Answer
$F = 106~N$
Work Step by Step
We can find the tension in the cord connected to box $A$:
$F_A = m_A~g~sin~\theta$
$F_A = (11.0~kg)(9.8~m/s^2)~sin~30.0^{\circ}$
$F_A = 53.9~N$
We can find the tension in the cord connected to box $B$:
$F_B = m_B~g$
$F_B = (7.00~kg)(9.8~m/s^2)$
$F_B = 68.6~N$
We can find the horizontal component of tension in the upper cord:
$F_x = F_A~cos~\theta$
$F_x = (53.9~N)~cos~30.0^{\circ}$
$F_x = 46.68~N$
We can find the vertical component of tension in the upper cord:
$F_y = F_A~sin~\theta+F_B$
$F_y = (53.9~N)~sin~30.0^{\circ}+(68.6~N)$
$F_y = 95.55~N$
We can find the tension in the upper cord:
$F = \sqrt{(46.68~N)^2+(95.55~N)^2} = 106~N$
