AVR-Chat

That is correct.  A cubic foot of water weighs considerably more than  
10 oz.  You could take that thing to the bottom of the Gulf and let it  
go and it'll shoot straight back up.  This does assume that it doesn't  
get crushed somewhere along the way.  Pressure does not matter.  The  
assumption that pressure is only acting on the bottom is a bad one.   
The pressure pushes up, down sideways and every direction all at  
once.  The only pressure difference is the difference in the head  
pressure of the water from the top of the object to the bottom of the  
object and this is actually what causes it to be buoyant in the first  
place.  This pressure difference is constant at all depths (Assuming  
the density change of the water is negligible.  Which is a good  
assumption at 15m but probably not at the bottom of the Gulf).  The  
force pushing up on the object is the difference between the objects  
weight and the weight of the fluid it displaces.  This is true of any  
object in any fluid, including air.  Yes you are being lifted up by  
the atmosphere, but you're just too darn heavy to notice.

Phil

On May 15, 2009, at 8:10 PM, Brian wrote:

> I might be wrong but what you are saying is if I have a non- 
> compressable ball weighing 10 oz displacing 1 cubic foot of water it  
> would take the same amount of force to hold it under 50 foot of  
> water as it would at 100 feet? If weight and volume didnt change  
> what would keep it from going deeper? I think I might be missing  
> something.
>
> Brian
>
> --- In AVR-Chat@yahoogroups.com, "Graham Davies" <Yahoo37849@...>  
> wrote:
>>
>> --- In AVR-Chat@yahoogroups.com, "Brian" <blue_eagle74@> wrote:
>>>
>>> From what I know, as the water
>>> level rises the pressure at the
>>> bottom increases causing more
>>> of an uplift of an object held
>>> at the bottom.
>>
>> OK, forget Archimedes, have a word with your tenth-grade physics  
>> teacher.
>>
>> Bouyancy has nothing to do with pressure.  The upward force on an  
>> object imersed in a liquid is the weight of the liquid displaced  
>> minus the object's weight.  To a first approximation, this is  
>> constant with depth of imersion, which is why your idea is a bad one.
>>
>> Arguing about second order effects is fine, but it's still a bad  
>> idea to use second order effects to make a measurement when sensors  
>> are available for the first order effect itself (pressure).  As  
>> others have posted, there will be a very small variation of the  
>> force of bouyancy depending on the relative compressibility of the  
>> liquid and the object.
>>
>> Graham.
>>
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