Determine the water pressure force on a vertical

Determine the water pressure force on a vertical triangular plate whose base $a$ lies at the water surface and whose apex points straight down, at depth (height) $h$.

Step 1: Width as a function of depth

\[w(y)=a\left(1-\frac{y}{h}\right),\qquad 0\le y\le h.\]

Step 2: Force on a thin strip and total force

\[F=\int_0^h\rho gy\cdot a\left(1-\frac{y}{h}\right)dy=\rho ga\int_0^h\left(y-\frac{y^2}{h}\right)dy=\rho ga\left(\frac{h^2}{2}-\frac{h^2}{3}\right)=\frac{\rho gah^2}{6}.\]

\[\boxed{F=\frac{\rho gah^2}{6}}\]

Step 3: Cross-check via the centroid

The triangle's centroid (average of its three vertices $(\pm a/2,0)$ and $(0,h)$) lies at depth $h/3$. With area $A=ah/2$, the average-pressure method gives $F=\rho g(h/3)\cdot(ah/2)=\rho gah^2/6$, matching Step 2.

Step 4: Center of pressure

\[y_{\text{cp}}=\frac{\int_0^hy^2(1-y/h)\,dy}{\int_0^hy(1-y/h)\,dy}=\frac{h^3/12}{h^2/6}=\frac{h}{2}.\]

\[\boxed{y_{\text{cp}}=\frac{h}{2}}\]

Note that this is deeper than the triangle's own centroid at $h/3$ — as expected, since pressure increases with depth and so weights the lower part of the plate more heavily, pulling the center of pressure below the geometric centroid (it is not itself the centroid, though for this particular triangle it happens to land exactly at the plate's mid-depth $h/2$).