My information is very limited. All I know is that there is a normal force and gravity acting on the car. I know what a banked road is, I know a centripetal force is a force that tries to pull the car towards the center. However, I have tried researching and I can't seem to understand so please make me comprehend on this topic. I need it to be simple otherwise it'll be futile. English isn't my native language. Thanks in advance.

Can someone just tell me why friction isn't needed?

## Best Answer

The frictionless banked curve exerts a normal force $F_{n}$ perpendicular to its surface. The downward force of the gravity

$F_{g}$is present. The two forces add as vectors and the resultant or net force $F_{net}$ points toward the center of the circle. This is the centripetal force.When the forces are resolved into their components, you will find that $F_{net,y}=F_{n}\cos\theta-F_{g}=0$. Hence, $F_{n}=\frac{F_{g}}{\cos\theta}$. You will also find that $F_{net,x}=F_{n}\sin\theta$. You know that this is equal to the radial force, $F_{r}=m\frac{v^2}{r}$.

$$F_{r}=F_{net,x}\Longrightarrow\frac{mv^2}{r}=F_{n}\sin\theta\Longrightarrow\frac{mv^2}{r}=\frac{F_{g}}{\cos\theta}\sin\theta\Longrightarrow\frac{mv^2}{r}=mg\tan\theta\Longrightarrow\frac{v^2}{r}=g\tan\theta$$

After simplifying, you will find that $v=\sqrt{rg\tan\theta}$, where $\theta$ is the angle that will allow a car to travel on a frictionless curve of radius $r$ with constant speed $v$. A banked curve is designed for one specific speed. Traveling at a speed higher than $v$ means the car will slide out, up, and over the edge. Traveling at a speed lower than $v$ means the car will slide in, down, and off the bank.