
#1
Jan3112, 05:48 AM

P: 122

1. The problem statement, all variables and given/known data
Mercury is poured into a Utube. The left arm has a crosssectional area A1=.001 m and the right has A2=.0005m. One hundred grams of water are then poured into the right arm. Determine the length of the water column in the right arm of the Utube; given the density of mercury is 13.6 g/cm^3, what distance h does the mercury rise in the left arm? 2. Relevant equations Just volumes of cylinder and pressure variations with depth: [tex]P=P_o + ρgh[/tex] 3. The attempt at a solution I’m not sure how to deal with this problem. I found the height in the column fairly easily since we know the density and mass and cross sectional area, it’s pretty easy to solve for h of the water column, it is .2 meters. I’m not sure how to set up equations to solve for the distance the mercury has risen though. I don’t know how to apply the principles of static fluids here since there are two different types of fluid. Pressures are not equal at equal heights I presume. Both surfaces of the water and the mercury must be at the same pressure, atmospheric pressure, yet are at different heights. However I do think the pressures at points A and B (labeled in the crude scetch) are equal because below the horizontal line I draw there is only mercury. 



#2
Jan3112, 06:21 AM

Sci Advisor
HW Helper
Thanks
P: 26,167

Hi AdkinsJr!
now combine that with … 



#3
Jan3112, 06:38 AM

P: 122

Ok; I think I'm making progress here. I just need to create another length "d" in the drawing and express the height of the mercury column as a sum h+d...
[tex]P_A=P_B[/tex] [tex]P_A=P_{atm}+ρ_{h_2O}g(.2 m)[/tex] [tex]P_B=P_{atm}+ρ_{Hg}g(h+d)[/tex] The three equations above give: [tex]ρ_{h_2O}(.2 m )=ρ_{Hg}(h+d)[/tex] So the 2 unknowns are h and d. So if I just come up with another equation with them it should get things moving. I think the weights of the columns have to be equal (above B for the mercury and above A for the water). So the mass of the mercury above the point B should be the same as the mass of the water poured in. I have the mass and the density, and thus the volume of mercury above B... which I can write as: [tex]V=(h+d)A_1[/tex] Since the areas are given I think this should work; two equations two unknowns. 



#4
Jan3112, 07:13 AM

P: 122

Fluid Statics in UTubes
Actually I think I'm wrong about the weights; they shouldn't be equal because of that principal of the hydraulic press...different cross sectional areas.
The pressures are equal, so I'll have to say that F1/A1=F2/A2 and this should give the mass of the mercury column. Then I can find that volume and use V=(h+d)A1 


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