Free body diagram of a wheel driven by a motor

  • #76
A.T.
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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?
 
  • #77
212
49
The summation of forces causes acceleration.

##∑F = ma##

It turns out in this case that ##∑F = T##
 
  • #78
A.T.
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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..
 
  • #79
212
49
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.
 
  • #80
jbriggs444
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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?
 
  • #81
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49
torque from hub applied to the rim: ##\frac{1}{24}##th

torque (or force) from friction applied to contact patch: the entire frictional force
 
  • #82
jbriggs444
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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.
 

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