Engineering Mechanics: Statics & Dynamics (14th Edition)

$N=277lb$, $F=13.4lb$
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: $F=150sin60-(\frac{150}{32.2})[\frac{(20)^2}{8}]cos60$ $\implies F=13.44lb$