Sliding, rolling and overall friction

In summary, the conversation discusses the concept of friction forces on a wheel in motion on a flat surface. The reaction force at the contact point of the wheel with the ground can be decomposed into horizontal and vertical components, with the vertical component passing through the center of the wheel and the horizontal component pushing the wheel to the right. When including rolling friction, the horizontal component is applied at the contact point while the vertical component is applied at a point slightly to the right. The idea of combining the two frictions into a single force is explored, but it is suggested to approach it by finding the net force and its point of intersection with the road. The use of a brake is also discussed, with it being equivalent to adding a torque opposing the
  • #1
barzi2001
10
0
Hi all. I'm struggling to understand how and where the friction forces that appear on the motion of a wheel on a flat surface, but I don't know if I understand it properly. I hope that you can help me. I try to describe the problem as follows.

When a wheel is not moving, and there are no forces/torques applied on that, at the contact point of the wheel with the ground (let's call it P) the reaction force points "upwards", it passes for the center of the wheel and it has the same magnitude of the weight force. Hence, the resulting force is zero as well as the resulting torque.

When a torque is applied, let's say clockwise, it generate a horizontal force F that is applied to the contact point P, and such force F points toward left. The magnitude of such force F is equal to T/r, where T is the applied torque and r is the wheel radius. The reaction force at the contact point P in this case will point toward up-right. This means that the reaction force can be decomposed in two components: one horizontal and one vertical. The vertical force (which points upwards) passes again for the center of the wheel, while the horizontal force (which points toward right) "push" the wheel toward right (I assumed that F is smaller that the static sliding friction). In this case, both the horizonal and vertical components of the reaction force are applied at the point P.

By including also the rolling friction, we have, like before, that the reaction force at the point P points again in up-right direction. Again, this force can be decomposed in a horizontal and vertical component. The difference with respect to the previous case is that now the horizontal component is applied to the point P while the vertical component is applied in a point P' that is located slightly to the right of P. Since the vertical component of the reaction force is applied at P', it does not passes for the center of the wheel, and then it generates a resistance torque, while the horizontal force, being applied to P, does the same job as the previous case (pushes the wheel toward right).

In synthesis, by combining the two frictions in one shot, can I say that it can be represented as a reaction force that points up-right, but the horizontal component is applied to P while the vertical component is applied to P'? But where it is applied such reaction force that include both the rolling and the sliding frictions?

And moreover, if the force F in the contact point P due to the applied torque is greater than the static friction, I imagine that the wheel would rolls without translating, but after some time it will it stop because of a vertical component applied to P'?

Finally, when I use a brake (for example a disk brake), is equivalent to say that I'm adding an additional friction that points toward left and thus it opposes to the motion of the wheel?

I hope I provided a good description of the example. Looking forward for your answer.
 
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  • #2
barzi2001 said:
By including also the rolling friction,
This is usually referred to as rolling resistance.
The difference with respect to the previous case is that now the horizontal component is applied to the point P while the vertical component is applied in a point P' that is located slightly to the right of P.
I've never thought of it like that, but I believe that works.
In synthesis, by combining the two frictions in one shot, can I say that it can be represented as a reaction force that points up-right, but the horizontal component is applied to P while the vertical component is applied to P'?
I don't see any benefit in trying to combine them as a single force, but you could approach it like this:
Find the line of action of the forces at P
Find the line of action of the forces at P'
Find the point of intersection, Q.
Find the direction, D, of the net force.
Find the point on the road s.t. a line through it in direction D passes through Q.
And moreover, if the force F in the contact point P due to the applied torque is greater than the static friction, I imagine that the wheel would rolls without translating,
i.e. skidding, not rolling
but after some time it will it stop because of a vertical component applied to P'?
I didn't understand that part. Are you saying it would stop skidding? I don't see why, as long as F exceeds static friction.
Finally, when I use a brake (for example a disk brake), is equivalent to say that I'm adding an additional friction that points toward left and thus it opposes to the motion of the wheel?
It's a torque opposing the original torque.
I hope I provided a good description of the example. Looking forward for your answer.[/QUOTE]
 
  • #3
Hi! Thanks for the answer.
I agree with you that I should use the term "skidding" and "rolling resistance" :)

I didn't understand that part. Are you saying it would stop skidding? I don't see why, as long as F exceeds static friction.

Yes, I mean to stop skidding (F exceeds the static friction, but what about the rolling resistance? I guess there is not rolling resistance because the wheel is skidding and not rolling, right?)

However, to try to clarify once and for ever these concepts, I have tried to write down the equations of the motion a system composed by two wheels connected with a rigid bar, and where an external torque is applied to the front wheel (when developing this example I was thinking about the longitudinal dynamics of a car). But I am not sure that i did it properly. Could you take a look at that? I attached a pdf. Thanks again! :smile:
 

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  • Pure rolling model.pdf
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  • #4
A few things I don't understand in your equations.
Front wheel torque: Shouldn't there be a term Nf r sin(θf) somewhere?
Rear wheel forces: You show a θf. Should be θr?
Rear wheel torque: Judging from the RHS, you've taken moments about the wheel centre (as with front wheel). So first term on LHS should read r Nr sin(θr). Fr has no moment about the centre.
 
  • #5



Hello there,

It seems like you have a good understanding of the concept of friction and its effects on a wheel's motion. Let's break down your questions one by one:

1. Where is the reaction force applied in the case of sliding and rolling friction?

In both cases, the reaction force is applied at the contact point P between the wheel and the ground. However, in the case of rolling friction, there is also a small vertical component of the reaction force applied at a point P' slightly to the right of P. This is due to the fact that the wheel is not perfectly rigid and there is some deformation happening at the contact point.

2. Can we represent the combination of sliding and rolling friction as a single reaction force?

Yes, we can represent the combination of sliding and rolling friction as a single reaction force that points up-right. The horizontal component of this reaction force is applied at P, while the vertical component is applied at P'.

3. What happens if the applied torque is greater than the static friction?

If the applied torque is greater than the static friction, the wheel will start to roll without translating. However, after some time, the vertical component of the reaction force at P' will cause the wheel to stop. This is because the vertical component is creating a resistance torque that opposes the motion of the wheel.

4. Is using a brake equivalent to adding an additional friction force?

Yes, using a brake is equivalent to adding an additional friction force that opposes the motion of the wheel. This is why using a brake can slow down or stop a moving wheel.

I hope this helps clarify your understanding of sliding, rolling, and overall friction. If you have any further questions, don't hesitate to ask. Keep up the good work in your studies!
 

Related to Sliding, rolling and overall friction

What is the difference between sliding and rolling friction?

Sliding friction occurs when two surfaces slide against each other, while rolling friction occurs when an object rolls over a surface. Sliding friction is generally greater than rolling friction because there is more surface contact and therefore more resistance.

How does the type of surface affect friction?

The type of surface can greatly affect the amount of friction. Rough surfaces have more irregularities and therefore create more friction, while smooth surfaces have less friction. Additionally, the type of material can also affect friction, as some materials have more or less resistance to sliding or rolling.

What are some real-life examples of sliding and rolling friction?

Sliding friction can be seen when pushing a heavy object across the floor, or when a car's brakes are applied to stop it. Rolling friction is present when a ball is rolling across the ground or when a tire is rolling on pavement.

How is friction related to energy?

Friction converts kinetic energy (motion) into thermal energy (heat). This means that when two surfaces rub against each other, some of the energy is lost as heat due to friction. This is why surfaces can become warm or even hot when there is a lot of friction present.

How can friction be reduced?

Friction can be reduced through the use of lubricants, such as oil or grease, which create a layer between the two surfaces and reduce the amount of direct contact. Additionally, smoother surfaces can also help reduce friction, as there are less irregularities to create resistance.

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