## Engineering Mechanics: Statics & Dynamics (14th Edition)

$200lb$
We know that $F_B=\mu_s N_B$ $\implies F_B=0.35N_B$ We also know that $\Sigma M_B=0$ $\implies N_A(3)-P(3-1)=0$ $\implies N_A(3)-P(2)=0$ This simplifies to: $N_A=200lb$