Answer
$42KN$
Work Step by Step
We can find the required force as follows:
$\Sigma F_y=2Tsin22^{\circ}-mg=0$
This simplifies to:
$T=\frac{mg}{2sin22^{\circ}}$
We plug in the known values to obtain:
$T=\frac{(3500)(9.81)}{2sin22^{\circ}}$
$T=45.8KN$
Now the force in the forward direction is given as
$T_x=Tcos22^{\circ}$
$T_x=45.8cos22^{\circ}$
$T_x=42KN$
