Hydrostatic force on submerged rectangle

1. The problem statement, all variables and given/known data



(Referring to attached diagram)



Find the hydrostatic force F2 on the rectangular part and the point of action of this force on rectangular area.

The triangular part is submerged in oil of specific gravity 0.8, while the rectangular part is submerged in water. The whole structure is upright with no inclination with fluid surface.



2. Relevant equations



Hydrostatic force= ρghA, where h=perpendicular distance of centroid of object from fluid surface.

ρ= water density

A= area of surface submerged





3. The attempt at a solution



According to me:

F2= ρxgx(height of triangle+further distance to get centroid of rectangle)x(area of rectangle)

= 1000x9.81x(3+1)x(2x4)



However the solution in the same book gives

F2= 9.81x1000x(3x0.8 + 1)x(2x4)



I don't understand why 3 is multiplied by 0.8.



For point of action of force F2,

According to me,

y_p (vertical distance from fluid surface of point of action)

= ∫(integral between 0 to 5m) ρgz(0.8xarea of triangle+4x(z-3))



where z= total vertical distance downward from fluid surface



In the solutions on my book,

it gives the right answer as:

y_p (vertical distance from fluid surface of point of action)

= ∫(integral between 3 to 5m) ρgz(3x0.8+(z-3))(4dz)z



Again, there's that 0.8x3...Also, it's as if they're ignoring the triangle since the integral is between 3 to 5 m...but then why would the 3x0.8 be in there?

Attached Images

IMG_0570.jpg (85.0 KB)