= Solution
For an <inclined viscous film with opposing surface shear>, the steady parallel velocity automatically satisfies the <continuity equation> and the advective acceleration vanishes. The two components of the <Navier-Stokes equations> reduce to
$$
0=-p_x+\mu u''+\rho g\sin\alpha,\qquad0=-p_z-\rho g\cos\alpha.
$$
The uniform thickness and constant atmospheric pressure make $p_x=0$. The boundary conditions are
$$
u(0)=0,\qquad\mu u'(h)=-S,\qquad p(h)=p_{\rm atm}.
$$
These are respectively <no-slip boundary condition>, imposed tangential <traction> on the surface with outward normal $+\mathbf e_z$, and the normal stress condition on the flat free surface. The normal viscous stress is zero because the normal velocity vanishes. Integrating gives
$$
\boxed{u(z)=\frac{\rho g\sin\alpha}{\mu}\left(hz-\frac{z^2}{2}\right)-\frac S\mu z,\qquad p(z)=p_{\rm atm}+\rho g\cos\alpha(h-z)}.
$$
In particular $\mu u'(0)=\rho gh\sin\alpha-S$ is the downslope tangential <traction> exerted by the fluid on the wall. The wall's tangential <traction> on the fluid has the opposite sign, because the outward normal of the fluid at its bottom is $-\mathbf e_z$. This fixes the potentially confusing wall sign convention.
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