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Static weight transfer of car with two wheels elevated

  1. Mar 28, 2015 #1
    I am trying to figure out what the weight transfer, CoG transfer, and actual weight for each wheel is, for the vehicles (specifically the Subaru wagon at 2:00 minutes) undergoing this test:

    My friend argues the test is flawed because so much of the weight is on the tires contacting the pavement. I don't think it makes enough of a difference. The results would be the same on flat ground in my opinion.

    I am curious as to how much weight is actually transferred? If the test was conducted on flat ground, would the AWD Subaru fare any better?

    I have drawn a picture of the problem. Using numbers from Subaru automobiles for dimensions and CoG. If you have a moment, I would really appreciate your solving of this problem.

    http://[ATTACH=full]199720[/ATTACH]

    Thanks for your time!
     
    Last edited by a moderator: May 7, 2017
  2. jcsd
  3. Mar 28, 2015 #2

    haruspex

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    The video link doesn't work on my iPad - might try later on another device - so I'm uncertain what the question is. I assume you want to determine the normal force on each tire from the ground.
    In the drawing, the tires should be tilted at the same angle as the vehicle, and the weight should act vertically through the mass centre.
    Just treat the lowest points of the tires and CoG as the three points of a rigid isosceles triangle. Figure out the horizontal distances between them when tilted, and take moments about one of the contact points. That will tell you the normal force at the other.
     
  4. Mar 28, 2015 #3
    Essentially, yes, I would like to determine the normal force on each tire. If you had a scale underneath each tire, what would it read? Thanks for your reply!
     
  5. Mar 28, 2015 #4

    billy_joule

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    PF won't do the work for you, see the forum rules for more info.
    Follow Haruspex method below and you will find the normal force.

     
  6. Mar 28, 2015 #5
    My apologies. Thanks for your help.
     
  7. Mar 28, 2015 #6

    billy_joule

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    If you have any trouble there are many people here who can help.

    My immediate thought is look at the spring compression as it's proportional to applied force. Ie if the wheels are different distances to the arches they are supporting a different mass.
    You can find the spring rate online and calc. The difference in mass. compare it to the theoretical result found by haruspex method.
     
  8. Mar 28, 2015 #7
    Some basic geometry to find the perpendicular distance from the point at which you are taking the moment at to the force applied on the tire. Then take sum of the moments either at the left tire or right tire depending on which lever arm you chose. If you have any questions I'd be happy to help.
     
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