Answer
(a) $F = \frac{mg~\mu_k}{cos(\theta)- ~sin(\theta)~\mu_k }$
(b) $\mu_s = cot(\theta)$
Work Step by Step
(a) Since the crate is moving with constant velocity, the horizontal component of the force is equal in magnitude to the force of kinetic friction.
$F~cos(\theta) = F_N~\mu_k$
$F~cos(\theta) = (mg+F~sin(\theta))~\mu_k$
$F~cos(\theta)- F~sin(\theta)~\mu_k = mg~\mu_k$
$F = \frac{mg~\mu_k}{cos(\theta)- ~sin(\theta)~\mu_k }$
(b) $F = \frac{mg~\mu_s}{cos(\theta)- ~sin(\theta)~\mu_s }$
If the denominator of the fraction goes to zero, then the required force becomes infinitely large.
$cos(\theta)- ~sin(\theta)~\mu_s = 0$
$\mu_s = cot(\theta)$
The critical value is $\mu_s = cot(\theta)$
