Answer
$N_{A/leg}=22.1lb$
$N_{B/leg}=30.6lb$
and $a=13.58 ft/s^2$
Work Step by Step
The required normal reaction and acceleration can be determined as follows:
We know that equilibrium in x-direction is given as
$\Sigma F_x=0$
$\implies Fcos30-\mu_s N_A-\mu_s N_B=0$
$\implies Fcos30-0.5N_A-0.5N_B=0$ [eq(1)]
Similarly, the equilibrium in y-direction is given as
$\Sigma F_y=0$
$\implies N_A+N_B-W-Fsin30=0$
$\implies N_A+N_B-F=75$ [eq(2)]
After solving eq(1) and eq(2), we obtain:
$F=60.874lb$
We know that
$\Sigma F_x=ma$
$\implies Fcos30-\mu_k N_A-\mu_k N_B=ma$
$\implies 60.874cos30-0.2N_A-0.2N_B=\frac{75}{32.2}a$ [eq(3)]
and $\Sigma F_y=0$
$\implies N_A+N_B-W-Fsin30=0$
$\implies N_A+N_B=75+60.874sin30$ [eq(4)]
After solving eq(3) and eq(4), we obtain:
$a=13.58 ft/s^2$
Now, we sum the moments about point A
$M_A=\Sigma M_{kA}$
$\implies (-2)ma=-3Fcos30-4N_B-2W$
$\implies 2(\frac{-75}{32.2})=-3(60.874 cos30)-4\times N_B-2(75)$
This simplifies to:
$N_B=61.2lb$
We plug in this value in eq(4) to obtain:
$N_A=44.2lb$
$\implies N_{A/leg}=\frac{44.2}{2}=22.1lb$
and $N_{B/leg}=\frac{61.2}{2}=30.6lb$
