= Solution
The compressive tip <force> is limited by <Euler buckling of an elastic filament>. With unsupported length $\ell$ and <filament bending modulus> $B=EI$,
$$
\boxed{F_c\sim\frac B{\ell^2}\sim\frac{Er^4}{\ell^2}.}
$$
For a hollow <microtubule>, use $I=\pi(r_o^4-r_i^4)/4$ rather than the solid-cylinder value. For a clamped seed and a moment-free tip free to slide along the wall, linearized angle balance is $B\theta''+F\theta=0$, with $\theta(0)=0$ and $\theta'(\ell)=0$. Its first nonzero solution has $\sqrt{F/B}\,\ell=\pi/2$, hence $F_c=\pi^2B/(4\ell^2)$. A different tip constraint changes this coefficient.
Increasing the unsupported length decreases that threshold. For growth against a wall at fixed projected gap $d$, <polymerization-driven cantilever postbuckling> gives a more direct quantitative statement. The inextensible elastica obeys $B\theta''+F\sin\theta=0$. Multiplying by $\theta'$ and using the zero tip moment gives
$$
\frac B2(\theta')^2=F(\cos\theta-\cos\theta_t).
$$
Put $k=\sin(\theta_t/2)$ and substitute $\sin(\theta/2)=k\sin\phi$. Integrating arc length and its horizontal projection gives
$$
\ell=\sqrt{B/F}\,K(k),\qquad d=\sqrt{B/F}\,[2E(k)-K(k)],
$$
where the complete <elliptic integrals> are $K(k)=\int_0^{\pi/2}(1-k^2\sin^2\phi)^{-1/2}d\phi$ and $E(k)=\int_0^{\pi/2}(1-k^2\sin^2\phi)^{1/2}d\phi$. Using $K=(\pi/2)(1+k^2/4+\cdots)$ and $E=(\pi/2)(1-k^2/4+\cdots)$ gives $\ell/d=1+k^2+O(k^4)$, and therefore
$$
\boxed{F=\frac{\pi^2B}{4d^2}\left[1-\frac32\frac{\ell-d}{d}+O\left(\frac{(\ell-d)^2}{d^2}\right)\right].}
$$
Thus the initial postbuckling <force> decreases as more contour is inserted. This result describes the weakly buckled branch before overhang, further wall contact, or altered tip constraints invalidate the model. It is a mechanical <force> law; a chemical <force> limit from <polymerization> can impose a smaller <force>.
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