Hydrostatics on plane surface (basic fluid mechanic Q)

[uzman1243]

I don't understand how they got the "position of individual forces".

For F1, I'm guessing they used a pressure diagram and that's why it's 2/3 from the tip of the triangle. But why wouldn't the force act on the centroid for this square section (thus its 3.5/2)?

But for position of F2 (y2), they used the equation. Can you explain this?

 

[SteamKing]

I don't understand how they got the "position of individual forces".

For F1, I'm guessing they used a pressure diagram and that's why it's 2/3 from the tip of the triangle. But why wouldn't the force act on the centroid for this square section (thus its 3.5/2)?

The hydrostatic force F1 doesn't act at the centroid of the square because it's not constant with respect to depth. You must integrate the pressure distribution acting over the area to find the magnitude of the force F1, and then calculate the first moment of this force with respect to the waterline in order to locate the point at which the hydrostatic pressure acts.

Since the pressure distribution is triangular w.r.t. depth, it's easy to calculate the total hydrostatic force and to find the center of pressure, knowing some geometric facts about triangles.

But for position of F2 (y2), they used the equation. Can you explain this?

Since the top of the square is at the waterline for the body, the pressure distribution is pretty simple: it's a triangle. No complicated formulas are required to find the center of pressure.

For the bottom of the body, the bottom is sloping and the pressure distribution is trapezoidal, so there are no easy formulas to apply except the one shown in the solution.