# Static pressure on the bottom of cylinder [closed]

i have a little doubt about a very simple problem regarding static pressure on the bottom of two different cylinder. We have the same quantity of liquid inside two cylinder with different section, as displayed in figure that follows The questions are:

• Since there is the same amount of liquid in the two cylinders, the force exerted by the liquid on the bottom of the two cylinders is the same, right?
• Since the section of the cylinders is different the pressure will be different, right?
• In particular in the cylinder having section $$S_1$$, with $$S_1 >S_2$$, the pressure will be lower, correct?

I know that is a very stupid problem, but some colleagues of mine says that on the bottom of the two cylinder we have the same static pressure but i was not agree with that.

If the density of the liquid is $$\rho$$ then the weight of the liquid in the first cylinder is $$W_1 = \rho h_1 S_1 g$$ and the pressure at the bottom of the first cylinder (neglecting atmospheric pressure) is $$P_1=W_1 / S_1=\rho h_1 g$$. For the second cylinder we have $$W_2 = \rho h_2 S_2 g$$ and $$P_2=W_2 / S_2=\rho h_2 g$$.

So, yes, if $$h_1 then $$P_1. Note that this applies even if the volume of liquid inside the two cylinders is not the same - pressure is always proportional to depth.

Assuming that only gravity is at work here and that the gravitational acceleration is the same on all the liquid, the force exerted on the bottom of the cylinder will be the same, independently of the section of the cylinder. That force will just be $$mg$$ where $$m$$ is the mass of water inside the cylinder.

On the other hand, the static pressure won't be the same being the pressure a measure of oh much force is exerted per unit area. In fact the two pressures will be $$P_1 = \frac{mg}{S_1}<\frac{mg}{S_2} = P_2$$

This is why we can produce diamonds using not so big of a force by concentrating that force on a really small surface area. On the tip of the apparatus which is pressing on carbon like object, the pressure will be so high that a diamond can be created. This is an oversimplification but you get the gist. If now the same force is spread over a bigger surface, at a certain point, you won't get enough pressure to make a diamond.

Both volumes weigh the same. The pressure due to these weights differ though. The pressure on the small surface is higher than that on the big surface.

The same holds for the hydrostatic pressure. The pressure on the water due to the air in the closed space is of course dependent on how much air is contained in that closed space. This pressure can be less, equal, or higher than the pressure imparted by the mass of the water. And again, if the air pressures are equal in both containers, the force on the surfaces will be distributed so that the pressure on the small surface is higher than that on the large surface.

You are correct in all your answers. Since the pressure is force/area you get: $$\begin{equation} P_1 = \frac{\rho_{liquid}h_1S_1g}{S_1} = \rho_{liquid}gh_1\text{,} \end{equation}$$

which is a very well known formula for the pressure of a liquid column. The same stands for $$P_2 = \rho_{liquid}gh_2$$. So it can be seen, using the $$P=\rho gh$$ formula, that your answers are right.