Standart

Task. Rectangular surface. If a rectangle of height H and base B is vertical and submerged in liquid with its vertex at the liquid surface (see Fig. A-1.00a), derive an expression for the depth to its center of pressure.

Calculation algorithm

1. depth of immersion of the center of gravity of the surface \(h_{C}=\frac{h}{2} [m]\);
2. hydrostatic pressure in the center of gravity of the surface \(p=\rho gh_{C} [Pa]\);
3. wetted surface area \(w=b\cdot h [m^2]\);
4. the force of hydrostatic pressure on the wetted surface \(P=p\cdot w [N]\);
5. the moment of inertia of the surface relative to the horizontal axis passing through the center of gravity \(I_{0}=\frac{b \cdot h^{3}}{12} [m^4]\);
6. depth of immersion of the pressure center \(h_{D}=h_{C}+\frac{I_{0}}{h_{C}\cdot w} [m]\);
7. to build a diagram of hydrostatic pressure.

Calculation

1.Depth of immersion of the center of gravity:

\[h_{C}=\frac{h}{2} [m],\]

\[h_{C}=\frac{0.3}{2}=0.15 [m].\]

2.Hydrostatic pressure in the center of gravity of the surface:

\[p=\rho gh_{C} [Pa],\]

\[p=1000 \cdot 9.81 \cdot 0.15=1471.5 [Pa].\]

3.Wet surface area:

\[w=b\cdot h [m^2],\]

\[w=0.1\cdot 0.3=0.03 [m^2].\]

4.The force of hydrostatic pressure on the surface:

\[P=p\cdot w [N],\]

\[P=1471.5\cdot 0.03=44.145 [N].\]

5.The moment of inertia of the surface relative to the horizontal axis passing through the center of gravity:

\[I_{0}=\frac{b \cdot h^{3}}{12} [m^4],\]

\[I_{0}=\frac{0.1 \cdot0.3^{3}}{12}=0.000225 [m^4].\]

6.Depth of immersion of the pressure center:

\[h_{D}=h_{C}+\frac{I_{0}}{h_{C}\cdot w} [m],\]

\[h_{D}=0.15+\frac{0.000225}{0.15\cdot 0.03}=0.2 [m].\]

7.Diagram of hydrostatic pressure

Fig. A.1.00b