Question:
(The other axis is $y$, got cut-off as I was cropping the image in Paint)
Determine the total force on the vertical water gate in the above image as well as the center of pressure.
The density of water is $\rho ~kg/m^3$ (1000 $kg/m^3$), all measurements are in meter and the acceleration due to gravity is $g~m/s^2$.
Attempted answer:
The pressure in any point $(x,y)$ at a right angle to the water gate is
$$\rho gh = 1000gh$$
The force on a horizontal segment with the area $dA$ at depth $y$ is therefore:
$$F_{total} \cdot y_{center~of~pressure} = \int y~dF$$
$$dF = \rho gh \cdot dA = 1000gh \cdot dA$$
$$F_{total} = \int dF = 1000gh \cdot \int dA$$
The pressure center is then:
$$y_{center~of~pressure} = \frac{1}{F_{total}} \int y \cdot dF$$
However, I am not quite sure how to get $dA$, but I assume it comes from the formula for $y$. Is this just the integral of $y$ from $-a$ to $a$? Let us give this a try:
$$dA = \int_{-a}^{a} h – \frac{h}{a^2} x^2 ~dx = \bigg[ hx – \frac{h}{3a^2} x^3 \bigg]_{-a}^{a} = ha – \frac{ha}{3} – \big( -ha + \frac{ha}{3} \big) = -\frac{2ah}{3}$$
Not sure if this makes sense (both a and h are positive numbers, so $dA$ should not be able to be negative?) or if I have lost the variable to integrate $dF$ for $F_{total}$.
So I am stuck here. Since I have the large-scale path for what to calculate to get $F_{total}$ and $y_{center~of~pressure}$, I wonder how this can be finished off?
The expected answer is that the total force is:
$$\frac{8\rho g}{15}ah^2 N$$
…and the center of pressure is $\frac{4h}{7} m$ below the water surface.
Best Answer
Recall the equation for hydrostatic force on a submerged object
$$F = \rho ghA$$
where $\rho = \text{density}$, $g = \text{gravity} \approx 9.81 \text{ } m/s^2$, $h = \text{depth}$, and $A = \text{area}$.
In order to use the definition of definite integrals to calculate the total force, we can split up the vertical watergate into $n$ horizontal thin strips such that each strip has a base of length $2\sqrt{\dfrac{a^2}{h}(h-y_i)}$ (where $y_i$ is the distance from the bottom of a strip to the surface of the water) and a height of small $\Delta y$. We derive the length from
$$y=h-\frac{x^{2}}{a^{2}}h \iff x=\pm\sqrt{\frac{a^{2}}{h}\left(h-y\right)}$$
and noticing that the length of the base is always $2\sqrt{\dfrac{a^2}{h}(h-y_i)}$.
Enforcing the definition of definite integrals, we will add the sum of all forces, define it $F_{\text{total}}$ (in Newtons), on the watergate and derive the desired answer:
$$ \eqalign{ \lim_{n\to\infty}\sum_{i=1}^{n}\rho gy_{i}2\sqrt{\frac{a^{2}}{h}\left(h-y_{i}\right)} \Delta yN &= \int_{0}^{h}\rho gy2\sqrt{\frac{a^{2}}{h}\left(h-y\right)}dyN \cr &= 2\rho g\sqrt{\frac{a^2}{h}} \int_0^h y\sqrt{h-y}dyN \cr &= 4\rho g\sqrt{\frac{a^2}{h}}\int_{0}^{\sqrt{h}}y^{2}\left(h-y^{2}\right)dyN \cr &= 4\rho g\sqrt{\frac{a^2}{h}}\left(\frac{2h^{\frac{5}{2}}}{15}\right) \cr &= \frac{8\rho g}{15}ah^2 N. } $$
To get the center of pressure, define it $F_{\text{CoP}}$, we can first calculate this:
$$ \eqalign{ \int ydF &= \int y(\rho ghdA) N \cr &= \int y(\rho gydA) N \cr &= \rho g \int_0^h y\left(y2\sqrt{\frac{a^{2}}{h}\left(h-y\right)}\right)dy N \cr &= \frac{32}{105}agh^{3}\rho N, \cr } $$ using the same kind of substitution from the previous integral.
The desired answer for the center of pressure is
$$F_{\text{CoP}} = \frac{1}{F_{\text{total}}} \int ydF = \frac{1}{\frac{8\rho g}{15}ah^2 N} \cdot \frac{32}{105}agh^{3}\rho N = \frac{4}{7}h.$$