Coefficient of static friction and banking of roads

[hyunxu]

Homework Statement

Why banking of roads or tracks depend on the co-efficient of static friction?

Relevant Equations

co-efficient of static friction = ?

Could you please explain the term 'co-efficient of static friction'?

why do the banking of roads or tracks depend of co-efficient of static friction?and not on the co-efficient of kinetic friction?

 

[haruspex]

Problem Statement: Why banking of roads or tracks depend on the co-efficient of static friction?
Relevant Equations: co-efficient of static friction = ?

Could you please explain the term 'co-efficient of static friction'?

why do the banking of roads or tracks depend of co-efficient of static friction?and not on the co-efficient of kinetic friction?

It generally takes a greater lateral force to start an object sliding across a surface than it does to maintain a constant motion. Thus, the coefficient (ratio of required force to normal force) of static friction is usually greater than that of kinetic friction.
In the case of a car on a banked turn, we don't want the car to start skidding. As long as the tyres are making rolling contact there is no relative motion between the surfaces and static friction applies.

 

[doggydan42]

Problem Statement: Why banking of roads or tracks depend on the co-efficient of static friction?
Relevant Equations: co-efficient of static friction = ?

Could you please explain the term 'co-efficient of static friction'?

why do the banking of roads or tracks depend of co-efficient of static friction?and not on the co-efficient of kinetic friction?

On the banking road, if there was no friction it would start moving outwards, towards the outer ends of the circle since there is no force pushing against it's motion. Since the car is not moving outwards, it is static in that direction.

So, the friction has to do with the centripital force for banked turns. The coefficient of static friction applies since the banked turn is to keep the vehicle from changing it's radius while in the turn. It is static in the radial direction.

 

[haruspex]

On the banking road, if there was no friction it would start moving outwards,

Not necessarily. It depends on the speed, radius of turn and angle of bank. It could go the other way.

 

[doggydan42]

Not necessarily. It depends on the speed, radius of turn and angle of bank. It could go the other way.

Yes that definitely makes since. I think I simplified it when I was answering it for myself before typing my answer. Typically, I think of a situation when you are moving at a high enough speed and does not banked too steeply, that the car goes outwards. That's just the first situation I think of.

Thanks for pointing out my mistake. I need to work on conceptualizing more possibilities.

 

[hyunxu]

It depends on the speed, radius of turn and angle of bank. It could go the other way.

why does static friction involve in this?since the object is moving.Why it doesn't have kinetic friction?

 

[hyunxu]

It is static in the radial direction.

Please explain this. How could you say it is static?
So far I can imagine that the car is moving , it isn't at rest.Hence there's kinetic friction.
From which frame of reference it is static?

 

[doggydan42]

Please explain this. How could you say it is static?
So far I can imagine that the car is moving , it isn't at rest.Hence there's kinetic friction.
From which frame of reference it is static?

Maybe that photo makes it more clear?
It has to do with the direction of the centripetal acceleration, but along the surface of the banked turn.

 

[hyunxu]

View attachment 247489
Maybe that photo makes it more clear?
It has to do with the direction of the centripetal acceleration, but along the surface of the banked turn.

My question is simple. What's the role of static friction here?

 

[haruspex]

Please explain this. How could you say it is static?
So far I can imagine that the car is moving , it isn't at rest.Hence there's kinetic friction.
From which frame of reference it is static?

As I wrote in post #2, sliding is relative motion of surfaces in contact. In rolling motion, the point where the wheel contacts the other surface is instantaneously stationary relative to that surface.

 

[hyunxu]

As I wrote in post #2, sliding is relative motion of surfaces in contact. In rolling motion, the point where the wheel contacts the other surface is instantaneously stationary relative to that surface.

Thank you I got it!

 

[haruspex]

It is static in the radial direction.

Friction is either static or kinetic. It cannot be static in one direction and kinetic in another. That is why it is dangerous to brake hard while taking a corner - the combination of the braking force vector (in the direction of travel) and the (lateral) cornering force exceeds the maximum static frictional force; skidding begins, and the kinetic frictional force is directly opposed to the relative velocity of tyre on road, so has little or no radial component.

 

[hyunxu]

so has little or no radial component.

Could you explain me? What is this radial component?(sorry because I'm absolute beginner but I want to learn it anyway)

 

[haruspex]

Could you explain me? What is this radial component?(sorry because I'm absolute beginner but I want to learn it anyway)

It might help to plug in some variables.
Suppose you are turning a corner radius r at steady speed v on a level surface. The acceleration you need is centripetal, v²/r. Suppose the coefficient of static friction is sufficient to handle this, i.e. is at least v²/(rg).
You now apply the brake to the extent that, if going straight, you would achieve acceleration a (negative) in the direction of velocity. If still no skidding, the frictional force on the tyres needs to be mv²/r laterally (i.e. normal to the velocity) and ma parallel to the velocity. If the net force, ##m\sqrt{a^2+\frac{v^4}{r^2}}## exceeds ##\mu mg## you will skid (and the wheels are likely to lock).
So now you have kinetic friction, and that acts opposite to the direction of relative motion of the surfaces. If the wheels have locked then that means opposite to the velocity of the vehicle. There's a bit of moderation in direction at individual wheels since the vehicle is still rotating, but that cancels out for the car as a whole. So the kinetic frictional force does decelerate the car, but there is no longer a lateral component, so you travel in a straight line.