Answer
a) $\mu_s\frac{l}{(1.99h)}$
Work Step by Step
a) $(mg\sin(45^o))(0.71l)>Fh$
$m>\frac{0.203Fh}{l}$
$F_y=F_N-F_G=0$
$F_N=F_G$
$F_x=F-F_f=F-\mu_sF_N$
$\mu_s=\frac{F}{mg}=\frac{l}{(0.203h)g}=\frac{l}{(1.99h)}$
Using the inequality, $\mu_s\frac{l}{(1.99h)}$
