# I Limiting friction

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1. Aug 19, 2016

### 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

Last edited by a moderator: Aug 19, 2016
2. Aug 19, 2016

### Staff: Mentor

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.)

3. Aug 19, 2016

### Faiq

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

4. Aug 19, 2016

### Staff: Mentor

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.

5. Aug 19, 2016

### Faiq

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

6. Aug 19, 2016

### Staff: Mentor

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

7. Aug 19, 2016

### Faiq

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

8. Aug 19, 2016

### 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.

9. Aug 19, 2016

### Staff: Mentor

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.

Last edited: Aug 21, 2016
10. Aug 19, 2016

### Faiq

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?

11. Aug 19, 2016

### Staff: Mentor

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

12. Aug 19, 2016

### Faiq

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

13. Aug 19, 2016

### Kajal Sengupta

The concept is that for the body to turn with out 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.

14. Aug 19, 2016

### Kajal Sengupta

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.

15. Aug 19, 2016

### Faiq

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.

16. Aug 19, 2016

### sophiecentaur

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.

17. Aug 19, 2016

### Kajal Sengupta

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

18. Aug 19, 2016

### pgardn

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

19. Aug 19, 2016

### pgardn

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

20. Aug 19, 2016

### sophiecentaur

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.