A cylindrical tank with horizontal axis is filled to half

A cylindrical tank with horizontal axis is filled to half with a liquid of specific gravity $0.9$. Find the pressure force of the liquid on each circular end, given the radius $R=2$ m.

Step 1: Coordinates and width

Put the origin at the circle's center with $y$ measured upward, so the liquid surface is $y=0$ and the liquid fills $-R\le y\le0$. The circular boundary $x^2+y^2=R^2$ gives width $w(y)=2\sqrt{R^2-y^2}$, and the depth below the surface at height $y\le0$ is $-y$.

Step 2: Force integral

\[F=\int_{-R}^0\rho g(-y)\cdot2\sqrt{R^2-y^2}\,dy=-2\rho g\int_{-R}^0y\sqrt{R^2-y^2}\,dy.\]

With $u=R^2-y^2$: at $y=-R$, $u=0$; at $y=0$, $u=R^2$, giving

\[F=\rho g\int_0^{R^2}\sqrt u\,du=\rho g\cdot\frac{2R^3}{3}=\frac{2\rho gR^3}{3}.\]

\[\boxed{F=\frac{2\rho gR^3}{3}}\]

Step 3: Numerical value

With $\rho=0.9\times1000=900\ \text{kg/m}^3$, $g=9.8\ \text{m/s}^2$, $R=2$ m:

\[F=\frac{2\times900\times9.8\times8}{3}=\frac{141{,}120}{3}=47{,}040\ \text{N}=47.04\ \text{kN}.\]

Step 4: Cross-check via the centroid

The submerged semicircle has area $A=\pi R^2/2$ and centroid at depth $4R/(3\pi)$ below the surface, giving $F=\rho g\cdot\dfrac{4R}{3\pi}\cdot\dfrac{\pi R^2}{2}=\dfrac{2\rho gR^3}{3}$, matching Step 2.

Step 5: Center of pressure

Carrying out the same trigonometric-substitution computation as in Problem 191 (mirrored below the surface here) gives

\[\boxed{y_{\text{cp}}=-\frac{3\pi R}{16}\approx-1.178\ \text{m}\ \left(\text{i.e. depth }\frac{3\pi R}{16}\approx1.178\ \text{m below the surface}\right)}\]