Limiting Friction & Centripetal Force: Explained

[Faiq]

What does the line in the rectangle box means? (What is the difference between limiting friction and centripetal frictional force?)
What type of motion is meant by skidding?
http://prnt.sc/c7ptm8

 

[NascentOxygen]

Friction completely cancels the applied force up until the point where the applied force is great enough to overcome friction---and at this point movement occurs.

Skidding is sliding perpendicular to the direction of rolling. (A wheel locked during braking can also skid.)

 

[Faiq]

But if friction is itself the applied force, how can applied force> friction?

 

[NascentOxygen]

Friction has a maximum value, the limiting value. Once the inertial force of the turning cycle equals the maximum friction then the cycle can't be steered in any sharper angle.

 

[Faiq]

What do you mean by "inertial force of the turning cycle"?Perhaps the centripetal force?

 

[NascentOxygen]

Yes, the force needed to overcome inertia and compel the body to follow a certain speed and path.

 

[Faiq]

I repeat my question
Friction = Centripetal force (in this case)
How can centripetal force> friction if Friction = Centripetal force?

 

[Kajal Sengupta]

First you need to understand the origin of frictional force. The origin of frictional force between two surfaces is due to the inter-atomic attractions of the two surfaces in contact. They are electromagnetic in nature. When we try to move an object over another surface then this frictional force tend to oppose it. If we increase the applied external force then the frictional force has the property to increase itself and hence the body would not move. Due to this property frictional force is said to be 'self-adjusting". There is a limit upto which this force can increase and if the external force is greater than that force then the body starts moving. Limiting friction is that maximum value of frictional force upto which it can be increased. It depends on nature of surface and normal reaction. When a person is taking a turn a centripetal force is necessary. In this case it is frictional force which provides for it. The person will not skid provided the force required ( mv^2 /r , determined by mass, velocity and radius of the path) to go round the bend is less than or equal to the force of limiting friction ( determined by nature of surfaces and normal reaction ) . That is why on a slippery road it is difficult to take a turn because the force of friction is less than the centripetal force required.

 

[NascentOxygen]

If you wanted a body to follow a path where the required centripetal force exceeded the available friction then you'd need to supply some extra force by some other means, e.g., a sloping floor so gravity assisted, or rocket side thrusters, or a cable tethered at the radius centre of curvature of the bend, etc.

 

[Faiq]

If you wanted a body to follow a path where the required centripetal force exceeded the available friction then you'd need to supply some extra force by some other means, e.g., a sloping floor so gravity assisted, or rocket thrusters, or a cable tethered at the radius of curvature, etc.

Yes that was what I was suggesting. So in this example are they hinting towards some extra force which is also contributing to centripetal force?

 

[NascentOxygen]

The situation with bicycles is that you can't turn more sharply than friction allows. The maximum friction dictates your maximum cornering speed.

 

[Faiq]

How is tilting angle related to centripetal force? The only force providing centripetal force is friction, which is independent on tilt angle.

 

[Kajal Sengupta]

I repeat my question
Friction = Centripetal force (in this case)
How can centripetal force> friction if Friction = Centripetal force?

The concept is that for the body to turn without skidding the centripetal force required should be less than or equal to frictional force. In the expression they are equated to calculate the maximum velocity with which you can take the turn in a given condition because all other factors like mass, radius of the road nature of surfaces are fixed.

Centripetal force depends on Mass of the person, velocity and radius of curvature ( mv^2 / r). Whereas the force of friction depends on nature of surface. Imagine if the radius of the path is too small ( a sharp bend) then the centripetal force increases. It may happen that it becomes more than frictional force. In that case the person will not be able to take the turn safely.

 

[Kajal Sengupta]

How is tilting angle related to centripetal force? The only force providing centripetal force is friction, which is independent on tilt angle.

Force of friction depends on normal reaction. In this case you have to take normal reaction as the normal component of the weight which contains angle of tilt.

 

[Faiq]

Force of friction depends on normal reaction. In this case you have to take normal reaction as the normal component of the weight which contains angle of tilt.

I think you are a little bit wrong here. Normal reaction and friction forces are themselves the component of contact force. So Normal force doesn't have a friction component. Moreover weight doesn't have a normal component. Weight has a reaction force which is contact force which has a normal component.

 

[sophiecentaur]

But if friction is itself the applied force, how can applied force> friction?

I am repeating what has been written in parts of the above posts but the following is an attempt to bring it all together:
It is important to get the cause and effect in the right order and to appreciate that 'the equation' represents the limiting case. The maximum applied force cannot be greater than the limiting friction force. If, instead of friction on the tyres, there was a piece of string, then the breaking stress of the string would define the fastest rotation speed. that could be maintained. at that radius. Limiting friction is the equivalent to the strength of the string and that relates to the weight and coeff of friction. The angle of lean is determined by the coeff of friction. (Weight force and limiting friction force) The angle of lean is also related to the centripetal force at any speed (if the bike is in equilibrium). That gives you two sides of a limiting case equation which can be solved for the maximum possible bike speed.

 

[Kajal Sengupta]

I think you are a little bit wrong here. Normal reaction and friction forces are themselves the component of contact force. So Normal force doesn't have a friction component. Moreover weight doesn't have a normal component. Weight has a reaction force which is contact force which has a normal component.

Yes I may skipped elaborating it. The weight mg acts vertically downwards. the way the normal force R has been shown is simplified version. We take component of mg along the axis of the cycle

 

[pgardn]

Why in the diagram does it appear show mg causing a torque (thus rotation indicated CCW)? Mg is a force "through" the CM?

 

[pgardn]

First you need to understand the origin of frictional force. The origin of frictional force between two surfaces is due to the inter-atomic attractions of the two surfaces in contact. They are electromagnetic in nature.

In the model I have static friction is a repulsive force? I am a bit confused here.

 

[sophiecentaur]

Why in the diagram does it appear show mg causing a torque (thus rotation indicated CCW)? Mg is a force "through" the CM?

You can ignore the torque because the bike is in equilibrium. The only relevant forces when the bike is at the correct angle are acting where the tyre makes contact with the road.
The confusion about the direction of any force can probably be down to the choice of the 'wrong' one of the N3 pair. Of course, the friction force on the bike is towards the centre of the circle because it is producing circular motion.
If you are trying to look at this in a way that includes the inter molecular effects then you are making life hard for yourself, I think. I would think that the sideways force of friction would be due to both attractive and repulsive effects between neighbouring tyre and road molecules.

 

[pgardn]

You can ignore the torque because the bike is in equilibrium. The only relevant forces when the bike is at the correct angle are acting where the tyre makes contact with the road.
The confusion about the direction of any force can probably be down to the choice of the 'wrong' one of the N3 pair. Of course, the friction force on the bike is towards the centre of the circle because it is producing circular motion.
If you are trying to look at this in a way that includes the inter molecular effects then you are making life hard for yourself, I think. I would think that the sideways force of friction would be due to both attractive and repulsive effects between neighbouring tyre and road molecules.

Thanks.

The friction part is my bad habit of trying to model things too simply. If I might draw a picture of a tire of how static friction might start and stop a car maybe you could elaborate on it? I don't wish to hijack the original question. Just trying to understand.

 

[pgardn]

Comments?

 

[sophiecentaur]

Comments?

Your 'peg in hole' model could perhaps be improved by making the cross section a V shape in a V shaped slot (many of them side by side). The relationship between weight force and the lateral force to balance it, would be Tan(slope angle) and that would represent the coeff of friction (∞ for vertical sides). If the lateral force exceeds this, the V will slide up out of the slot and you would have slipping. The nice thing about that simple model is that it follows the law of friction and gives a constant value of μ for all loads.
The force is always against the desired displacement, which provides a braking or driving force, as required. Sometimes you can sort this thing out in your head by drawing diagrams (like above) of all different situations and 'believing' that the forces are in the right direction - then fitting your in-brain model to that. (The conceptual leap is never half way. It's one way or the other)

 

[pgardn]

Your 'peg in hole' model could perhaps be improved by making the cross section a V shape in a V shaped slot (many of them side by side). The relationship between weight force and the lateral force to balance it, would be Tan(slope angle) and that would represent the coeff of friction (∞ for vertical sides). If the lateral force exceeds this, the V will slide up out of the slot and you would have slipping. The nice thing about that simple model is that it follows the law of friction and gives a constant value of μ for all loads.
The force is always against the desired displacement, which provides a braking or driving force, as required. Sometimes you can sort this thing out in your head by drawing diagrams (like above) of all different situations and 'believing' that the forces are in the right direction - then fitting your in-brain model to that. (The conceptual leap is never half way. It's one way or the other)

Actually the model (teeth being triangles) is the model I use to provide for static and kinetic friction.

But here is where I run into problems with that model:

If a ball rolls up a rough surface while slipping and slowing down I see the static frictional force acting up the hill and the kinetic friction acting down? And this confuses me. Like the triangular teeth slip back (the surface of the tooth parallel to the plane of the hill) and then the part of the tooth that is perpendicular to the plane of the surface then acts to keep it from slipping further providing a static frictional force up? Can you make sense of this?

Thanks for your help.

I am more of a Biochem guy trying to get rid of my misconceptions of which I have found many and this requires people like you walking me through. A book with just math problems does not always help. I need models and then proper application and then math that fits the model. Thanks for your patience.

 

[sophiecentaur]

If a ball rolls up a rough surface while slipping and slowing down I see the static frictional force acting up the hill and the kinetic friction acting down?

The thing that's making it slow down is the downhill weight component . There can only be one Net force due to friction which, in this case, is pushing the ball uphill - just not enough to counteract the downhill weight component

 

[pgardn]

The thing that's making it slow down is the downhill weight component . There can only be one Net force due to friction which, in this case, is pushing the ball uphill - just not enough to counteract the downhill weight component

I get the weight component.

But if the ball was slipping while slowing down going up the hill kinetic friction would act down as well? Yes?

 

[sophiecentaur]

I get the weight component.

But if the ball was slipping while slowing down going up the hill kinetic friction would act down as well? Yes?

The logic has to tell you that there is only one force at work, due to friction. Whichever direction the wheels are moving against the ground then the friction force on the wheels must be against that direction. if the wheels are driving you uphill then friction is acting uphill on the wheels (you can't have dynamic and static friction at the same time and, with the wheels moving, if you have to give it a name, it's kinetic friction that's involved - personally I would call it 'slipping friction')
There are two forces acting on the vehicle: friction drives it uphill and weight component drives it downhill. Depending on which is greater, it can be going up or down the slope.
But we (you) are after a simple model for a very complex situation and there isn't one.
The driven wheels of a car are subject to a forward force (horizontal reaction against the ground) which would be much the same as the static friction, if the wheels were stationary, with the same torque applied. There is significant loss of power due to the constant deformation of the tyre (constantly climbing up the leading edge and the road surface which involved hysteresis and also the grinding of grit and dust in the footprint. That Lost Power can be looked upon as an effective Force times the Speed of the car over the road. This Force is subtracted from the Force from the drive and will be reducing the acceleration or top speed of the car. Rather than calling it 'Friction', it would normally be called Rolling Resistance. That avoids your problem of reconciling two forces of 'friction', acting in two different directions.
PS The only times that friction would be acting downhill would be if the brakes were applied and no power was being applied to the wheels OR if the car were actually being driven downhill. In both cases, there would be an Acceleration downhill.

 

[pgardn]

The logic has to tell you that there is only one force at work, due to friction. Whichever direction the wheels are moving against the ground then the friction force on the wheels must be against that direction. if the wheels are driving you uphill then friction is acting uphill on the wheels (you can't have dynamic and static friction at the same time and, with the wheels moving, if you have to give it a name, it's kinetic friction that's involved - personally I would call it 'slipping friction')
There are two forces acting on the vehicle: friction drives it uphill and weight component drives it downhill. Depending on which is greater, it can be going up or down the slope.
But we (you) are after a simple model for a very complex situation and there isn't one.
The driven wheels of a car are subject to a forward force (horizontal reaction against the ground) which would be much the same as the static friction, if the wheels were stationary, with the same torque applied. There is significant loss of power due to the constant deformation of the tyre (constantly climbing up the leading edge and the road surface which involved hysteresis and also the grinding of grit and dust in the footprint. That Lost Power can be looked upon as an effective Force times the Speed of the car over the road. This Force is subtracted from the Force from the drive and will be reducing the acceleration or top speed of the car. Rather than calling it 'Friction', it would normally be called Rolling Resistance. That avoids your problem of reconciling two forces of 'friction', acting in two different directions.
PS The only times that friction would be acting downhill would be if the brakes were applied and no power was being applied to the wheels OR if the car were actually being driven downhill. In both cases, there would be an Acceleration downhill.

So my physical model which allows for both static and kinetic friction is wrong. This is the problem I have. I have a model which makes physical sense to me. Now what I would normally try to do is fit the model in with the tire changing shape, the ground beneath actually changing composition, etc... (I can also see there might also be a bit of adhesion on the molecular level thinking about dragsters spinning their tires to make a sticky surface)

But I won't.

Thanks for your time.

 

[sophiecentaur]

So my physical model which allows for both static and kinetic friction is wrong.

My view is that the tyre is either static with respect to the ground or it's moving. If you just take a block and try to pull it along, you have static friction until it breaks away and then, once it is free, you have kinetic friction (often less). As the speed increases, there can be other losses, due to deformation of block and ground. With a driven wheel, you are mostly dealing with simple static friction Until it slips - when it's kinetic. If the wheel is slipping, how can you say that there is a static condition?
If your model makes physical sense to you then you should try to test it for credibility. Is it really consistent with experience?
ABS braking improves the retardation by ensuring that no wheel goes faster than the others (showing it is slipping.) The static friction force is greater than the force after is slips. That applies on your sloping situation.
When dragsters melt their tyres the situation is non-linear and there is no simple model to describe the phenomenon.

 

[pgardn]

My view is that the tyre is either static with respect to the ground or it's moving. If you just take a block and try to pull it along, you have static friction until it breaks away and then, once it is free, you have kinetic friction (often less). As the speed increases, there can be other losses, due to deformation of block and ground. With a driven wheel, you are mostly dealing with simple static friction Until it slips - when it's kinetic. If the wheel is slipping, how can you say that there is a static condition?
If your model makes physical sense to you then you should try to test it for credibility. Is it really consistent with experience?
ABS braking improves the retardation by ensuring that no wheel goes faster than the others (showing it is slipping.) The static friction force is greater than the force after is slips. That applies on your sloping situation.
When dragsters melt their tyres the situation is non-linear and there is no simple model to describe the phenomenon.

It makes physical sense with the triangular teeth model. The diagonal teeth can slip and at the same time can catch.

Again if you model the vertical component of the teeth you get some static friction (which really is a lot like a normal force like in the peg picture I drew. It's possible the peg like tooth could just sheer after exerting a static force. Like if one puts a more massive load on plane of wood till it breaks. Even when something slips this could still exist with the model I already presented using diagonal teeth.The horizontal component of the tooth would allow for the classic slip. I can see different parts of a larger body slipping while others remain static.

Now I may to read more about shearing, compression etc... as I may have ideas about this that are horribly wrong.

But if my simplified model does not fit experiment and observation then it's no good. It's much simpler actually to use your idea of you are static or you are not.

PS

I realize I am not very good at expressing my ideas with English. I'm much better with a face to face sit down while drawing pictures. I tend to think visually before mathematically. So sorry for this.