I Apparent mass loss due to air

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We were discussing apparent mass loss in water in class yesterday. A student asked if apparent mass loss happens in water, does it also happen in air? And if it does happen in air then why do we not compensate for that when determining the mass of an object in air.

I did some rough math trying to solve for the mass of an object in a vacuum assuming that the object would weigh more in the vacuum versus air because air does provide a buoyant force. My envelope math shows an apparent mass loss of .0001 g where the object is experiencing mass loss due to the buoyant force of air. However, I don't think its right. I would image that there is conceptual problems associated with what I calculated.

My question is the same question that my student had, does an object experience apparent mass loss due to the buoyant force of air and if so is it ever taken into account?
 

HallsofIvy

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I am not happy with the phrase "apparent mass loss". It really should be "apparent weight loss". There loss of mass either real or apparent. In any case, the apparent weight loss in water is due to the buoyant property- immersing an object in water displaces a quantity of water equal to the volume of the object and reduces the apparent weight of the object by the weight of the water displaced. The same thing is true of air but with two differences. First, the density of air is very much less than that of water so the effect is much less. Second, we normally weigh things in air so that effect is already accounted for. But, yes, if you were to weigh an object in vacuum it would weigh very slightly more than if you weigh it in air.
 
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I am not happy with the phrase "apparent mass loss". It really should be "apparent weight loss". There loss of mass either real or apparent. In any case, the apparent weight loss in water is due to the buoyant property- immersing an object in water displaces a quantity of water equal to the volume of the object and reduces the apparent weight of the object by the weight of the water displaced. The same thing is true of air but with two differences. First, the density of air is very much less than that of water so the effect is much less. Second, we normally weigh things in air so that effect is already accounted for. But, yes, if you were to weigh an object in vacuum it would weigh very slightly more than if you weigh it in air.
Thats what I told him before I attempted to crunch some numbers. That because the density of air is so much less than that of water, that the effect would very small.

I also thought "mass loss" was a weird phrase because that's not what they (we) are taught about mass. However, the textbook refers to it as "mass loss". The textbook is my guide this year so I used the same terminology that the book used.
 

anorlunda

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I also thought "mass loss" was a weird phrase because that's not what they (we) are taught about mass. However, the textbook refers to it as "mass loss". The textbook is my guide this year so I used the same terminology that the book used.
What book is that?
 
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My question is the same question that my student had, does an object experience apparent mass loss due to the buoyant force of air and if so is it ever taken into account?
Yes and yes. If the buoyant force doesn't remain below the required accuracy it must be corrected or the measurement must be performed under vaccum. Here is a description of buoyancy correction (chapter 4):

https://www.npl.co.uk/special-pages/guides/gpg71_mass
 

anorlunda

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You misquoted the book. Here is what it says:
Archimedes' principle states that the buoyant force on the object equals the weight of the fluid displaced. This, in turn, means that the object appears to weigh less when submerged; we call this measurement the object's apparent weight. The object suffers an apparent weight loss equal to the weight of the fluid displaced. Alternatively, on balances that measure mass, the object suffers an apparent mass loss equal to the mass of fluid displaced. That is apparent weight loss = weight of fluid displace (11.44) or apparent mass loss = mass of fluid displaced (11.45) The next example illustrates the use of this technique.
The word apparent changes the whole meaning.
 
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You misquoted the book. Here is what it says:


The word apparent changes the whole meaning.
I don't know what I misquoted. The example right below that statement and some of the problems use "apparent mass loss."
 

Bystander

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Whenever masses of different densities are compared, all buoyancy corrections must be considered. If found to be negligible, it is then permissible to "ignore" them. Brass weights and gas bottles are two obviously different densities. NPL has already been mentioned.
Yes and yes. If the buoyant force doesn't remain below the required accuracy it must be corrected or the measurement must be performed under vaccum. Here is a description of buoyancy correction (chapter 4):

https://www.npl.co.uk/special-pages/guides/gpg71_mass
 

anorlunda

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I don't know what I misquoted. The example right below that statement and some of the problems use "apparent mass loss."
"apparent mass loss" means that it appears to be a mass loss but it really isn't.

As others said, it is a weight loss. In zero gravity, there is no such thing as buoyant force.
 

DaveC426913

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Yes. Air creates buoyancy - albeit small.

In theory, if you can make a container large enough, rigid enough and light enough, you can "fill" it with vacuum, and it will float away.
 

DaveC426913

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Fun fact:

One cubic metre of air weighs 1.293kg*.
A human body is about 0.071 cubic metres.

So that means we have 91g of buoyancy. We actually weigh 91g less* than we would if we were in vacuum.

*at sea level
 
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Scales are typically calibrated (assuming they're are calibrated at all) for masses with a density of around 8 g/ml. So if you're weighing a water based solution or a gas or something styrofoamy you'll read a value that is lower than the actual mass due to the buoyancy. Whether or not you care about the the discrepancy depends on the context. But it's there.
 

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