- #1

Urmi Roy

- 753

- 1

## Homework Statement

__(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.

## Homework Equations

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

ρ= water density

A= area of surface submerged

## 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?**

#### Attachments

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