a vertical dam has a semicircular gate as shown in the picture below. find the hydrostatic force against the…

Explanation:

Step1: Determine the density of water

The density of water $\rho = 1000\ kg/m^{3}$, and the gravitational - acceleration $g=9.8\ m/s^{2}$.

Step2: Set up a coordinate system

Place the origin at the center of the semi - circular gate. The radius of the semi - circular gate $r = 3\ m$. The depth of a horizontal strip of width $\Delta y$ at a distance $y$ from the origin is $h=(11 + y)$ (where $y$ ranges from $- 3$ to $3$). The length of the horizontal strip (chord) of the semi - circle at a distance $y$ from the origin is $l = 2\sqrt{r^{2}-y^{2}}=2\sqrt{9 - y^{2}}$ (using the equation of a circle $x^{2}+y^{2}=r^{2}$ and the fact that the length of the chord is $2x$).

Step3: Calculate the area of the horizontal strip

The area of the horizontal strip $\Delta A=l\Delta y=2\sqrt{9 - y^{2}}\Delta y$.

Step4: Calculate the pressure on the horizontal strip

The pressure on the horizontal strip $P=\rho gh=\rho g(11 + y)$.

Step5: Calculate the force on the horizontal strip

The force on the horizontal strip $\Delta F=P\Delta A=\rho g(11 + y)\times2\sqrt{9 - y^{2}}\Delta y$.

Step6: Integrate to find the total force

The total hydrostatic force $F=\int_{-3}^{3}\rho g(11 + y)\times2\sqrt{9 - y^{2}}dy$. Since $\int_{-3}^{3}y\sqrt{9 - y^{2}}dy = 0$ (because the integrand is an odd function), we only need to calculate $\int_{-3}^{3}11\times2\sqrt{9 - y^{2}}dy$. We know that $\int_{-a}^{a}\sqrt{a^{2}-y^{2}}dy=\frac{1}{2}\pi a^{2}$. Here $a = 3$. So $F = 22\rho g\int_{-3}^{3}\sqrt{9 - y^{2}}dy$. Substitute $\rho = 1000\ kg/m^{3}$, $g = 9.8\ m/s^{2}$ and $\int_{-3}^{3}\sqrt{9 - y^{2}}dy=\frac{9\pi}{2}$ into the formula. $F=22\times1000\times9.8\times\frac{9\pi}{2}$. $F = 22\times1000\times9.8\times4.5\pi$. $F=970200\pi\ N\approx 3.04\times 10^{6}\ N$.

Answer:

$970200\pi\ N\approx3.04\times 10^{6}\ N$