# Pressure-related problem with Mercury and Water in a U-tube

1. May 13, 2016

### horsedeg

1. The problem statement, all variables and given/known data
Mercury is poured into a U-tube as shown in Figure a. The left arm of the tube has cross-sectional area A1 of 11.0 cm2, and the right arm has a cross-sectional area A2 of 4.60 cm2. One hundred grams of water are then poured into the right arm as shown in Figure b.

The length of the column of water is 21.7 cm.

Given that the density of mercury is 13.6 g/cm3, what distance h does the mercury rise in the left arm?
2. Relevant equations
V1 = V2
P = P0 + pgh

3. The attempt at a solution
I honestly have no clue where to start. Finding the length of the column of water was easy, but I don't really know what to do from here.

2. May 13, 2016

### billy_joule

let h' be the height of the water above the dotted line. Pressure at the dotted line is the same on both sides of the tube so we know that: ρmercurygh = ρwatergh'

We also know the volume of mercury above the line is the same as the volume of water below the line so we can express h in terms of h', L (length of water column) A1 and A2. then substitute in the first equation and solve for h.

3. May 13, 2016

### Staff: Mentor

I don't think that this is correct. In figure B, the pressures are the same in both tubes only at the level of the water-mercury interface in the right hand tube, (and below). This equation is correct for figure B only if the dotted line were moved to the level of the water-mercury interface in the figure.

Last edited: May 13, 2016
4. May 13, 2016

### billy_joule

Ah, of course.

Corrected:

5. May 13, 2016

### horsedeg

Clearly I'm lacking some sort of basic concept here. Why is pressure the same on both sides at that level? Also, what level would that be on the left? Obviously you mean the level on the right where the mercury and water meet, but on the left side what level does that correspond to? The surface of the mercury above the line or some imaginary line at that same exact height (I'd assume it's the former)?

6. May 13, 2016

### billy_joule

The pressure in a fluid is the same at all points with same elevation, which is what P = ρgh states mathematically.
My mistake in #2 was ignoring that there were two different fluids (with different densities).

The same elevation, so, it's a horizontal line.

7. May 13, 2016

### Staff: Mentor

Let's take a step back. Imagine (or actually draw) a second horizontal dashed line across the U tube at the level of the interface between the mercury and the water in the right arm. Beneath this dashed line is only mercury in both arms. I think we can all agree that the highest pressure in the system is going to be at the very bottom of the U tube. Call this elevation z = 0, and call the pressure at this location $p_B$. Now consider the pressure in either arm at an elevation z that is less than the elevation of the interface $z_I$. In the right arm at elevation z, the pressure will be $p =p_B-\rho_mgz$, where $\rho_m$ is the density of mercury. What would be the pressure in the left arm at elevation z?

8. May 15, 2016

### horsedeg

Even this is confusing me. The highest pressure is at the lowest point, I suppose. "In the right arm at elevation z," but didn't we just declare the pressure at elevation z to be pB? Or am I misunderstanding what z is?

9. May 15, 2016

### Staff: Mentor

Where else would you think the highest pressure would be? When you swim to the bottom of a swimming pool, is the pressure highest at the surface or is it highest at the bottom?

10. May 15, 2016

### horsedeg

The pressure would be highest at the bottom.
So that equation, p=pB−ρmgz, would represent the difference in pressure between the lowest point and the elevation at z = zi, the imaginary line?

11. May 15, 2016

### Staff: Mentor

No. It would be the pressure at arbitrary elevation $0\leq z\leq z_I$. The difference in pressure between the lowest point and the elevation $z_I$ would be $p(z_I)-p_B=-\rho_m g z_I$, and the difference in pressure between the lowest point and the arbitrary elevation $0\leq z\leq z_I$ would be $p(z)-p_B=-\rho_m g z$.

Can you see why the pressure in the left arm at elevation $0\leq z\leq z_I$ would be exactly the same as the pressure in the right arm at elevation $0\leq z\leq z_I$?

12. May 15, 2016

### horsedeg

Yes, because they have only mercury at those elevations, right?

13. May 15, 2016

### Staff: Mentor

Excellent.

So, now we're in agreement that the pressures in the two arms of the U tube at the elevation $z=z_I$ (i.e., at the interface in the right arm) are equal, with the value equal to $p(z_I)$. We obtained this result by working our way up from the bottom of the U to elevation $z=z_I$.

Now, we are going to go in the reverse direction by working our way down from the top. Let

$h_w$ equal the height of the water column in the right arm above elevation $z_I$

$h_m$ equal the height of the mercury column in the left arm above elevation $z_I$

$p_a$ = atmospheric pressure prevailing at the top of both columns

So, in the right column, the pressure at elevation $z_I$, $p(z_I)$, will be given by:
$$p(z_I)=p_a+\rho_w g h_w$$where $\rho_w$ is the density of the water.

By this same rationale, what would be the equation for the pressure at elevation $z_I$, $p(z_I)$ in the left column?

14. May 15, 2016

### horsedeg

I think it would be $$p(z_I)=p_a+ρ_mg(h+h_2)$$

h being the given height in the diagram that we're trying to find, and h2 being the height between the two dotted lines, right?

15. May 15, 2016

### Staff: Mentor

Yes, that is correct. What do you think should be done next?

16. May 15, 2016

### horsedeg

I think you could compare the volumes, so that V1 = V2? Not sure if that would be right. If it's true, you could do A1(h+h2) = A2(height of water)

17. May 15, 2016

### Staff: Mentor

You have to set the left arm pressure equal to the right arm pressure at the lower dotted line. That is, you have to set the pressures in our last two equations equal to one another. This gives you an equation for $h+h_2$ in terms of $h_w$ (i.e., your 21.7 cm of water). So, you now have an algebraic equation involving the two unknowns h and h2. You need another equation.

In figure B, in terms of h, what is the volume of mercury above the upper dotted line? In terms of $h_w$ and $h_2$, what is the volume of water above the upper dotted line? The sum of these two volumes must be 100 cc, the total volume of water added. What do you get when you add these two volumes and set them equal to 100? This gives you your second algebraic equation involving the two unknowns h and h2.

18. May 16, 2016

### horsedeg

Okay, so from that equation, the P's cancel out because at the surface they have atmospheric pressure. Solving for h+h2, I get that h+h2=1.5956 cm.

In terms of h, the volume of mercury above the dotted line would be A1h. Not sure what the volume of water above the dotted line would be.

19. May 16, 2016

### Staff: Mentor

The height of the water column above the dotted line is $h_w-h_2$.