Vertical component of friction

If we have a car of mass $$m$$ sliding around a banked curve of angle $$\phi$$ with some coefficent of friction $${\mu}$$, then the normal force will be

$$Ncos\phi= mg + {\mu}F_{g}\sin\phi$$

But this would mean that there is a frictional force acting pressing directly against the plane, I was of that understanding that friction only happens horizontally. Is my understanding in-correct or have I missed some important point in my study? edit:http://www.batesville.k12.in.us/physics/phynet/mechanics/circular%20motion/banked_with_friction.htm (see when they resolve for vertical forces)

• Assuming that the ramp is making an angle of $\phi$ with horizontal ground. The normal force will be balancing only the $F_g cos\phi$. The $sin\phi$ component would be parallel to the surface of the ramp, why are you equating it with the Normal force? – orion Jul 14 at 16:05
• edited question – DDD4C4U Jul 14 at 16:29
• Your equation seem to be incorrect, please add diagram and also how did you get this equation. It is not matching with any equation from the link you have shared. – orion Jul 14 at 17:22
• sry fixed it, Check again! – DDD4C4U Jul 14 at 17:30
• If your understanding is that friction force only acts horizontally than it is incorrect. If your understanding is that friction force act parallel to the surface than it is correct. I think you are mixing two things here, friction is parallel to the surface but as the surface here is making an angle $\phi$ with horizontal, hence it has a vertical component. – orion Jul 14 at 18:14

"friction only happens horizontally" -- this in incorrect.

The force of friction is applied in the direction that opposes the relative motion between two bodies.

Take a book and press it against a wall. The friction force will act vertically in that scenario. It will try to keep the book in the same place and will act against gravity.

Now, start sliding that book in upwards direction. The friction force will still act vertically but now it will work in the opposite direction i.e. work with the gravity. This is all for the same reason which is to oppose the relative motion between two bodies.

All the directions (horizontal, vertical, diagonal...) are relative and are based on frame of reference. Based on the question in your case of this ramp, the force of gravity will act downward, hence mg. Then the force of friction will be parallel to the road. To make the calculation easier, we use a bit of math for the "resolution of vectors" and break the force of friction to two components; therefore, you got the sine and cosine in your equation.

• It might help here to show how the given equation relates to looking at the vertical force components specifically. – Aaron Stevens Jul 14 at 22:15

Assuming the angle is the incline with the horizontal only the first term is the normal force to the plane. I don’t know what your second term is. Change it to the cosine and then it is the friction force acting up the plane. The force acting down the plane is $$F_{g}\sin\phi$$

Hope this helps

• I edited the question check at it again thanks – DDD4C4U Jul 14 at 16:29