A right circular cone of base radius R and height H is

A right circular cone of base radius $R$ and height $H$ is submerged apex-down in water, with its flat base exactly at the water surface. Find the total (vertical) hydrostatic force on the cone.

Step 1: Geometry

Put the base (radius $R$) at the surface $z=0$ and the apex at depth $z=H$; by similar triangles the radius at depth $z$ is $r(z)=R(1-z/H)$.

Step 2: Vertical force via horizontal projection

As $z$ increases by $dz$, the radius shrinks by $|dr|=(R/H)\,dz$, so the thin conical band between $z$ and $z+dz$ projects onto a flat annulus of area $2\pi r(z)\cdot(R/H)\,dz$. The pressure there is $p(z)=\rho gz$, so

\[dF=p(z)\cdot2\pi r(z)\cdot\frac{R}{H}\,dz=2\pi\rho g\,\frac{R}{H}\,z\,R\left(1-\frac{z}{H}\right)dz=\frac{2\pi\rho gR^2}{H}\left(z-\frac{z^2}{H}\right)dz.\]

Step 3: Integrate over the height

\[F=\frac{2\pi\rho gR^2}{H}\int_0^H\left(z-\frac{z^2}{H}\right)dz=\frac{2\pi\rho gR^2}{H}\left(\frac{H^2}{2}-\frac{H^2}{3}\right)=\frac{2\pi\rho gR^2}{H}\cdot\frac{H^2}{6}=\frac{\pi\rho gR^2H}{3}.\]

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

Step 4: Confirm via Archimedes' principle

Since the base sits exactly at the surface (zero gauge pressure there, so it contributes no force), the entire vertical hydrostatic force on the cone comes from the lateral surface, and by Archimedes' principle it must equal the weight of the water displaced by the whole submerged solid:

\[F=\rho g\cdot V_{\text{cone}}=\rho g\cdot\frac{\pi R^2H}{3}=\frac{\pi\rho gR^2H}{3},\]

exactly matching Step 3 — a clean and reassuring independent check.

Step 5: Numerical example

For $R=2$ m, $H=6$ m, $\rho=1000\ \text{kg/m}^3$, $g=9.8\ \text{m/s}^2$:

\[F=\frac{\pi\times1000\times9.8\times4\times6}{3}=\pi\times78400\approx246{,}300\ \text{N}\approx246.3\ \text{kN},\]

which indeed equals $\rho g$ times the cone's volume $\tfrac13\pi(2)^2(6)=8\pi\approx25.13\ \text{m}^3$.