How to calculate how much foam is needed to float something of a certain mass? I have some object (300g) that I want to keep above the water. I could run experiments to figure out how much foam I need but I'd have to do this for every object with a different mass. What equation should I use that will get me close to what I need without trial and error?
 A: According to the Archimedes principle, you need to displace the amount of water equal to the weight of the object plus the weight of the foam. Neglecting the weight of the foam, you need $300\,cm^3$ or $0.3\,liters$ of foam to hold an object of $300\,g$.
To account for the weight of the foam, you need to increase this amount slightly. The density of styrofoam is about 20 times less than water, so you need to increase the volume by about 5% to $315 cm^3$ as the minimum, give or take. A third of a liter would do. This is roughly a $6\text{-}cm$ or 2.7" cube.
The equation would be to use the same amount of cubic centimeters of foam, as the number of grams in the object weight and then increase the result by at least 5%. This would position your object at the surface. To avoid the object getting wet, just slightly increase the amount of foam.
A: I suppose you can use specific gravity, which is $\frac{\text{density of object}}{\text{density of water}}$. This ratio will give you what percent of the entire volume of the object will be underwater. For example, if my object's density is 700 $kg/m^3$, since the density of water is 1000 $kg/m^3$, 70% of the object's volume will be underwater.
A: The boundary conditions would be :
$$ F_b = -W $$
This gives :
$$ \rho_{_f} g \left(V_{foam}+\frac {m}{\rho_{_b}}\right) = -mg $$
Where $\rho_{_f}$ is fluid density, $m$ - body mass, $\rho_{_b}$ body density & (assuming massless foam).
From there we can get required foam amount (neglecting signs):
$$ \boxed {V_{foam} = \frac{m}{\phi}} $$
Where $\phi$ is reduced density of body + fluid :
$$ \phi = \frac{\rho_{_f} \rho_{_b}}{\rho_{_f} + \rho_{_b}} $$
Nice question by the way, I have not expected such nice and short expression.
