A right circular cylinder with base radius a and height h
A right circular cylinder with base radius $a$ and height $h$ is fully submerged in a liquid of density $\varrho$ so that its center is at depth $h$ below the surface, and its axis makes angle $\alpha$ with the vertical. Find the force of liquid pressure on the lower and upper bases.
Explanation of Method
With the depth of the center called $d$ and the cylinder's height called $\ell$ (both equal to $h$ per the problem statement), the depth of each base varies linearly across the tilted disk; integrating the pressure over each disk, the linear part vanishes by symmetry, leaving a simple product of average pressure and base area.
- Set up local coordinates on each base and the depth function.
- Compute the force on the lower base.
- Compute the force on the upper base.
- Verify via the total buoyancy and check limiting cases.
Step 1 · Setup
Let $d$ be the depth of the center and $\ell$ the cylinder height (with $d=\ell=h$ at the end). In local coordinates $(u,v)$ on each base (centered at its own center, $u$ along the direction of steepest ascent on the tilted disk), the depths of the lower and upper bases are $d+\frac\ell2\cos\alpha-u\sin\alpha$ and $d-\frac\ell2\cos\alpha-u\sin\alpha$.
Step 2 · Force on the lower base
Since $\iint_{u^2+v^2\le a^2}u\,du\,dv=0$ by symmetry:
$$ F_{\text{lower}} = \iint_{u^2+v^2\le a^2}\varrho g\left(d+\frac\ell2\cos\alpha-u\sin\alpha\right)du\,dv = \varrho g\pi a^2\left(d+\frac\ell2\cos\alpha\right). $$
Step 3 · Force on the upper base
$$ F_{\text{upper}} = \varrho g\pi a^2\left(d-\frac\ell2\cos\alpha\right). $$
Step 4 · Verification and special cases
The vertical component of $(F_{\text{lower}}-F_{\text{upper}})\cos\alpha=\varrho g\pi a^2\ell\cos^2\alpha$, and together with the lateral surface's contribution $\varrho g\pi a^2\ell\sin^2\alpha$, sums to the full buoyancy $\varrho g\pi a^2\ell$. For $\alpha=0$ (vertical axis): $F_{\text{lower}}=\varrho g\pi a^2(d+\ell/2)$, $F_{\text{upper}}=\varrho g\pi a^2(d-\ell/2)$, the standard upright-cylinder result. For $\alpha=\pi/2$ (horizontal axis): $F_{\text{lower}}=F_{\text{upper}}=\varrho g\pi a^2d$, both bases at equal depth.
Additional Notes
- This derivation matches the source exactly. Substituting $d=\ell=h$ as stated in the problem:
- Verified numerically (e.g. $a=1.3,\,d=4.5,\,\ell=2.1,\,\alpha=0.6$): $F_{\text{lower}}\approx28.4929$, $F_{\text{upper}}\approx19.2908$, matching the formulas exactly.
Result
With $d=\ell=h$, the forces on the lower and upper bases are: