= Solution
Choose tangent and normal vectors
$$
\widehat{\mathbf t}=(\sin\theta,\cos\theta),
\qquad
\widehat{\mathbf n}=(-\cos\theta,\sin\theta),
$$
so $\widehat{\mathbf t}_s=-\kappa\widehat{\mathbf n}$, $\widehat{\mathbf n}_s=\kappa\widehat{\mathbf t}$, and $\kappa=\theta_s$. Then
$$
\mathbf F_s=(N_s-T\kappa)\widehat{\mathbf n}
+(T_s+N\kappa)\widehat{\mathbf t},
$$
while the drag components are $f_n=-\zeta_\perp U\cos\theta$ and $f_t=(\zeta_\perp U/2)\sin\theta$.
Let
$$
\xi=\frac sL,
\quad K=L\kappa,
\quad \mathcal N=\frac{L^2N}{A},
\quad \mathcal T=\frac{L^2T}{A}.
$$
Since $N=A\kappa_s$, one has $\mathcal N=K'$. Force balance becomes
$$
\boxed{\mathcal N'-\mathcal TK-\operatorname{Sp}\cos\theta=0},
$$
$$
\boxed{\mathcal T'+\mathcal NK
+\frac{\operatorname{Sp}}2\sin\theta=0},
\qquad
\boxed{K=\theta',\quad\mathcal N=K'}.
$$
At the free end, $\mathcal T(1)=\mathcal N(1)=K(1)=0$; at the held midpoint, position and orientation impose $\theta(0)=0$ in this convention. Reversing the normal reverses the corresponding signs without changing the shape.
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