Consider a plate submerged in fluid with an inclination of θ with free surface.

Now consider small strip of are a da lying at a depth of ‘x’ from ‘O’.

Pressure intensity on the strip,

dp  =  γx

Total pressure force on strip,

dp  =  Pressure x Area

      = γ . x. dA

but x =  Y sin θ

dp =  γ . y. sin θ. dA

Total pressure force on the whole area,

Integrate Equation ,  ∫ dp =  ∫ γ. y sin θ. dA

=  γ. sin θ ∫ y. dA

But ∫ is the sum of first moments of the area of the strip, which is equal to A¯y

∫ ydA  =  A¯y

P = γ. sin θ. A¯y

From figure,  ¯x / ¯ y = sin θ

¯x = ¯ y sin θ

P =  γ. A. ¯x

Hence there is no change in equation for total pressure force weather it is vertical or inclined.

Centre of pressure can be calculated by using law of moment.

Total pressure force on strip,

dp = γ. y sin θ dA

Moment of force on strip, dM  =  dp. y = ( γ. y. sinθ dA) y

= γ. sinθ y² dA

Now, sum of moment of all such pressure forces on strip about free surface.

Integrate Equation

∫ dM =  ∫ γ. sin θ. y² dA

M = γ. sin θ ∫ y² dA

But we know  ∫ y² dA = Io = moment of inertia about free surface

M = γ. sinθ.  Io

By parallel axis theorem

Io =  I_G + A ( ¯ y )² = x / sin θ but  ¯ y = ¯x / sin θ and y_c = h / sin θ

But, moment of total pressure from free surface,

M = P ¯ y_c

Equate Equations and we get,

P ¯ y_c  = γ sinθ Io

γ A¯x (h / sin θ)  = γ sin θ [ I_G + A (¯x² / sin θ)² ] = γ  sin θ I_G + A¯x² / sin² θ

¯ h  =  sin² θ I_G / A¯x + sin θ. A ¯x² / A ¯x sin θ

¯ h  = I_G sin² θ / A ¯x + ¯x

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