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This question already has an answer here:

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 :

tanθ = v^2/rg

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

N*sinθ = mv^2/r

Then, the y component is set equal to mg:

N*cosθ = mg =>N = mg/cosθ

Dividing Nsinθ by Ncosθ, we get :

tanθ = 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 is:

mg*sinθ.

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

mg*cosθ.

The normal force is equal to this component into the plane by Newton's 3rd Law, so,

N =mg*cosθ.

Why in the first scenario (banked curve) is

N=mg/cosθ,

HOWEVER, in the second (inclined plane)

N=mg*cosθ ?

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

Is the normal force in the first scenario (banked curve) greater than mgcosθ because:

a) the normal force also is responsible for the centripetal acceleration, so it needs to be greater?

Or,

b) the car is not sliding down the curve, so the normal force is greater because of the translational equilibrium requirement? The textbook that I have (Resnick, Fundamentals of Physics) seems to suggest (b), since they use the equation for equilibrium in the Y direction. But if that is the case, then what is PHYSICALLY CREATING this "extra" normal force as compared to the second scenario (inclined plane)?

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marked as duplicate by Michael Seifert, user259412, honeste_vivere, David Hammen, Jon Custer Jul 5 '17 at 4:27

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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In the banked track scenario you are interested in the horizontal component of the normal force which will contribute to the net horizontal force which provides the centripetal acceleration.
In the inclined plane scenario you are interested in the component of the normal force parallel to the slope which will contribute to the net force along the slope which provides the acceleration along the slope.

So for the banked track there is no motion in the vertical plane so the vertical component of the normal reaction is related to the weight whilst for the slope there is no motion at right angles to the slope so the normal force is related to the component of the weight at right angles to the slope.

Update in response to a comment

Ignoring friction, here are the force diagrams for the two situations.

enter image description here

For the slope the weight does two things.
One component is equal in magnitude to the normal reaction and the other component provides the force which accelerates the body down the slope.
So the normal reaction must be smaller than the weight.

For the banked track the normal reaction does two things. One component is equal to the magnitude of the weight and the other component provides the force which causes the centripetal acceleration.
So the normal reaction must be greater than the weight.
The banked track (normal reaction) is not only trying to prevent the body go vertically downwards it is also making the body change its direction of motion.

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  • $\begingroup$ I asked what is PHYSICALLY CREATING this "extra" normal force as compared to the second scenario (inclined plane)? $\endgroup$ – Hisab Jul 3 '17 at 16:16
  • $\begingroup$ On the banked curve, the normal force must be greater than the weight of the car in order to support the weight of the car AND provide the centripetal force to carry the car around in a circle. For a car that is merely on an inclined plane, the normal force is less than the weight of the car. This means that the two situations, which look similar, are actually quite different from a physics standpoint. $\endgroup$ – David White Jul 3 '17 at 17:18
  • $\begingroup$ The object must exert a larger force in order to get a larger normal reaction force. So, why does the object exert more force on a banked surface than an inclined plane? $\endgroup$ – Hisab Jul 3 '17 at 23:49
  • $\begingroup$ Think of the object trying to move in a straight line and the banked track as well pushing the object up has to push the object inwards to make its path an arc of a circle. $\endgroup$ – Farcher Jul 4 '17 at 5:28
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You ask what's "physically creating" the "extra" normal force. This question doesn't really have anything to do with banked curves & inclines per se. You could ask the same question as follows: if I place a block on a level table, the normal force on the block is a particular magnitude. If I then place another block on top of the first block, the normal force on the bottom block increases. What "physically creates" this extra normal force?

In intro mechanics, we normally think of the normal force as being as big or as small as it needs to be to accomplish the goal of not having an object accelerated through the surface of a table, or a ramp, or a road. In reality, the surface we place the objects on will deform ever so slightly; if we push more on the object, the surface will deform a little bit more. In other words, the surface acts a little bit like a spring; and a spring can generate any magnitude of force you ask it to, depending on how far it's stretched. (At least until it breaks — but then, you can break a table by applying too much weight to it, so the analogy still holds there.) These deformations are usually so small that we can treat the surface as though it's still flat, but they're the reason that the normal force can vary in magnitude.

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  • $\begingroup$ The object must exert a larger force in order to get a larger normal reaction force. So, why does the object exert more force on a banked surface than an inclined plane? $\endgroup$ – Hisab Jul 3 '17 at 23:40
  • $\begingroup$ The object exerts a larger force on the surface because the surface is exerting a larger force on it (by Newton's Third Law.) As to why there must be a greater force on the object, that follows from Newton's Second Law and the geometry of the situation, as explained in Farcher's answer (as well as the duplicate question I linked above.) $\endgroup$ – Michael Seifert Jul 4 '17 at 12:41

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