# [Physics] Can a car move on a banked road without friction

centripetal-forcefree-body-diagramfrictionhomework-and-exercisesnewtonian-mechanics

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?

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.