Lateral force of water on the walls of a tank

In summary, the lateral force on a vertical wall of a tank filled with fluid can be calculated by multiplying the height and width of the wall by half the density of the fluid and the acceleration due to gravity. The force on the base of the tank can be calculated by multiplying the height of the tank by the total mass of the fluid and the acceleration due to gravity. To convert from mass to weight, the weight density of the fluid should be used.
  • #1
jimhebert
29
0
I want to build a plywood fish tank 16' long x 4' wide x 3.5' high. I need to know how to calculate the lateral force at any point on the walls in order to build with strong enough lumber to support the lateral pressure and also the weight pushing down. I know that one gallon weighs 8.345 lbs. and one cubic foot has 7.48 gallons, so it's not too difficult to calculate the weight pushing down. The lateral force is where comes the problem.
 
Physics news on Phys.org
  • #2
jimhebert said:
I need to know how to calculate the lateral force at any point on the walls.
You don't have a force acting at a point, you have a pressure:
http://en.wikipedia.org/wiki/Fluid_statics#Hydrostatic_pressure

pressure = depth * gravitational_acceleration * water_densityThe force is the integral of pressure over an area of the wall. To get the total force, you can take the average pressure. Assuming a rectangle:

pressure_avg = 0.5 * total_depth * gravitational_acceleration * water_density

force_total = wall_area * pressure_avg

But the force is not evenly distributed. The wall must be stronger at the bottom.
 
Last edited:
  • #3
I think that the formula is to calculate the average pressure pushing down on the floor, not any point on the wall which will decrease as you go up the wall.
 
  • #4
jimhebert said:
I think that the formula is to calculate the average pressure pushing down on the floor, not any point on the wall which will decrease as you go up the wall.
This applies to any point at the wall (depth = depth of the point):

pressure = depth * gravitational_acceleration * water_density
 
  • #5
So let me see if I got this correctly. If I have a tank that is 16’ x 4’ x 3.5 ‘ and I want to know the pressure at a point that is at 3’ below the top of any of the four walls, then I would calculate the following: 3’ x 32 ft/sec/sec x 62.22 lb/cubic foot = 5973 pounds.
Is that correct? Is it 5973 lb/square foot or per square inch or is it on an area the size of a pencil point?
 
  • #6
Pressure = height times density

Pressure = 3 feet x 62 lb/cu ft = 186lbs/sq ft

So it's like, roughly, somebody standing on a patch 1 ftx 1ft

[the formula in the previous posts above has the wrong units...]
 
Last edited:
  • #7
Naty1 said:
Pressure = height times density

Pressure = 3 feet x 62 lb/cu ft = 186lbs/sq ft

So it's like, roughly, somebody standing on a patch 1 ftx 1ft

[the formula in the previous posts above has the wrong units...]

The formula in the previous posts is correct if we are using the usual definition of density, which is mass per unit volume. It will give correct results if we are using SI units, or any other absolute system of units.

Your formula is only correct if we define density as weight per unit volume. It will work in a system where "pound" is used both as a unit of mass and of force, so it's better suited to jimhebert's needs.

To jimhebert: you have worked out how to calculate the total weight on the floor of the aquarium. Divide this by the area of the floor in feet and you will get the pressure, in pounds per square foot. The pressure at a point on the wall just near the floor will be almost the same as this. The pressure on a point halfway up the wall will be half this pressure and the pressure at the very top will be zero.
 
  • #8
Michael C said:
The formula in the previous posts is correct if we are using the usual definition of density, which is mass per unit volume. It will give correct results if we are using SI units, or any other absolute system of units.

Your formula is only correct if we define density as weight per unit volume. It will work in a system where "pound" is used both as a unit of mass and of force, so it's better suited to jimhebert's needs.

To jimhebert: you have worked out how to calculate the total weight on the floor of the aquarium. Divide this by the area of the floor in feet and you will get the pressure, in pounds per square foot. The pressure at a point on the wall just near the floor will be almost the same as this. The pressure on a point halfway up the wall will be half this pressure and the pressure at the very top will be zero.

No, I haven’t calculated the total weight on the floor. The way to do that would have been 16x4x3.5 = 224 cu.ft.

Then 224x62.22 lb/cu.ft = 13,937 lbs. Total

Therefore, 13937 divided by the area of the floor (64) =218 lb/sq ft.

So based on the last part of your explanation, the lateral force pushing out from a point 2 feet deep 2/3.5 x 218 = 125 pounds/sq ft.

Those are the correct units as far as the math is concerned, but not the correct units for for the answer to the problem. The answer has to be in pounds.
 
  • #9
The pressure in a fluid at depth [itex]z[/itex] from the surface is [itex]P(z) = \rho g z[/itex], where [itex]\rho[/itex] is the density of the fluid, and [itex]g[/itex] is the acceleration due to gravity.

The get the total force on a surface, the pressure would need to be integrated across the surface.

[tex]F = \int \limits_S P da[/tex]

For a vertical surface (like a wall of the tank) the force is:

[tex]F_{wall} = \int \limits_S P da =\rho g \int_0^w \int_0^h z dz dx [/tex]
[tex]F_{wall} = \frac{1}{2} \rho g w h^2[/tex]

where [itex]w[/itex] is the width of the wall and [itex]h[/itex] is the height of the wall. For the base:

[tex]F_{base} = \int \limits_S P da =\rho g h \int_0^w \int_0^l dy dxl[/tex]
[tex]F_{base} = m g[/tex]

where [itex]m[/itex] is the total mass of the fluid.

To get from mass to weight, replace [itex]m g[/itex] with the weight (i.e. pounds) and replace [itex]\rho g[/itex] with the weight density (i.e. pounds/foot^3).
 
  • #10
You have the total weight right, but you still aren't getting the lateral force right. At a given depth, the pressure on the wall at that point is:
[tex]P = \rho g h[/tex]

Where:
[itex]P[/itex] is pressure (force per area)
[itex]\rho[/itex] is density (mass per volume)
[itex]g[/itex] is the acceleration due to gravity (length per time per time)
[itex]h[/itex] is depth (length)

You have to make sure that since you insist on using Imperial units, that your units remain consistent. If you are taking your density in pounds per cubic foot, that is essentially the [itex]\rho g[/itex] term.

Now, that gives you the pressure at a given depth. Force is pressure times area, but you know that the pressure varies with depth, so you have to integrate through the depth to get the total force.

[tex]dF = P\;dA = P\;dx\;dh[/tex]

Where:
[itex]dA[/itex] is a differential area
[itex]dx[/itex] and [itex]dh[/itex] are differential lengths that describe the area [itex]dA[/itex]

Now let's say the x-direction just the other coordinate for a given wall. You need to integrate the pressure over the area of the wall to get the force.

[tex]F = \int\limits_0^D \int\limits_0^W \rho g h\;dx\;dh[/tex]

Where:
[itex]F[/itex] is force
[itex]D[/itex] is the total depth
[itex]W[/itex] is the width of the side
and the rest have already been defined.

For vertical, rectangular walls, this simplifies to:
[tex]F = \rho g W \int\limits_0^D h\;dh[/tex]
[tex]F = \rho g W \left[\frac{h^2}{2}\right]_0^D[/tex]

So, that leaves you with:
[tex]F = \frac{\rho g W D^2}{2}[/tex]

Just make sure your units are consistent.

---

Looks like Jasso beat me to it. Just see whose description makes more sense to you, haha.
 
Last edited:
  • #11
Jasso and boneh3ad have given the formula for calculating the total force on one side. In your case, since you're using imperial units (pounds for both mass and force), the [itex]\rho g[/itex] term is 62.42 lb per cubic foot. So, for example, the total force on one of the longer walls would be:

62.42 x 16 x 3.52/2 lb ≈ 6117 lb

But remember that that force is spread over the whole side, and not evenly spread: there is more pressure at the bottom than at the top.
 
  • #12
Michael C said:
Jasso and boneh3ad have given the formula for calculating the total force on one side. In your case, since you're using imperial units (pounds for both mass and force), the [itex]\rho g[/itex] term is 62.42 lb per cubic foot. So, for example, the total force on one of the longer walls would be:

62.42 x 16 x 3.52/2 lb ≈ 6117 lb

But remember that that force is spread over the whole side, and not evenly spread: there is more pressure at the bottom than at the top.

Is it correct to say that the total force acting on the upper rectangular section of one of the 16' walls that is 2' from the top would be 62.42 x 16 x 2 squared divided by 2 = 1997 lb ?
 
  • #13
jimhebert said:
Is it correct to say that the total force acting on the upper rectangular section of one of the 16' walls that is 2' from the top would be 62.42 x 16 x 2 squared divided by 2 = 1997 lb ?

Yes, that's correct. By subtracting this result from the previous one that gives the total force on the whole of that wall, you can also calculate that the total force on the rest of the wall is 6117 lb - 1997 lb = 4120 lb.
 
  • #14
We know that the force per square inch on the wall will increase as we go from top to bottom. Also, if my logic is correct, the force per square inch should be the same along a horizontal line drawn across the wall. With this information that we have previously calculated, how do we now calculate the force per square inch at any given point on that wall. Let's say, what would be the force per square inch at any point on a horizontal line 2 feet below the top edge of the wall?

I think I figured it out. The force per square inch would be the same or very near the force or weight pushing down. I figured .8 lb. but doesn't sound reasonable.
 
Last edited:
  • #15
jimhebert said:
I think I figured it out. The force per square inch would be the same or very near the force or weight pushing down. I figured .8 lb. but doesn't sound reasonable.

That sounds reasonable enough to me.
 
  • #16
OK, now I've got to find the supports that will keep the tank from rupturing due to pressure. Thanks for the help.
 
  • #17
http://www.fnzas.org.nz/?p=1732: it gives a detailed method for calculating the glass thickness for aquarium walls.
 
  • #18
Appreciate it, Michael. We'll give a try.
 

1. What is the lateral force of water on the walls of a tank?

The lateral force of water on the walls of a tank is the sideways pressure exerted by the water on the walls of the tank. This force is caused by the weight of the water and can be significant, especially in large tanks or when the water level is high.

2. How is the lateral force of water on the walls of a tank calculated?

The lateral force of water on the walls of a tank can be calculated using the formula F = ρgh, where F is the force in Newtons, ρ is the density of water (1000 kg/m³), g is the acceleration due to gravity (9.8 m/s²), and h is the height of the water above the wall in meters. This formula assumes the tank is completely filled with water.

3. How does the shape of the tank affect the lateral force of water?

The shape of the tank can affect the lateral force of water in two ways. First, a tank with curved walls will distribute the force more evenly compared to a tank with straight walls. Second, a tank with a larger surface area will have a higher lateral force compared to a tank with a smaller surface area.

4. What factors can affect the lateral force of water on the walls of a tank?

Apart from the shape and size of the tank, the lateral force of water can also be affected by the density of the water, the acceleration due to gravity, and the height of the water above the wall. Additionally, any external forces such as wind or earthquakes can also increase the lateral force on the walls of the tank.

5. How can the lateral force of water on the walls of a tank be reduced?

To reduce the lateral force of water on the walls of a tank, the tank can be designed with curved walls or additional reinforcement can be added to the walls. Additionally, reducing the water level in the tank can also decrease the lateral force. Proper maintenance and inspection of the tank can also help prevent any potential damage from the lateral force of water.

Similar threads

Replies
12
Views
772
  • Mechanical Engineering
Replies
11
Views
2K
  • Engineering and Comp Sci Homework Help
Replies
2
Views
2K
Replies
8
Views
3K
  • Introductory Physics Homework Help
Replies
7
Views
1K
  • Mechanical Engineering
Replies
5
Views
361
Replies
17
Views
2K
Replies
2
Views
9K
Replies
7
Views
3K
Back
Top