# Fluid pressure problem

1. Apr 5, 2013

### wcase

1. The problem statement, all variables and given/known data

The L-shaped tank shown in the figure is filled with water and is open at the top. If d = 4.81 m, what is the total force exerted by the water (a) on face A and (b) on face B?

image: http://edugen.wileyplus.com/edugen/courses/crs4957/art/qb/qu/c14/fig14_33.gif
2. Relevant equations

F=A*p

3. The attempt at a solution

For A I believe the solution should be 1000*9.81*2d*d^2 = 2183404N
and B should be 1000*9.81*2.5*d*d^2 = 2729255.8N
but these both seem to be wrong. Based on other solutions to the same question found around the internet, this should be the solution as well, but apparently it is not. Unless atmospheric pressure is supposed to be factored in, which I doubt it is, I have no idea why this is wrong.

2. Apr 5, 2013

### TSny

The tank is open at the top. Does your calculation take that into account?

[EDIT: Note that you want to find the total force exerted by the water alone on the surfaces. So, you don't want to include the atmospheric pressure acting on the outside of the surfaces. Maybe the solutions you found on the net are finding the total force on the surfaces from both the water and the outside atmospheric pressure.

But, you will need to take into account the fact that the tank is open to the atmosphere. So you will start with atmospheric pressure at the top of the tank, and than as you go down into the water, the pressure increases by ρgh.]

Last edited: Apr 5, 2013
3. Apr 5, 2013

### wcase

I doubt it, but the question only refers to the pressure from the water. And even if atmospheric pressure was factored in, assuming the tank is completely exposed to the atmosphere, shouldn't any effects cancel out since it would be acting on the top and bottom of the tank?

And when I did factor in the atmospheric pressure (d*1e5) my answers were still wrong.

4. Apr 5, 2013

### TSny

Sorry, I edited by first post while you were posting your second post. Do the remarks in my edit help?

5. Apr 5, 2013

### wcase

I understand, but I did not add any atmospheric pressures initially.
If I search the problem on the internet, and change the d value to whatever the alternate problem is, and do 1000*(2*d)*d^2, with 1000 being the density of water, 2d being the height of water above face A, and d^2 being the Area of the face, I get the answer that those problems claim is the answer, but when I do the same calculation with this problem it is wrong.

6. Apr 5, 2013

### TSny

I guess it depends on the interpretation of the phrase "force exerted by the water". If you want just the force due to the pressure of the water, then you need to add the pressure at the top of the tank (atmospheric pressure) to your calculations. If you want the net total force on a surface due to the water and the outside atmospheric pressure acting on the surface, then the net force would be what you calculated.

Since you are not getting the correct answer by ignoring the pressure at the top of the water in the tank, maybe you should include it and see.

7. Apr 5, 2013

### wcase

I just add d*atmospheric pressure to the previous calculations, right?

(1000*(2*d)*d^2)+(d*1e5)

8. Apr 5, 2013

### TSny

No, first find the pressure of the water at the point you are working with. The pressure at a point in a static fluid is Po + ρgh where Po is the pressure at the top of the fluid. In your case, Po is atmospheric pressure since the top of the water is exposed to the atmosphere. After getting the pressure, use it to find the force.

9. Apr 5, 2013

### SteamKing

Staff Emeritus
For surface A, what is the area? What is the pressure at the top? What is the pressure at the bottom? For pressure, ignore atmospheric pressure.