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With positive opposing load magnitude $L$ and positive viscous drag coefficient $\gamma$, steady contraction requires $L=N⟨f⟩-\gamma u$. A decreasing motor force curve and an increasing drag-magnitude line have a unique intersection for each load below the limiting stall force. Using $+\gamma u$ as the signed drag would instead make drag assist forward motion, reversing its physical role.

Put $V=k_1{x}_1$ and $F_{\ast }=N{f}_0{x}_2/\Delta$. The constant-force spatially gated motor producesThe limit is approached after the periodic steady state is established; at exactly $u=0$ arbitrary initially bound heads in inactive phase intervals need not equilibrate. The curve decreases from $F_{\max }$ and has formal large-$u$ tail $F_{\max }V/u$. Extrapolation of that tail beyond the small-$u$ detachment assumptions is not controlled.

Sign inconsistency in the printed force balance. For a positive viscous drag coefficient $\gamma$, the signed drag force is $-\gamma u$. The printed $F_{\mathrm{ext}}+F_m+\gamma u=0$ cannot give the requested positive no-load contraction speed: both last terms are positive when $u>0$. With $F_{\mathrm{ext}}=-F_{\max }/2$, the left side starts at $F_{\max }/2>0$ and may have zero, one or two positive roots depending on parameters; the requested usual load construction is not implied by that equation.

The physical linear-drag load curve for a spatially gated motor usesOn the same axes plot the positive drag magnitude $D(u)=\gamma u$ and the decreasing motor force $F_m(u)$. The no-load speed $u_0$ is their intersection. For opposing load $F_{\max }/2$, intersect $F_m(u)$ with the parallel raised line $D(u)+F_{\max }/2$; equivalently intersect $D(u)$ with $F_m(u)-F_{\max }/2$. Drop verticals to the speed axis. The loaded speed obeys $0<u_{1/2}<u_0$. Each intersection is unique because $F_m$ decreases and $D$ increases. The signed drag is also indicated below the axis in the sketch.

Take an overdamped particle with viscous drag coefficient $\gamma$, so its deterministic drift in the harmonic potential $U=k{x}^2/2$ is $\dot{x}=-kx/\gamma$. The one-dimensional Fokker-Planck equation has probability currentAt thermal equilibrium, $J=0$ and the Boltzmann distribution is $p(x)=Z^{-1}{e}^{-k{x}^2/(2k_{B}T)}$, with $p^{\prime }/p=-kx/(k_{B}T)$. Inserting this into the zero-current condition givesThe result is the Einstein diffusion relation between diffusion and mobility. It assumes a bath at one temperature, linear drag and thermal equilibrium. The same Gaussian probability density gives $⟨x^2⟩=k_{B}T/k$, providing an independent calibration of trap stiffness.