So there were two methods presented: the first one was using the $F = PhA$ where $P$ is density, $h$ is the height, $A$ is the area of the strip. The second one was using the center of mass which is $F = PHA$, where $H$ is the center of mass. I really couldn’t upload any pics for the visuals but I’ll try to explain as much as I could. You can find the problem on the book entitled 10th Edition Calculus by Ron Larson and Bruce H. Edwards on page 501, exercises 7.7 number 14. The problem is:
Find the fluid force on the vertical plate submerged in water, where the dimensions are given in meters and the weight-density of water is $9800~\frac{\text{N}}{\text{m}^3}$.
It only gave a square illustration. Take not that the square isn’t positioned properly but it is tilted 45 degress do it looks like a diamond but it is really a square. All of its sides is 3m. The square is also 1 m deep.
Using method 1: $$ h = 1+ \frac{3}{\sqrt{2}}-y$$, the 1 is taken from the depth. $A = x$, I divided the square into 4 traingles thus solving for one but multiplied by 4. Since the the height of the element is $dy$, that means we should change x in terms of y. Using the slope which turned out to be $y=-x + (3\sqrt{2}/2)$. So $$ x = -y + \frac{3 \sqrt{2}}{2}. $$ $$ F = 9800(4) \int _0^{\frac{3 \sqrt{2}}{2}} \left(1+3\sqrt{2}-y \right)\left(-y+\frac{3\sqrt{2}}{2}\right)\, dy $$ The answer becomes $212~933.6362~\text{Nm}$
Method 2
Using method 2 was a lot easier but the answer was different. So using F = PHA $$ \left\{ \begin{array}{cc} H =& 1+ \frac{3\sqrt{2}}{2} \\ A =& \frac{1}{2} bh\\ \end{array} \right. $$ where the base is 3 and the height is 3 so it becomes 1/2 * 3 * 3. Thus
$$ F = 9800(1+\frac{3 \sqrt{2}}{2})(3)(3)$$. The answer is $275~300.45~\text{Nm}$.
Is there anything wrong with the first method? I really hope someone could help to clear my confusion. Thanks in advance.
