How does placing objects in liquids affect the mass?

Question

I was dazing off in my physics class when I came up with this question and I was wondering about it all day. I could not provide myself with an adequate solution, so here I am asking the forum about it!

I know the community prefers generalizations, but for the sake of not being too confusing, I must be a bit specific for now.

We have a liquid of mass $M_l$ and density $D_l$ laying on a scale (which obviously reads "$M_l$"). A solid of mass $M_s$ and density $D_s$ is hanging by another scale (which again obviously reads "$M_s$"). We move the solid down so that it is halfway submerged in the liquid. (all masses in kg, densities in g/cm^3)

How will the masses read by the two scales change, if at all? Intuitively, I would say that the sum of the masses read by both scales would have to be the same as the original sum ($M1 +M2$) but I could be wrong. Density probably plays a key role.

Any help is appreciated :)

Answer 1

Submerging objects in a liquid does not change the mass of those objects. It does effect the weight they would register on a scale, though. The bouyant force a fluid exerts upwards on a body submerged in it, $$F=\rho Vg$$ where $\rho$ is the density of the fluid, $V$ is the volume of the fluid displaced, and $g$ is the acceleration due to gravity.

The liquid pushes up on the solid, which means, by Newton's third law, that there is a force of equal magnitude acting downward on the liquid. The force of the liquid up on the solid is

$$F_{ls}=D_l\frac{M_sg}{2 D_s}$$

by the previous equation, so the scales would read that the solid weighs

$$M_s- F_{ls}/g$$

and the liquid scale would read

$$M_l+ F_{ls}/g $$

Note that the actual mass of the solid and liquid would be unaffected, only the reading on the scale would change.