Force in connected tubes with unequal water heights

[Rockdog]

See the picture.
1 atm = 101300 Pa = 101300 N/m2
density of water = 1000 kg/m3

The tube is filled with water. A1 = 0.05m^2 and A2 = 0.08m^2. Two pistons apply different forces to the water in the tube so that the water in the right side of the tube is a height h = 0.43m above the height of the water in the left side of the tube. If F2 = 138 N what is F1?
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I know general equation is F/A equals pressure.

Well, I wanted to do F1/A1=F2/A2... but the fact that both sides of the tube are not level gives me a problem.

It's probably a simple concept, but I'm not seeing it. HOw do I set it up to account for the height=.43m factor?

 

[Chi Meson]

The pressure is equal in both tubes at exactly the same height. THe pressure in the higher tube decreases according to the static fluid pressure formula, P=(rho)gh .

So the pressure at the surface of the right tube is (rho)gh less than the other.

 

[Doc Al]

Originally posted by Rockdog
Well, I wanted to do F1/A1=F2/A2... but the fact that both sides of the tube are not level gives me a problem.

It's probably a simple concept, but I'm not seeing it. HOw do I set it up to account for the height=.43m factor?

The pressure in the fluid is the same at all points with the same height. So... set the pressure at the left side equal to the pressure on the right at depth h. Make sense?

 

[HallsofIvy]

What both Rockdog and Doc Al are saying is that the total force on the right (higher) side is F₂+ the weight of the additional water. Since the additional height is h and the area is A₂ the volume is hA₂ and the weight is that times g times the density of water (which I'll call δ): The total Force on the right is F₂+ gδhA₂. The pressure is then F₂/A₂+ gδh and that must equal F₁/A₁.