Thrust And Pressure

Variation of Pressure in a static fluid
Let us consider a container filled with water. Density of water can be taken as uniform through out the container. Let A and B be two points inside the water located at the same level. Consider a thin fluid element located between A and B as shown in the figure15.2(a). 
                     

Now, let us apply Newton's law of motion to the elements. It is clear that the acceleration of the element  is zero. From the surrounding fluid, the net vertical fluid contact force must balance its weight. 
                   

PAS' and 'PBS' act on the element as shown; where 'S' is the area of cross-section of the element. As the net horizontal force on the element is also zero, 
           \( P_A S = P_B S\,\,\,\,or\,\,\,\,\,P_A = P_B \).
 Therefore, when the fluid is at rest, the points located on the same horizontal level and within a given fluid, must have equal pressures.
Now, let us take A & B  on a vertical line as shown. Separation between A&B is 'h'. Again, consider a thin water element located between A&B and apply Newton's law of motion to it.
                         

The vertical pressure forces 'P_AS' and 'P_BS' must balance the weight of the element.
    where, \( m = Sh\rho \) 
      \( \therefore P_A S + mg = P_B S \)
    or  \( \therefore P_A S + (Sh\rho )g = P_B S \)  
       \( \therefore P_B = P_A + h\rho g \)  

The above expression can also be written as pressure gradient \( \frac{{P_B - P_A }} {h} = \rho g \), or in differential from \( \frac{{dp}} {{dh}} = \rho g = \) constant. 
    \( (\rho \to \)density assumed to be constant)
In other words, when density is constant, pressure increases linearly with depth. 
          

 Now, let h_A & h_B be the depths of A and B from the free surface. Then, applying the above relation, 
        \( P_A = P_0 + h_1 \rho g \)  and \( P_B = P_0 + h_2 \rho g \)  
    where P₀ is atmospheric pressure.
The pressure difference between hydrostatic pressure and atmosphenic pressure is called gauge-pressure, 
   \(\therefore\) gauge pressure = \( P - P_0 = h\rho g \)

Regarding pressure in a static fluid it is worth noting that:
(i) At a point, pressure acts in all direction. If a pressure measuring device is placed at a point within the fluid, the reading of the device is same for all its orientations.
(ii) Pressure always acts normal to the fluid boundary because a static fluid cannot sustain a tangential force.
                      

(iii) In a uniform fluid, pressure depends on .
 a) atmospheric pressure (P₀)
 b) depth of the desired point below the free surface (h)
 c) density of the fluid \((\rho)\), and 
 d) acceleration due to gravity (effective acceleration of the fluid element)
It is independent of : a) amount of the fluid b) shape of the container etc., So, if a given liquid is filled in vessels of different shapes up to same height, the pressure at the base in each vessel will be same, though the volume or weight of the liquid will be different in different vessels.
              

At the base,  P_A = P_B = P_C but w_A < w_B < w_C (weight)
(iv)  in a static liquid at same level, the pressure will be same at all points provided, the points are in communication with one another. This is why, the height of liquid is the same in vessels of diffeent shapes
    Here P_A = P_B = P_C = P_D = P_E 
    and  h_A = h_B = h_C = h_D = h_E  
    If the weight of the fluid c

Note: 1. Pressure increases linearly with depth, if  \(\rho\) and g are uniform. A graph between P and h is shown below.
                  

Further, the pressure is the same at any two points at the same level in the fluid. The shape of the container does not matter. 
2.  At same point n a fluid pressure is same in all directions. In the figure,  P1 = P2 = P3 = P4 
                    

3. Forces acting on a fluid in equilibrium have to be perpendicular to its surface. 
4. In the same liquid pressure will be same at all points at the same level. 
    For example, in the figure.   
              

   \( P_1 \ne P_2 \) , P3 = P4  and P5 = P6 
    Further, P3 = P4
    \( \therefore P_0 + \rho _1 gh_1 = P_0 + \rho _2 gh_2 \)  or \( p_1 h_1 = \rho _2 h_2 \)  or  \( h \propto \frac{1} {\rho } \)