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Lateral force of water on the walls of a tank

  1. Feb 26, 2012 #1
    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.
  2. jcsd
  3. Feb 27, 2012 #2


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    You don't have a force acting at a point, you have a pressure:

    pressure = depth * gravitational_acceleration * water_density

    The 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: Feb 27, 2012
  4. Feb 27, 2012 #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.
  5. Feb 27, 2012 #4


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    This applies to any point at the wall (depth = depth of the point):

    pressure = depth * gravitational_acceleration * water_density
  6. Feb 27, 2012 #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?
  7. Feb 27, 2012 #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: Feb 27, 2012
  8. Feb 27, 2012 #7
    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.
  9. Feb 27, 2012 #8
    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.
  10. Feb 27, 2012 #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).
  11. Feb 27, 2012 #10


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    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]

    [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]

    [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]

    [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: Feb 27, 2012
  12. Feb 27, 2012 #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.
  13. Feb 27, 2012 #12
    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 ?
  14. Feb 28, 2012 #13
    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.
  15. Feb 28, 2012 #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: Feb 28, 2012
  16. Feb 28, 2012 #15
    That sounds reasonable enough to me.
  17. Feb 28, 2012 #16
    OK, now I've got to find the supports that will keep the tank from rupturing due to pressure. Thanks for the help.
  18. Feb 29, 2012 #17
  19. Feb 29, 2012 #18
    Appreciate it, Michael. We'll give a try.
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