 Solution

Configurational entropy counts the arrangements compatible with a macroscopic configuration: $S=k_B\ln \Omega$. Let $x=L/(Na)$, $N_{\downarrow }=N(1+x)/2$ and $N_{\uparrow }=N(1-x)/2$. The multiplicity is $N!/(N_{\downarrow }!N_{\uparrow }!)$. Stirling's approximation therefore givesWith downward $L$ positive, the gravitational potential energy is $-mgL$. Minimizing the Helmholtz free energy $F=-mgL-TS$ gives $\mathrm{artanh}x=mga/(k_{B}T)$, henceRaising temperature reduces the extension: the entropic preference for many up/down arrangements opposes alignment by the load. For small loads, $L\simeq N{a}^{2}mg/(k_{B}T)$.

For the positive-feedback entropic chain with replacement force $f(L)=f_{0}L/a$, $f_0>0$, the conservative potential is $U(L)=-{\int }_0^{L}f(s)ds=-f_0{L}^2/(2a)$. Using $-f(L)L$ would overcount it by a factor two. The stationary equation becomesExpansion about $x=0$ givesFor $T>T_c$ the stable extension is zero. Below $T_c$, nonzero minima appear, with $x\simeq \pm \sqrt{3(T_c-T)/T}$ close to the transition and $|x|\to 1$ as $T\to 0$. Restricting to downward extensions selects the positive branch. The onset is continuous in this model, with a nonanalytic mean-field change at $T_c$; a finite jump in $L$ does not follow from the printed force law, despite the question's discontinuity wording. A finite chain has rounded fluctuations rather than a literal thermodynamic singularity.

This is the same self-consistency equation as the zero-field mean-field treatment of the Ising model: the segment orientation corresponds to a spin, extension to magnetization, and the extension-dependent force to the self-consistent molecular field. The competition between alignment energy and binary configurational entropy explains the shared temperature dependence.