The Pressure in a Fluid Under Gravity Forces

Consider an element of a fluid in the form of a vertical cylinder, located as shown in Fig. 1.

If we draw a free-body diagram of this element in equilibrium, we will have acting the forces shown; a force $pdA$ acting vertically downward on the top of the element, an upward force $(p+dp)dA$ acting on the bottom of the element, and a gravity force equal to the weight of the fluid in the element. Using the notation:

$\rho =$ the density of the fluid $=$ the mass per unit volume $\gamma =\rho g=$ the specific weight of the fluid $=$ the weight per unit volume, where $g$ is the acceleration of gravity

the weight of the fluid in the element is: $$\gamma dAdh.$$ For equilibrium, we have: $$\begin{array}{c}\sum F_y=0=(p+dp)dA-pdA-\gamma dAdh\\ dp-\gamma dh=0\\ \frac{dp}{dh}=\gamma \end{array}$$ From this relationship we can determine the variation of pressure with depth in an incompressible fluid, for which $\gamma$ is a constant: \begin{aligned} \int_{p_{1}}^{p_{1}} d p & =\gamma \int_{n_{1}}^{h_{2}} d h \\ \left(p_{2}-p_{1}\right) & =\gamma\left(h_{2}-h_{1}\right) \end{aligned} hence the difference in pressure at two points varies directly with the difference in the depth of the two points.