# Buoyancy calculating unknown density

1. Feb 29, 2012

### masterburn

1. The problem statement, all variables and given/known data

A cube of wood, of mass 0.50 kg and side length 10.80 cm, is floating in a beaker containing alcohol, as shown.
a) What is the density of the wood?
b) Ifthe wood is exactly half submerged, find the density ofthe alcohol
2. Relevant equations

not sure how to go about to solve density of alcohol
3. The attempt at a solution

A) p=m/v = 0.5/ (0.1080^3) = 396.92 kg/m^3

B) density of alcohol = height of box/ height of water * known p
= 0.05/ height * 369.92

not sure how to do B at all ? am i even close

2. Feb 29, 2012

### Staff: Mentor

Draw the diagram. When encountering occasional exercises of this type, I prefer to go back to basics and derive the equations in situ.

Imagine there is an ant (kitted out in scuba gear) walking around the bottom of the beaker. He doesn't feel any extra pressure on his body when he is in the shadow of the floating block compared to when he is out from under it. At any given depth, the pressure is the same everywhere at that level. (If it wasn't, then some fluid would rearrange itself until the pressure was equalised.)

So when the ant is under the floating block, pressing down on his body is a height of alcohol + 10.8cm of wood. When he is not directly under the block, pressing down on him is a certain height of alcohol. Formalize this into volumes, densities and pressures.

Can you continue from there?

(Acting above both of these is atmospheric pressure, but because it acts uniformly over both the block and the alcohol, it need not enter into the computations for density here.)

3. Feb 29, 2012

### masterburn

no i cannot continue from there not sure how ? since half the block is submerged and, the volume displaced is the amount from the submerged block therefore density unknown = h of block/ h of liquid * density known but since the h is equal wouldn't it cancel out and the density of the liquid = density of block ?

4. Feb 29, 2012

### Staff: Mentor

Notice that I didn't mention anything about volume displaced because I was not using that approach.

5. Feb 29, 2012

### masterburn

I'm not sure what way to approach ? it was the only way I was taught ? the only thing i know following your logic is density avg > density fluid since it only half sunk, so the free diagram is mg going down and fb going up then it would be Fb>mg, how so i have buoyancy force of FB but i think it wrong since it doesn't take into account of the block being half sunk