= Solution
The fluid's tangential traction on the lower stationary surface is $\tau=\mu u_y|_{y=0}$ in the positive $x$ direction. Using the <pressure> gradient found above gives
$$
\tau=\frac{\mu U}{h}-\frac h2p_X=4\mu U\left(\frac1h-\frac{h_0}{h^2}\right).
$$
The <drag on a plane beneath a translating cylinder> follows by integrating along the wall:
$$
\boxed{\mathbf e_x\cdot\mathbf F=4\mu U\frac\ell{h_0}\left(\pi-\frac\pi2\right)=2\pi\mu U\sqrt{\frac{2a}{h_0}}}
$$
per unit cylinder-axis length. This is the force on the lower surface, so its direction follows the cylinder's translation.
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