Answer
$F=6KN$
Work Step by Step
As we know that the car is in equilibrium so
$\sum T_p=0$
$\implies F\times 1.8sin34^{\circ}-Mg\times (2.4-1.8)cos34^{\circ}=0$
This simplifies to:
$F=\frac{Mg\times 0.6cos34^{\circ}}{1.8sin34^{\circ}}$
We plug in the known values to obtain:
$F=\frac{1250\times 9.8\times cot34^{\circ}}{3}$
$F=6kN$
