Re: In a fish tank how many pound of force is applied on each wall of the tank.

Hi Scott,

There are a couple of different possible quantities of interest here. There is the total force exerted on the walls of the tank. Then there is the pressure (the force exerted per unit area) exerted on the walls, which varies with depth in the tank.

For a reference, what I have handy is the fifth edition of "Fundamentals of Physics," by Halliday, Resnick and Walker. The same information can, however, be found in essentially any introductory physics text, usually in a chapter on fluids.

Rather than pounds, I will use metric units, in which force is measured in Newtons, and pressure in Pascals (1 Pascal = 1 Newton per square meter). As you may realize, pressure increases with depth in a fluid. The rate of increase is given by P = P_ext+r*g*h. P _ext is the external pressure, above the surface of the liquid, r is the density of the liquid. The term g is the acceleration of gravity, and h is the depth into the fluid (the letter d would be more obvious, but we prefer not to use it because of confusion when calculus gets involved). For water, P_ext is the pressure of the air at the surface (about 10^{5 Pascals at sea level), r is about 1000 kg/m3 (kilograms per cubic meter), and the acceleration of gravity is 9.8 m/s2 (meters per second per second). Any physics text can give you more accurate numbers, as well as the conversions between Newtons and pounds.}

You may perhaps know of the concerns faced by scuba divers after they dive to depths that may not seem so far to us (200 feet is a pretty deep dive). The pressure at a given depth in a fluid, however, varies directly with the density of that fluid. The density of air is only about 1.2 kg/m³ , and the pressure change from a 200 foot ocean dive is equivalent to that from an altitude change in air of almost 170,000 feet! Much thicker than the dense region of Earth's atmosphere.

The maximum pressure in your fish tank will therefore occur at the bottom of the tank, and will be given by the equation above, using the height of the tank for h.

To get the total force exerted on the walls of the tank, you'll need to do a simple integral. If you already know basic calculus, this won't be a problem. If not, then imagine dividing the walls of the tank into horizontal strips. Make the vertical height of the strips so small that the pressure doesn't change noticeably from top to bottom. The total force exerted outward by the water on that strip is then P(h)*L*dh. By P(h) I am indicating that P varies with depth, h (we read this as, "P is a function of h"). The term L is the horizontal length of the strip (the width of the face of the tank), while dh is its vertical extent (we use dh to indicate a really tiny change in h). The total force is then just the sum of the force on each strip.

As dh becomes smaller and smaller, the sum turns into an integral (an integral is really just a fancy way of doing a sum). I can't represent it easily in html, but it looks like
F=Integral[P(h)*L*dh]
where F is the total force, and the limits of h for the integral are at 0 (top of the tank) and H (the total depth of the tank). P(h) is given from the first equation above, and so
F=Integral[(P_ext+r*g*h)*L*dh]
The pressure is a sum of two terms, one that is a constant, and one that increases with depth into the tank. The sum means that we can write this as two integrals like this:
F=Integral[P_ext*L*dh]+Integral[r *g*h*L*dh].

Without going into too much detail, the first integral gives the simple result (because nothing is changing with h) of P_ext*L*H. This is just the pressure that would be exerted by air on the inside of the tank. Because the term in the second integral is increasing with h, it gives the slightly more complicated result of 0.5*r*g*L*H². The total force exerted is then just the sum of the two terms:
F= P_ext*L*H + 0.5*r*g*L*H².

The next question is whether you want the total pressure exerted by the water, or the net pressure. Remember that the air outside is also pressing inwards. The inward pressure is just P_ext*L*H (because air has such a low density, we can treat the air pressure as constant over a vertical distance equal to the height of the tank, and so the total force is just this constant pressure times the area of the wall of the tank). This cancels the first term above, leaving the net force as
F_net=0.5*r*g*L*H².