water in reservoir

[logistic_guy]

The water in a \(\displaystyle 25\)-\(\displaystyle \text{m}\)-deep reservoir is kept inside by a \(\displaystyle 150\)-\(\displaystyle \text{m}\)-wide wall whose cross section is an equilateral triangle, as shown in the figure. Determine \(\displaystyle \bold{(a)}\) the total force (hydrostatic + atmospheric) acting on the inner surface of the wall and its line of action and \(\displaystyle \bold{(b)}\) the magnitude of the horizontal component of this force. Take \(\displaystyle P_{\text{atm}} = 100 \ \text{kPa}\).

 

[logistic_guy]

Determine \(\displaystyle \bold{(a)}\) the total force (hydrostatic + atmospheric) acting on the inner surface of the wall

\(\displaystyle F_R = (\rho g h_c + P_\text{atm})A = \left(\rho g\frac{l}{2}\sin \theta + P_\text{atm}\right)lw = \left(\rho g\frac{h}{2} + P_\text{atm}\right)\frac{hw}{\sin \theta}\)

 

[logistic_guy]

\(\displaystyle F_R = (\rho g h_c + P_\text{atm})A = \left(\rho g\frac{l}{2}\sin \theta + P_\text{atm}\right)lw = \left(\rho g\frac{h}{2} + P_\text{atm}\right)\frac{hw}{\sin \theta}\)

Plug in numbers.

\(\displaystyle F_R = \left(\rho g\frac{h}{2} + P_\text{atm}\right)\frac{hw}{\sin \theta} = \left(1000 \times 9.8\frac{25}{2} + 100 \times 10^3\right)\frac{25 \times 150}{\sin 60^{\circ}} = \textcolor{blue}{9.635 \times 10^8 \ \text{N}}\)

 

[logistic_guy]

Determine \(\displaystyle \bold{(a)}\) its line of action

\(\displaystyle y_p = y_c + \frac{I_{xx}}{\left(y_c + \frac{P_0}{\rho g \sin\theta}\right)A} = \frac{h}{2\sin\theta} + \frac{\frac{wh^3}{12\sin^3\theta}}{\left(\frac{h}{2\sin\theta} + \frac{P_0}{\rho g \sin\theta}\right)\frac{hw}{\sin\theta}}\)


\(\displaystyle = \frac{25}{2\sin 60^{\circ}} + \frac{\frac{(150)25^3}{12\sin^3 60^{\circ}}}{\left(\frac{25}{2\sin 60^{\circ}} + \frac{100 \times 10^3}{1000 \times 9.8 \sin 60^{\circ}}\right)\frac{25(150)}{\sin 60^{\circ}}} = \textcolor{blue}{17.1 \ \text{m}}\)

 

[logistic_guy]

\(\displaystyle \bold{(b)}\) the magnitude of the horizontal component of this force.

\(\displaystyle F_x = F_R\sin 60^{\circ} = 9.635 \times 10^8 \ \frac{\sqrt{3}}{2} = \textcolor{blue}{8.344 \times 10^8 \ \text{N}}\)