A substance that can flow is called fluid. Unlike solids, liquids and gases can flow and hence they are fluids. Liquids are said to be nearly incompressible while gases are easily compressible.
When you immerse a body in a fluid such as water, the fluid exerts normal force on the surface of the body. In Figure 1 a body is immersed in a fluid and the fluid exerts the force perpendicular to the surface of the body. The force exerted by the fluid on the body at a point on its surface is always perpendicular to the surface at that point regardless of the body's shape.
If an infinitesimally small normal force of magnitude \(d{F_n}\) acts over an infinitesimally small surface area \(dA\), the pressure is the force \(d{F_n}\) divided by \(dA\). Therefore, the pressure \(P\) over the small area is
\[P = \frac{d{F_n}}{{dA}} \tag{2} \label{2}\]
But if the normal force \(F_n\) is uniform over an area \(A\), the pressure due to the force on the area is \(P = F_n/A\) and therefore the pressure on the surface is the normal force per unit area of the surface.
If the normal force is not uniform, you should use Eq. \eqref{2} for infinitesimally small normal force \(dF_n\) acting over an infinitesimally small area \(dA\). The SI unit of pressure is \(N/{m^2}\). You can also call the SI unit of pressure as \(\text{Pa}\) from the original name Pascal so, \(1{\rm{Pa}} = {\rm{1}}N/{m^2}\).
Note that pressure is not a vector quantity even if the force is a vector. The air in the atmosphere exerts normal force on the surface of the Earth and also on every object on its surface.
Note that the force exerted by a fluid on a body is always perpendicular to the surface of the body. The normal atmospheric pressure denoted by \(\text{atm}\) at sea level is \(1{\rm{atm}} = 101.3{\rm{kPa}}\) or \(1{\rm{atm}} = {\rm{1}}{\rm{.013}} \times {\rm{1}}{{\rm{0}}^5}{\rm{Pa}}\).
The fluid exerts force on the immersed body which in turn results pressure. We derive an expression for the pressure of a fluid at a particular depth here. We consider a fluid of uniform density \(\rho\) in a vessel as shown in Figure 2.
We consider an imaginary fluid element as a box having imaginary boundary. The imaginary fluid element is at rest that is the fluid element is in hydrostatic equilibrium. If the net force on a particular portion of the fluid is zero, the portion of the fluid is said to be in hydrostatic equilibrium.
We distinguish the force acting on the top and bottom surfaces of the fluid element as vertical force while the force acting on the other sides as horizontal force. Only the vertical forces are shown in Figure 2.
Since the fluid element is in hydrostatic equilibrium, the net force acting on the element is zero. Which concludes that the horizontal forces should cancel each other and the net vertical force should also be zero.
Let the area of the top or bottom surface of the fluid element be \(A\). Suppose pressure on the top surface of the element is \(P_1\) and the pressure on the bottom surface is \(P_2\). Note that we choose the positive y-axis to be upwards and therefore the upward force is positive while the downward force is negative.
The downward force on the top surface due to the pressure \(P_1\) is \(-P_1A\) while the upward force on the bottom surface is \(P_2A\). Since the weight of the fluid element is also vertically downwards, the pressure on the bottom surface of the fluid element must be greater than the pressure on the top surface.
The height of the fluid element is \((y_2-y_1)\) where \(y_1\) and \(y_2\) are the heights of the bottom and top surfaces of the fluid element from the bottom of the vessel respectively. If \(V\) is the volume of the fluid element, its mass is \(m = \rho V = \rho A(y_2 - y_1)\). The fluid element is at rest so the sum of all vertical forces must be zero:
\[\begin{align*} - P_1A + P_2A - \rho A(y_2-y_1){\kern 1pt} g &= 0\\ {\rm{or,}}\quad P_2 - P_1 &= \rho g{\kern 1pt} (y_2 - y_1) \tag{3} \label{3} \end{align*}\]
As in Figure 3 let \({P_0}\) be the pressure on the top surface of the fluid in the vessel at height \(y_2\) and \(P\) be the pressure at height \(y_1\), the above equation can be rewritten as
\[P - {P_0} = \rho g({y_2} - {y_1})\]
Let the depth of any point where the pressure is \(P\) be \(h = y_2 - y_1\). And hence the pressure due to the fluid at any depth \(h\) from the top surface is given by
\[P = P_0 + \rho gh \tag{4} \label{4}\]
Suppose the vessel in Figure 3 is open to the atmosphere and the pressure on the top surface of the fluid is the atmospheric pressure. Then the total pressure at the depth \(h\) from the top surface is the sum of the atmospheric pressure and the quantity \(\rho gh\) as given by Eq. \eqref{4}.
The excess pressure greater than the atmospheric pressure is called gauge pressure and the total pressure is called absolute pressure. Therefore, if \(P_0\) in Eq. \eqref{4} is the atmospheric pressure, the gauge pressure is \(P - P_0 = \rho gh\) and \(P\) is the absolute pressure.
