Little confused regarding centirpedal force

This question combines a combination of questions.
See below: I understand how friction acts on a slope when a car is driving up or down an incline but in centripetal force, friction varies based on the velocity of the car.
This wasn’t explained in my material.
I am wondering if anyone can confirm if I am on the correct path.
I have provided all parts to the question, so that you get a better understanding of question.
Thanks in advance

not quite sure what you mean. Friction is like a sponge in that, up to its limiting value it will simply absorb enough force to ensure balance. So it doesnt really depend on velocity, but if going faster means that more (frictional) force is required to ensure circular motion, then it will increase (up to the limiting value).
The last couple of slides in
https://www.sfu.ca/~mxchen/phys1011101/Lecture06B.pdf
demonstrate how banking provides the centripetal motion without friciton (zero frictional force). Then obviously for diferent speeds about this value (either greater or smaller) friction will ensure that circular motion still occurs up to the limiting value.

Sorry, I meant friction acts different directions depending on the velocity.

Dont really understand that either. If the track is banked, there is a velocity such that the banking solely provides the centripetal force, as shown in the slides. Then friction will mean that circular motion still occurs for a range of velocities about that value. The frictional force required to ensure ciricular motion will obviously depend on the velocity and if it exceeds the limiting value, then circular motion will not occur.
Is that what you are asking?

Yes, exactly that. This concept is new to me in this regard, so I am trying apply to apply it in a real world scenario. From what I understand, is, friction acts in the opposing direction in a linear concept. Driving forwards and backward. In a circle, it acts differently in order to maintain the circular motion of the vehicle.
What has confused me is. If the vehicle drives slow, almost straight forward around the circle, does the friction still act towards the centre or? Just trying to make sense of these different scenarios is confusing me a little.

A banked curve sliding up and sliding down should be covered in
https://www.youtube.com/watch?v=i6saYsd1qwE
https://www.youtube.com/watch?v=VBNn4soa0_U
but there are a few others (just google if necessary). Its one of the reasons why banking gets steeper close to the top and shallower close to the bottom.

A banked curve sliding up and sliding down should be covered in
https://www.youtube.com/watch?v=i6saYsd1qwE
https://www.youtube.com/watch?v=VBNn4soa0_U
but there are a few others (just google if necessary). Its one of the reasons why banking gets steeper close to the top and shallower close to the bottom.

I have been doing this all day, watching various videos to try and grasp what's happening.
Firstly, my first thought about the car facing towards the centre doesn't exist unless it is drifting ii guess.
So, on an incline where a car is driving towards you as you're looking at this screen.
From what I understand, due to the normal being perpendicular to the incline, to find this resultant force, we must consider the normal as two components: one point in the y direction and one in the x direction, sohcahtoa, is applied to find the formulas for these.
Let Fn= normal force, so Fnx = horizontal normal force and Fny = vertical normal force.
If I consider a frictionless scenario, up is positive, left is positive, right is negative and down is negative
Fx has just one force, and Fy has two forces.
Fx = Fn sin theta and Fy = Fn cos theta - mg
so because there's no upward force in Fny = zero,
Fn cos theta - mg = Fny
Fn cos theta - mg = 0,
mg = Fn cos theta,
Fn = mg/ cos theta.....
This bit is where I might be wrong.
resultant Fx = centripetal force,
mv^2/r = Fn sin theta
Introduce Fn here:
Mv^2/r = (Mg/cos theta) x Sin theta
v^2/r = g x tan theta
V^2 = gr tan theta
V = root gr tan theta...
So, when friction is involved, like the Normal force, is it also split into two components, and this must be introduced?
Am I right as of so far?
Additionally, if I know the max friction from when the track was horizontal and not banked, would this friction remain the same, or in my head, I think that this would increase due to the greater likelihood of the car slipping and supporting the increased velocity.

The main part looks correct / like the last couple of slides in the previous post.
Friction acts down the plane/banking (rough surface) and the limiting value depends on the normal which is perpendicular to the plane/banking. So to adapt the previous to the friction case, you need to resolve both and balance horizontally (centripetal) and vertically like the previous case and as explained in the videos in the previous post.
The limiting frictional force is a function of the normal reaction, which will depend on the banking. So if the banking was vertical (stupid/extreme case), then the limiting friciton would be zero.

I finished my studies when I received your message to allow what I had learned to absorb.
So, we know the friction is acting down the slope because the car asks us to find the maximum velocity required to prevent the car from flying off the track?
Well, if a car could drive at a 90-degree angle, this would defy physics as we know it, and I imagine the car would roll and fall due to gravity.

Friction acts parallel to the floor which in this case is the slope. Limiting friction is determined by the force pressing down perpendicularly onto the floor as the videos show.
It helps to have an idea of what friction is. Imagine the floor is like a stationary saw (with teeth) and you have another saw with teeth that wants to move across / along it. The v ^ shapes in the two saws mesh together and theyll resist moving the larger the perpendicular force. So here the stationary saw is the banking and the other saw that wants to slide along it is the wheels and therefore friction acts along the banking and the greater the perpendicular force pushing the teeth together the harder will be the motion along the plane/bank.
I think thats something like what you were asking.

Yeah, I understand what friction is and how it acts, except for bizarre situations when two smooth plates become stuck together. I assume this is more related to the atoms in water sticking, lol.
I mean that when talking about centripetal force,
At Maximum Velocity: When the vehicle is at or above the ideal speed for the banked curve, the frictional force acts downwards, opposing the car's tendency to slide up the bank due to the higher centripetal force required.
At Lower Velocity: When the vehicle is below the ideal speed, the frictional force acts upwards, opposing the car's tendency to slide down the bank because the centripetal force required is less than what the banking angle alone can provide.

Greater than ideal velocity (for the radius). There would be a tendancy to move along the bank/slope upwards. Friction would act against that, down the slope/bank until limiting friction is reached.
Obv the less than ideal velocity is the flip.
It seems to be roughly what youre saying, but uploading a pic might help.

okay, so I think I am just about there; the problem is regarding the coefficient of friction. from the previous scenario, this was 0.2 when the track was horizontal, but I assume this would change on a banked surface.
Once I clarify this, I think I would have solved it . i can then share with you my working and you can tell me off where I have gone wrong lol

coeff of friction would be the same, the "roughness" of the two saws would be the same,

Mmm, it makes sense in that scenario but on a banked curved, I feel like if friction=μN, where N=mg, obviously gravity stays the same. In the scenario, max friction is 20% of the weight of man and kart x gravity in the horizontal track.
In my mind, I feel like the friction value would reduce.
Any how, I found the max value to be 9.62 approx. I can upload the pictures after. I’m just grocery shopping

The normal reaction would reduce because of the banking (its not mg as its perpendicular to the banking), not the coefficient.

Perhaps, I have made an error then. Because, I tried Fn based on the angle and it seems to be higher than my original normal.
Here are my workings:I know there's probably a more direct route, but this for me, makes sure that I don't make any mistakes.

Perhaps, I have made an error then. Because, I tried Fn based on the angle and it seems to be higher than my original normal.
Here are my workings:

I know there's probably a more direct route, but this for me, makes sure that I don't make any mistakes.