U-Tube Fluid Statics: Solving for Water Column Length and Mercury Rise

[AdkinsJr]

Homework Statement

Mercury is poured into a U-tube. The left arm has a cross-sectional 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 U-tube; given the density of mercury is 13.6 g/cm^3, what distance h does the mercury rise in the left arm?

Homework Equations

Just volumes of cylinder and pressure variations with depth:

P=P_o + ρgh

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.

 

[tiny-tim]

Hi AdkinsJr!

Both surfaces of the water and the mercury must be at the same pressure, atmospheric pressure …

that's right!

now combine that with …

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.

(or alternatively, the principle that if you start at the bottom, and deduce the pressure at the top of each tube, they have to come out the same)

 

[AdkinsJr]

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...

P_A=P_B

P_A=P_{atm}+ρ_{h_2O}g(.2 m)

P_B=P_{atm}+ρ_{Hg}g(h+d)

The three equations above give:

ρ_{h_2O}(.2 m )=ρ_{Hg}(h+d)

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:

V=(h+d)A_1

Since the areas are given I think this should work; two equations two unknowns.

 

[AdkinsJr]

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

 

[tiny-tim]

that looks better!