Question

Calculate the hydrostatic force against the gate. The depth of vertical dam is\(12m\), the water level of vertical dam is \(10m\) and the radius of the semi circle gate is \(2m\)

Short answer

The hydrostatic force against the gate is\(5.63 \times {10^5}\;{\rm{N}}\).

Step-by-step solution

Pythagoras theorem and draw a semi-circular gate.

Pythagoras Theorem:\(A{B^2} + B{C^2} = A{C^2}\)

The depth of vertical dam is\(12m\).

The water level of vertical dam is\(10m\).

The radius of the semi circle gate is\(2m\).

Draw a semicircular gate as shown in Figure 1.

Use Pythagoras theorem and refer to figure 1 for calculation.

Refer to figure 1.

Calculate the width of the \({i^{th}}\)strip as follows:

\({w_i} = 2\sqrt {4 - {x^2}} \)

Calculate the depth of the \({i^{th}}\)strip as follows:

\({\rm{di = 10 - x}}\)

Consider the acceleration due to gravity as\(g = 9.8\;{\rm{m}}/{{\rm{s}}^2}\).

Consider the density of water as\(\rho = 1,000\;{\rm{kg}}/{{\rm{m}}^3}\).

Calculate the area of the \({i^{th}}\)strip as follows:

\({\rm{Ai = wi dx}}\) …… (1)

Substitute \(2\sqrt {4 - {x^2}} {\rm{ for }}{w_i}\) in equation (1).

\({A_i} = 2\sqrt {4 - {x^2}} dx\)

Calculate the pressure of the \({i^{th}}\) rectangular strip:

\({P_i} = \rho g{d_i}\) …… (2)

Substitute \(1,000\;{\rm{kg}}/{{\rm{m}}^3}{\rm{ for }}\rho ,9.8\;{\rm{m}}/{{\rm{s}}^2}{\rm{ for }}g{\rm{,}}\)and \((10 - x){\rm{ for }}{d_i}\) in equation (2).

\(\begin{aligned}{}{P_i} = 1,000 \times 9.8 \times (10 - x)\\ = 9,800(10 - x)\end{aligned}\)

Calculate the hydrostatic force on one end of the tank:

\(F = \int_a^b {{P_i}} {A_i}\) …… (3)

Substitute \(0{\rm{ for }}a,2{\rm{ for }}b,9,800(10 - x){\rm{ for }}{P_i}\) and \(2\sqrt {4 - {x^2}} dx{\rm{ for }}A\)in equation (3).

\(\begin{aligned}{}F = \int_0^2 9 ,800(10 - x)2\sqrt {4 - {x^2}} dx\\ = 19,600\int_0^2 {\left( {10\sqrt {4 - {x^2}} - x\sqrt {4 - {x^2}} } \right)} dx\\ = 196,000\int_0^2 {\sqrt {4 - {x^2}} } dx - 19,600\int_0^2 x \sqrt {4 - {x^2}} dx\end{aligned}\) …… (4)

Consider \({F_1}{\rm{ and }}{F_2}\) as follows:

\({F_1} = 196,000\int_0^2 {\sqrt {4 - {x^2}} } dx\) …… (5)

\({F_2} = 19,600\int_0^2 x \sqrt {4 - {x^2}} dx\) …… (6)

Substitute \({F_1}{\rm{ for }}19,6000\int_0^2 {\sqrt {4 - {x^2}} } dx\) and \({\rm{ }}{F_2}{\rm{ for }}19,600\int_0^2 x \sqrt {4 - {x^2}} dx\)in equation (4).

\({\rm{F = F1 - F2}}\) …… (7)

Consider \(x = 2\sin \theta \) …… (8)

Differentiate both sides of the equation (8).

\(dx = 2\cos \theta d\theta \)

Substitute \((2\sin \theta ){\rm{ for }}x\) and \((2\cos \theta d\theta ){\rm{ for }}dx\) in equation (5).

\(\begin{aligned}{}{F_1} = 196,000\int_0^2 {\sqrt {4 - {{(2\sin \theta )}^2}} } (2\cos \theta d\theta )\\ = 196,000 \times 2\int_0^2 {\sqrt {4 - 4{{\sin }^2}\theta } } \cos \theta d\theta \\ = 392,000\int_0^2 {\sqrt {4\left( {1 - {{\sin }^2}\theta } \right)} } \cos \theta d\theta \\ = 392,000\int_0^2 2 \sqrt {{{\cos }^2}\theta } \cos \theta d\theta \\ = 392,000 \times 2\int_0^2 {\cos } \theta \cos \theta d\theta \\ = 392,000 \times 2\int_0^2 {\cos } \theta d\theta \\ = 392,000 \times 2\int_0^2 {\frac{1}{2}} (1 + \cos 2\theta )d\theta \\ = 392,000\int_0^2 {(1 + \cos 2\theta )} d\theta \end{aligned}\) …… (9)

Integrate equation (8).

\(\begin{aligned}{}{F_1} = 392,000\left( {\theta + \frac{1}{2}\sin 2\theta } \right)_0^2\\ = 392,000\left( {\theta + \frac{1}{2}(2\sin \theta \cos \theta )} \right)_0^2\\ = 392,000(\theta + \sin \theta \cos \theta )_0^2\end{aligned}\) …… (10)

Rearrange equation (8).

\(\begin{aligned}{}\sin \theta = \frac{x}{2}\\\theta = {\sin ^{ - 1}}\left( {\frac{x}{2}} \right)\end{aligned}\) …… (11)

Represent Equation (11) with Pythagorean Theorem as shown in Figure 2.

Refer to Figure 2.

\(\begin{aligned}{}A{B^2} + B{C^2} = A{C^2}\\A{B^2} + {x^2} = {2^2}\\A{B^2} = 4 - {x^2}\\AB = \sqrt {4 - {x^2}} \end{aligned}\)

Calculate the value of \(\cos \theta \) using Figure 2.

\(\begin{aligned}{}\cos \theta = \frac{{AB}}{{AC}}\\ = \frac{{\sqrt {4 - {x^2}} }}{2}\end{aligned}\)

Substitute \({\sin ^{ - 1}}\left( {\frac{x}{2}} \right){\rm{ for }}\theta ,\frac{x}{2}{\rm{ for }}\sin \theta {\rm{, }}\) and \({\rm{ }}\frac{{\sqrt {4 - {x^2}} }}{2}{\rm{ for }}\cos \theta \) in equation (10).

\(\begin{aligned}{}{F_1} = 392,000\left( {{{\sin }^{ - 1}}\left( {\frac{x}{2}} \right) + \frac{x}{2}\frac{{\sqrt {4 - {x^2}} }}{2}} \right)_0^2\\ = 392,000\left( {{{\sin }^{ - 1}}\left( {\frac{x}{2}} \right) + \frac{{x\sqrt {4 - {x^2}} }}{4}} \right)_0^2\\ = 392,000\left\{ {\left( {{{\sin }^{ - 1}}\left( {\frac{2}{2}} \right) + \frac{{2\sqrt {4 - {{(2)}^2}} }}{4}} \right) - \left( {{{\sin }^{ - 1}}\left( {\frac{0}{2}} \right) + \frac{{0\sqrt {4 - {{(0)}^2}} }}{4}} \right)} \right\}\\ = 392,000\left( {\frac{\pi }{2} - 0} \right)\\\;\; = \;392,000\left( {\frac{\pi }{2} - 0} \right)\\ = 196,000\pi \end{aligned}\)

Consider \(u = 4 - {x^2}\) …… (12)

Differentiate both sides of equation (12).

\(\begin{aligned}{}du = - 2xdx\\ - \frac{1}{2}du = xdx\end{aligned}\)

Calculate the lower limit value \(u\)using equation (12).

Substitute \(2{\rm{ for }}x\) in equation (12).

\(\begin{aligned}{}u = 4 - {(2)^2}\\ = 0\end{aligned}\)

Apply lower and upper limits for \(u\) in equation (6).

\(\begin{aligned}{}{F_2} = 19,600\int_4^0 {\sqrt u } \left( { - \frac{1}{2}du} \right)\\ = - 19,600 \times \frac{1}{2}\int_4^0 {{u^{\frac{1}{2}}}} du\\ = - 9800\int_4^0 {{u^{\frac{1}{2}}}} du\end{aligned}\) …… (13)

Integrate equation (13).

\(\begin{aligned}{}{F_2} = - 9800\left( {\frac{{{u^{\frac{3}{2}}}}}{{\frac{3}{2}}}} \right)_4^0\\ = - 9800\left( {\frac{{2{u^{\frac{3}{2}}}}}{3}} \right)_4^0\\ = - 9800\left( {\frac{{2{{(0)}^{\frac{3}{2}}}}}{3} - \frac{{2{{(4)}^{\frac{3}{2}}}}}{3}} \right)_4^0\\ = - 9800\left( {0 - \frac{{16}}{3}} \right)\\ = \frac{{156,800}}{3}\end{aligned}\)

Substitute \(196,000\pi {F_1}{\rm{ and }}\frac{{156,800}}{3}{\rm{ for }}{F_2}\) in equation (7).

\(\begin{aligned}{}F = 196,000\pi - \frac{{156,800}}{3}\\ = 563485.5\;{\rm{N}}\\ \approx 5.63 \times {10^5}\;{\rm{N}}\end{aligned}\)

Hence, the hydrostatic force against the gate is\(5.63 \times {10^5}\;{\rm{N}}\).