Sliding, rolling and overall friction

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Discussion Overview

The discussion revolves around the mechanics of friction forces acting on a wheel in motion, particularly focusing on the distinctions between static friction, rolling resistance, and the effects of applied torque. Participants explore the implications of these forces on the wheel's motion, including scenarios of skidding versus rolling, and the impact of braking.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant describes the forces acting on a stationary wheel and how they change when a torque is applied, suggesting that the reaction force can be decomposed into horizontal and vertical components.
  • Another participant introduces the concept of rolling resistance and questions the benefit of combining the two friction forces into a single representation.
  • There is a discussion about the conditions under which a wheel would skid rather than roll, with some participants suggesting that if the applied force exceeds static friction, the wheel would skid.
  • Concerns are raised about the role of rolling resistance when a wheel is skidding, with one participant questioning whether rolling resistance is relevant in that scenario.
  • Participants discuss the effect of braking on the wheel's motion, with one asserting that braking introduces an opposing torque to the original torque.
  • One participant attempts to clarify their understanding by writing equations for a system of two wheels connected by a rigid bar, seeking feedback on their approach.
  • Another participant critiques the equations presented, pointing out potential errors and suggesting corrections related to torque and forces acting on the wheels.

Areas of Agreement / Disagreement

Participants express differing views on the representation of friction forces and the conditions under which a wheel skids versus rolls. There is no consensus on the implications of rolling resistance during skidding or the correctness of the equations presented.

Contextual Notes

Participants acknowledge the complexity of the forces involved and the need for precise definitions and conditions in their discussions. Some equations presented may lack clarity or contain errors that have not been resolved.

barzi2001
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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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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]
 
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:
 

Attachments

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.
 

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