## College Physics (4th Edition)

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.
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.