## Essential University Physics: Volume 1 (3rd Edition)

$F=6KN$
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: $F=\frac{1250\times 9.8\times cot34^{\circ}}{3}$ $F=6kN$