# Pressure on the wall of aquarium

1. Nov 6, 2015

### RyanTAsher

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

A large aquarium of height 5.00 m is filled with fresh water to a depth of 2.00 m. One wall of the aquarium consists of thick plastic 8.00 m wide. By how much does the total force on that wall increase if the aquarium is next filled to a depth of 4.00m?

2. Relevant equations

$p_1 + \rho g h_1 + \frac {1} {2} \rho v_{1}^{2} = p_2 + \rho g h_2 + \frac {1} {2} \rho v_{2}^{2}$ (Bernoulli's Equation)

$\Delta p = \frac {F} {A}$

3. The attempt at a solution

Since the pressure at any level of the fluid is constant at that height level, the pressure on the wall must be the sum of all the forces at each level of the water filled in the aquarium. The P is the change of the pressure at the point of depth, and the pressure at the top of the water.

1)

$p = \frac {dF} {dA}$

$dF = \Delta p dA$

$\int dF = \int \Delta p dA$

$F = \int \Delta p dA$

2)

For this particular scenario, we can use bernoulli's equation to extract a general equation for the change of pressure. Since the system is at rest, the velocity term of Bernoulli's equation will be removed.

$p_1 + \rho g h_1 = p_2 + \rho g h_2$

We will use $h_2$ as our reference point for height, so that will be a zero term as well.

$p_1 + \rho g h_1 = p_2$

Rearranging for $\Delta P$

$\rho g h_1 = p_2 - p_1 = \Delta p$

3)

We can now proceed to use this $\Delta p$ for the integral we devised above to sum the forces on the wall.

$F = \int \Delta p dA$

$F = \int \rho g h dA$

$F = \rho g h \int dA$

$F = \rho g h (h * w)$

$F = \rho g h^{2} w$

This is the answer I get (not for the final answer, I know how to do that part, just having trouble on the integral part).

They are getting $\frac {1} {2} \rho g h^{2} w$, while I'm not getting the $\frac {1} {2}$ part.

What am I doing wrong either in the formulations of my equations, or in my taking of the integral?

Last edited: Nov 6, 2015
2. Nov 6, 2015

### Staff: Mentor

You haven't taken into account the fact that the pressure is gradually increasing with depth, from a value of zero at the surface to a value of ρgh at the bottom. Your equations assume that the pressure is constant at ρgh over the entire depth.

3. Nov 6, 2015

### RyanTAsher

Okay, how would I take that into account in an integral? Isn't the sum already taking that into consideration?

Also, I'm still confused as to where they are getting that 1/2 in front of the $\rho g h^{2} w$

4. Nov 6, 2015

### RyanTAsher

Sorry, but, so in regards to your original statement. Would I have to take it into account by making $\Delta p = \rho g \Delta h = \rho g (h_1 - h_2)$, then proceed to take the integral of that?

5. Nov 6, 2015

### SteamKing

Staff Emeritus
If you draw the pressure profile along one wall of the aquarium, you'll see that the hydrostatic pressure p = 0 at the surface and p = ρgh at a depth of h meters below the surface.

The hydrostatic pressure on the sides of the aqarium would look like so:

Does this give you a hint as to where the factor of 1/2 comes from, when evaluating the force due to hydrostatic pressure on one wall of the aquarium?

6. Nov 6, 2015

### RyanTAsher

Ohhhh, so I was evaluating it at 2 walls essentially, since I was just doing the whole pressure going across one level?

7. Nov 6, 2015

### SteamKing

Staff Emeritus
I think in your original analysis, you overlooked some things.

For one wall of the aquarium, the hydrostatic force along one horizontal strip will be dF = P dA, where P = ρg h. For dA, since it is a horizontal strip, we can say dA = w * dh, where w is the width of the side and dh is a small height.

Plugging into dF = P dA,

dF = ρg h * w * dh = w * ρg * h dh

Integrating both sides of the equation above:

∫ dF = ∫ w * ρg * h dh

F = w * ρg ∫ h dh

F = w * ρg h2 / 2

which is basically saying that the hydrostatic force is equal to the volume of a triangular prism w units long × h units deep × ρgh units wide at the base.

8. Nov 6, 2015

### RyanTAsher

Oh! It makes so much more sense now that I saw you split the dA into it's width and height components! Thank you!