Fluid Mechanics - Pressure - U-Tube Manometer

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

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 Recognitions: Homework Help Science Advisor 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
 Here's a scan of the piston I'm talking about. Attached Thumbnails

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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.
 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 Blog Entries: 1 Don't forget that the tube is open to the air. (But you are on the right track.)
 Hmmm. I did take that into account when I did -$$\gamma$$(Hg)*h

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 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?
 well, isn't the pressure at the open end supposed to be equal to zero?

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 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.
 hmmm then what would it be? because my book says "at the open end the pressure is zero."

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