Car on Banked Curve: Centripetal Force?

[starstruck_]

Does a car on a banked curve have a component of gravity that accounts for the centripetal force ( along with the normal force)?

 

[CWatters]

Gravitational force acts vertically. Centripetal force acts horizontally. Does that help answer your question?

 

[starstruck_]

Gravitational force acts vertically. Centripetal force acts horizontally. Does that help answer your question?

If it's banked is there no component of gravity inwards?

 

[kuruman]

To rephrase the answer by @CWatters, gravity will contribute to the centripetal force only if the centripetal force has a vertical component. Is that the case here?

 

[starstruck_]

To rephrase the answer by @CWatters, gravity will contribute to the centripetal force only if the centripetal force has a vertical component. Is that the case here?

No it isn't. This is what my FBD looks like:
That component of gravity has no affect to the center of the circle right? It's just opposing friction?

 

[starstruck_]

 

[sophiecentaur]

This thread could be interpreted as suggesting something that isn't right. If the track were frictionless, there would still be a centripetal force to allow circular motion at (just one) speed. There will be a component of the weight force down hill and this will provide a component the required centripetal force to support circular motion. The presence of the inclined plane gives a situation where the 'effect' of gravity can be horizontal. Resolving vectors doesn't always produce obvious results.

 

[kuruman]

The presence of the inclined plane gives a situation where the 'effect' of gravity can be horizontal.

I am not sure what you mean by this 'effect'. Gravity is strictly a vertical vector while the centripetal (or net) force is horizontal (assuming that motion takes place in a horizontal circle.) A vertical vector cannot have a component in the horizontal direction, therefore gravity does not contribute to the acceleration. In the case of the frictionless banked curve, the contact force with the plane is normal to the plane and has a horizontal component that provides the centripetal acceleration while its vertical component keeps the mass in the horizontal circle. In other words, being a contact force it adjusts itself to provide the observed acceleration.

 

[sophiecentaur]

A vertical vector cannot have a component in the horizontal direction

A vertical vector can have a component that isn't vertical. Eliminate the 'other' vector in that arbitrary resolution, by interposing a sloping surface, for instance. That gives you a vector that acts along the surface of the slope. That vector can be subsequently resolved, if you choose . . . .and so on.
Consider three linear polarisers. Cross two of them and you see nothing. Insert the third at 45° and lo!, you see light through all three.
Let's face it, the message I was getting from the thread was actually that banking would achieve nothing. That would be daft, wouldn't it?

 

[kuruman]

The polarizer example is not germane to this example. When the electric field vector goes through the polarizer, only the component parallel to the polarizer's axis makes it through while the perpendicular component is extinguished resulting in a loss of intensity. Here there is no loss of vector magnitude, the equivalent of intensity. Nevertheless, I went along with your suggestion and did a vector decomposition of the weight. Please see figures below, they are drawn to scale.
Left to right:
The weight vector.
The weight vector decomposed into components parallel and perpendicular to the incline.
Each of the two vectors in the previous diagram decomposed into horizontal and vertical components.

Yes, indeed, there are horizontal arrows in the third figure, but their sum is zero. I think it is inappropriate to pick and choose one of them and say that it contributes to the centripetal force whilst ignoring the other. Note that the sum of the vertical arrows gives the weight as expected. Also note that if the weight were a polarized electric field vector and the incline line were the axis of the polarizer, then in the second figure there would be no $W \cos \theta$ term. That would eliminate the horizontal component pointing to the right in the third figure, which is probably what you had in mind.

 

[A.T.]

A vertical vector can have a component that isn't vertical.

No.

Eliminate the 'other' vector in that arbitrary resolution, by interposing a sloping surface, for instance.

That is just an obfuscatory way to say "apply a second force" (here normal force), which has an horizontal component. But this is completely different from the claim, that a single vertical force (gravity) has a horizontal component.

 

[sophiecentaur]

the perpendicular component is extinguished

Yes. And the component of weight that's normal to the slope ceases to have any effect because of the equal and opposite normal reaction from the slope. The third diagram is not relevant ; all you have done is to recompose the weight force from its two components and ignored the effect of the Normal force from the slope. What is relevant is the component of the W sinθ that acts toward the centre of the track (w sinθcos θ). That is the centripetal force needed for circular motion.
If you can't accept that then you imply that gravity has no place in the function of the slope. Do you think that the car could not have a curved path without lateral tyre friction? Look in this link (ignore the gravitational analogy model aspect of it - just look at the motion of the balls) There is no frictional resistance keeping the balls from falling down the slope or flying over the top. Can you suggest any force, other than the weight component down the slope that can be identified with the centripetal force?

 

[kuruman]

What is relevant is the component of the W sinθ that acts toward the centre of the track (w sinθcos θ).

Please explain why the component of W cosθ that also has a horizontal component in the opposite direction becomes irrelevant.

If you can't accept that then you imply that gravity has no place in the function of the slope. Do you think that the car could not have a curved path without lateral tyre friction?

We are discussing the no friction case here.

? Look in this link (ignore the gravitational analogy model aspect of it - just look at the motion of the balls) There is no frictional resistance keeping the balls from falling down the slope or flying over the top. Can you suggest any force, other than the weight component down the slope that can be identified with the centripetal force?

I looked at the link and it's not relevant to this case. The balls are in orbits that are not horizontal circles as is the case here.

Can you suggest any force, other than the weight component down the slope that can be identified with the centripetal force?

Yes I can and I did so in post #8:

In the case of the frictionless banked curve, the contact force with the plane is normal to the plane and has a horizontal component that provides the centripetal acceleration while its vertical component keeps the mass in the horizontal circle. In other words, being a contact force it adjusts itself to provide the observed acceleration.

 

[sophiecentaur]

Please explain why the component of W cosθ that also has a horizontal component in the opposite direction becomes irrelevant.

The only forces that are relevant to the motion of the car up and down the slope are components in that direction. The weight force has a component in that direction. The centripetal force is supplied by the horizontal component of that force. Alternatively, you could say that the component of the centripetal force parallel to the slope will be equal to the weight component down the slope. The other component of the required centripetal force is provided by the normal force from the slope.

I'm not sure what it is that you are arguing about. If there's no friction and there is circular motion then there must be a centripetal force. That force can only be due to the weight of the car - it's the only force that's available. Your second diagram shows one force (the one parallel to the slope - to the left) and the other force is provided by the normal reaction component, also towards the left. They add up to the centripetal force.

The balls are in orbits that are not horizontal circles as is the case here.

Orbits can be circular but the orbital analogy is not the point. They are balls on a circular sloping track, to all intents and purposes. They do not escape or fall directly to the centre.
The guy did not have the time to get it just right so that the paths were circles (it was not relevant to his demo) but they could easily have been, apart from the losses. Over a short time the motion is precisely the same as our ideal car on a track.

But this is completely different from the claim, that a single vertical force (gravity) has a horizontal component.

I really didn't put that in a good way. The horizontal force 'arises' due to the presence of the vertical force. It's only there when there is a weight force. The presence of the slope causes one component to be unchanged and the effect of the other component is reversed. Verbal descriptions of vectors are a bit fraught. I could have said "gives rise to" rather than "has".

 

[kuruman]

I quit this thread without prejudice.

 

[A.T.]

The horizontal force 'arises' due to the presence of the vertical force... I could have said "gives rise to" rather than "has".

This fuzzy causation philosophy is irrelevant to the vector analysis of forces.

 

[CWatters]

Does a car on a banked curve have a component of gravity that accounts for the centripetal force ( along with the normal force)?

My answer is no because you said "along with the normal force"...

In this example:
http://hyperphysics.phy-astr.gsu.edu/hbase/Mechanics/carbank.html

The centripetal force is given by

∑ Fx = mv²/r = Nsinθ + μ_sNcosθ

where
Nsinθ is the horizontal component of the Normal force and
μ_sNcosθ is the horizontal component of the friction force

Note there is no additional term containing gravity.

 

[sophiecentaur]

Note there is no additional term containing gravity.

The "Frictionless Case" in that link is given as V_max = √(rg tanθ) . That's the case we're dealing with, I think.
Are we talking at cross purposes?
With no friction Vmax and Vmin are 'almost' equal and lie on either side the required speed. (Forgive the imprecise mathematical description; there is sure to be a better one.)

This fuzzy causation philosophy is irrelevant to the vector analysis of forces.

OK, that's fair enough; I was still being sloppy. How about "The horizontal centripetal force is a function of gravity, a vertical force"? We can agree that it's only there because of the reaction from the sloping surface.

 

[A.T.]

We can agree that it's only there because of the reaction from the sloping surface.

Yes, the normal force from the surface accounts for all the centripetal force. No need to look for additional contributions from gravity.

 

[jbriggs444]

How about "The horizontal centripetal force is a function of gravity, a vertical force"?

Not until you specify what other things are being held constant while you fiddle with gravity.

 

[CWatters]

The "Frictionless Case" in that link is given as Vmax = √(rg tanθ) . That's the case we're dealing with, I think.

The OP doesn't mention friction but the frictionless case is essentially just a special case.

The equation for the centripetal force becomes
∑ Fx = mv2/r = Nsinθ

still no additional term for gravity needed. In effect it's already included in the normal force N.

 

[CWatters]

Are we talking at cross purposes?

I'm trying to stick to the OP's question.

The OP asked if a component of gravity contributes to the centripetal force in addition to the normal force. The answer to this is no because the effect of gravity is already included in the normal force.

 

[sophiecentaur]

The answer to this is no because the effect of gravity is already included in the normal force.

I'm having a problem with that, actually. If you put a ball on a slope, it will roll downhill. The Normal Force won't make it do that. Doesn't that Force Vector count towards the centripetal force? I realize that I may be suggesting double accounting but . . . .

EDIT: I re-read your post and you seem to have explained that.

 

[A.T.]

If you put a ball on a slope, it will roll downhill.

Along the net force.

Doesn't that Force Vector count towards the centripetal force?

The net force is already the vector sum of all forces. You want to add it again?

I realize that I may be suggesting double accounting

In an endless recursion.

 

[CWatters]

kuruman was right earlier. You can resolve gravity into two components with one down the slope but you can't ignore the other one. If you could ignore one then it would be possible to prove gravity had a component that acted upwards.