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I've been trying to derive this equation for about two hours:

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But I can't seem to get it. In my class, we are using it to accurately measure the volume of irregulary-shaped objects, but I completely missed the derivation.

I'm not looking to leech off of somebody's proof, but any help with this would be greatly appreciated. Thank you in advance!

Just in case anyone would like to see my implementation of Mark's answer:

$$ w_{dry} = m_{dry}g $$ $$ m_{dry}g = \rho_{o}Vg $$ $$ \rho_{o} = \frac{m_{dry}}{V} $$

The forces acting on the same object, but submerged fully in a fluid are:

$$ w_{wet} = m_{dry}g - B $$ $$ = m_{dry}g - \rho_{f}gV $$ $$ m_{dry} - m_{wet} = \rho_{f}V $$ $$ \rho_{f} = \frac{m_{dry} - m_{wet}}{V}$$

The specific gravity of the object in relation to the fluid it is submerged in is therefore:

$$ S_{g} = \frac{\rho_{o}}{\rho_{f}} $$ $$ = \frac{\frac{m_{dry}}{V}}{\frac{m_{dry} - m_{wet}}{V}} $$ $$ = \frac{m_{dry}}{m_{dry} - m_{wet}} $$

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1 Answer 1

up vote 1 down vote accepted

If you place an object with volume $V$ in the water, the buoyant force on it is $\rho_w g V$, with $\rho_w$ the density of water and $g$ gravitational acceleration. This is because the force on the body from the water is the same as it would be on a thin bag of the same shape but filled with water; the surrounding water will push on anything the same way. The bag wouldn't go anywhere since water doesn't move on average, so the buoyant force would have to equal the weight of the bag. This buoyant force is the difference between the weight of the body and the "weight" of the immersed body.

The force on the body when out of the water is $mg$, which is the same as $\rho_b V g$, with $\rho_b$ the density of the body.

Therefore, the right hand side of your equation is

$$\frac{\rho_b V g}{\rho_w V g}$$

canceling $Vg$ we have


However, the equation you wrote up is a little imprecise. The "weight" of something is the gravitational force on it. That's unchanged as you put it underwater, so "weight of immersed body" is not really the right term.

Additionally, the most important correction to the formula comes from our failure to account for the buoyancy of the atmosphere, which is about .1% the density of water, so we'll be off by that much (depending on how we measure the density of water).

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Thank you for the answer. I didn't know I could use $\LaTeX$ in posts here ;) The $g$ would cancel out of the whole right side of that equation, leaving me with just masses (which is what was being measured in the experiment). I'll see what I can write up, so thank you once again! –  Blender Nov 21 '11 at 5:25
Thanks for your help, once again. I've included my proof in my answer, so thank you for pointing me in the right direction! –  Blender Nov 21 '11 at 6:33

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