Answer to Question #86587 in Mechanics | Relativity for dddd

Question

Q1. A dam has its water face as a quarter circle arc of radius of 4m submerged under water surface so that the total dam depth is 9 m. The width (span) of the dam is 150 m.

Calculating:
(i) Horizontal force component acting on the dam.

Answer 1

Let $y$ be the total height of the water, $r$ - radius of the arc, $b$ - span of the dam.

Split the total force into two forces: horizontal component acting on the flat vertical side $F_1$ and the component acting on the curved part $F_2$:

$$F_1=\gamma h_{C1}A_1=\rho g \cdot \frac{y-r}{2}\cdot b(y-r)=$$ $$=1000\cdot 9.8\cdot\frac{9-4}{2} \cdot150\cdot (9-4)=18375000\text{ N}.$$

$$F_2=\gamma h_{C2}A_2=\rho g \cdot (y-\frac{r}{2})\cdot br=$$ $$=1000\cdot 9.8\cdot(9-4/2)\cdot 150\cdot 4=41160000\text{ N}.$$

$$F_h=F_1+F_2=18375000+41160000=59535000\text{ N}.$$