= Solution
Write $u$ for the effective unloaded addition rate and $w$ for the removal rate of the same elementary tip step $a$. If the supplied on-rate is a bimolecular rate constant, then $u=k_{\rm on}c_m$, with monomer concentration $c_m$; if it already denotes an event rate, use $u=k_{\rm on}$. Assume gap <diffusion> is much faster than chemical events, removal is <force>-independent, and the monomer bath is maintained.
Under a positive constant resisting <force>, the load equilibrates in the potential $Fh$ above a reflecting tip. The normalized <Boltzmann distribution> is $p(h)=\beta F e^{-\beta Fh}$ for $h\ge0$, where $\beta=1/(k_BT)$. The fraction of time with room to insert is
$$
\int_a^\infty p(h)\,dh=e^{-\beta Fa}.
$$
Steric gating reduces addition to $ue^{-\beta Fa}$; each addition advances by $a$ and each removal retreats by $a$. The <reaction-limited polymerization Brownian ratchet> therefore has
$$
\boxed{V(F)=a\left(ue^{-Fa/(k_BT)}-w\right),\qquad V(0)=a(u-w).}
$$
For $u>w>0$, the stall <force> is \b[$F_s=(k_BT/a)\log(u/w)$]. Equivalently, the available chemical free <energy> per effective step is $\Delta\mu=k_BT\log(u/w)$, and stall balances $F_sa=\Delta\mu$. A half-monomer step requires rates defined for that half-step, not automatically single-strand monomer rates with a gratuitous factor of two.
At exactly zero <force> a load on an unbounded half-line has no normalizable equilibrium gap density. The unloaded expression is the $F\downarrow0$ limit, or the usual freely available-gap approximation with a weak positional confinement. For a positive load the equilibration time must remain short compared with reaction times. Load <inertia>, filament bending, monomer depletion, and <force>-dependent bond rupture have been excluded. For arbitrary finite reaction and <diffusion> rates one must solve the joint gap dynamics rather than attach $-aw$ to the <diffusion>-limited first-passage formula: a removal changes the gap and the distribution from which the next addition starts.
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