
lower surface of the bubble pressure is equilibrated with the
surrounding water, at the top surface the air pressure is higher than
that of the surrounding water. This pressure differential at the top
of the bubble accounts for net "bouyant force on the jar" in this
question. And, as the trapped air bubble is shrinks in height, this
pressure differential, hence the buoyant force, decreases.

And that's how the buoyant force on the jar itself, strictly speaking,
descreases as the jar is pushed down in the water, even though, as you
correctly note, the volume of the jar itself is not changing.

If you speak more loosely of "the jar" as comprising the jar and the
trapped air -- which does change in volume -- you reach a similar
conclusion more simply, though not, therefore, more correctly.

From: Hexenmeister on 2 Apr 2006 15:11

"Bell Foster" <bell3774(a)gmail.com> wrote in message
news:1143969800.842841.263490(a)u72g2000cwu.googlegroups.com...
| Here is a test question that I didn't get and don't understand:
| "An empty jar is pushed open-side downward into water so that air
| trapped in it cannot get out. As the jar is pushed deeper, the buoyant
| force on the jar...decreases."
|
| I don't understand the answer of "decreases" (as opposed to say "stays
| the same" or "increases") because I understood buoyant force to be
| related to the weight of the displaced fluid, and pushing the jar
| further down displaces more fluid, which weighs more. So it seems to me
| it should be "increases"...maybe someone can explain to me why/how I'm
| wrong.
|
When I was a child I obtained a small plastic "diver", a figure of a
man in diving gear, a free gift from a Kellogg's cornflakes packet
I think it was.
At that time soda pop came in bottles with screw caps, and by
filling the bottle with water, placing the diver inside and screwing
down on the cap, compressing the air inside, one could cause the diver
to sink and then rise again by unscrewing the cap. There was a tiny
bubble of air in the diver's "helmet" and you could see this compress
if you looked carefully. The density of the plastic was only slightly
greater than that of water, the air bubble providing the buoyancy.
It has since occurred to me that this could make a sensitive barometer.

Androcles.






From: tadchem on 2 Apr 2006 16:27

Bell Foster wrote:
> Here is a test question that I didn't get and don't understand:
> "An empty jar is pushed open-side downward into water so that air
> trapped in it cannot get out. As the jar is pushed deeper, the buoyant
> force on the jar...decreases."
>
> I don't understand the answer of "decreases" (as opposed to say "stays
> the same" or "increases") because I understood buoyant force to be
> related to the weight of the displaced fluid, and pushing the jar
> further down displaces more fluid, which weighs more. So it seems to me
> it should be "increases"...maybe someone can explain to me why/how I'm
> wrong.

As the jar is pushed down the hydraulic pressure increases.

For an incompressible fluid of density rho, the pressure P at a depth
of h is
P = rho * g * h
where g is the local acceleration due to gravity.

Increased pressure compresses the trapped air.

The compressed trapped air takes up less volume.

Less gas volume means less displaced fluid (once the entire jar is
submerged).

Less displaced fluid means less buoyancy.

Google "Cartesian Diver"
I have a Cartesian diver on my desk at work made with a medicine
dropper and a glass 'hip flask' with a rubber stopper. The diver is so
sensitive that I can squeeze the glass flask with my fingers and make
it sink, or relax and let it rise.


Tom Davidson
Richmond, VA

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