Physics > Fluid Mechanics > 2.0 Fluid pressure
Fluid Mechanics
1.0 Introduction
1.1 Ideal liquid
1.2 Density of a liquid $\left( \rho \right)$
1.3 Relative density of a liquid $(RD)$
1.4 Density of a mixture of two or more liquid
1.5 Density variation with temperature
1.6 Density variation with pressure
2.0 Fluid pressure
2.1 Atmospheric pressure
2.2 Pressure variation with depth
2.3 Measurement of pressure
2.4 Pressure difference in accelerating fluids
3.0 Pascal's law
4.0 Buoyant force
5.0 Flow of fluids
6.0 Viscosity
7.0 Stoke's law
8.0 Intermolecular forces
9.0 Angle of contact
2.4 Pressure difference in accelerating fluids
1.2 Density of a liquid $\left( \rho \right)$
1.3 Relative density of a liquid $(RD)$
1.4 Density of a mixture of two or more liquid
1.5 Density variation with temperature
1.6 Density variation with pressure
2.2 Pressure variation with depth
2.3 Measurement of pressure
2.4 Pressure difference in accelerating fluids
1. Horizontal acceleration
2. Circular motion
2.4.1 Horizontal acceleration
Consider a liquid kept at rest in a container as shown in the figure.
Suppose the container is accelerated with an acceleration $a$ as shown in the figure.
The surface of the liquid becomes incline due to the pressure difference.
Now, the pressure is different at same level..
As we know, \[\left. \begin{gathered} {P_1} = {P_O} + \rho g{h_1} \hspace{1em} \\ {P_2} = {P_O} + \rho g{h_2} \hspace{1em} \\ \end{gathered} \right\}\quad ...(i)\]
Consider a section of liquid of cross-sectional area $A$ as shown in the figure.
From Newton's law of motion we get, $$\begin{equation} \begin{aligned} {P_1}A - {P_2}A = ma \\ \left( {{P_1} - {P_2}} \right)A = \left( {\rho Ax} \right)a \\ {P_1} - {P_2} = \rho xa\quad ...(ii) \\\end{aligned} \end{equation} $$
From equation $(i)$ and $(ii)$ we get, $$\begin{equation} \begin{aligned} \rho g\left( {{h_1} - {h_2}} \right) = \rho xa \\ \left( {\frac{{{h_1} - {h_2}}}{x}} \right) = \frac{a}{g} \\ \tan \theta = \frac{a}{g} \\\end{aligned} \end{equation} $$
2.4.2 Circular motion
Consider a container filled with liquid of density $\rho $ which is rotated with an angular velocity $\omega $ as shown in the figure
Let $P_O$ be the atmospheric pressure, 
Consider an infinitesimally small section of liquid of cross-sectional area $A$ and length $dr$ at a distance $r$ from the axis of rotation as shown in the figure.
Centripetal acceleration of the liquid section, $$a = {\omega ^2}r$$
Mass of the liquid section, $$m = \rho Adr$$
As we know, centripetal acceleration is towards the centre as shown in the figure. So, $$\begin{equation} \begin{aligned} (P + dP)A - PA = ma \\ AdP = \rho A{\omega ^2}rdr \\ dP = \rho {\omega ^2}rdr \\\end{aligned} \end{equation} $$
Integrating with proper limits we get, $$\begin{equation} \begin{aligned} \int\limits_{{P_O}}^P {dP} = \rho {\omega ^2}\int\limits_0^r {rdr} \\ P - {P_O} = \rho {\omega ^2}\left[ {\frac{{{r^2}}}{2}} \right]_0^r \\ P = {P_O} + \frac{{\rho {\omega ^2}{r^2}}}{2} \\\end{aligned} \end{equation} $$The above equation gives the variation of pressure with distance $r$ from the axis of rotation.
