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

$154.89N$
The required minimum force $P$ can be determined as follows: We know that $\Sigma M_B=0$ $\implies W(2)-N_A(3.0)-F_A(4.0)=0$ $\implies W(2)-N_A(3.0)-3.8N_A=0$ This simplifies to: $N_A=154.89N$ The sum of forces in the x-direction is zero, so: $\implies \Sigma F_x=0$ $\implies P-N_A=0$ $\implies P-154.89=0$ $\implies P=154.89N$