# Homework Help: Archimedes/Buoyancy Question

1. May 2, 2010

### patricks

1. The problem statement, all variables and given/known data
A cube of iron 60 cm on each side is suspended between layers of mercury and water. The density of mercury is 13.6 g/cm^3. What height of the cube is in each liquid? What if the cube is made of lead?

2. Relevant equations
h=$$h_1$$ + $$h_2$$=0.6 m
$$\rho_{Fe}$$=7880 kg/m^3
$$\rho_{Pb}$$=11300 kg/m^3
$$\rho_{Hg}$$=13600 kg/m^3
$$\rho_{H2O}$$=1000 kg/m^3

Where $$h_1$$ = height of the block in the Mercury, and $$h_2$$ = height of the block in the Water.

3. The attempt at a solution
F=0N=F_b - F_g
F_g=$$\rho_{Pb}$$*A*h*g
F_b=$$\Delta P$$ * A=(P_f - P_i)*A
P_f = P_i + $$\rho_{H2O}$$*h_2*g + $$\rho_{Hg}$$*h_1*g
F_b=A*g*($$\rho_{H2O}$$*h_2 + $$\rho_{Hg}$$*h_1)
0N=A*g*($$\rho_{H2O}$$*h_2 + $$\rho_{Hg}$$*h_1) - $$\rho_{Pb}$$*A*h*g
$$\rho_{H2O}$$*h_2 + $$\rho_{Hg}$$*h_1 = $$\rho_{Pb}$$*h
h_2+h_1=0.6m
h_1=0.49m
h_2=0.11m

The way I thought through the problem is any pressure at the top of the block would be canceled by that same amount of pressure at the bottom of the block, so all I needed to calculate was the addition pressure at the bottom of the block (gauge pressure, I believe it is called), which I took to be the weight of the water displaced plus the weight of the mercury displaced.

This is NOT the solution my teacher came up with. My teachers solution requires the weight of water displaced to act in the same direction as the gravitational force on the block itself which results in a solution of $$h_1=0.506m, h_2=0.094m$$, which does not seem to make any sense to me. If this is indeed true, why is it true?

Also, how exactly do I do a new line in the Latex? Google suggests \\ or \newline, but neither of those seem to work in the "Preview Post" thing. Maybe if I submit it will fix itself, if not, I apologize for the broken Latex.

Okay submitting didn't work. Not sure what to try. Edit: deleted most of the tex stuff.

Last edited: May 2, 2010
2. May 2, 2010

### willem2

Your teacher was probably led astray by the fact that the water does press down on the block. We are only interested in the difference in pressure between the top and the bottom, and that difference is equal to a column of height h_2 of water on top of a column of height h_1 of mercury.

I think it's somewhat simpler to keep pressure out of it and use Archimedes law.

weight of the block = weight of displaced mercury + weight of displaced water

$$h \rho_{Pb} A g = h_1 \rho_{Hg} A g + h_2 \rho_{H20} A g$$ wich is the same as you have.

Your Latex looked fine. There seems to be a problem with updating of the preview. I get to see a preview of a reply of mine in another thread. I hope it comes out right if I press "Submit Reply"