Answer
The magnitude of the force on the car from the rope is $~~8255~N$
Work Step by Step
We can find the angle $\theta$ that the rope makes with the horizontal:
$tan~\theta = \frac{0.30~m}{9~m}$
$\theta = tan^{-1}~(\frac{0.30~m}{9~m})$
$\theta = 1.9^{\circ}$
A component of the tension in each side of the rope opposes the applied force $F$
We can use the sum of the forces to find the tension $F_T$ in the rope:
$F_{net} = 0$
$2~F_T~sin~\theta-550~N = 0$
$2~F_T~sin~\theta = 550~N$
$F_T = \frac{550~N}{2~sin~\theta}$
$F_T = \frac{550~N}{2~sin~1.9^{\circ}}$
$F_T = 8255~N$
The tension in the rope is $~~8255~N$
The magnitude of the force on the car from the rope is $~~8255~N$.
