= Solution
{c}
Since $b>1$, the denominator in
$$
F_x=\frac{2\pi\rho}{b^2(b^2-1)}\,[b-U(b^2-1)]
$$
is positive. The positive root of $Ub^2-b-U=0$ is
$$
\boxed{b_* =\frac{1+\sqrt{1+4U^2}}{2U}>1}.
$$
The other root is negative, so the horizontal <force> points in the positive $x$ direction for $1<b<b_*$, vanishes at $b=b_*$, and points in the negative $x$ direction for $b>b_*$. In particular, the nearby <line source> pulls the <cylinder> towards itself as $b\downarrow1$, while at large $b$ the horizontal <force> is $F_x\sim-2\pi\rho U/b^2$.
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