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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 :)

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It doesn't, unless they dissolve. – Keith Thompson Jan 19 '12 at 1:21
I think the question is about measured mass, not acutal mass. – Dan Jan 19 '12 at 1:54
A scale doesn't measure mass. It measures force. It is calibrated in units of mass, but the calibration assumes that the scale is measuring the force due to gravity on some object at or near sea-level on Earth. A balance doesn't have to assume a particular gravity field, but it still measures force. A balance works by comparing the force exerted by gravity on an unknown mass vs. the force exerted by a known mass. – james large Mar 23 at 1:54

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.

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Well you are given the original mass of the solid ($M_S$) as well as its density ($D_S$)... same with the liquid.... can you not use this information to find the volume? And the solid is place halfway in, so the submerged volume would have half the volume of the solid. Using this information, can you complete the problem? I am having trouble understanding the concept. Thanks :) Also, if the mass of the objects do not change, and the scale registers in KG, wouldnt the sum of the masses not change either? The scales register mass, not weight, correct? – Raymond Jan 18 '12 at 22:38
Also - scales always measure weight. They work by measuring the force that gravity applies to the object. – Mark Beadles Jan 19 '12 at 0:18
Ahh ok... Wait I think I did give the density of the liquid ($D_l$). How would this then effect the weight of the objects measured by the scales? Sorry for all the questions, i'm just trying to get this straight :) – Raymond Jan 19 '12 at 0:38
Ah!, I missed that the density of the solid was given as well. – Dan Jan 19 '12 at 0:39
I've edited my answer to reflect my error. – Dan Jan 19 '12 at 0:49

protected by Qmechanic Mar 23 at 7:14

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