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Physics behind a powered tire

  1. Jun 21, 2011 #1
    Diagram = http://i.imgur.com/LaUO4.jpg

    The tire is powered by a motor of torque T. It's weight is W and the static friction force is Fs. Consider that the tire is moving upwards in a constant speed of V.

    I would like to understand the relations between forces on a model like above.
    Also how can I show the rolling resistance? What's the direction?
    For example what would be the value of Fs?

    What resources/books can you recommend to better understand this?

    Thank you
     
  2. jcsd
  3. Jun 21, 2011 #2
  4. Jun 21, 2011 #3
    Ok from what i understand rolling resistance creates a torque opposite to the motor but where from the force is applied? How can i include in my calculations?
     
  5. Jun 21, 2011 #4

    jack action

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    See http://hpwizard.com/car-performance.html" [Broken] (at the bottom of the page, Theory »» Longitudinal acceleration)

    rolling-resistance.gif
     
    Last edited by a moderator: May 5, 2017
  6. Jun 22, 2011 #5
    Ok I think I cracked it.
    According to wiki page rolling resistance is applied in the reverse direction of the motion from the center of the wheel (somehow I missed it the first time sorry)

    Since there is no acceleration the overall force should be 0 thus I can write this equation

    [itex]F_s = F_{rr} + W.sin(\alpha)[/itex] where [itex]F_s = T/r[/itex] and [itex]F_{rr} = W.cos(\alpha).C_{rr}[/itex] so the equation becomes:

    [itex]T/r = W.cos(\alpha).C_{rr} + W.sin(\alpha)[/itex]

    Am I on the right track here?
     
    Last edited: Jun 22, 2011
  7. Jun 29, 2011 #6
    How would I have to change the above equation if let's say there is an obstacle in front of the wheel by the dimensions LxL like in below
    QUc0r.jpg

    thank you
     
  8. Jul 1, 2011 #7
    Ok I now know that the above diagram and formula is false.
    The force needed to move the tire is equal to the static friction between the tire and the road right. So [itex]F_s = W.cos(\alpha).\mu_s[/itex]
    But there is also [itex]W.sin(\alpha)[/itex] and [itex]F_{rr}[/itex] working in the opposite direction of [itex]F_s[/itex] so the force needed to move the tire upwards is [itex]F = F_s + F_{rr} + W.sin(\alpha)[/itex] am I correct? anyone know any books that explains the tractive forces and rolling resistance on a car wheel?
    Or can you at least provide me with a free body diagram of a car wheel at constant speed?

    thank you
     
    Last edited: Jul 1, 2011
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