Free body diagram of a wheel driven by a motor

[EM_Guy]

Again, you come back to the model with tennis shoes on spokes. If that is the model you want, just say so.

That is not the model I want. How did I "come back" to this model?

 

[jbriggs444]

That is not the model I want. How did I "come back" to this model?

You have mass units on the end of spokes supporting shear stresses and unaffected by the rim. That is precisely the situation for tennis shoes on spokes.

 

[EM_Guy]

You have mass units on the end of spokes supporting shear stresses and unaffected by the rim.

I didn't say that they are unaffected by the rim. Can you not have mass units on the end of spokes supporting shear stresses that are also affected by the rim?

 

[jbriggs444]

I didn't say that they are unaffected by the rim. Can you not have mass units on the end of spokes supporting shear stresses that are also affected by the rim?

But you do not seem to want to accept that the rim does anything. If it does nothing, it might as well not be there.

If we take a model where each spoke transmits part of the torque from hub to rim and where the contact force from the ground exerts a force at a single point, what can we say about the change in tension or compression of the rim across the point of contact with the ground?

 

[EM_Guy]

But you do not seem to want to accept that the rim does anything.

I didn't say that. I acknowledged that we have two tangential forces of the rim on the infinitesimal section of wheel touching the ground as well as two normal forces of the rim on the section of wheel touching the ground.

If we take a model where each spoke transmits part of the torque from hub to rim and where the contact force from the ground exerts a force at a single point, what can we say about the change in tension or compression of the rim across the point of contact with the ground?

I don't know. First, it is not clear to me - even in free space (no ground) whether the torque on the axle causes the rim to be in greater compression or tension? On one hand the torque transmitted through each spoke would seem to compress (push) the section of rim ahead of it, while also pulling the section of rim behind it. Therefore, without considering the ground, I don't think the torque on the axle changes the compression / tension of the rim.

But for the section of wheel touching the ground, right as that section starts to touch the ground, I think that there is increased compression (torque through spoke pushing the rim, while static friction is pushing the rim in the opposite direction). But then as that section of rim leaves the ground, the spoke seems to be pulling that section of wheel, and - in a sense - the static friction force is also pulling that section of wheel - causing increased tension (or decreased compression) in that part of the rim. Meanwhile the normal component of the force of the rim (not yet touching the ground) on the point touching the ground would be downwards, while the normal component of the force of the rim (just previously touching the ground) on the point touching the ground would be upwards.

 

[A.T.]

It still remains true that T = ma.

And that means T causes the acceleration?

Lets have:

F1 = 1 N
F2 = 1 N
F3 = -2 N
F4 = 1 N

F1 = ma
F2 = ma
F4 = ma

So which of the 3 forces causes the acceleration?

 

[EM_Guy]

The summation of forces causes acceleration.

$∑F = ma$

It turns out in this case that $∑F = T$

 

[A.T.]

The summation of forces causes acceleration.

That is also my interpretation. Although I would rather say "determines", instead of "causes".

It turns out in this case that ∑F=T

It doesn't "turn out". You simply choose to decompose a physical force such that one component is equal to the net force. There is nothing significant about the fact that you can find such a decomposition, because as you correctly noted yourself: Any given vector (representing a physical force) can be decomposed into any number of components..

 

[EM_Guy]

That is also my interpretation

Well, that's good! I'm glad we agree on Newton's 2nd Law! :)

It doesn't "turn out". You simply choose to decompose a physical force such that one component is equal to the net force. There is nothing significant about the fact that you can find such a decomposition, because as you correctly noted yourself: Any given vector (representing a physical force) can be decomposed into any number of components..

But I did so - taking two FBDs into account. I could have decomposed the Fspoke arbitrarily into any number of components. But my choice was not arbitrary.

 

[jbriggs444]

First, it is not clear to me - even in free space (no ground) whether the torque on the axle causes the rim to be in greater compression or tension? On one hand the torque transmitted through each spoke would seem to compress (push) the section of rim ahead of it, while also pulling the section of rim behind it. Therefore, without considering the ground, I don't think the torque on the axle changes the compression / tension of the rim.

Yes, we agree.

But for the section of wheel touching the ground, right as that section starts to touch the ground, I think that there is increased compression (torque through spoke pushing the rim, while static friction is pushing the rim in the opposite direction).

If there are (for instance) 24 spokes attached to the rim, what fraction of the total torque from the hub would you expect the spoke nearest the bottom to be applying to the rim?

If there is one contact patch of the tire on the road, what fraction of the total torque from the road would you expect from the contact patch?

 

[EM_Guy]

torque from hub applied to the rim: $\frac{1}{24}$^th

torque (or force) from friction applied to contact patch: the entire frictional force

 

[jbriggs444]

So if there happens to be a spoke near the contact patch, one would expect a difference between compression/tension before and after the contact patch/spoke area roughly equal to 23/24 of the frictional force. And one would expect a delta of 1/24 of the frictional force around each of the other 23 spokes.