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**Normal force on a banked curve????**

In every textbook I've looked at, I can't find an answer to this question. When determining the centripetal force on an object on a banked curve, it is stated that the banking angle for a given speed and radius is found by tan θ = v^2/rg

It is found as follows (there is an attachment as well):

The normal force on the object is resolved into components. The x-component (the one providing the centripetal force) is:

Fn * sin θ = mv^2/r (1)

Then, the y component is set equal to mg and it is found that the normal force:

Fn = mg / cos θ (2)

Substitute Fn from (2) into (1) and get:

tan θ = v^2/rg

I'm fine with that.

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Here is where I'm confused...When resolving the forces of an object resting on an inclined plane:

The component down and parallel to the plane due to gravity, Fp, is as follows:

Fp = mg * sin θ

The component representing the force of gravity into (perpendicular) to the plane is:

mg * cos θ

The normal force is equal to this component into the plane by Newton's 3rd Law, so Fn = mg * cos θ

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Why in the first scenario (banked curve) is Fn = mg / cos θ, HOWEVER, in the second (block on plane) Fn = mg * cos θ ????

How can this be? There are two different values for Fn

I'm stumped. I have no solution.

Thank you for any help.