= Solution
With endpoints held at zero displacement but free to rotate, variation permits arbitrary endpoint $\delta\psi_x$ and gives the natural zero-moment condition $\psi_{xx}=0$ there. These are simply supported endpoints. The <normal modes> are
$$
\psi_n(x,t)=A_n(t)\sin(n\pi x/l),\qquad k_n=n\pi/l.
$$
The first mode becomes unstable when $F+B(\pi/l)^2<0$, so
$$
\boxed{l>\pi\sqrt{\frac{B}{-F}}\quad(F<0).}
$$
Equivalently the compressive load exceeds $|F|_{\rm crit}=\pi^2B/l^2$. At equality the linear mode is neutral and nonlinear terms are needed to determine its post-buckling amplitude. The quadratic energy of $A\sin(\pi x/l)$ is $A^2l[F(\pi/l)^2+B(\pi/l)^4]/4$, giving the same sign test.
The endpoint displacement condition alone does not impose clamping. The quoted threshold uses the natural free-slope <boundary condition> of the stated variational problem. If rotations were also fixed, the admissible modes and threshold would change; one must not import a clamped-end formula into this result.
\Image[/past-exam-of-the-theoretical-physics-i-course-of-the-university-of-cambridge/2015/paper-1-compressed-rod.png]
{title=Original bending-versus-compression dispersion and the first simply supported buckling mode}
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