Answer
$F = 798~N$
Work Step by Step
In the final position, we can find the angle $\theta$ the rope makes with the vertical:
$sin~\theta = \frac{4.00~m}{12.0~m}$
$\theta = sin^{-1}(\frac{4.00~m}{12.0~m})$
$\theta = 19.5^{\circ}$
The vertical component of the tension in the rope is equal in magnitude to the gravitational force on the crate.
We can find the force of tension $F_T$ in the rope:
$F_T~cos~\theta = mg$
$F_T = \frac{mg}{cos~\theta}$
$F_T = \frac{(230~kg)(9.8~m/s^2)}{cos~19.5^{\circ}}$
$F_T = 2391~N$
In the final position, the force $F$ is equal in magnitude to the horizontal component of the tension in the rope.
We can find the magnitude of $F$:
$F = F_T~sin~\theta$
$F = (2391~N)~sin~19.5^{\circ}$
$F = 798~N$
