[Physics] Force on the bottom of a tank full of liquid – Hydrostatic Pressure or Gravity

fluid-staticsforcespressure

Imagine a tank filled with water that has some height $h$ and at the bottom area $A$ but as it goes up, for example at height $h/2$, it's area is now $A/2 $. What's the correct way to calculate the force at the bottom of the tank? (Let's ignore atmospheric pressure for now)

If I use $W=mg$, we get $F=W=ρVg=ρ(\frac{Ah}{2}+\frac{Ah}{4})g=\frac{3}{4}ρghA$

If I calculate the hydrostatic pressure at the bottom, it's $p=ρgh$, and then $F=pA=ρghA.$

Which one is the correct one and why?

Best Answer

As @Berend mentioned you are calculating two different things. The first calculation gives you the weight of all water in the tank which is what an scale would read.(internal forces explained in the second part cancel out.)
In the second case though you are calculating hydrostatic pressure of water. Theses answers are different because water in the tank applies force $f$ to the tank upwards as shown in the figure and according to Newton's laws the tank applies force $f$ downwards so hydrostatic pressure gets bigger than water weight and their difference is as much as the weight of water in the stripped region.

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.

The pressure at D is not significantly greater than the surrounding atmosphere. The jet of fluid is surrounded by the atmosphere, and the pressure within the jet (like at point D) is virtually identical to atmospheric pressure.

What you are missing is what is happening inside the tank in the immediate vicinity of the exit hole. The hydrostatic pressure distribution inside the tank is disturbed by the fluid flow converging toward the exit hole. The effective region where this disturbance occurs is within just a few hole diameters of the exit. Thus, the point A in the figure is just about close enough to the exit hole for this disturbance to begin to be felt. In the region near the exit hole, the pressure rapidly decreases from the hydrostatic pressure variation away from the hole to atmospheric pressure at the exit hole. This localized pressure gradient provides the driving force for the flow out the hole. The surfaces of constant pressure are roughly hemispherical, with their center at the hole. The streamlines near the hole are all converging radially toward the hole. As the flow converges toward the hole, the fluid velocity increases, and the pressure decreases.

## Best Answer

As @Berend mentioned you are calculating two different things. The first calculation gives you the weight of all water in the tank which is what an scale would read.(internal forces explained in the second part cancel out.) In the second case though you are calculating hydrostatic pressure of water. Theses answers are different because water in the tank applies force $f$ to the tank upwards as shown in the figure and according to Newton's laws the tank applies force $f$ downwards so hydrostatic pressure gets bigger than water weight and their difference is as much as the weight of water in the stripped region.