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. ---------------- 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 ---------- 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.