(^730) „„„„„ A Textbook of Engineering Mechanics
M=wlbdxsinθ∫^2
But ∫lbdx^2 =I 0
∴ M =w sin θ I 0 ...(i)
where I 0 = Moment of inertia of the immersed surface about the point O,
or second moment of area.
We know that the sum of the moments of all such pressures

sin
Ph
θ
...(ii)
where P= Total pressure of the surface, and
h = Depth of centre of pressure from the liquid surface.
Equating equations (i) and (ii),
sin
Ph
θ =w sin θ I^0
sin
wA x×h
θ
=w sin θ I 0 ...(QPwAx= )
h =
2
I 0 sin
Ax
θ
...(iii)
We know from the Theorem of Parallel Axis that
I 0 =IG + Ah^2
where IG= Moment of inertia of the figure about horizontal axis through
its centre of gravity.
h= Distance between O and the centre of gravity of the figure
(l =
sin
x
θ
in this case).
Rearranging equation (iii),
h =
()sin^22
IAhG
Ax
+θ

2
2
sin^2 sin sin
G
A x
I
Ax Ax
⎛⎞
⎜⎟θ
θ ⎝⎠θ

• =
  sin^2
  IG x
  Ax
  θ


• 
  

  Thus the centre of pressure is always below the centre of gravity of the area by a distance
  equal to
  sin^2
  G.
  I
  Ax
  θ
  Example 35.10. A circular plate 3 metres dia. is submerged in water, with its greatest and
  least depths below the surface being 2 metres and 1 metre respectively. Find (i) the total pressure
  on one face of the plate, and (ii) the position of the centre of pressure.
  Solution. Given: Diameter of the circular plate (d) = 3 m; Greatest depth = 2 m and least
  depth = 1 m.