Consider a plance plane vertical surface of arbitrary shape immmersed in a liquid as shown in Figure 1.

$$ Let\ A = Total\ area\ of\ surface. $$ $$ \overline{h} = Distance\ of\ C.G\ of\ the\ area\ from\ free\ surface\ $$ $$ G = centre\ of\ gravity\ of\ plane\ surface\ $$ $$ P = centre\ of\ pressure\ $$ $$ h^* = Distance\ of\ centre\ of\ pressure\ from\ free\ surface\ of\ liquid $$

$$ Figure\ 1 $$

1. Consider a strip of thickness dh and width b at a depth of h from free surface of liquid as shown in figure 1.

   Pressure intensity on the strip $$ p = \rho g h $$ Area of the strip, $$ dA = b \times dh $$ Total pressure force on the strip, $$ dF = p \times Area = \rho g h \times b \times dh $$ ∴ Total pressure force on the whole surface $$ F = \int dF = \int \rho g h \times b \times dh = \rho g \int b \times h \times dh $$ but $$ \int b \times h \times dh = \int h \times dA $$ $$ = Moment\ of\ surface\ area\ about\ the\ free\ surface\ of\ liquid\ $$ $$ = Area\ of\ surface \times Ditance\ of\ C.G\ from\ free\ surface $$ $$ A \times \overline{h} $$ $$ F = \rho g A \overline{h} \ \ \ \ \ \ \ \ \ \ \ \ -> equation\ 1 $$ ∴ For water the value of ρ = 100 kg/m³ and g = 9.81 m/sec². The force will be in Newtons.

   
   

2. The resultant force F is acting at P, at a distance h^* from free surface of the liquid as shown in figure 1. Hence moment of the force F about free surface of the liquid

   $$ = dF \times h \ \ \ \ \ \ \ (because\ dF = \rho g h \times b \times dh) \ \ \ \ -> equation\ 2 $$ $$ = \rho g h \times b \times dh \times h $$ Sum of moments of all such forces about free surface of liquid $$ = \int \rho g h \times b \times dh \times h = \rho g \int b \times h \times hdh $$ $$ = \rho g \int bh^2 dh = \rho g \int h^2dA \ \ \ \ \ \ \ \ \ (because\ bdh = dA) $$ but $$ \int h^2dA = \int bh^2 dh $$

   $$ = Moment\ of\ Intertia\ of\ the\ surface\ about\ free\ surface\ of\ liquid\ $$ $$ = I_o $$

   ∴ Sum of moments about free surface $$ = \rho g I_o \ \ \ \ \ \ \ \ \ -> equation\ 3 $$

   equating equation 2 and 3, we get $$ F \times h^* = \rho g I_o $$ but $$ F = \rho g A \overline{h} $$ therefore $$ \rho g h A \overline{h} \times h^* = \rho g I_o $$ or $$ h^* = \frac{\rho g I_o}{\rho g A \overline{h}} = \frac{I_o}{A\overline{h}} \ \ \ \ \ \ \ \ \ -> equation\ 4 $$ by the theorem of prallel axis, we have $$ I_o = I_G + A \times \overline{h}^2 $$ where I_G = Moment of Intertia of area about an axis passing through the C.G of the area and parallel to the free surface of the liquid

   Subtitution I_o in equation 4, we get $$ h^* = \frac{I_G + A\overline{h}^2}{A\overline{h}} = \frac{I_G}{A\overline{h}} + \overline{h} \ \ \ \ \ \ \ \ \ -> equation\ 5 $$

1. Center of pressure (i.e h<sup>*</sup>) lies below the center of gravity of the vertical surface
          2. The distance of centre of pressure from free surface of liquid is independent of the density of the liquid.