Answer
$F=6.37N$
Work Step by Step
We can determine the required force as follows:
We apply Newton's second law
$\Sigma F_y=0$
$\implies N_A-W_Acos\theta=0$
$\implies N_A=W_Acos\theta$
$\implies N_A=(10)(9.81)cos 30=84.96N$
and $\Sigma F_x=ma_x$
$\implies Wsin\theta-\mu_AN_A=ma$
We plug in the known values to obtain:
$40.55-F=10a$.eq(1)
Now we apply Newton's second law for block $B$
$\Sigma F_y=0$
$\implies N_B-W_B cos\theta=0$
$\implies N_B=(6)(9.81)cos30=50.97N$
and $\Sigma F_x=ma_x$
$Wsin\theta-\mu_BN_B+F=ma$
$\implies 14.14+F=6a$..eq(2)
Solving eq(1) and eq(2), we obtain:
$F=6.37N$
