#### Answer

$2.76KN$

#### Work Step by Step

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:
$F=2.76KN$