# Fluid Mechanics - Pressure - U-Tube Manometer

by Aerospace
Tags: fluid, manometer, mechanics, pressure, utube
 P: 39 A piston having a cross-sectional area of 0.07 m^2 is located in a cylinder containing water as shown in the figure. An open u-tube manometer is connected to the cylinder as shown. For h1=60 mm and h=100 mm, what is the value of the applied force, P, acting on the piston? The weight of the piston is negligible. I'm confused because there is more than one "u"-shape in the figure, so, it's unclear to me how to proceed. Help would be appreciated! legend for figure : p = piston first weird looking h = h1 second weird looking h = h piston is present on top of the water. Force P downwards on piston. Attached Thumbnails
 Sci Advisor HW Helper P: 3,033 The U on the right is the manometer. I've never seen a manometer filled with water on one side, but that's the way your diagram looks. Is there any air between the water and the mercury? It's not clear to me why the 60mm is given, but the height of the water relative to the top of the mercury is important
 P: 39 Here's a scan of the piston I'm talking about. Attached Thumbnails
Mentor
P: 40,689

## Fluid Mechanics - Pressure - U-Tube Manometer

This fact may help: Since the fluid is connected (and in hydrostatic equilibrium) the pressure at any height within the same fluid must be the same throughout.
 P: 39 Ok, so far this is what I've got. Let me know if I'm on the right track. If you take the pressure at the piston as P1 then, P1 + $$\gamma$$H2O*h1 - $$\gamma$$Hg*h = 0 Solve for p1 and then P(Force)= p1*Area
 Mentor P: 40,689 Don't forget that the tube is open to the air. (But you are on the right track.)
 P: 39 Hmmm. I did take that into account when I did -$$\gamma$$(Hg)*h
Mentor
P: 40,689
 Quote by Aerospace Hmmm. I did take that into account when I did -$$\gamma$$(Hg)*h
That's the pressure due to the column of mercury. But what about the air on top of the mercury?
 P: 39 well, isn't the pressure at the open end supposed to be equal to zero?
Mentor
P: 40,689
 Quote by Aerospace well, isn't the pressure at the open end supposed to be equal to zero?
Only if it's in a vacuum.
 P: 39 hmmm then what would it be? because my book says "at the open end the pressure is zero."
HW Helper
P: 3,033
 Quote by Aerospace hmmm then what would it be? because my book says "at the open end the pressure is zero."
This is a matter of interpretation. If the book says zero, they intend for you to find only the extra force applied to the piston that is added to the force of air pressure already acting. Air pressure is the same on both sides, so it can be neglected if all you need is the extra force.

Don't forget to calculate the force. If all you have done is caclulate the added pressure, you are not finished.
 P: 39 so the force P = p1*A, correct? I stated that above in my post...is this incomplete? And in any case, if Pair was not neglected, how would you write it into the equation?
 Sci Advisor HW Helper P: 3,033 Sorry. I had missed your force calculation. That is OK. If you had to include air pressure in the force you would need to look up the normal pressure of air (1 atmosphere) and multiply by the area of the piston to find the force the air exerts on the piston.
Mentor
P: 40,689
 Quote by Aerospace hmmm then what would it be? because my book says "at the open end the pressure is zero."
I was interpreting the problem as the piston being enclosed but the tube open to air. (But there's no good reason to think that the piston is enclosed.) I think OlderDan is correct, that they just want you to ignore the equal air pressure on both sides.

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