# Homework Help: Fluid mechanics and buoyancy

1. Nov 17, 2004

### thunderbug

i seem to be stuck on a problem which involves fluids of 2 different densities, and an object floating in/on them. A block of wood is floating in a jar of water. oil is then poured onto the water. how can i calculate the mass of "the fluid displaced" when it is composed of 2 different fluids, and i have no idea how large the jar is? i don't think i should average their densities, but i keep getting a wrong result.

for a floating (partially submerged) object, i can see that
FB = mg = (fraction submerged) (weight of fluid displaced by entire object)

= (fraction) (rho of fluid * Volume of object * g)

i have used this to find the fraction of an object which is above or below the surface. how can i change it to apply to 2 fluids with 2 different densities?

the specifics given:
rho(wood) = 500 kg/m^3
rho(oil) = 600 kg/m^3
v(wood) = (.01m)^3
question asks how deep oil layer is when it is 4 cm below the top of the block.[/CODE]
PHP:

2. Nov 17, 2004

### DaveC426913

I am not an engineering whiz, but isn't the shape of the block an essential element? Normally, the shape would be irrelevant, since bouyancy acts on the volume/mass, and is independent of shape. But you need to know when it is 4cm out of the water - a linear measurement.

If the block were 1 cm^2 and 1m tall, you would have a very different answer than if it were 100cm^2 and 5cm tall.

Do we assume it is a cube?

Also, when you say the block of wood has a volume of (.01m)^3, I can interpret that two ways:
.01m is equal to 1cm; 1cm^3 is 1 cubic cm
or
.01m^3 could be read as 1/100th of a cubic meter; which is 10000 cubic cm.

Something tells me this isn't about decimals. I'm going to assume the block is a cube.

I'm also going to go out on a limb and propose that the cube is not supposed to be .01m^3 (this would make the cube 21.544379972138894748806069367288cm on a side).
Nor it is supposed to be 1cm^3, since it could never float 4cm out of the water.

I'm going to say the cube is supposed to be 0.001m^3 - making it 10 cm on a side (and incidentally having a volume of 1 liter and a mass of 0.5kg).

Last edited: Nov 17, 2004
3. Nov 17, 2004

### thunderbug

sorry if i was unclear, yes, the block is .01m per side, and a cube. and yes, the mass would be 0.5 kg. am i correct in saying that if this block were floating only on the water, that it would be submerged 50% ?

4. Nov 17, 2004

### thunderbug

wow. sorry. 0.1 m per side, yes 10 cm per side, etc.