possible to levitate the object into a stable equilibrium away from the boundaries. For this one needs (at least) two polymers. This more difficult problem was solved when the two polymers start at the same point or nearby on the real axis and both go to infinity [@GamsaCardy2005]. We use their extension of Schramm’s formula to two SLEs conditioned not to merge before reaching infinity. One defines ${\cal P}_l$, ${\cal P}_m$ and ${\cal P}_r=1-{\cal P}_m-{\cal P}_l$ the relative weights of configurations such that the object is constrained to lie on the left of both polymers (l), in the middle (m) or to the right (r). Then ${\cal P}_m= \frac{4}{5} \sin^2(\theta)$, hence the free energy is: $${\cal F}_m = - k T \left[ 2 \ln (\sin \theta) + \ln(4/5)\right]\ .$$ More complicated formula hold for ${\cal P}_r$ and ${\cal F}_r$. We obtain for the force exerted on a point which remains to the left of the two polymers (l), in the middle (m) or to the right (r) as $\vec f= f^\theta \vec n_\theta$ with: $$\begin{aligned}
f^\theta_{{l}} &=& -\frac{k T}{r} 
\frac{8 \sin (\theta ) [-12 \theta  \cos (\theta )+9 \sin (\theta )+\sin (3 \theta )]}{-24 \cos (2 \theta ) \theta -36 