Answer
The magnitude of the total force exerted on the car by the track is $3910~N$ and this force is directed at an angle of $53.3^{\circ}$ above the horizontal.
Work Step by Step
The horizontal component $F_x$ of the force exerted on the car by the track provides the centripetal force to keep the car moving in a circle. We can find $F_x$:
$F_x = \frac{mv^2}{r}$
$F_x = \frac{(320~kg)(16~m/s)^2}{35~m}$
$F_x = 2340.6~N$
The vertical component $F_y$ of the force exerted on the car by the track is equal in magnitude to the car's weight. We can find $F_y$:
$F_y = mg$
$F_y = (320~kg)(9.80~m/s^2)$
$F_y = 3136~N$
We can find the magnitude of the total force:
$F = \sqrt{F_x^2+F_y^2}$
$F = \sqrt{(2340.6~N)^2+(3136~N)^2}$
$F = 3910~N$
We can find the angle $\theta$ above the horizontal that this force is directed:
$tan~\theta = \frac{F_y}{F_x}$
$tan~\theta = \frac{3136~N}{2340.6~N}$
$\theta = tan^{-1}(\frac{3136~N}{2340.6~N})$
$\theta = 53.3^{\circ}$
The magnitude of the total force exerted on the car by the track is $3910~N$ and this force is directed at an angle of $53.3^{\circ}$ above the horizontal.
