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I am trying to derive a formula to calculate the density of a irregulary shaped object.

I can measure the (false) weight of the object in pure air (of known density), and the (false) weight of the object in water (of known density).

I cannot measure (directly) the volume of the object, nor its real weight (in vaccum).

I tried this approach: $$G_{air} = G_{vaccum} - B_{air} $$ $$G_{water} = G_{vaccum} - B_{water} $$ where $B$ is the buoyant force.

Solving both equations for $G_{vaccum}$ gives $$G_{vaccum} = G_{air} + B_{air}$$ $$G_{vaccum} = G_{water} + B_{water}$$ so $$G_{air} + B_{air} = G_{water} + B_{water}$$

Because $B_{air} = \rho_{air} V g$ resp. $B_{water} = \rho_{water} V g$ and $$V = \frac{m_{obj}}{\rho_{obj}}$$ I can write:

$$ G_{air} - G_{water} = B_{water} - B_{air}$$ $$ G_{air} - G_{water} = Vg \left(\rho_{water} - \rho_{air}\right)$$ $$ \frac{G_{air} - G_{water}}{\rho_{water} - \rho_{air}} = \frac{m_{obj}g}{\rho_{obj}} $$

And this is where I am stuck, as I cannot measure the real weight of the object $m_{obj}$. It feels like I am missing something here ...

So, how do I solve this equation to finally get an equation for $\rho_{obj}$? Tell me if I forgot something I could measure.


Edit: Of course I can assume that the influence of the buoyant force in air is negligible and write: $$ m_{obj}g \approx G_{air} $$ Then I can also assume that the density of air is much less than the density of water and write: $$ \rho_{water} - \rho_{air} \approx \rho_{water} $$ Using those two assumpions I could write: $$ \rho_{obj} = \rho_{water} \frac{G_{air}}{G_{air} - G_{water}} $$

Can I only assume the first simplification and not the second and get a better result?

And still: How do I solve the problem without those simplifications?

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The mass/volume terms can be cancelled out.

You know $G_{water}=G_w$ and $G_{air}=G_a$, $\rho_w$, $\rho_a$, and want to know $\rho$.

$$G_w=mg-\rho_w V g= Vg(\rho-\rho_w)$$ $$G_a=mg-\rho_a V g= Vg(\rho-\rho_a)$$

Dividing, $$\frac{G_w}{G_a}=\frac{\rho-\rho_w}{\rho-\rho_a}$$

You can rearrange the terms after this to get $\rho$.

Remember, when solving such equations, dividing them makes more sense than adding.

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Awesome, thank you! –  Daniel Oertwig Mar 14 '13 at 13:38
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