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

$2.76KN$
We know that $\Sigma M_A=0$ $\implies N_B(2.5)-G(1.0)-F(cos 30^{\circ}(0.3)+sin30^{\circ}(0.75))=0$ $\implies N_B=0.04608F+7.848$ Now the friction forces under B are $F_B=0.3N_B$ $\implies F_B=(0.3)(0.04608F+7.848)$ $\implies F_B=2.3544+0.013824F$ As the sum of forces in the x-direction is zero, then: $\implies \Sigma F_x=0$ $\implies F.cos 30^{\circ}-F_B=0$ $\implies F.cos 30^{\circ}-(2.3544+0.013824F)=0$ This simplifies to: $F=2.76KN$