# Resultant force on the side of a tank of water

A closed tank rectangular in plane with vertical sides is 1.8m deep and contains water to a depth of 1.2m.

Air is pumped into the space above the water until the air pressure is 35kN/m2.

If the length of one wall of the tank is 3m, determine the resultant force on this wall and the height of the centre of pressure above the base.

So far my calculations have been to calculate the pressure due to the water which is: rho(g)(h) = 1000(9.81)(1.2) = 11,772 N/m^2. I then multiply this by the arc of the side of the tank covered by water which is: 1.2(3) = 3.6m^2.

Multiplying these gives 42379.2N.

I am not entirely sure what to do with the air pressure. Whether to multiply it again by 3.6 (the area of the side of the tank covered by water) or to multiply it by the area of the top of the water but I would not have enough details to calculate this.

• Is the air pumped into the tank resulting in pressure that is above atmospheric pressure? There are two ways to specify pressure: 1) absolute pressure; 2) gauge pressure, which is everything above or below ambient pressure. – David White Feb 20 '17 at 2:57

Above the water line, the pressure on the wall is that of air pressure, so you can calculate the area of the wall above the water line and multiply it by the air pressure to get the force on that section. Below the water line, the pressure varies with depth as $$P=\rho g h +P_0$$ where $P_0$ is the air pressure above the water and $h$ is the distance below the water. To get the force below the water line you need to integrate the pressure. For each section of the wall with height $dh$, the force is given by $$dF = (\rho g h +P_0)Ldh$$ You then need to integrate this force to the bottom of the tank and add it to the force from the top of the tank.