# Cube Density

1. Mar 10, 2016

### vinamas

1. The problem statement, all variables and given/known data
A wooden cube is floating in a cup containing drinking water and an unidentified fluid.0.5 of the cube's volume is submerged in water and 0.4 of the cube volume is floating on the unidentified fluid
find the density of the cube and the density of the unidentified fluid

2. Relevant equations
density=mass/volume
FB=density*volume*gravity

3. The attempt at a solution
well I try to solve it with a system of equations but am having no luck my attemp:
FB=1000*0.5*V*9.81 and FB=density*0.4*V*9.81 you can see that I have 3 variable as a ressult I cant solve it

2. Mar 10, 2016

### Orodruin

Staff Emeritus
Your equations do not make sense. What you have quoted (I am going to make a qualified guess, you really should use symbols throughout and explain what you mean by them) is the buoyancy force for the water and unknown fluid, respectively. This is not going to get you anywhere on its own and the buoyant force from the water need not be equal to that of the unknown fluid. You need to consider the force equilibrium equation for the cube.

Edit: Is this the problem statement exactly as given? Is there both water and an unknown fluid in the glass at the same time or do they refer to different attempts?

3. Mar 10, 2016

### haruspex

I assume these two FBs represent different variables. As Orodruin says, that is confusing for the reader, especially since you showed no further working. I also assume that you wrote an equation containing these two variables.
The volume and g should cancel out, leaving you only the two densities as the unknowns in the equation. But that is still not enough information. Are you sure this is the complete word-for-word statement of the problem?

4. Mar 10, 2016

### Merlin3189

Putting equations aside for the moment, I'm having trouble understanding this situation.
I am assuming, 50% of the cube is in the water and 40% in the other liquid, leaving 10% in the air.
But "0.4 of the cube volume is floating on the unidentified fluid" could mean, 40% in the air and only 10% in the fluid.

So why is the container not filled further with water and other fluid?
Whichever liquid is on top, if more of that liquid is added, then less will be submerged in the other liquid and less in the air.

The problem seems to depend on just how deep the top layer is:
if the second liquid had negligible density (but greater than air), the block would have a density of about 0.5, but the 40% is arbitrary - the block could be submerged (50%) or any lesser % according to the depth of this liquid.
If it had a density very slightly less than water, the block would have a density of about 0.9, but again, if this liquid were deeper, more would be submerged (up to 90%) If the liquid layer were thinner, less would be submerged down to any lesser % and more submerged in the water up to 90%.
Fall all liquid densities in between, we can find a block density and a range of submersion %'s depending on the depth of the liquid.
There doesn't seem to be any density combination which would require the %'s to be as specified, nor vice versa.

And maybe the liquid could be more dense than water (depending on the interpretation in para 1) giving a whole new range of possible densities and submersion %'s depending on the depth of the water.

Edit: Perhaps my second interpretation is more sensible, as it limits the density of the block to 0.5 to 0.6, so we could just guess 0.55 and be near enough?

5. Mar 10, 2016

### Orodruin

Staff Emeritus
This was my initial thought. Upon rereading the OP, I instead get the impression that it is two different tests, one in water and one in an unknown liquid. We cannot know until the OP returns and clarifies.

6. Mar 10, 2016

### Merlin3189

It would be most sensible (& soluble) if that were the case.
But using all info provided, I looked for woods around RD = 0.55 , Maples look to be a plausible candidate.

7. Mar 10, 2016

### vinamas

Thanks for your help guys and sorry for making things a bit vague for you but I solved this question yesterday like this:
FB=1000*0.5*V*9.81=Density of cube*V*9.81
which resulted in the density being half of that of water so 500
FB=Density of fluid*0.4*V*9.81=9.81*500*V
which resulted in the density of the unknown fluid being 1.25 times of that of water so 1250

8. Mar 10, 2016

### haruspex

Good, but do you understand that what you originally posted suggested the two liquids were in the same cup at the same time?

9. Mar 11, 2016

### Merlin3189

If 0.4 is floating, isn't 0.6 submerged? Which would make the fluid $RD = \frac{0.5}{0.6} = 0.833$ instead of $\frac{0.5}{0.4} = 1.25$