Outward radial motion in unbanked turn

[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.