# Physics of Banked Curves at an angle [closed]

When calculating the normal force of an object on a ramp inclined at an angle, the force of gravity is broken into components: $f_{gx}$ and $f_{gy}$, where $f_{gx} = f_g \sin (\theta)$ and $f_{gy} = f_g \cos (\theta)$ and the normal force is equal to $f_{gy}$ (since there's no acceleration in the $y$ direction).

However, when dealing with banked curves $F_n$ is not equal to $f_{gy}$, rather $F_{n}$ is equal to $f_g$. Why is this? Normal force is the opposite force of any force exerted on the surface. Since $f_{gy}$ is exerted on the surface, shouldn't $F_n$ be equal to $f_{gy}$ ? I am aware of centripetal force, but shouldn't that be taken care of by the horizontal component of $f_{gx}$?

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Can you provide some reference where they say $F_n = f_{g}$ and not $f_{gy}$. Wikipedia says you need to take the component of the weight (i.e. force due to gravity). – mythealias Nov 10 '12 at 3:40
A sketch would be helpful for you and for us to understand better. – ja72 Feb 17 at 2:51
Echoing mythealias's comment, I'm closing this question as not a real question. – Qmechanic May 18 at 10:55

## closed as not a real question by Qmechanic♦May 18 at 10:55

It's difficult to tell what is being asked here. This question is ambiguous, vague, incomplete, overly broad, or rhetorical and cannot be reasonably answered in its current form. For help clarifying this question so that it can be reopened, see the FAQ.

$$\sum F_x = N\sin\theta = ma \\ \sum F_y = N\cos\theta - mg = 0$$
where $a$ is the horizontal acceleration (see figure). If the vehicle is to stay on the ground, then it cannot be that the normal force is equal to the gravitational force unless $\theta = 0$.