Pressure

Normal force per unit cross-sectional area

$$\begin{align*} p &= \frac{F}{A} \\ \text{pressure} &= \frac{\text{normal force}}{\text{cross-sectional area}} \end{align*}$$

Unit of $p$ = $\mathrm{Pa}$

Pressure in a fluid produces a perpendicular force onto any surface

Hydrostatic pressure

Pressure in fluids

$$\begin{align*} \Delta p &= \rho g \Delta h \\ \text{change in pressure} &= \text{density} \cdot \text{gravitational acceleration} \cdot \text{change in depth} \end{align*}$$

Pressure at the same depth is the same regardless of the shape of the water column above

Derivation

$$\begin{align*} \text{volume of liquid ~} V &= A \Delta h \\ \text{mass of liquid ~} m &= \rho V = \rho A \Delta h \\ \text{weight of liquid ~} F &= m g = \rho A \Delta h g \\ \text{change in pressure ~} \Delta p &= \frac{F}{A} \\ &= \frac{\rho A \Delta h g}{A} \\ &= \rho g \Delta h \\ \end{align*}$$