x/L)-\cos (2\pi y/L)}} \right]$$ On the symmetric line $y=L/2$ the force is directed along $x$ and equal to $
f_x = \frac{k T}L  \frac{2 \pi }{1+e^{2 \pi  x/L}}, 
$ which has a finite limit at large negative $x$.

(100,41) (0,38) (0,0)[[![Left: Strip geometry ${\cal D}$: A self-avoiding polymer constrained to depart from $x=0$, passing through $x=i L$, and staying left of point $z$. Right: Mapping of the strip to the plane.[]{data-label="f:3"}](./figures/geom3 "fig:"){width="0.44\columnwidth"}]{}]{} (48,30)[$\to$]{} (48,9)[$\to$]{}

[![Top: Strip geometry ${\cal E}$: A self-avoiding polymer constrained to depart from $x=-\infty$, going to $x=+\infty$, and passing at the top of point $z$. Bottom: Mapping of the strip to the plane.[]{data-label="f:5"}](./figures/geom5 "fig:"){width="0.75\columnwidth"}]{}

$\to$$\to$

In the second strip geometry, $\cal E$ on figure \[f:5\], the polymer is attached infinitely far away on each side and the object is below it. Using the map $w=e^{\pi z/L}$, one finds the free energy and force (with $f_x=0$): $$\begin{aligned}
 {\cal F}_{\cal E} &=& - kT \left[\ln \left( 1 + \cos(\pi y/L)\right) - \ln2\right] \\
 f^{\cal E}_{y} &=& - \frac{\pi}{L} \tan\!\left({\pi y\over 2 L}\right)\ .  \label{strip2} \end{aligned}$$ In all cases considered above the force tends to bring the object towards a portion of the boundary. One can ask whether it is 