So I understand that if we have a system that involves a container of water the pressure will equal atmospheric pressure at the top and as we go further down the container the pressure will increase with depth until it is at its maximum pressure at the bottom of the container. In terms of a 1-D fluid equation (neglecting viscosity) we have

$$\frac{\partial v}{\partial t} = \frac{1}{\rho} \frac{\partial p}{\partial z} – \frac{\partial v}{\partial z} + g$$

$$0 = \frac{1}{\rho} \frac{\partial p}{\partial z} – 0 + g$$

$$\frac{\partial p}{\partial z} = \rho g$$

$$p = \rho g z + const$$

$$p = \rho g z $$

where we have taking the constant to be $0$.

Now say you removed the bottom of this container. Obviously the water starts to flow out the bottom. So the velocity changes with time. As the fluid is incompressible we still have $\frac{\partial v}{\partial z} = 0$.

So

$$\frac{\partial v}{\partial t} = \frac{1}{\rho} \frac{\partial p}{\partial z} + g$$

But if pressure still balances with gravity we have no fluid flow! But the fluid does flow so the pressure must not balance with gravity anymore. So what is the pressure in the system now that the bottom of the container has been removed?

## Best Answer

Take a 1-cm square tube and place it vertically in the container from top to bottom, touching the bottom so that the bottom of the container is the bottom of the tube.

The pressure at the bottom of the tube is nothing but the weight of water it is supporting - the water in the tube. Supporting means to keep from falling. (Forget the air pressure - that's just confusing the issue.)

If you take away the bottom of the container, you are taking away the bottom of the tube, so it is no longer supporting the column of water, so the weight of water above goes to zero because it is not being supported. So the pressure goes to zero because it's nothing more than the weight of water being supported.

You don't need equations to understand this.