= Equilibrium peeling force of an adsorbed Gaussian chain
{title2=$f=\sqrt{6k_BT\mu}/b$}
Let an adsorbed <Gaussian chain> gain <free energy> $\mu>0$ per bound <monomer>. A pulled segment of $n$ <Kuhn length> steps spanning height $z$ has elastic <free energy> $3k_BTz^2/(2nb^2)$, giving $F=-(N-n)\mu+3k_BTz^2/(2nb^2)$. Minimizing over $n$ yields $n=(z/b)\sqrt{3k_BT/(2\mu)}$ and $F_{\min}=-N\mu+z\sqrt{6k_BT\mu}/b$. Its height <derivative> is the constant peeling <force> displayed above. The plateau lasts while $0<n<N$ and the <Gaussian chain> approximation remains valid. In the slab model $F_{\mathrm{ads}}=Nk_BT[(b/\xi)^2-\epsilon b/\xi]$, optimization gives $\mu=k_BT\epsilon^2/4$ and $f=\sqrt{3/2}\,\epsilon k_BT/b$.
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