#### Answer

We should pull with 4.3 units of force at an angle of $83.4^{\circ}$ below the negative x-axis.

#### Work Step by Step

$F_x+3.0-(5.0)~sin(30^{\circ})= 0$
$F_x = -3.0+(5.0)~sin(30^{\circ})$
$F_x = -0.5~units~of~force$
$F_y + (5.0)~cos(30^{\circ}) = 0$
$F_y = -(5.0)~cos(30^{\circ})$
$F_y = -4.3~units~of~force$
We can find the magnitude of the applied force $F$;
$F = \sqrt{(F_x)^2+(F_y)^2}$
$F = \sqrt{(-0.5)^2+(-4.3)^2}$
$F = 4.3~units~of~force$
We can find the angle $\theta$ below the negative x-axis;
$tan(\theta) = \frac{4.3}{0.5}$
$\theta = arctan(\frac{4.3}{0.5})$
$\theta = 83.4^{\circ}$
We should pull with 4.3 units of force at an angle of $83.4^{\circ}$ below the negative x-axis.