Answer
$N=277lb$, $F=13.4lb$
Work Step by Step
We know that
$\Sigma F_y=ma_n$
$N-Wcos\theta=ma_nsin\theta$
$\implies N=Wcos\theta+m\frac{v^2}{\rho}sin\theta$
We plug in the known values to obtain:
$N=150cos60+(\frac{150}{32.2})[\frac{(20)^2}{8}]sin60$
$\implies N=277lb$
We also know that
$\Sigma F_x=ma_x$
$\implies Wsin\theta-F=ma_ncos\theta$
$\implies F=Wsin\theta-m\frac{v^2}{\rho}cos\theta$
We plug in the known values to obtain:
$F=150sin60-(\frac{150}{32.2})[\frac{(20)^2}{8}]cos60$
$\implies F=13.44lb$