= Solution
The leading <shear stress> exerted by the fluid on the lower surface is
$$
\tau(x)=\mu u_y(x,0)
=-\frac{\mu U}{h}-\frac h2p_x.
$$
The horizontal force per unit span is $\int_0^L\tau\,dx$. Substituting the pressure gradient from part (c) gives
$$
\frac1{\mu L}\int_0^L\tau\,dx
=-4U\langle h^{-1}\rangle
+6V\langle xh^{-2}\rangle
+6C\langle h^{-2}\rangle.
$$
Using the value of $C$ and setting this force to zero yields
$$
U\left(
4\langle h^{-1}\rangle
-3\frac{\langle h^{-2}\rangle^2}{\langle h^{-3}\rangle}
\right)
=6V\left(
\langle xh^{-2}\rangle
-\frac{\langle h^{-2}\rangle\langle xh^{-3}\rangle}
{\langle h^{-3}\rangle}
\right).
$$
Consequently the <zero-shear lower wall in a translating and squeezing finite gap> condition is
$$
\boxed{
U=6V\,
\frac{\displaystyle
\langle xh^{-2}\rangle
-\langle h^{-2}\rangle\langle xh^{-3}\rangle/
\langle h^{-3}\rangle}
{\displaystyle
4\langle h^{-1}\rangle
-3\langle h^{-2}\rangle^2/\langle h^{-3}\rangle}.}
$$
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