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Homework Statement
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?
Homework Equations
h=[tex]h_1[/tex] + [tex]h_2[/tex]=0.6 m
[tex]\rho_{Fe}[/tex]=7880 kg/m^3
[tex]\rho_{Pb}[/tex]=11300 kg/m^3
[tex]\rho_{Hg}[/tex]=13600 kg/m^3
[tex]\rho_{H2O}[/tex]=1000 kg/m^3
Where [tex]h_1[/tex] = height of the block in the Mercury, and [tex]h_2[/tex] = height of the block in the Water.
The Attempt at a Solution
F=0N=F_b - F_g
F_g=[tex]\rho_{Pb}[/tex]*A*h*g
F_b=[tex]\Delta P[/tex] * A=(P_f - P_i)*A
P_f = P_i + [tex]\rho_{H2O}[/tex]*h_2*g + [tex]\rho_{Hg}[/tex]*h_1*g
F_b=A*g*([tex]\rho_{H2O}[/tex]*h_2 + [tex]\rho_{Hg}[/tex]*h_1)
0N=A*g*([tex]\rho_{H2O}[/tex]*h_2 + [tex]\rho_{Hg}[/tex]*h_1) - [tex]\rho_{Pb}[/tex]*A*h*g
[tex]\rho_{H2O}[/tex]*h_2 + [tex]\rho_{Hg}[/tex]*h_1 = [tex]\rho_{Pb}[/tex]*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 [tex]h_1=0.506m, h_2=0.094m[/tex], 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.
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