1
$\begingroup$

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?

$\endgroup$
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.