What is the Relationship Between Pressure and Height in Fluid Mechanics?

[Tekneek]

The correct answer to this problem is h2 = h1 but I don't get it. According to what I understand,

F=p/A

So for the Tube on the left:

F1 = P/A = ($\rho$*g*h1)/A1

For tube on the right:

F1 = P/A = ($\rho$*g*h1)/3*A1

I get h2=3h1 (i know if you think about it this does not make sense, but i don't what i am not getting).

 

[1994Bhaskar]

I think your solution is wrong... it should come h1=3h2...but it's the last line of the question that's bugging me... i think we got to understand h2 relative to h1...

 

[paisiello2]

This problem illustrates the concept of a hydraulic piston.

First of all, a small minor correction in your formula where F=P*A, not F=P/A.

Next, you have to accept the principle that for a given fixed depth a body of water will always have the same pressure everywhere. This means the pressure where the force F is applied in the left column is the same pressure at the same elevation in the right column. I think you made the mistake of equating forces which is not correct. If you equate the pressures then I think you'll see h1 = h2.

A hydraulic piston or cylinder is a way to use a small force over a large distance to move a large force over a small distance.

 

[Tekneek]

This problem illustrates the concept of a hydraulic piston.

First of all, a small minor correction in your formula where F=P*A, not F=P/A.

Next, you have to accept the principle that for a given fixed depth a body of water will always have the same pressure everywhere. This means the pressure where the force F is applied in the left column is the same pressure at the same elevation in the right column. I think you made the mistake of equating forces which is not correct. If you equate the pressures then I think you'll see h1 = h2.

A hydraulic piston or cylinder is a way to use a small force over a large distance to move a large force over a small distance.

Yeah I agree the equation i had was wrong..but how can the height of the water level on system B be equal to System A when the area on the system B is 3 times the area of system A? I mean the volume of water on both system is same..where would the extra water for system B come from?

Doing math i do get h2=h1

For system B:

Fout = 3F1
P=(3F1)/(3A1) = F1/A1
ρgh1=F1/A1
h1=F1/(A1*ρ*g)

For system A:
Fout=Fin
ρgh2=F1/A1
h2=F1/(A1*ρ*g)

So h1=h2

 

[paisiello2]

The volume of water is irrelevant in both systems. It could be the same or they could have added more water, it makes no difference to the answer.

I think you might still be trying to equate forces so starting with Fout and Fin seems confusing to me and also I think you mixed up h1 and h2 between the two systems.

Just equate the pressures P1 = P2 and I think it would answer your question immediately.