In the book Serway-Jewett pag. 420 (file is available via the following URL https://ufile.io/p6qru1g0, free on the net), for the Stevin'law I read:
Now consider a liquid of density $\rho$ at rest as shown in figure. We assume $\rho$ is uniform throughout the liquid, which means the liquid is incompressible. Let us select a parcel of the liquid contained within an imaginary block of cross-sectional area $A$ extending from depth $d$ to depth $d + h$. The liquid external to our parcel exerts forces at all points on the surface of the parcel, perpendicular to the surface. The pressure exerted by the liquid on the bottom face of the parcel is $p$, and the pressure on the top face is $p_0$. Therefore, the upward force exerted by the outside fluid on the bottom of the parcel has a magnitude $pA$, and the downward force exerted on the top has a magnitude $p_0A$. The mass of liquid in the parcel is $M=\rho V$; therefore, the weight of the liquid in the parcel is $Mg=\rho Ahg$. Because the parcel is at rest and remains at rest, it can be modeled as a particle in equilibrium, so that the net force acting on it must be zero. Choosing upward to be the positive $y$ direction, we see that:
$$\sum F_y= pA-p_0A-Mg=0$$
hence
$$p=p_0+\rho gh \tag 1$$
That is, the pressure $p$ at a depth $h$ below a point in the liquid at which the pressure is $p_0$ is greater by an amount $\rho gh$. If the liquid is open to the atmosphere and $p_0$ is the pressure at the surface of the liquid, then $p_0$ is atmospheric pressure.
Question:
I can't quite compute the heights to put in Stevino's law because looking at the Serway has me confused. Can you give me a preferably comprehensive explanation or answer of how to compute the actual pressures?
I think that if the barrel have height $w$ and it is open in top there is the atmosferic pressure $p_{\mathrm {at}}$ plus the pressure in a point $B$ in the depth $w$ at the bottom.
Hence: $$p=p_{\mathrm {at}}+\rho gw$$
If point $B$ is not in the bottom of the barrel called $h$ the distance between the water surface and a point in the fluid then
$$p=p_{\mathrm {at}}+\rho gh$$
If the barrel have a mobile piston and there is a weight on it, three pressures act:
$$p=(p_{\text{at}}+p_{\text{weight}})+\rho gh=p_{\text{ext}}+\rho gh$$
$$p_{\text{at}}+p_{\text{weight}}=p_{\text{ext}}$$
As far as I can increase the acting force (within certain limits) it does not produce compression, and the liquid remains in static equilibrium.
This should be correct, I think, looking at the Halliday-Resnick figure.
