- Variant 1. Standart
- Variant 2. Analitics
- Variant 3. Num. methods
- Variant 4. Plot
APPENDIXES
The hydrostatic pressure. The method of three commands K123
Variant 4. Plot
Force on rectangular surface
Force
\[H=7 m,h=7 m, B=3 m, \rho = 1000 \frac{kg}{m^{3}}, g=9.81 \frac{m}{s^{2}} \]
\[P=\int\rho \cdot g \cdot \left ( H -h\right )\cdot Bdh=29430.0\,\left(7\,h-{{h^2}\over{2}}\right) [N]\]
Moment of Force
\[mP=\int_{0}^{H}\rho \cdot g \cdot \left ( H -h\right )\cdot h\cdot Bdh [m \cdot N]\]
Center of pressure
\[h_D=\frac{mP}{P} [m]\]
Problem A-1.
The force of hydrostatic pressure on a rectangular surface
For a rectangular surface:
\[H=7 m,h=7 m, B=3 m, \rho = 1000 \frac{kg}{m^{3}}, g=9.81 \frac{m}{s^{2}} \]
Counting
\[P=\int_{0}^{H}\rho \cdot g \cdot \left ( H -h\right )\cdot Bdh [N]\]
\[P(\rho ,g,H,h,B)=\int_{0}^{7}1000 \cdot 9.81 \cdot \left ( 7 -h\right )\cdot 3dh = 721035 N \]
\[mP=\int_{0}^{H}\rho \cdot g \cdot \left ( H -h\right )\cdot h\cdot Bdh [m \cdot N]\]
\[mP=\int_{0}^{7}1000 \cdot 9.81 \cdot \left ( 7 -h\right )\cdot h\cdot 3dh = 1680011.55[m \cdot N]\]
\[h_D=\frac{\int_{0}^{H}\rho \cdot g \cdot \left ( H -h\right )\cdot h\cdot Bdh}{\int_{0}^{H}\rho \cdot g \cdot \left ( H -h\right )\cdot Bdh} [m]\]
\[h_D=\frac{mP}{P} [m]\]
\[h_D=\frac{\int_{0}^{7}1000 \cdot 9.81 \cdot \left ( 7 -h\right )\cdot h\cdot 3dh}{\int_{0}^{7}1000 \cdot 9.81 \cdot \left ( 7 -h\right )\cdot 3dh} = 2.33 m \]
\[h_D=\frac{1680011.55 \cdot m \cdot N}{721035 N} = 2.33 m \]
Variant 4. Plot
Facts:
\[P(\rho ,g,H,h,B)=\int_{0}^{7}1000 \cdot 9.81 \cdot \left ( 7 -h\right )\cdot 3dh = 721035 N \]
\[mP=\int_{0}^{7}1000 \cdot 9.81 \cdot \left ( 7 -h\right )\cdot h\cdot 3dh = 1680011.55[m \cdot N]\]
\[h_D=\frac{1680011.55 \cdot m \cdot N}{721035 N} = 2.33 m \]