Answer
$F=960lb$
Work Step by Step
We can find the required maximum force as follows:
We know that the sum of forces in the y-direction is given as
$\Sigma F_y=0$
$\Sigma F_y=-Fsin(30)+T_B(\frac{4}{5})=0$
$\Sigma F_y=-Fsin(30)+(600)(\frac{4}{5})=0$..eq(1)
Now the sum of forces in the x-direction is given as
$\Sigma F_x=-Fcos(30)+T_B(\frac{3}{5})=0$
$\Sigma F_x=-Fcos(30)+(600)(\frac{3}{5})=0$..eq(2)
Solving eq(1) and eq(2), we obtain:
$F=960lb$
