Outward radial motion in unbanked turn

[jason.farnon]

Sorry for not following the template, this is a follow up to a few other old posts.

In solutions to problems where a car negotiates an unbanked turn, I usually see that the centripetal force is supplied by friction. My understanding is that friction applies in the direction opposite motion. So there is some outward radial motion the friction is counteracting? What force supplies this or is my analysis incorrect? Is it the wheels of the car?

 

[Ericv_91]

These centripetal force questions are rather difficult to visualize, so here maybe one of the diagrams here will help you a bit: http://hyperphysics.phy-astr.gsu.edu/hbase/corf.html

 

[PeterO]

Sorry for not following the template, this is a follow up to a few other old posts.

In solutions to problems where a car negotiates an unbanked turn, I usually see that the centripetal force is supplied by friction. My understanding is that friction applies in the direction opposite motion. So there is some outward radial motion the friction is counteracting? What force supplies this or is my analysis incorrect? Is it the wheels of the car?

Your mind is [like most people] stuck in the inertial frame - one which is not accelerating.

When the car travels in a circle, it is accelerating towards the centre of the circle. Even if it travels an constant speed [note speed has no direction; it is velocity when we add direction]

In order to accelerate towards the centre, something has to push it that way. it is the friction force that provides that.

The force is achieved by the driver turning the steering wheel. This angles the front wheels, meaning they would tend to slide across the surface, rather than merely roll along. If friction is large enough, instead of sliding across the surface, that friction results in circular motion.

Note that if the friction is not big enough, the car instead goes straight ahead or, at best, travels in a circle of greater radius than is necessary to follow the desired path.
Such a motion is generally referred to as "a skid".

If you want to see the effect of such a skid, draw the following on some graph paper:

Using a compass, with the origin as centre, draw a circle of radius 5, and another of radius 6 units. That gives you the "road" you wish to negotiate.

Now using (-2,0) as the centre, draw a circle of radius 7, starting at (5,0) and going anticlockwise.
That path shows the car "skidding wide" on the corner due to in sufficient friction.

 

[PeterO]

Sorry for not following the template, this is a follow up to a few other old posts.

In solutions to problems where a car negotiates an unbanked turn, I usually see that the centripetal force is supplied by friction. My understanding is that friction applies in the direction opposite motion. So there is some outward radial motion the friction is counteracting? What force supplies this or is my analysis incorrect? Is it the wheels of the car?

Lets look at a straight line example of basically the same thing.

A brief-case is placed on the floor of a train, then the train moves off from the station.
The friction [between the case and the floor] supplies the force to accelerate the case so that it "keeps up with" with the accelerating train.

That frictional force clearly acts forward - but that seems to be in the direction of motion!

That friction force does act in the direction the case would move [relative to the train] if there was no friction - like for example a skate board placed on the floor of the train beside the brief-case.

NOTE: The train provides an accelerated frame of reference [it is, after all, accelerating].
In that accelerated frame of reference it is common to "find" a fictitious force the one accelerating the skateboard towards the rear of the train.
The friction "overcomes" that fictitious force.
Of course there is no fictitious force: that friction force is merely supplying the required force to accelerate the brief-case.

EXAMPLE 2: When you are a passenger in a very powerful car, and the driver guns the motor, you are "pushed back into the seat" as the car accelerates away.
Look carefully and you will find there is nothing there to actually push you into the seat.

what actually happens is that the seat - attached to the car - accelerates down the road. You inertia means that you briefly do not accelerate.
That mans you stay behind, and the seat back gets squashed between a stationary you, and a moving seat frame.
Once squashed, the springs/padding/frame of the seat NOW apply sufficient force to you, to make you ALSO accelerate with the car.
The only option was for you to break through the seat, through the back structure of the car and end up sitting on the road, as the (now damaged) car accelerates away from you.

 

[jason.farnon]

PeterO,

Thanks for the explanation and examples. I think you got me close but not quite there. I can see the operation of friction in your briefcase on a train example. In the absence of friction the briefcase moves to the back of the train relative to the train's motion, so the direction of friction must be in the same direction as the train. I can follow your example about the passenger in the car. Still, I don't see the relevance to the question of a car going in a circle. My understanding of inertial/non-inertial frames may be weak. But I can see that those examples involve objects in accelerating objects, so the coordinate system set in that object is accelerating. How does that apply to the radial motion of a car going in a circle? I'm not talking inside the car, but isn't the car going in a circle on a speedway, with the speedway assumed fixed in space, have the speedway as an inertial frame?

I think what you said about the wheels agrees with my sense of the answer. If I turn the wheels theta>0 degrees off the path I'm going in, which is tangential to the circular path, friction will act in that direction theta off the path, so it will have an inward radial component. Then the answer to my question, what is the outward radial motion friction is counteracting, would be, it is the relative motion of the turned tires to the ground. Those tires, relative to the ground, move outward. Does that sound right?

 

[PeterO]

PeterO,

Thanks for the explanation and examples. I think you got me close but not quite there. I can see the operation of friction in your briefcase on a train example. In the absence of friction the briefcase moves to the back of the train relative to the train's motion, so the direction of friction must be in the same direction as the train. I can follow your example about the passenger in the car. Still, I don't see the relevance to the question of a car going in a circle. My understanding of inertial/non-inertial frames may be weak. But I can see that those examples involve objects in accelerating objects, so the coordinate system set in that object is accelerating. How does that apply to the radial motion of a car going in a circle? I'm not talking inside the car, but isn't the car going in a circle on a speedway, with the speedway assumed fixed in space, have the speedway as an inertial frame?

I think what you said about the wheels agrees with my sense of the answer. If I turn the wheels theta>0 degrees off the path I'm going in, which is tangential to the circular path, friction will act in that direction theta off the path, so it will have an inward radial component. Then the answer to my question, what is the outward radial motion friction is counteracting, would be, it is the relative motion of the turned tires to the ground. Those tires, relative to the ground, move outward. Does that sound right?

The friction with the circular motion is a parallel to the brief case in the train.

In the train, without friction, the case would slide straight back along the floor. Friction makes it accelerate forward with the train.

On the flat track, the car would proceed straight ahead. Friction makes it accelerate towards the centre of the circle - and thus travel in circular motion.
Note:
If the acceleration was too small, the car would spiral out.
If the acceleration was too big, it would spiral in.

Like all good friction forces in action, the friction is just the right size to achieve the desired circular motion [unless to alter the steering wheel position, and thus the size of the friction force].

REMEMBER: CIRCULAR MOTION IS ACCELERATED MOTION.
An INWARDS force causes that accleeration. [Centripetal Force]
There is no outward force to balance the inward force.
If there was, the net Force would be zero, and the motion would be in a straight line.

 

[tiny-tim]

welcome to pf!

hi jason! welcome to pf!

… where a car negotiates an unbanked turn, I usually see that the centripetal force is supplied by friction. My understanding is that friction applies in the direction opposite motion …

this really is only to do with the difference between sliding and rolling

if we ignore air resistance etc, the car will keep going in a circle (even on an unbanked surface) without any power from the engine

no forward friction is needed

and there's no forward relative motion between the rolling wheel and the road …

the point of contact is stationary …

(nor is there any tendency to forward relative motion)

so there's no forward friction

there is a tendency to move sideways (the wheel can't roll sideways! ), and so the friction is opposing that ​

 

[cmb]

Which comes first - the slide or the friction?

Actually, in automotive chassis-dynamics it is often said there is no grip without slide. I don't buy that, I think there is such a thing as purely static friction, but I don't think it is a completely unreasonable argument to make either way.

 

[PeterO]

Which comes first - the slide or the friction?

Actually, in automotive chassis-dynamics it is often said there is no grip without slide. I don't buy that, I think there is such a thing as purely static friction, but I don't think it is a completely unreasonable argument to make either way.

Based on the example of applying an ever increasing force to a block on a rough surface, I would say the friction comes first.

Suppose μN for a block is 10N [μN is the maximum possible friction]

Apply a force of 1N, you get Friction of 1N
Incresase to 2N - friction is 2N
keep going to 10N - friction becomes 10N

Finally get to 11N and we get the slide.

 

[cmb]

I think I should've used the word 'slip' rather than 'slide'.

'Slip' is the term the tyre dynamicists use to describe this concept. I'm with you, I'd say 'the friction' comes before 'the slip'.

 

[jason.farnon]

Hi Tiny-tim, Thanks I had not even considered the difference between rolling and sliding. But I am still not clear.

this really is only to do with the difference between sliding and rolling

if we ignore air resistance etc, the car will keep going in a circle (even on an unbanked surface) without any power from the engine

You mean no additional power, after whatever gives the car an initial velocity? I can visualize that, e.g., a satellite rotating around a planet, where gravity replaces the friction of our example. But for gravity there is a law that says it always acts radially between two bodies in the direction of the segment between their centers. With friction it only acts in the direction opposite motion. This is my question, what motion is it opposite to.

Let me try this way. Tell me where I mess up. 1. When a car is driving in a circle with uniform speed, friction supplies the centripetal acceleration. 2. Friction is a force between two surfaces in contact that acts in the direction opposite to the direction of motion between the two surfaces that would exist in the absence of friction. 3. Therefore in the car example friction opposes motion in the outward radial/centrifugal direction. 4. Therefore there is a centrifugal force.

 

[PeterO]

You mean no additional power, after whatever gives the car an initial velocity? I can visualize that, e.g., a satellite rotating around a planet, where gravity replaces the friction of our example. But for gravity there is a law that says it always acts radially between two bodies in the direction of the segment between their centers.
Gravity only acts radially on a body orbiting in a circle! Similarly the friction you are encountering her acts radially because the car is traveling in a circle.
Gravity doesn't "look" at you, when you leave a diving board and say "Ah, he is just going to fall so I will act down - or look at a rocket being launched and say "Oh, that rocket is trying to get away, so I will pull down and at least make it little bit harder for it" or look at a satellite in orbit and think " I may as well be kind here and supply a radial force to keep it in orbit - gee, I hope that is what it wanted!"
As common, there is an attraction - centre of mass to centre of mass regardless

With friction it only acts in the direction opposite motion.

Opposed to motion or opposed to a potential motion.
When a box slides across the floor, the friction opposes that motion. When a box is placed on a gentle slope, friction opposes the potential for the box to slide down the slope


This is my question, what motion is it opposite to.

It is the potential motion that it is opposing.
When the front wheels of the car are angled (you turn the wheel) there is the potential for the tyres to now slide sideways (to some extent - we usually don't turn them through 90^o
Friction opposes that potential to slide sideways by creating a force at right angles to the motion - and so an acceleration at right angles to the motion - a centripetal acceleration


Let me try this way. Tell me where I mess up.
1. When a car is driving in a circle with uniform speed, friction supplies the centripetal acceleration. Yes, because the front wheels have been angled
2. Friction is a force between two surfaces in contact that acts in the direction opposite to the direction of motion between the two surfaces that would exist in the absence of friction. That's right - the tyres are trying to slip sideways; but remember, sideways to an angled tyre may be straight ahead to a moving car. Many drivers discover this problem each winter when a snap freeze means ice on the road and the car can no longer turn or stop
3. Therefore in the car example friction opposes motion in the outward radial/centrifugal direction. The motion is not radially outward. When the driver in winter slips off the road, it is only that the road was constructed with a radius of, say, 50m but the limited friction means the car travels in a circle of , say, 100m.
When suddenly placed in an unexpected non-inertial (accelerating) frame - the human will immediately invent a fictitious force to "explain" the problem. You don't need understanding to make things up. The fictitious force everyone invents even has a name - centrifugal force - and once a force gets a name it must be real - NOT
4. Therefore there is a centrifugal force.

Exactly - as soon as everyone agrees on the same name for a piece of fiction it must exist, mustn't it? That's what physics is for - to show that there is no such thing!

 

[tiny-tim]

(PeterO, do you know you're typing in red? )​

hi jason!

(just got up :zzz:)

You mean no additional power, after whatever gives the car an initial velocity?

assuming no air resistance, friction in the bearings, rolling resistance, etc … yes

(of course, the front wheels must be angled, or the whole vehicle must be leaning, like a motorbike)

… With friction it only acts in the direction opposite motion. This is my question, what motion is it opposite to.

it's static friction, so it's opposite the way the car "wants" to slide, ie outwards …

you've got that anyway, in 2 and 3 below! ​

1. When a car is driving in a circle with uniform speed, friction supplies the centripetal acceleration.
2. Friction is a force between two surfaces in contact that acts in the direction opposite to the direction of motion between the two surfaces that would exist in the absence of friction.
3. Therefore in the car example friction opposes motion in the outward radial/centrifugal direction.

yes

4. Therefore there is a centrifugal force.

no, no such thing as centrifugal force for an inertial observer

the friction can be called a centripetal force

(though if the road was banked, the friction wouldn't be centripetal, would it? )

 

[PeterO]

(PeterO, do you know you're typing in red? )​

hi jason!
I type in red when I respond within the previous post rather than after it - so it is obvious which part I am commenting on
(just got up :zzz:)


assuming no air resistance, friction in the bearings, rolling resistance, etc … yes

(of course, the front wheels must be angled, or the whole vehicle must be leaning, like a motorbike)



it's static friction, so it's opposite the way the car "wants" to slide, ie outwards …

you've got that anyway, in 2 and 3 below! ​

yes

With description given, I am afraid it is NO

no, no such thing as centrifugal force for an inertial observer

the friction can be called a centripetal force

(though if the road was banked, the friction wouldn't be centripetal, would it? )

But would the friction by up the bank or down the bank ?
Perhaps no fiction at all, even if the track was "rough"!

 

[PeterO]

(PeterO, do you know you're typing in red? )​

hi jason!

(just got up :zzz:)assuming no air resistance, friction in the bearings, rolling resistance, etc … yes

(of course, the front wheels must be angled, or the whole vehicle must be leaning, like a motorbike)
it's static friction, so it's opposite the way the car "wants" to slide, ie outwards …

And importantly the car will NEVER slide out, it just may turn in less sharply than the road/planned route turns in. ie the car may be traveling in a circle of radius 100m when the road was constructed with radius 70m. Under no circumstances, on a flat surface, will a car go right when you turn left - it just may not go left as sharply as you hoped for! It may even continue straight ahead if the surface is ice.

you've got that anyway, in 2 and 3 below! ​

yes no, no such thing as centrifugal force for an inertial observer

the friction can be called a centripetal force

(though if the road was banked, the friction wouldn't be centripetal, would it? )

...

 

[tiny-tim]

I type in red when I respond within the previous post rather than after it - so it is obvious which part I am commenting on

but then it appears in italics

it's much easier to read (and to quote!) if you simply divide the previous post into separate quote-blocks (eg as in my last post) ​

 

[PeterO]

but then it appears in italics

it's much easier to read (and to quote!) if you simply divide the previous post into separate quote-blocks (eg as in my last post) ​

What is the problem with italics - I would concentrate more on correcting your suggestion there is a radially outward motion - even a potential for it - in the situation mentioned in the OP.

 

[tiny-tim]

What is the problem with italics

i find italics difficult to read

… in the situation mentioned in the OP.

i'm confused …

can you quote it?​

 

[PeterO]

i find italics difficult to read


i'm confused …

can you quote it?​

The original post was about a car traveling in a circle on a flat track.
OP pondered
"So there is some outward radial motion the friction is counteracting? What force supplies this or is my analysis incorrect?"

any explanation that involves some idea of " some outward radial motion " is wrong, and your suggestion "so it's opposite (to) the way the car "wants" to slide, ie outwards …" just re-inforces that error.

Not wanting to turn in is a far cry from sliding outwards.

I think what OP, and you, think of as outward motion is simply insufficient inward motion. Like when you watch video of sky-divers from a head-camera. Another diver deploys their 'chute and is seen to disappear upwards. However they have not gone up, they have merely changed to traveling more slowly down.
A racing car "running wide" on a bend - even NASCAR oval tracks - is a case of the car not turning left as sharply as the track - they don't actually turn right.

 

[tiny-tim]

OP pondered
"So there is some outward radial motion the friction is counteracting? What force supplies this or is my analysis incorrect?"

uhh? …

that was in his original post …

So there is some outward radial motion the friction is counteracting? What force supplies this or is my analysis incorrect?

i never replied to that!

i replied to …

hi jason! welcome to pf!

… where a car negotiates an unbanked turn, I usually see that the centripetal force is supplied by friction. My understanding is that friction applies in the direction opposite motion …

anyway, you seem to be ignoring the fact that if the road suddenly became fricitionless (ie the car hits an ice patch), then the wheels would continue to roll forwards, but would slide sideways ​

 

[PeterO]

uhh? …

that was in his original post … that is what OP stands for!i never replied to that!

i replied to …

anyway, you seem to be ignoring the fact that if the road suddenly became fricitionless (ie the car hits an ice patch), then the wheels would continue to roll forwards, but would slide sideways ​

NO! the car would cease traveling in a circle and travel in a straight line - it would not go "sideways" to the direction it was traveling at the time. We are supposed to be discussing the motion of the car.

EDIT: and the way most drivers react, the tyres would stop rolling too, since the driver will have applied the brakes in (confused) panic.

 

[tiny-tim]

NO! the car would cease traveling in a circle and travel in a straight line - it would not go "sideways" to the direction it was traveling at the time. We are supposed to be discussing the motion of the car.

you discuss the car, i'll discuss the tyres ​

the front tyres certainly will not be going directly forward …

they were at an angle to forward before the ice, and they'll remain at that same angle …

there will be a combination of forward (relative to the tyre) rolling and sideways sliding

(the rear tyres, I'm not sure about … i think the answer probably is that they were never completely rolling in the first place … it's a lot easier to analyse if we pretend that all four tyres are angled so as to be moving tangentially along the original circle! )

 

[PeterO]

you discuss the car, i'll discuss the tyres ​

the front tyres certainly will not be going directly forward …

they were at an angle to forward before the ice, and they'll remain at that same angle …

there will be a combination of forward (relative to the tyre) rolling and sideways sliding

(the rear tyres, I'm not sure about … i think the answer probably is that they were never completely rolling in the first place … it's a lot easier to analyse if we pretend that all four tyres are angled so as to be moving tangentially along the original circle! )

Even the tyres don't slide radially out. The tyres are on the wheels, the wheels are attached to the car, and the car does not slide radially out.

 

[tiny-tim]

Even the tyres don't slide radially out. The tyres are on the wheels, the wheels are attached to the car, and the car does not slide radially out.

(the car doesn't slide at all)

the (front) tyres are at an angle to the car …

if the car is moving forward, the tyres (which are still rolling) aren't rolling forward, they're rolling at an angle, and sliding sideways

 

[PeterO]

(the car doesn't slide at all)

the (front) tyres are at an angle to the car …

if the car is moving forward, the tyres (which are still rolling) aren't rolling forward, they're rolling at an angle, and sliding sideways

But as I said - the tyres do not slide radially out ?
The OP was wondering if there was some outward radial motion.

 

[tiny-tim]

The OP was wondering if there was some outward radial motion.

he was wondering why the friction was not opposite to the forward motion …

My understanding is that friction applies in the direction opposite motion.

… which is what i set out to answer

 

[jason.farnon]

he was wondering why the friction was not opposite to the forward motion …


… which is what i set out to answer

I was wondering, if friction is centripetal and if friction opposes motion that is opposite the motion that would occur in its absence, what is that outward motion.

I think we may be agreed it is just the tangential motion being opposed. Let's say I have a switch that can make friction on the track exist or not. First it exists, and the car is traveling in a circle. Then I turn it off. The car will travel tangential to the circle. This is the motion the car travels in in the absence of friction. And relative to the circle, the tangential direction has an outward radial component. Friction is opposing this.

 

[tiny-tim]

hi jason!

And relative to the circle, the tangential direction has an outward radial component. Friction is opposing this.

no, that makes no sense …

tangential is always perpendicular to radial, it cannot have a radial component ​

imagine that all four wheels can be steered, and that initially they are all pointing tangentially along the circle as they go round it, at angles of ±θ to the car itself …

so the static friction force is exactly sideways to each wheel

(anything else is too complicated to analyse exactly)

then you "turn the friction off" …

the car will go in a straight line, but the wheels will still be at ±θ to the car …

they will still be rolling at the same speed, but they will be sliding sideways (to the wheel) also

 

[jason.farnon]

hi jason!


no, that makes no sense …

tangential is always perpendicular to radial, it cannot have a radial component ​

yes I guess it is, at the point of tangency. Then I still don't see why there is a tendency to move in the outward radial direction, which is needed for friction to act inward?

What is the relevance of the wheels continuing to roll at an angle at the same speed, if friction is "turned off"? The car is stil moving along the tangent at the point that friction was turned off. The definition of friction requires that it opposes the direction of motion that exists in its absence. As you pointed out, at the point of tangency, there is no inward radial component to the tangent.

 

[tiny-tim]

What is the relevance of the wheels continuing to roll at an angle at the same speed, if friction is "turned off"? The car is stil moving along the tangent at the point that friction was turned off. …

but the car isn't in contact with the road!

friction is between two surfaces in contact

static friction is between two surfaces in contact with no relative motion at the point(s) of contact​

friction in this case is static, and is between the tyre and the road

if the wheel is tending to move forward, it will tend to roll, which involves no relative motion at the point of contact, so no friction in that direction

if the wheel is tending to move sideways, it will tend to slide, which does involve relative motion at the point of contact, so there is friction in that direction

 

[PeterO]

Then I still don't see why there is a tendency to move in the outward radial direction, which is needed for friction to act inward?

Jason, Newtons first law reminds us that in the absense of an unbalanced external force, a moving object will continue in a straight line.
In this case you want the car to travel in a circle, which is definitely not a straight line, so an external force is required.
Friction is supplying that force.
If you want to look at straight line motion as an example of outward radial motion, as if circular motion was the normal condition, then that may satisfy you, but will be foreign to the thinking of most people. Mind you, millions of people cling to centrifugal force when explaining the sensation they feel when a car moves around a circle too.
Deep down, I think you want the friction to be opposing the centrifugal force.
Problem there is: there is no centrifugal force, that is the imaginary force needed to make an accelerated (non-inertial) frame of reference seem to be a non-accelerating (inertial) frame. The belief that "no change of speed" constitutes no acceleration is very strong in the world. The fact that "no change of velocity" is the real "no acceleration" example confuses non-physics people - especially when merely changing direction constitutes a change in velocity.

From a physics point of view a car is equipped with 3 accelerating devices: the pedal every one calls the accelerator, the pedal everyone calls the brake, and the wheel in front of the driver. Alter the position of anyone of those and the car should change its velocity. {there are a few more if you count the hand-brake, friction in the bearings etc}

 

[jason.farnon]

Deep down, I think you want the friction to be opposing the centrifugal force.

I don't personally. I just interpret the definition of friction as requiring it.

Perhaps a related question I have had. In the usual example of a block resting on an incline, with gravity, the normal force is less than the weight. The component of the weight perpendicular to the incline alone equals the normal. If you have a car going in a circle on a banked track, the normal force exceeds the weight in the analyses I encounter. Eg http://www.batesville.k12.in.us/physics/phynet/mechanics/circular motion/banked_no_friction.htm. Now the vertical component of the normal equals the weight, and the normal also has a centripetal component. Why the difference in relative magnitude of normal and weight?

I should also add, thanks for your patience.

 

[tiny-tim]

hi jason!

In the usual example of a block resting on an incline, … The component of the weight perpendicular to the incline alone equals the normal. If you have a car going in a circle on a banked track, … the vertical component of the normal equals the weight …

that's because with a well-behaved planar surface, the normal acceleration is zero whatever the block is doing,

and so we can say 0 = ma = components of N + W in the normal direction​

with an essentially conical surface, the normal acceleration is not zero, so we can't …

but if the block stays at the same height (and if the friction is zero), then the vertical acceleration is zero, and we can say 0 = ma = components of N + W in the vertical direction

 

[PeterO]

I don't personally. I just interpret the definition of friction as requiring it.

Perhaps a related question I have had. In the usual example of a block resting on an incline, with gravity, the normal force is less than the weight. The component of the weight perpendicular to the incline alone equals the normal. If you have a car going in a circle on a banked track, the normal force exceeds the weight in the analyses I encounter. Eg http://www.batesville.k12.in.us/physics/phynet/mechanics/circular motion/banked_no_friction.htm. Now the vertical component of the normal equals the weight, and the normal also has a centripetal component. Why the difference in relative magnitude of normal and weight?

I should also add, thanks for your patience.

The short answer would be that when the car is traveling around a banked track, it is in an accelerated frame of reference, so the forces will appear different to when they are in a frame of reference that is not accelerating.

Now how to show the answers:
Remember, Friction opposes actual motion as well as potential motion.
Remember a mass placed on a gentle rough slope, such that friction prevents it slipping down the slope.
Friction is not opposing an actual slip down the slope, it is opposing a potential slip - the slip that would occur if the surface was smooth.

Case 1, the car is parked on the banked track: This is just like the block on the slope I referred to above - let's just model the car as a block.

If the slope was smooth, the block (car) would slide down the slope, accelerating as it went - like a bob-sled.

Draw the actual forces acting on the block - there are only 2.
Weight force - straight down - represent it as an arrow straight down from the centre of mass. make it as long as you like - but I would draw it about 2 cm long as that is a pleasant size to work with.
Normal Reaction Force - perpendicular to the slope - represent that as an arrow perpendicular to the slope, from a point where the block touches the slope.

To find the net force, we add the two vectors by connecting those arrows head to tail.

Translate the Normal Force arrow so that it starts at the end of the weight vector.
The obvious question is "How long should the Normal Force vector be"
Again the short answer "Just long enough!"
Reaction forces are always "Just strong enough"

If the Reaction Force is too short, the net force will be angled down into the slope - suggesting the block will slide down and dig into the surface [like a box placed on the side of a sand dune] The block is not going to "dig in" so the Normal force can't be that small.

If the Reaction Force is too long, the net force will be angled up away from the slope (perhaps only slightly) suggesting the block will "lap off" the slope as it begins to move. We know that never happens either.

The Reaction force has to be just the right size so that the net force is parallel to the slope - because that is the direction the block is about to move.

If you resolved the weight force into components perpendicular and parallel to the surface, you would get the same sized Normal Reaction Force - but you don't have to resolve it, and this method is more useful when the block (car) starts traveling around a banked track.

Note that if the slope was rough, and this block didn't slide down the slope, there has to be a friction force directed UP the slope so that weight, Normal and Friction arrows form a closed triangle, with a net force of zero.

banked track next post.

 

[PeterO]

I don't personally. I just interpret the definition of friction as requiring it.

Perhaps a related question I have had. In the usual example of a block resting on an incline, with gravity, the normal force is less than the weight. The component of the weight perpendicular to the incline alone equals the normal. If you have a car going in a circle on a banked track, the normal force exceeds the weight in the analyses I encounter. Eg http://www.batesville.k12.in.us/physics/phynet/mechanics/circular motion/banked_no_friction.htm. Now the vertical component of the normal equals the weight, and the normal also has a centripetal component. Why the difference in relative magnitude of normal and weight?

I should also add, thanks for your patience.

Now the banked track.

When the car (block) is traveling around the circular track, we draw the free body diagram looking from behind/in front as the car travels in the middle of the bend.

The picture actually looks exactly like the block on the inclined plain I described earlier - but in this case represents a moving object - moving into or out of the page.

Since the block is moving i a circle, the net force is directed horizontally towards the centre of the circle. I can't say left or right, since I am not sure which way you drew your banking.

This is too hard.
If you give me an email address - open up a new temporary hotmail address if you like - I can send you a Power Point presentation that shows clearly what I want to show.
Once you have the PowerPoint you can cancel that temporary hotmail address.

I am in Australia, where it is 1:40 am at the moment [GMT +11] so I wouldn't send a file for about 8 hours (going to bed) so if you don't want an email address sitting about on here don't post it immediately. I might try a message.