Normal Force: Inclined Plane vs. Banked Turn

[jon4444]

I"m wondering, at the abstract level, why different mathematics is used to calculate the Normal Force in an inclined plane versus a banked turn (which a vehicle is driving around).
For an inclined plane, the standard approach is take weight and resolved into parallel and perpendicular vectors, giving mg* cos for normal force. For banked turn, the standard approach is take the Normal force, and calculate the component that's anti-parallel to weight, and then "infer" the Normal Force is mg / cos.
I'm assuming the Physics behind this is that the Normal Force in incline plane is solely due to gravity, while in banked turn it changes due to the circular motion of the vehicle.
I'm wondering if this is correct and how one justifies the different mathematical approaches? (I.e., how does one decide which vector to resolve into components and why wouldn't the vertical component of the Normal Force in a banked turn just be mg--what other tangible forces could one speak of in that situation)?

 

[kuruman]

You are asking a very good question. First let me say that the math is not different. In both cases trigonometry and vectors are used. The difference lies in the choice of coordinate system. These problems involve acceleration vectors in two dimensions. It is easier to write the acceleration vector in a frame where one of its components is zero. Then the other component is just the magnitude $a$. Thus, in the inclined plane, the x-axis is chosen down the incline, so that the acceleration is parallel to the surface, and has zero component normal to the surface. This makes the calculation of the normal force easy. In the banked curve, the acceleration is perpendicular to the direction of gravity and in a horizontal plane. Therefore the x-axis is chosen so the acceleration is horizontal and towards the center. Of course in this case the normal force has two components as you remarked which makes the analysis slightly more involved. So the upshot is that, if you choose the x-axis along the acceleration, then you have Newton's 2nd law in the form $F_{net,x} = ma$ and $F_{net,y}=0$ which makes the algebra easier to handle.

 

[jon4444]

Thanks--could you comment on the Physics behind it. The normal forces are, in fact, different. Is this due to the circular motion of the vehicle, as I presume?

 

[CWatters]

A stationary car on a banked track is the same as an object on an incline. Only gravity contributes to the normal force.

A moving car on a banked track has two forces that contribute to the normal force (gravity and centripetal).

A racing car on a banked track has three force that contribute to the normal force (gravity, centripetal and aerodynamic down force).

There might be other cases but I can't think of them at the moment.

 

[A.T.]

why wouldn't the vertical component of the Normal Force in a banked turn just be mg?

Who says it isn't?

 

[kuruman]

A moving car on a banked track has two forces that contribute to the normal force (gravity and centripetal).

I prefer to think of it slightly differently. The normal force is a contact force and, as such, it adjusts itself to provide the observed acceleration. In the case of the inclined plane, it adjusts itself to provide zero acceleration in a direction perpendicular to the incline. In the case of the banked curve, it adjusts itself to do two things, (a) provide zero acceleration in the vertical direction (by balancing the acceleration of gravity) and (b) provide the centripetal acceleration v²/r.

 

[Mister T]

For an inclined plane, the standard approach is take weight and resolved into parallel and perpendicular vectors, giving mg* cos for normal force. For banked turn, the standard approach is take the Normal force, and calculate the component that's anti-parallel to weight, and then "infer" the Normal Force is mg / cos.

In both cases the standard approach is to solve Newton's 2nd Law $\vec{F}_{net}=m \vec{a}$.

One chooses the coordinate axes such that the acceleration vector $\vec{a}$ has a direction parallel to the x-axis. In the cases of the inclined plane $\vec{a}$ is directed down the incline, whereas in the case of the banked turn $\vec{a}$ is directed towards the the center of the circle.

I'm assuming the Physics behind this is that the Normal Force in incline plane is solely due to gravity, while in banked turn it changes due to the circular motion of the vehicle.

The normal force is the normal component of the force exerted by the surface. It's caused by the deformation of the surface which in turn is caused by both gravity and the motion. You should try working the problem where the inclined plane is formed by a wedge that's free to move in a horizontal direction.

I'm wondering if this is correct and how one justifies the different mathematical approaches? (I.e., how does one decide which vector to resolve into components and why wouldn't the vertical component of the Normal Force in a banked turn just be mg--what other tangible forces could one speak of in that situation)?

The vertical component is indeed equal to $mg$. You can choose a coordinate system where $\vec{a}$ is not directed along one of the coordinate axes. For example, choose a horizontal x-axis for the inclined plane and an x-axis parallel to the banked turn.

 

[jon4444]

"The vertical component is indeed equal to mg"

I see now, the increase in the Normal force is due entirely to the increase from the contact force which operates only in the x direction (creating the centripetal force)--correct?

 

[Mister T]

I see now, the increase in the Normal force is due entirely to the increase from the contact force which operates only in the x direction (creating the centripetal force)--correct?

The normal force has both a vertical and a horizontal component. The horizontal component is centripetal (directed towards center) and since it's the only force with a component in that direction it's equal to $\frac{mv^2}{r}$. The vertical component is equal to $mg$.

 

[CWatters]

The normal force has both a vertical and a horizontal component. The horizontal component is centripetal (directed towards center) and since it's the only force with a component in that direction it's equal to $\frac{mv^2}{r}$. The vertical component is equal to $mg$.

That's in the frictionless case or when the car is going at the design speed for the banking.

If there is friction the velocity of the car determines if friction is zero or acts up or down the slope - so it may also have a component acting towards the centre.

 

[adomanim1]

1
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.

 

[adomanim1]

1
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 snaptube telegram web 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.

the slope which will contribute to the net force along the slope which provides the acceleration along the slope. thanks

 

[annamal]

I believe it is because in a banked curve, the car's weight is not fully on the slope as it is on an incline plane. Therefore, on a banked curve we are dependent on the normal for the car's weight. Nbanked*cos(theta) = weight --> Nbanked = weight/cos(theta)

In an inclined plane we can calculate the normal from the weight, so N = weight*cos(theta)

 

[kuruman]

I believe it is because in a banked curve, the car's weight is not fully on the slope as it is on an incline plane. Therefore, on a banked curve we are dependent on the normal for the car's weight. Nbanked*cos(theta) = weight --> Nbanked = weight/cos(theta)

In an inclined plane we can calculate the normal from the weight, so N = weight*cos(theta)

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