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- Thread starter oreo
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- #2

Chestermiller

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Chet

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Okay but the object at greater depth is due to its greater density.

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BruceW

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Thanks

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BruceW

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Chestermiller

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The buoyant force is determined by the density of the fluid that the object is immersed in, not the density of the object.Okay but the object at greater depth is due to its greater density.

Chet

- #9

Doug Huffman

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lots of fancy maths there for so simple an hypothesis, div and del and surface integrals.

- #10

RonL

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If you like to read, here are two books about events that took place in the past, very interesting IMO, :)

https://www.amazon.com/dp/B0007DP27Q/?tag=pfamazon01-20

https://www.amazon.com/dp/0195061918/?tag=pfamazon01-20&tag=pfamazon01-20

two links from wiki,......

http://en.wikipedia.org/wiki/Bathyscaphe_Trieste

http://en.wikipedia.org/wiki/DSV_Alvin

The Trieste has lots of information about overcoming problems encountered relating to buoyancy.

https://www.amazon.com/dp/B0007DP27Q/?tag=pfamazon01-20

https://www.amazon.com/dp/0195061918/?tag=pfamazon01-20&tag=pfamazon01-20

two links from wiki,......

http://en.wikipedia.org/wiki/Bathyscaphe_Trieste

http://en.wikipedia.org/wiki/DSV_Alvin

The Trieste has lots of information about overcoming problems encountered relating to buoyancy.

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- #11

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I am talking about the depths achieved by two objects that displace equal volumes of fluid.The buoyant force is determined by the density of the fluid that the object is immersed in, not the density of the object.

Chet

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No. Not necessarily. If the fluid is incompressible (what is usually implied by the word "liquid") then the density is constant regardless of depth/pressure.Okay but the object at greater depth is due to its greater density.

You seem to be thinking of a gas, not a liquid.

- #13

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I have another mathematical confusion. As the Bouyant force is equal to weight of water displaced which is equal to weight of object and both have same volumes, therefore g and Volume are canceled out which leaves: density of water= density of object, which is not possible. Mathematically,

ρ(water)V(fluid)g=ρ(object)V(object)g

ρ(water)=ρ(object)

Please clarify it.

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Doug Huffman

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For a submerged object this is only true if the object is neutrally buoyant.the Bouyant force is equal to weight of water displaced which is equal to weight of object

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russ_watters

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Well, for air (for example), density increases with "depth" in the atmosphere, so yes, pressure increases with the density of the fluid, as the object sinks. But if you're trying to get back to pressure as the driver for buoyancy change in general, no, it isn't correct. Higher density and therefore buoyancy can also be compared between different fluids, regardless of the pressure.I am talking about the depths achieved by two objects that displace equal volumes of fluid.

- #17

BruceW

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yeah, DaleSpam has the right idea. The Buoyant force on the object equals the weight of water displaced, not the weight of the object. These things are only equal in the special case when the (average) density of the object is the same as the density of the water, i.e. neutrally buoyant.I have another mathematical confusion. As the Bouyant force is equal to weight of water displaced which is equal to weight of object and both have same volumes, therefore g and Volume are canceled out which leaves: density of water= density of object, which is not possible. Mathematically,

ρ(water)V(fluid)g=ρ(object)V(object)g

ρ(water)=ρ(object)

Please clarify it.

edit: and Chestermiller mentioned this earlier.

- #18

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Thanks for the clarification.For a submerged object this is only true if the object is neutrally buoyant.

- #19

RonL

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Were you water pressure testing a tank ?

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Doug Huffman

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- #21

RonL

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I had the similar experience at a much smaller and lower pressure scale, what I thought would be a simple unloading of pressure produced a bit of surprise (no damage or injury) I'm always very Leary of pressure and spin speed

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