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Why does a less dense object rise in a dense fluid surrounding?

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  Why does a less dense object rise in a dense fluid surrounding? Arindam Banerjee <banerjeeadda1234@gmail.com> - 2019-01-09 21:12 -0800
    Re: Why does a less dense object rise in a dense fluid surrounding? Sylvia Else <sylvia@email.invalid> - 2019-01-10 23:20 +1100
      Re: Why does a less dense object rise in a dense fluid surrounding? Arindam Banerjee <banerjeeadda1234@gmail.com> - 2019-01-10 05:10 -0800
        Re: Why does a less dense object rise in a dense fluid surrounding? Odd Bodkin <bodkinodd@gmail.com> - 2019-01-10 13:32 +0000
          Re: Why does a less dense object rise in a dense fluid surrounding? benj <benj.fake@gmail.com> - 2019-01-10 13:13 -0500
        Re: Why does a less dense object rise in a dense fluid surrounding? Sylvia Else <sylvia@email.invalid> - 2019-01-13 16:31 +1100
      Re: Why does a less dense object rise in a dense fluid surrounding? poraty6@gmail.com - 2019-01-10 16:11 -0800
    Re: Why does a less dense object rise in a dense fluid surrounding? Odd Bodkin <bodkinodd@gmail.com> - 2019-01-10 13:00 +0000
      Re: Why does a less dense object rise in a dense fluid surrounding? Arindam Banerjee <banerjeeadda1234@gmail.com> - 2019-01-10 05:04 -0800
        Re: Why does a less dense object rise in a dense fluid surrounding? benj <benj.fake@gmail.com> - 2019-01-10 13:12 -0500
      Re: Why does a less dense object rise in a dense fluid surrounding? benj <benj.fake@gmail.com> - 2019-01-10 13:11 -0500
        Re: Why does a less dense object rise in a dense fluid surrounding? Odd Bodkin <bodkinodd@gmail.com> - 2019-01-10 19:14 +0000
          Re: Why does a less dense object rise in a dense fluid surrounding? "reber G=emc^2" <herbertglazier0@gmail.com> - 2019-01-10 12:28 -0800
            Re: Why does a less dense object rise in a dense fluid surrounding? benj <benj.fake@gmail.com> - 2019-01-10 19:36 -0500
              Re: Why does a less dense object rise in a dense fluid surrounding? "reber G=emc^2" <herbertglazier0@gmail.com> - 2019-01-11 09:40 -0800
          Re: Why does a less dense object rise in a dense fluid surrounding? mitchrae3323@gmail.com - 2019-01-10 13:21 -0800
          Re: Why does a less dense object rise in a dense fluid surrounding? benj <benj.fake@gmail.com> - 2019-01-10 19:32 -0500
            Re: Why does a less dense object rise in a dense fluid surrounding? Odd Bodkin <bodkinodd@gmail.com> - 2019-01-11 16:20 +0000
              Re: Why does a less dense object rise in a dense fluid surrounding? poraty6@gmail.com - 2019-01-12 23:56 -0800
                Re: Why does a less dense object rise in a dense fluid surrounding? Odd Bodkin <bodkinodd@gmail.com> - 2019-01-13 12:40 +0000
                  Re: Why does a less dense object rise in a dense fluid surrounding? poraty6@gmail.com - 2019-01-13 16:01 -0800
      Re: Why does a less dense object rise in a dense fluid surrounding? poraty6@gmail.com - 2019-01-10 16:13 -0800
    Re: Why does a less dense object rise in a dense fluid surrounding? mitchrae3323@gmail.com - 2019-01-10 10:46 -0800
      Re: Why does a less dense object rise in a dense fluid surrounding? Sergio <invalid@invalid.com> - 2019-01-13 15:40 -0600

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#722900 — Why does a less dense object rise in a dense fluid surrounding?

On Earth. Why does a less dense object rise up in a dense fluid surrounding medium?
So exactly why does a cork bounce up when submerged in water?
If the top of a hot air balloon was pointy instead of flat would it rise?

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#722922

On 10/01/2019 4:12 pm, Arindam Banerjee wrote:
> On Earth. Why does a less dense object rise up in a dense fluid surrounding medium?
> So exactly why does a cork bounce up when submerged in water?

The force on a part of an object object due to the water is always 
perpendicular to the surface of the object. Perform the vector sum of 
all those forces.

Now consider that the object was replaced by the same volume of water, 
with the same shape. It would have exactly the same forces acting on it 
due to the surrounding water.

Since that water would stay where it is, the vector sum of all the 
forces from the water surrounding it must be equal and opposite to its 
weight, which is to say, the force is upwards.

Nor put back the original object. The vector sum of all the forces 
acting on it still equals the weight of the water that would otherwise 
be there. But since the object is less dense than water, that force is 
greater than its weight. There is therefore an upwards force on the 
object that is greater than its weight, so it will accelerate upwards.

 > If the top of a hot air balloon was pointy instead of flat would it rise?
 >
Yes, the shape has nothing to do with it.

Now try figuring out where the forces act on a piece of cooked spaghetti 
when you suck it into your mouth.

Sylvia.

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#722925

The vector sum of forces is small for a very thin flat solid and is easily checked by the viscosity of the medium.

Your pretentious response is ridiculous. 

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#722926

Arindam Banerjee <banerjeeadda1234@gmail.com> wrote:
> The vector sum of forces is small for a very thin flat solid and is
> easily checked by the viscosity of the medium.
> 
> Your pretentious response is ridiculous. 
> 

What on earth are you talking about? Do you really not know how a boat
works, the second oldest invention behind controlled fire?

-- 
Odd Bodkin — Maker of fine toys, tools, tables

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#722943

On 1/10/2019 8:32 AM, Odd Bodkin wrote:
> Arindam Banerjee <banerjeeadda1234@gmail.com> wrote:
>> The vector sum of forces is small for a very thin flat solid and is
>> easily checked by the viscosity of the medium.
>>
>> Your pretentious response is ridiculous.
>>
> 
> What on earth are you talking about? Do you really not know how a boat
> works, the second oldest invention behind controlled fire?
> 
It's pretty clear you don't, Odd. You need to go to India and find your 
guru.

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#723234

On 11/01/2019 12:10 am, Arindam Banerjee wrote:
> The vector sum of forces is small for a very thin flat solid and is easily checked by the viscosity of the medium.

If a net force is applied to an object in a viscous medium, the object 
will move, regardless of how small the force, or how viscous the medium. 
Only the terminal velocity is affected, and it is never zero.

Sylvia.

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#723003

On Thursday, January 10, 2019 at 2:20:39 PM UTC+2, Sylvia Else wrote:
> On 10/01/2019 4:12 pm, Arindam Banerjee wrote:
> > On Earth. Why does a less dense object rise up in a dense fluid surrounding medium?
> > So exactly why does a cork bounce up when submerged in water?
> 
> The force on a part of an object object due to the water is always 
> perpendicular to the surface of the object. Perform the vector sum of 
> all those forces.
> 
> Now consider that the object was replaced by the same volume of water, 
> with the same shape. It would have exactly the same forces acting on it 
> due to the surrounding water.
> 
> Since that water would stay where it is, the vector sum of all the 
> forces from the water surrounding it must be equal and opposite to its 
> weight, which is to say, the force is upwards.
> 
> Nor put back the original object. The vector sum of all the forces 
> acting on it still equals the weight of the water that would otherwise 
> be there. But since the object is less dense than water, that force is 
> greater than its weight. There is therefore an upwards force on the 
> object that is greater than its weight, so it will accelerate upwards.
> 
>  > If the top of a hot air balloon was pointy instead of flat would it rise?
>  >
> Yes, the shape has nothing to do with it.
> 
> Now try figuring out where the forces act on a piece of cooked spaghetti 
> when you suck it into your mouth.
> 
> Sylvia
==================
less dense object   are pushed up
by more dense entities 
just as simple as that

==
Y.P
===================  

.

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#722923

Arindam Banerjee <banerjeeadda1234@gmail.com> wrote:
> On Earth. Why does a less dense object rise up in a dense fluid surrounding medium?

Because the weight of the object is less than the buoyant force of the
fluid, which in turn is due the differential in pressure with depth of the
surrounding fluid. High school physics. 


> So exactly why does a cork bounce up when submerged in water?
> If the top of a hot air balloon was pointy instead of flat would it rise?
> 



-- 
Odd Bodkin — Maker of fine toys, tools, tables

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#722924

Silly replies. 

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#722942

On 1/10/2019 8:04 AM, Arindam Banerjee wrote:
> Silly replies.
> 
YOU SAID IT BANJOJI!

Odd is especially western running dog silly.

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#722941

On 1/10/2019 8:00 AM, Odd Bodkin wrote:
> Arindam Banerjee <banerjeeadda1234@gmail.com> wrote:
>> On Earth. Why does a less dense object rise up in a dense fluid surrounding medium?
> 
> Because the weight of the object is less than the buoyant force of the
> fluid, which in turn is due the differential in pressure with depth of the
> surrounding fluid. High school physics.

Babbling jargon. Try explaining something in English for a change if you 
can understand things well enough. You are one of those WEstern Idiot 
dogs that Banjoji is always talking about.

>> So exactly why does a cork bounce up when submerged in water?
>> If the top of a hot air balloon was pointy instead of flat would it rise?
>>
> 
> 
> 

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#722956

benj <benj.fake@gmail.com> wrote:
> On 1/10/2019 8:00 AM, Odd Bodkin wrote:
>> Arindam Banerjee <banerjeeadda1234@gmail.com> wrote:
>>> On Earth. Why does a less dense object rise up in a dense fluid surrounding medium?
>> 
>> Because the weight of the object is less than the buoyant force of the
>> fluid, which in turn is due the differential in pressure with depth of the
>> surrounding fluid. High school physics.
> 
> Babbling jargon. Try explaining something in English for a change if you 
> can understand things well enough. You are one of those WEstern Idiot 
> dogs that Banjoji is always talking about.

I’m a little surprised you find this jargon. Let’s suppose that everyone
here knows what pressure is and what a fluid is. It is common experience
that pressure increases linearly with the depth in a fluid, in cases where
the density of the fluid is approximately constant. Anyone that has
snorkeled knows this. The pressure on any patch is in fact the weight of
the fluid sitting on top of that patch. 

To make this visually clearer, consider first a pool of water, and imagine
an upright cylindrical volume of this water. If it helps, wrap this volume
in a thin cylindrical can made of plastic of the same density as the water,
so that it has no effect on the following. There is pressure inwards on the
outside of this cylinder due to water pressure. From symmetry, the force on
the left side of the can balances the force on the right side of the can,
at any given depth. But the pressure increases with depth, so the pressure
in the flat top surface of the can is different than the pressure on the
flat bottom surface of the can. The net force exerted by the fluid outside
the can is just then the difference in the forces on top and bottom of the
can. This is called the buoyant force of the fluid on the can. Since the
pressure on the bottom of the can is greater than in the top, the buoyant
force is upward. 

This volume of water (with or without the can) is obviously in equilibrium.
It moves neither up nor down. Thus the downward force of gravity on that
volume of water (namely its weight) is therefore balanced by the upward
buoyant force. This is the statement of Archimedes’ Principle, that the
buoyant force is equal to the weight of the fluid displaced. 

Quantitatively, the weight of the water inside the can is the mass times
the acceleration of gravity. The mass of the water is the density times the
volume. If our van has ends of area A and the top of the can is at depth d1
and the bottom is at depth d2, then the volume of the can is A * (d2-d1),
and thus the weight of the water displaced is (density)*A*(d2-d1)*g. We
just said this is equal to the buoyant force (Archimedes’ principle), and
we said earlier that this was due to the difference in water pressure on
the top and bottom of the can. This is obviously satisfied if the pressure
(force per area) on the top is (density)*d1*g and the pressure on the
bottom is (density)*d2*g. This is the source of our earlier statement that
the pressure increases linearly with the depth in the fluid. 

As the last step in thinking this through, replace the contents of the can
with some substance other than water. The pressure from the fluid OUTSIDE
the can doesn’t change — it doesn’t care what’s inside the can. So the
buoyant force is still the same, equal to the weight of the water displaced
by the can. But the weight of the contents of the can has changed. If the
can is filled with styrofoam, the weight of the contents is now smaller,
and the buoyant force is relatively bigger, and so there is a net force
upward and the can filled with styrofoam rises. If the can is filled with
birdshot, the weight of the can contents is now larger than the buoyant
force, and thus can will sink. 

This is high school physics. 

> 
>>> So exactly why does a cork bounce up when submerged in water?
>>> If the top of a hot air balloon was pointy instead of flat would it rise?
>>> 
>> 
>> 
>> 
> 
> 



-- 
Odd Bodkin — Maker of fine toys, tools, tables

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#722965

On Thursday, January 10, 2019 at 11:14:42 AM UTC-8, Odd Bodkin wrote:
> benj <benj.fake@gmail.com> wrote:
> > On 1/10/2019 8:00 AM, Odd Bodkin wrote:
> >> Arindam Banerjee <banerjeeadda1234@gmail.com> wrote:
> >>> On Earth. Why does a less dense object rise up in a dense fluid surrounding medium?
> >> 
> >> Because the weight of the object is less than the buoyant force of the
> >> fluid, which in turn is due the differential in pressure with depth of the
> >> surrounding fluid. High school physics.
> > 
> > Babbling jargon. Try explaining something in English for a change if you 
> > can understand things well enough. You are one of those WEstern Idiot 
> > dogs that Banjoji is always talking about.
> 
> I’m a little surprised you find this jargon. Let’s suppose that everyone
> here knows what pressure is and what a fluid is. It is common experience
> that pressure increases linearly with the depth in a fluid, in cases where
> the density of the fluid is approximately constant. Anyone that has
> snorkeled knows this. The pressure on any patch is in fact the weight of
> the fluid sitting on top of that patch. 
> 
> To make this visually clearer, consider first a pool of water, and imagine
> an upright cylindrical volume of this water. If it helps, wrap this volume
> in a thin cylindrical can made of plastic of the same density as the water,
> so that it has no effect on the following. There is pressure inwards on the
> outside of this cylinder due to water pressure. From symmetry, the force on
> the left side of the can balances the force on the right side of the can,
> at any given depth. But the pressure increases with depth, so the pressure
> in the flat top surface of the can is different than the pressure on the
> flat bottom surface of the can. The net force exerted by the fluid outside
> the can is just then the difference in the forces on top and bottom of the
> can. This is called the buoyant force of the fluid on the can. Since the
> pressure on the bottom of the can is greater than in the top, the buoyant
> force is upward. 
> 
> This volume of water (with or without the can) is obviously in equilibrium.
> It moves neither up nor down. Thus the downward force of gravity on that
> volume of water (namely its weight) is therefore balanced by the upward
> buoyant force. This is the statement of Archimedes’ Principle, that the
> buoyant force is equal to the weight of the fluid displaced. 
> 
> Quantitatively, the weight of the water inside the can is the mass times
> the acceleration of gravity. The mass of the water is the density times the
> volume. If our van has ends of area A and the top of the can is at depth d1
> and the bottom is at depth d2, then the volume of the can is A * (d2-d1),
> and thus the weight of the water displaced is (density)*A*(d2-d1)*g. We
> just said this is equal to the buoyant force (Archimedes’ principle), and
> we said earlier that this was due to the difference in water pressure on
> the top and bottom of the can. This is obviously satisfied if the pressure
> (force per area) on the top is (density)*d1*g and the pressure on the
> bottom is (density)*d2*g. This is the source of our earlier statement that
> the pressure increases linearly with the depth in the fluid. 
> 
> As the last step in thinking this through, replace the contents of the can
> with some substance other than water. The pressure from the fluid OUTSIDE
> the can doesn’t change — it doesn’t care what’s inside the can. So the
> buoyant force is still the same, equal to the weight of the water displaced
> by the can. But the weight of the contents of the can has changed. If the
> can is filled with styrofoam, the weight of the contents is now smaller,
> and the buoyant force is relatively bigger, and so there is a net force
> upward and the can filled with styrofoam rises. If the can is filled with
> birdshot, the weight of the can contents is now larger than the buoyant
> force, and thus can will sink. 
> 
> This is high school physics. 
> 
> > 
> >>> So exactly why does a cork bounce up when submerged in water?
> >>> If the top of a hot air balloon was pointy instead of flat would it rise?
> >>> 
> >> 
> >> 
> >> 
> > 
> > 
> 
> 
> 
> -- 
> Odd Bodkin — Maker of fine toys, tools, tables

Air bubble coming up two miles down under water gets bigger and bigger going up.Also come up faster Is it surrounded by water's tension? I say yes.Some bugs live inside an air bubble.Bert

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#723008

On 1/10/2019 3:28 PM, reber G=emc^2 wrote:
> On Thursday, January 10, 2019 at 11:14:42 AM UTC-8, Odd Bodkin wrote:
>> benj <benj.fake@gmail.com> wrote:
>>> On 1/10/2019 8:00 AM, Odd Bodkin wrote:
>>>> Arindam Banerjee <banerjeeadda1234@gmail.com> wrote:
>>>>> On Earth. Why does a less dense object rise up in a dense fluid surrounding medium?
>>>>
>>>> Because the weight of the object is less than the buoyant force of the
>>>> fluid, which in turn is due the differential in pressure with depth of the
>>>> surrounding fluid. High school physics.
>>>
>>> Babbling jargon. Try explaining something in English for a change if you
>>> can understand things well enough. You are one of those WEstern Idiot
>>> dogs that Banjoji is always talking about.
>>
>> I’m a little surprised you find this jargon. Let’s suppose that everyone
>> here knows what pressure is and what a fluid is. It is common experience
>> that pressure increases linearly with the depth in a fluid, in cases where
>> the density of the fluid is approximately constant. Anyone that has
>> snorkeled knows this. The pressure on any patch is in fact the weight of
>> the fluid sitting on top of that patch.
>>
>> To make this visually clearer, consider first a pool of water, and imagine
>> an upright cylindrical volume of this water. If it helps, wrap this volume
>> in a thin cylindrical can made of plastic of the same density as the water,
>> so that it has no effect on the following. There is pressure inwards on the
>> outside of this cylinder due to water pressure. From symmetry, the force on
>> the left side of the can balances the force on the right side of the can,
>> at any given depth. But the pressure increases with depth, so the pressure
>> in the flat top surface of the can is different than the pressure on the
>> flat bottom surface of the can. The net force exerted by the fluid outside
>> the can is just then the difference in the forces on top and bottom of the
>> can. This is called the buoyant force of the fluid on the can. Since the
>> pressure on the bottom of the can is greater than in the top, the buoyant
>> force is upward.
>>
>> This volume of water (with or without the can) is obviously in equilibrium.
>> It moves neither up nor down. Thus the downward force of gravity on that
>> volume of water (namely its weight) is therefore balanced by the upward
>> buoyant force. This is the statement of Archimedes’ Principle, that the
>> buoyant force is equal to the weight of the fluid displaced.
>>
>> Quantitatively, the weight of the water inside the can is the mass times
>> the acceleration of gravity. The mass of the water is the density times the
>> volume. If our van has ends of area A and the top of the can is at depth d1
>> and the bottom is at depth d2, then the volume of the can is A * (d2-d1),
>> and thus the weight of the water displaced is (density)*A*(d2-d1)*g. We
>> just said this is equal to the buoyant force (Archimedes’ principle), and
>> we said earlier that this was due to the difference in water pressure on
>> the top and bottom of the can. This is obviously satisfied if the pressure
>> (force per area) on the top is (density)*d1*g and the pressure on the
>> bottom is (density)*d2*g. This is the source of our earlier statement that
>> the pressure increases linearly with the depth in the fluid.
>>
>> As the last step in thinking this through, replace the contents of the can
>> with some substance other than water. The pressure from the fluid OUTSIDE
>> the can doesn’t change — it doesn’t care what’s inside the can. So the
>> buoyant force is still the same, equal to the weight of the water displaced
>> by the can. But the weight of the contents of the can has changed. If the
>> can is filled with styrofoam, the weight of the contents is now smaller,
>> and the buoyant force is relatively bigger, and so there is a net force
>> upward and the can filled with styrofoam rises. If the can is filled with
>> birdshot, the weight of the can contents is now larger than the buoyant
>> force, and thus can will sink.
>>
>> This is high school physics.
>>
>>>
>>>>> So exactly why does a cork bounce up when submerged in water?
>>>>> If the top of a hot air balloon was pointy instead of flat would it rise?
>>>>>
>>>>
>>>>
>>>>
>>>
>>>
>>
>>
>>
>> -- 
>> Odd Bodkin — Maker of fine toys, tools, tables
> 
> Air bubble coming up two miles down under water gets bigger and bigger going up.Also come up faster Is it surrounded by water's tension? I say yes.Some bugs live inside an air bubble.Bert
> 
Bert you genius tell us why the air bubble gets bigger and bigger. And 
why does it come up faster and faster? or does it? Is there more 
bouyancy as it get higher in the water? Can the bugs explain it? What 
role does surface tension play in this phenomenon? Maybe Odd can explain 
it to you and help you get your Nobel?

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#723059

On Thursday, January 10, 2019 at 4:36:45 PM UTC-8, benj wrote:
> On 1/10/2019 3:28 PM, reber G=emc^2 wrote:
> > On Thursday, January 10, 2019 at 11:14:42 AM UTC-8, Odd Bodkin wrote:
> >> benj <benj.fake@gmail.com> wrote:
> >>> On 1/10/2019 8:00 AM, Odd Bodkin wrote:
> >>>> Arindam Banerjee <banerjeeadda1234@gmail.com> wrote:
> >>>>> On Earth. Why does a less dense object rise up in a dense fluid surrounding medium?
> >>>>
> >>>> Because the weight of the object is less than the buoyant force of the
> >>>> fluid, which in turn is due the differential in pressure with depth of the
> >>>> surrounding fluid. High school physics.
> >>>
> >>> Babbling jargon. Try explaining something in English for a change if you
> >>> can understand things well enough. You are one of those WEstern Idiot
> >>> dogs that Banjoji is always talking about.
> >>
> >> I’m a little surprised you find this jargon. Let’s suppose that everyone
> >> here knows what pressure is and what a fluid is. It is common experience
> >> that pressure increases linearly with the depth in a fluid, in cases where
> >> the density of the fluid is approximately constant. Anyone that has
> >> snorkeled knows this. The pressure on any patch is in fact the weight of
> >> the fluid sitting on top of that patch.
> >>
> >> To make this visually clearer, consider first a pool of water, and imagine
> >> an upright cylindrical volume of this water. If it helps, wrap this volume
> >> in a thin cylindrical can made of plastic of the same density as the water,
> >> so that it has no effect on the following. There is pressure inwards on the
> >> outside of this cylinder due to water pressure. From symmetry, the force on
> >> the left side of the can balances the force on the right side of the can,
> >> at any given depth. But the pressure increases with depth, so the pressure
> >> in the flat top surface of the can is different than the pressure on the
> >> flat bottom surface of the can. The net force exerted by the fluid outside
> >> the can is just then the difference in the forces on top and bottom of the
> >> can. This is called the buoyant force of the fluid on the can. Since the
> >> pressure on the bottom of the can is greater than in the top, the buoyant
> >> force is upward.
> >>
> >> This volume of water (with or without the can) is obviously in equilibrium.
> >> It moves neither up nor down. Thus the downward force of gravity on that
> >> volume of water (namely its weight) is therefore balanced by the upward
> >> buoyant force. This is the statement of Archimedes’ Principle, that the
> >> buoyant force is equal to the weight of the fluid displaced.
> >>
> >> Quantitatively, the weight of the water inside the can is the mass times
> >> the acceleration of gravity. The mass of the water is the density times the
> >> volume. If our van has ends of area A and the top of the can is at depth d1
> >> and the bottom is at depth d2, then the volume of the can is A * (d2-d1),
> >> and thus the weight of the water displaced is (density)*A*(d2-d1)*g. We
> >> just said this is equal to the buoyant force (Archimedes’ principle), and
> >> we said earlier that this was due to the difference in water pressure on
> >> the top and bottom of the can. This is obviously satisfied if the pressure
> >> (force per area) on the top is (density)*d1*g and the pressure on the
> >> bottom is (density)*d2*g. This is the source of our earlier statement that
> >> the pressure increases linearly with the depth in the fluid.
> >>
> >> As the last step in thinking this through, replace the contents of the can
> >> with some substance other than water. The pressure from the fluid OUTSIDE
> >> the can doesn’t change — it doesn’t care what’s inside the can. So the
> >> buoyant force is still the same, equal to the weight of the water displaced
> >> by the can. But the weight of the contents of the can has changed. If the
> >> can is filled with styrofoam, the weight of the contents is now smaller,
> >> and the buoyant force is relatively bigger, and so there is a net force
> >> upward and the can filled with styrofoam rises. If the can is filled with
> >> birdshot, the weight of the can contents is now larger than the buoyant
> >> force, and thus can will sink.
> >>
> >> This is high school physics.
> >>
> >>>
> >>>>> So exactly why does a cork bounce up when submerged in water?
> >>>>> If the top of a hot air balloon was pointy instead of flat would it rise?
> >>>>>
> >>>>
> >>>>
> >>>>
> >>>
> >>>
> >>
> >>
> >>
> >> -- 
> >> Odd Bodkin — Maker of fine toys, tools, tables
> > 
> > Air bubble coming up two miles down under water gets bigger and bigger going up.Also come up faster Is it surrounded by water's tension? I say yes.Some bugs live inside an air bubble.Bert
> > 
> Bert you genius tell us why the air bubble gets bigger and bigger. And 
> why does it come up faster and faster? or does it? Is there more 
> bouyancy as it get higher in the water? Can the bugs explain it? What 
> role does surface tension play in this phenomenon? Maybe Odd can explain 
> it to you and help you get your Nobel?

benji when scuber diving you could see bubbles getting smaller the as you went down.Water can not be pressured and made smaller.The spider bug uses the bubble as a diving bell.I can see water always having a tension on its surface strong binding molecular force.(it curves) If you hold in farts you float better.Bert

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#722973

On Thursday, January 10, 2019 at 11:14:42 AM UTC-8, Odd Bodkin wrote:
> benj <benj.fake@gmail.com> wrote:
> > On 1/10/2019 8:00 AM, Odd Bodkin wrote:
> >> Arindam Banerjee <banerjeeadda1234@gmail.com> wrote:
> >>> On Earth. Why does a less dense object rise up in a dense fluid surrounding medium?
> >> 
> >> Because the weight of the object is less than the buoyant force of the
> >> fluid, which in turn is due the differential in pressure with depth of the
> >> surrounding fluid. High school physics.
> > 
> > Babbling jargon. Try explaining something in English for a change if you 
> > can understand things well enough. You are one of those WEstern Idiot 
> > dogs that Banjoji is always talking about.
> 
> I’m a little surprised you find this jargon. Let’s suppose that everyone
> here knows what pressure is and what a fluid is. It is common experience
> that pressure increases linearly with the depth in a fluid, in cases where
> the density of the fluid is approximately constant. Anyone that has
> snorkeled knows this. The pressure on any patch is in fact the weight of
> the fluid sitting on top of that patch. 
> 
> To make this visually clearer, consider first a pool of water, and imagine
> an upright cylindrical volume of this water. If it helps, wrap this volume
> in a thin cylindrical can made of plastic of the same density as the water,
> so that it has no effect on the following. There is pressure inwards on the
> outside of this cylinder due to water pressure. From symmetry, the force on
> the left side of the can balances the force on the right side of the can,
> at any given depth. But the pressure increases with depth, so the pressure
> in the flat top surface of the can is different than the pressure on the
> flat bottom surface of the can. The net force exerted by the fluid outside
> the can is just then the difference in the forces on top and bottom of the
> can. This is called the buoyant force of the fluid on the can. Since the
> pressure on the bottom of the can is greater than in the top, the buoyant
> force is upward. 
> 
> This volume of water (with or without the can) is obviously in equilibrium.
> It moves neither up nor down. Thus the downward force of gravity on that
> volume of water (namely its weight) is therefore balanced by the upward
> buoyant force. This is the statement of Archimedes’ Principle, that the
> buoyant force is equal to the weight of the fluid displaced. 
> 
> Quantitatively, the weight of the water inside the can is the mass times
> the acceleration of gravity. The mass of the water is the density times the
> volume. If our van has ends of area A and the top of the can is at depth d1
> and the bottom is at depth d2, then the volume of the can is A * (d2-d1),
> and thus the weight of the water displaced is (density)*A*(d2-d1)*g. We
> just said this is equal to the buoyant force (Archimedes’ principle), and
> we said earlier that this was due to the difference in water pressure on
> the top and bottom of the can. This is obviously satisfied if the pressure
> (force per area) on the top is (density)*d1*g and the pressure on the
> bottom is (density)*d2*g. This is the source of our earlier statement that
> the pressure increases linearly with the depth in the fluid. 
> 
> As the last step in thinking this through, replace the contents of the can
> with some substance other than water. The pressure from the fluid OUTSIDE
> the can doesn’t change — it doesn’t care what’s inside the can. So the
> buoyant force is still the same, equal to the weight of the water displaced
> by the can. But the weight of the contents of the can has changed. If the
> can is filled with styrofoam, the weight of the contents is now smaller,
> and the buoyant force is relatively bigger, and so there is a net force
> upward and the can filled with styrofoam rises. If the can is filled with
> birdshot, the weight of the can contents is now larger than the buoyant
> force, and thus can will sink. 
> 
> This is high school physics. 
> 
> > 
> >>> So exactly why does a cork bounce up when submerged in water?
> >>> If the top of a hot air balloon was pointy instead of flat would it rise?


It would rise faster. Air friction with it is the factor...


God creates gravity.


Mitchell Raemsch
> >> 
> >> 
> >> 
> > 
> > 
> 
> 
> 
> -- 
> Odd Bodkin — Maker of fine toys, tools, tables

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#723007

On 1/10/2019 2:14 PM, Odd Bodkin wrote:
> benj <benj.fake@gmail.com> wrote:
>> On 1/10/2019 8:00 AM, Odd Bodkin wrote:
>>> Arindam Banerjee <banerjeeadda1234@gmail.com> wrote:
>>>> On Earth. Why does a less dense object rise up in a dense fluid surrounding medium?
>>>
>>> Because the weight of the object is less than the buoyant force of the
>>> fluid, which in turn is due the differential in pressure with depth of the
>>> surrounding fluid. High school physics.
>>
>> Babbling jargon. Try explaining something in English for a change if you
>> can understand things well enough. You are one of those WEstern Idiot
>> dogs that Banjoji is always talking about.
> 
> I’m a little surprised you find this jargon. Let’s suppose that everyone
> here knows what pressure is and what a fluid is. It is common experience
> that pressure increases linearly with the depth in a fluid, in cases where
> the density of the fluid is approximately constant. Anyone that has
> snorkeled knows this. The pressure on any patch is in fact the weight of
> the fluid sitting on top of that patch.
> 
> To make this visually clearer, consider first a pool of water, and imagine
> an upright cylindrical volume of this water. If it helps, wrap this volume
> in a thin cylindrical can made of plastic of the same density as the water,
> so that it has no effect on the following. There is pressure inwards on the
> outside of this cylinder due to water pressure. From symmetry, the force on
> the left side of the can balances the force on the right side of the can,
> at any given depth. But the pressure increases with depth, so the pressure
> in the flat top surface of the can is different than the pressure on the
> flat bottom surface of the can. The net force exerted by the fluid outside
> the can is just then the difference in the forces on top and bottom of the
> can. This is called the buoyant force of the fluid on the can. Since the
> pressure on the bottom of the can is greater than in the top, the buoyant
> force is upward.
> 
> This volume of water (with or without the can) is obviously in equilibrium.
> It moves neither up nor down. Thus the downward force of gravity on that
> volume of water (namely its weight) is therefore balanced by the upward
> buoyant force. This is the statement of Archimedes’ Principle, that the
> buoyant force is equal to the weight of the fluid displaced.
> 
> Quantitatively, the weight of the water inside the can is the mass times
> the acceleration of gravity. The mass of the water is the density times the
> volume. If our van has ends of area A and the top of the can is at depth d1
> and the bottom is at depth d2, then the volume of the can is A * (d2-d1),
> and thus the weight of the water displaced is (density)*A*(d2-d1)*g. We
> just said this is equal to the buoyant force (Archimedes’ principle), and
> we said earlier that this was due to the difference in water pressure on
> the top and bottom of the can. This is obviously satisfied if the pressure
> (force per area) on the top is (density)*d1*g and the pressure on the
> bottom is (density)*d2*g. This is the source of our earlier statement that
> the pressure increases linearly with the depth in the fluid.
> 
> As the last step in thinking this through, replace the contents of the can
> with some substance other than water. The pressure from the fluid OUTSIDE
> the can doesn’t change — it doesn’t care what’s inside the can. So the
> buoyant force is still the same, equal to the weight of the water displaced
> by the can. But the weight of the contents of the can has changed. If the
> can is filled with styrofoam, the weight of the contents is now smaller,
> and the buoyant force is relatively bigger, and so there is a net force
> upward and the can filled with styrofoam rises. If the can is filled with
> birdshot, the weight of the can contents is now larger than the buoyant
> force, and thus can will sink.
> 
> This is high school physics.

then why are you having so much trouble explaining it in words?

>>>> So exactly why does a cork bounce up when submerged in water?
>>>> If the top of a hot air balloon was pointy instead of flat would it rise?

Fact is I looked in the index in a number of College Freshman physics 
texts and none of them had the term "buoyant force" listed. Mostly they 
only mentioned in passing that under Archimedes principle, objects in 
fluid are "bouyed up". It's interesting that while you try to explain 
what is going on with words, you can't. You lapse into equations and 
jargon and end up chasing your own tail over and over.

You don't explain any general situation at all and then trying to use a 
single restricted example of a cylindrical "can" you don't even cover 
that case reasonably. if the can rises out of the liquid how does it 
have a force on it's top cover? Does the principle quit working them? Of 
course it still works. And even worse you don't even mention the problem 
that the can might flip over. Why? What are the conditions that control 
this?

Sylvia did much better than you on this. Maybe you should ahs Sylvia for 
some science lessons.

So If I take your mythical "can" and fill it completely with water 
sealed with it's lid and then toss it into a pond, will it sink or 
float? And if it floats in which orientation will it float? Explain 
using words with no math, please. This is high school physics.

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#723054

benj <benj.fake@gmail.com> wrote:
> On 1/10/2019 2:14 PM, Odd Bodkin wrote:
>> benj <benj.fake@gmail.com> wrote:
>>> On 1/10/2019 8:00 AM, Odd Bodkin wrote:
>>>> Arindam Banerjee <banerjeeadda1234@gmail.com> wrote:
>>>>> On Earth. Why does a less dense object rise up in a dense fluid surrounding medium?
>>>> 
>>>> Because the weight of the object is less than the buoyant force of the
>>>> fluid, which in turn is due the differential in pressure with depth of the
>>>> surrounding fluid. High school physics.
>>> 
>>> Babbling jargon. Try explaining something in English for a change if you
>>> can understand things well enough. You are one of those WEstern Idiot
>>> dogs that Banjoji is always talking about.
>> 
>> I’m a little surprised you find this jargon. Let’s suppose that everyone
>> here knows what pressure is and what a fluid is. It is common experience
>> that pressure increases linearly with the depth in a fluid, in cases where
>> the density of the fluid is approximately constant. Anyone that has
>> snorkeled knows this. The pressure on any patch is in fact the weight of
>> the fluid sitting on top of that patch.
>> 
>> To make this visually clearer, consider first a pool of water, and imagine
>> an upright cylindrical volume of this water. If it helps, wrap this volume
>> in a thin cylindrical can made of plastic of the same density as the water,
>> so that it has no effect on the following. There is pressure inwards on the
>> outside of this cylinder due to water pressure. From symmetry, the force on
>> the left side of the can balances the force on the right side of the can,
>> at any given depth. But the pressure increases with depth, so the pressure
>> in the flat top surface of the can is different than the pressure on the
>> flat bottom surface of the can. The net force exerted by the fluid outside
>> the can is just then the difference in the forces on top and bottom of the
>> can. This is called the buoyant force of the fluid on the can. Since the
>> pressure on the bottom of the can is greater than in the top, the buoyant
>> force is upward.
>> 
>> This volume of water (with or without the can) is obviously in equilibrium.
>> It moves neither up nor down. Thus the downward force of gravity on that
>> volume of water (namely its weight) is therefore balanced by the upward
>> buoyant force. This is the statement of Archimedes’ Principle, that the
>> buoyant force is equal to the weight of the fluid displaced.
>> 
>> Quantitatively, the weight of the water inside the can is the mass times
>> the acceleration of gravity. The mass of the water is the density times the
>> volume. If our van has ends of area A and the top of the can is at depth d1
>> and the bottom is at depth d2, then the volume of the can is A * (d2-d1),
>> and thus the weight of the water displaced is (density)*A*(d2-d1)*g. We
>> just said this is equal to the buoyant force (Archimedes’ principle), and
>> we said earlier that this was due to the difference in water pressure on
>> the top and bottom of the can. This is obviously satisfied if the pressure
>> (force per area) on the top is (density)*d1*g and the pressure on the
>> bottom is (density)*d2*g. This is the source of our earlier statement that
>> the pressure increases linearly with the depth in the fluid.
>> 
>> As the last step in thinking this through, replace the contents of the can
>> with some substance other than water. The pressure from the fluid OUTSIDE
>> the can doesn’t change — it doesn’t care what’s inside the can. So the
>> buoyant force is still the same, equal to the weight of the water displaced
>> by the can. But the weight of the contents of the can has changed. If the
>> can is filled with styrofoam, the weight of the contents is now smaller,
>> and the buoyant force is relatively bigger, and so there is a net force
>> upward and the can filled with styrofoam rises. If the can is filled with
>> birdshot, the weight of the can contents is now larger than the buoyant
>> force, and thus can will sink.
>> 
>> This is high school physics.
> 
> then why are you having so much trouble explaining it in words?
> 
>>>>> So exactly why does a cork bounce up when submerged in water?
>>>>> If the top of a hot air balloon was pointy instead of flat would it rise?
> 
> Fact is I looked in the index in a number of College Freshman physics 
> texts and none of them had the term "buoyant force" listed. 

Well, I’m not sure what ones you looked at. First three I pulled from my
shelf sampled different vintages: 

Halliday & Resnick (1977) pg 376

Giancoli (2000) pg 340-343

Mazur (2015) pg 469-471

Buoyant force called out (and carefully defined) there.


> Mostly they 
> only mentioned in passing that under Archimedes principle, objects in 
> fluid are "bouyed up". It's interesting that while you try to explain 
> what is going on with words, you can't. You lapse into equations and 
> jargon and end up chasing your own tail over and over.

Physics is a quantitative science. I added a paragraph about how to do
things quantitatively because it’s important to do so in physics. It also
served to justify the claim of linear dependence between pressure and depth
in a constant-density fluid. 

> 
> You don't explain any general situation at all 

Nor was that asked for. The question was Why does a less dense object rise
in a dense fluid surrounding? 

Which I answered. 

> and then trying to use a 
> single restricted example of a cylindrical "can" you don't even cover 
> that case reasonably. if the can rises out of the liquid how does it 
> have a force on it's top cover? 

I’m glad you asked! That’s an elaboration that can be easily explained in
much the same way. 

First note that the shape of the can doesn’t matter. The very same
arguments I used for a cylinder would also apply to a rectangular block,
for instance. One can also see that an assemblage of two stacked cylinders
or stacked rectangular blocks yields the same result. By extension, even
the same result can be found for a cone, either point up or point down, by
approximating it with a stack of concentric cylinders of different radii.
This last point also shows that the orientation of the shape doesn’t
matter. Thus we conclude that Archimedes’ principle applies regardless of
the submerged shape or orientation. Indeed, the story of Archimedes’
revelation had to do with a crown, quite a complicated shape. 

If the buoyant force is greater than the weight of the object, then it will
rise to the surface. At this point, neither the buoyant force nor the
weight have changed, so it will continue to rise so that some portion of
the body is above the fluid. When that happens, the amount of the fluid
displaced starts to drop, and so the buoyant force also drops (see
Archimedes’ principle). When the weight equals the buoyant force, then it
will be in equilibrium again. 

There are many applications that could be discussed here, over many many
pages of additional, plain English prose. 

> Does the principle quit working them? Of 
> course it still works. And even worse you don't even mention the problem 
> that the can might flip over. Why? What are the conditions that control 
> this?
> 
> Sylvia did much better than you on this. Maybe you should ahs Sylvia for 
> some science lessons.
> 
> So If I take your mythical "can" and fill it completely with water 
> sealed with it's lid and then toss it into a pond, will it sink or 
> float? And if it floats in which orientation will it float? Explain 
> using words with no math, please. This is high school physics.
> 
> 
> 
> 



-- 
Odd Bodkin — Maker of fine toys, tools, tables

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#723241

On Friday, January 11, 2019 at 6:21:03 PM UTC+2, Odd Bodkin wrote:
> benj <benj.fake@gmail.com> wrote:
> > On 1/10/2019 2:14 PM, Odd Bodkin wrote:
> >> benj <benj.fake@gmail.com> wrote:
> >>> On 1/10/2019 8:00 AM, Odd Bodkin wrote:
> >>>> Arindam Banerjee <banerjeeadda1234@gmail.com> wrote:
> >>>>> On Earth. Why does a less dense object rise up in a dense fluid surrounding medium?
> >>>> 
> >>>> Because the weight of the object is less than the buoyant force of the
> >>>> fluid, which in turn is due the differential in pressure with depth of the
> >>>> surrounding fluid. High school physics.
> >>> 
> >>> Babbling jargon. Try explaining something in English for a change if you
> >>> can understand things well enough. You are one of those WEstern Idiot
> >>> dogs that Banjoji is always talking about.
> >> 
> >> I’m a little surprised you find this jargon. Let’s suppose that everyone
> >> here knows what pressure is and what a fluid is. It is common experience
> >> that pressure increases linearly with the depth in a fluid, in cases where
> >> the density of the fluid is approximately constant. Anyone that has
> >> snorkeled knows this. The pressure on any patch is in fact the weight of
> >> the fluid sitting on top of that patch.
> >> 
> >> To make this visually clearer, consider first a pool of water, and imagine
> >> an upright cylindrical volume of this water. If it helps, wrap this volume
> >> in a thin cylindrical can made of plastic of the same density as the water,
> >> so that it has no effect on the following. There is pressure inwards on the
> >> outside of this cylinder due to water pressure. From symmetry, the force on
> >> the left side of the can balances the force on the right side of the can,
> >> at any given depth. But the pressure increases with depth, so the pressure
> >> in the flat top surface of the can is different than the pressure on the
> >> flat bottom surface of the can. The net force exerted by the fluid outside
> >> the can is just then the difference in the forces on top and bottom of the
> >> can. This is called the buoyant force of the fluid on the can. Since the
> >> pressure on the bottom of the can is greater than in the top, the buoyant
> >> force is upward.
> >> 
> >> This volume of water (with or without the can) is obviously in equilibrium.
> >> It moves neither up nor down. Thus the downward force of gravity on that
> >> volume of water (namely its weight) is therefore balanced by the upward
> >> buoyant force. This is the statement of Archimedes’ Principle, that the
> >> buoyant force is equal to the weight of the fluid displaced.
> >> 
> >> Quantitatively, the weight of the water inside the can is the mass times
> >> the acceleration of gravity. The mass of the water is the density times the
> >> volume. If our van has ends of area A and the top of the can is at depth d1
> >> and the bottom is at depth d2, then the volume of the can is A * (d2-d1),
> >> and thus the weight of the water displaced is (density)*A*(d2-d1)*g. We
> >> just said this is equal to the buoyant force (Archimedes’ principle), and
> >> we said earlier that this was due to the difference in water pressure on
> >> the top and bottom of the can. This is obviously satisfied if the pressure
> >> (force per area) on the top is (density)*d1*g and the pressure on the
> >> bottom is (density)*d2*g. This is the source of our earlier statement that
> >> the pressure increases linearly with the depth in the fluid.
> >> 
> >> As the last step in thinking this through, replace the contents of the can
> >> with some substance other than water. The pressure from the fluid OUTSIDE
> >> the can doesn’t change — it doesn’t care what’s inside the can. So the
> >> buoyant force is still the same, equal to the weight of the water displaced
> >> by the can. But the weight of the contents of the can has changed. If the
> >> can is filled with styrofoam, the weight of the contents is now smaller,
> >> and the buoyant force is relatively bigger, and so there is a net force
> >> upward and the can filled with styrofoam rises. If the can is filled with
> >> birdshot, the weight of the can contents is now larger than the buoyant
> >> force, and thus can will sink.
> >> 
> >> This is high school physics.
> > 
> > then why are you having so much trouble explaining it in words?
> > 
> >>>>> So exactly why does a cork bounce up when submerged in water?
> >>>>> If the top of a hot air balloon was pointy instead of flat would it rise?
> > 
> > Fact is I looked in the index in a number of College Freshman physics 
> > texts and none of them had the term "buoyant force" listed. 
> 
> Well, I’m not sure what ones you looked at. First three I pulled from my
> shelf sampled different vintages: 
> 
> Halliday & Resnick (1977) pg 376
> 
> Giancoli (2000) pg 340-343
> 
> Mazur (2015) pg 469-471
> 
> Buoyant force called out (and carefully defined) there.
> 
> 
> > Mostly they 
> > only mentioned in passing that under Archimedes principle, objects in 
> > fluid are "bouyed up". It's interesting that while you try to explain 
> > what is going on with words, you can't. You lapse into equations and 
> > jargon and end up chasing your own tail over and over.
> 
> Physics is a quantitative science. I added a paragraph about how to do
> things quantitatively because it’s important to do so in physics. It also
> served to justify the claim of linear dependence between pressure and depth
> in a constant-density fluid. 
> 
> > 
> > You don't explain any general situation at all 
> 
> Nor was that asked for. The question was Why does a less dense object rise
> in a dense fluid surrounding? 
> 
> Which I answered. 
> 
> > and then trying to use a 
> > single restricted example of a cylindrical "can" you don't even cover 
> > that case reasonably. if the can rises out of the liquid how does it 
> > have a force on it's top cover? 
> 
> I’m glad you asked! That’s an elaboration that can be easily explained in
> much the same way. 
> 
> First note that the shape of the can doesn’t matter. The very same
> arguments I used for a cylinder would also apply to a rectangular block,
> for instance. One can also see that an assemblage of two stacked cylinders
> or stacked rectangular blocks yields the same result. By extension, even
> the same result can be found for a cone, either point up or point down, by
> approximating it with a stack of concentric cylinders of different radii.
> This last point also shows that the orientation of the shape doesn’t
> matter. Thus we conclude that Archimedes’ principle applies regardless of
> the submerged shape or orientation. Indeed, the story of Archimedes’
> revelation had to do with a crown, quite a complicated shape. 
> 
> If the buoyant force is greater than the weight of the object, then it will
> rise to the surface. At this point, neither the buoyant force nor the
> weight have changed, so it will continue to rise so that some portion of
> the body is above the fluid. When that happens, the amount of the fluid
> displaced starts to drop, and so the buoyant force also drops (see
> Archimedes’ principle). When the weight equals the buoyant force, then it
> will be in equilibrium again. 
> 
> There are many applications that could be discussed here, over many many
> pages of additional, plain English prose. 
> 
> > Does the principle quit working them? Of 
> > course it still works. And even worse you don't even mention the problem 
> > that the can might flip over. Why? What are the conditions that control 
> > this?
> > 
> > Sylvia did much better than you on this. Maybe you should ahs Sylvia for 
> > some science lessons.
> > 
> > So If I take your mythical "can" and fill it completely with water 
> > sealed with it's lid and then toss it into a pond, will it sink or 
> > float? And if it floats in which orientation will it float? Explain 
> > using words with no math, please. This is high school physics.
> > 

============================================
much simpler 

and less    compose ass !!
things are ***moving from more dense environment
to less dance    one**!
there  is always inner   motion of particles   and objects  !!!!!
IN ALL   DIRECTINS (automatic   test )take for instance a  metaphor example
imitating the above situation



 you**  are in a dense crowded place 
and looking how to get   off it!!
 
looking   in   all  possible   directions 
than at last 
you see a direction in which it will look/
and not less ''feel' !!!!it will be the   easiest  
directions  to   escape it 
and bingo!!
there you  ''chose '   your way out !!!!
(automatic not automatic 
call    it as you like it 
nature as well   has that   ''wisdom !!!

=======================
TIA
Y.P
==================

 

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#723248

<poraty6@gmail.com> wrote:
> On Friday, January 11, 2019 at 6:21:03 PM UTC+2, Odd Bodkin wrote:
>> benj <benj.fake@gmail.com> wrote:
>>> On 1/10/2019 2:14 PM, Odd Bodkin wrote:
>>>> benj <benj.fake@gmail.com> wrote:
>>>>> On 1/10/2019 8:00 AM, Odd Bodkin wrote:
>>>>>> Arindam Banerjee <banerjeeadda1234@gmail.com> wrote:
>>>>>>> On Earth. Why does a less dense object rise up in a dense fluid surrounding medium?
>>>>>> 
>>>>>> Because the weight of the object is less than the buoyant force of the
>>>>>> fluid, which in turn is due the differential in pressure with depth of the
>>>>>> surrounding fluid. High school physics.
>>>>> 
>>>>> Babbling jargon. Try explaining something in English for a change if you
>>>>> can understand things well enough. You are one of those WEstern Idiot
>>>>> dogs that Banjoji is always talking about.
>>>> 
>>>> Im a little surprised you find this jargon. Lets suppose that everyone
>>>> here knows what pressure is and what a fluid is. It is common experience
>>>> that pressure increases linearly with the depth in a fluid, in cases where
>>>> the density of the fluid is approximately constant. Anyone that has
>>>> snorkeled knows this. The pressure on any patch is in fact the weight of
>>>> the fluid sitting on top of that patch.
>>>> 
>>>> To make this visually clearer, consider first a pool of water, and imagine
>>>> an upright cylindrical volume of this water. If it helps, wrap this volume
>>>> in a thin cylindrical can made of plastic of the same density as the water,
>>>> so that it has no effect on the following. There is pressure inwards on the
>>>> outside of this cylinder due to water pressure. From symmetry, the force on
>>>> the left side of the can balances the force on the right side of the can,
>>>> at any given depth. But the pressure increases with depth, so the pressure
>>>> in the flat top surface of the can is different than the pressure on the
>>>> flat bottom surface of the can. The net force exerted by the fluid outside
>>>> the can is just then the difference in the forces on top and bottom of the
>>>> can. This is called the buoyant force of the fluid on the can. Since the
>>>> pressure on the bottom of the can is greater than in the top, the buoyant
>>>> force is upward.
>>>> 
>>>> This volume of water (with or without the can) is obviously in equilibrium.
>>>> It moves neither up nor down. Thus the downward force of gravity on that
>>>> volume of water (namely its weight) is therefore balanced by the upward
>>>> buoyant force. This is the statement of Archimedes Principle, that the
>>>> buoyant force is equal to the weight of the fluid displaced.
>>>> 
>>>> Quantitatively, the weight of the water inside the can is the mass times
>>>> the acceleration of gravity. The mass of the water is the density times the
>>>> volume. If our van has ends of area A and the top of the can is at depth d1
>>>> and the bottom is at depth d2, then the volume of the can is A * (d2-d1),
>>>> and thus the weight of the water displaced is (density)*A*(d2-d1)*g. We
>>>> just said this is equal to the buoyant force (Archimedes principle), and
>>>> we said earlier that this was due to the difference in water pressure on
>>>> the top and bottom of the can. This is obviously satisfied if the pressure
>>>> (force per area) on the top is (density)*d1*g and the pressure on the
>>>> bottom is (density)*d2*g. This is the source of our earlier statement that
>>>> the pressure increases linearly with the depth in the fluid.
>>>> 
>>>> As the last step in thinking this through, replace the contents of the can
>>>> with some substance other than water. The pressure from the fluid OUTSIDE
>>>> the can doesnt change  it doesnt care whats inside the can. So the
>>>> buoyant force is still the same, equal to the weight of the water displaced
>>>> by the can. But the weight of the contents of the can has changed. If the
>>>> can is filled with styrofoam, the weight of the contents is now smaller,
>>>> and the buoyant force is relatively bigger, and so there is a net force
>>>> upward and the can filled with styrofoam rises. If the can is filled with
>>>> birdshot, the weight of the can contents is now larger than the buoyant
>>>> force, and thus can will sink.
>>>> 
>>>> This is high school physics.
>>> 
>>> then why are you having so much trouble explaining it in words?
>>> 
>>>>>>> So exactly why does a cork bounce up when submerged in water?
>>>>>>> If the top of a hot air balloon was pointy instead of flat would it rise?
>>> 
>>> Fact is I looked in the index in a number of College Freshman physics 
>>> texts and none of them had the term "buoyant force" listed. 
>> 
>> Well, Im not sure what ones you looked at. First three I pulled from my
>> shelf sampled different vintages: 
>> 
>> Halliday & Resnick (1977) pg 376
>> 
>> Giancoli (2000) pg 340-343
>> 
>> Mazur (2015) pg 469-471
>> 
>> Buoyant force called out (and carefully defined) there.
>> 
>> 
>>> Mostly they 
>>> only mentioned in passing that under Archimedes principle, objects in 
>>> fluid are "bouyed up". It's interesting that while you try to explain 
>>> what is going on with words, you can't. You lapse into equations and 
>>> jargon and end up chasing your own tail over and over.
>> 
>> Physics is a quantitative science. I added a paragraph about how to do
>> things quantitatively because its important to do so in physics. It also
>> served to justify the claim of linear dependence between pressure and depth
>> in a constant-density fluid. 
>> 
>>> 
>>> You don't explain any general situation at all 
>> 
>> Nor was that asked for. The question was Why does a less dense object rise
>> in a dense fluid surrounding? 
>> 
>> Which I answered. 
>> 
>>> and then trying to use a 
>>> single restricted example of a cylindrical "can" you don't even cover 
>>> that case reasonably. if the can rises out of the liquid how does it 
>>> have a force on it's top cover? 
>> 
>> Im glad you asked! Thats an elaboration that can be easily explained in
>> much the same way. 
>> 
>> First note that the shape of the can doesnt matter. The very same
>> arguments I used for a cylinder would also apply to a rectangular block,
>> for instance. One can also see that an assemblage of two stacked cylinders
>> or stacked rectangular blocks yields the same result. By extension, even
>> the same result can be found for a cone, either point up or point down, by
>> approximating it with a stack of concentric cylinders of different radii.
>> This last point also shows that the orientation of the shape doesnt
>> matter. Thus we conclude that Archimedes principle applies regardless of
>> the submerged shape or orientation. Indeed, the story of Archimedes
>> revelation had to do with a crown, quite a complicated shape. 
>> 
>> If the buoyant force is greater than the weight of the object, then it will
>> rise to the surface. At this point, neither the buoyant force nor the
>> weight have changed, so it will continue to rise so that some portion of
>> the body is above the fluid. When that happens, the amount of the fluid
>> displaced starts to drop, and so the buoyant force also drops (see
>> Archimedes principle). When the weight equals the buoyant force, then it
>> will be in equilibrium again. 
>> 
>> There are many applications that could be discussed here, over many many
>> pages of additional, plain English prose. 
>> 
>>> Does the principle quit working them? Of 
>>> course it still works. And even worse you don't even mention the problem 
>>> that the can might flip over. Why? What are the conditions that control 
>>> this?
>>> 
>>> Sylvia did much better than you on this. Maybe you should ahs Sylvia for 
>>> some science lessons.
>>> 
>>> So If I take your mythical "can" and fill it completely with water 
>>> sealed with it's lid and then toss it into a pond, will it sink or 
>>> float? And if it floats in which orientation will it float? Explain 
>>> using words with no math, please. This is high school physics.
>>> al
> 
> e ===========================================much simpler 
> 
> and less    compose ass !!
> things are ***moving from more dense environment
> to less dance    one**!

And WHY? 

Or are things just willful about needing room?

> 
> there  is always inner   motion of particles   and objects  !!!!!
> IN ALL   DIRECTINS (automatic   test )take for instance a  metaphor example
> imitating the above situation
> 
> 
> 
>  you**  are in a dense crowded place 
> and looking how to get   off it!!
>  
> looking   in   all  possible   directions 
> than at last 
> you see a direction in which it will look/
> and not less ''feel' !!!!it will be the   easiest  
> directions  to   escape it 
> and bingo!!
> there you  ''chose '   your way out !!!!
> (automatic not automatic 
> call    it as you like it 
> nature as well   has that   ''wisdom !!!
> 
> ======================TIA
> Y.P
> =================
>  
> 
> 



-- 
Odd Bodkin — Maker of fine toys, tools, tables

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