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Kinetic Energy & Forces

  1. Mar 3, 2008 #1
    1. The problem statement, all variables and given/known data
    Ok, well I feel this is a little beyond introductory physics, but maybe I'm wrong. I'm helping to create a program to simulate a car stopping based on input variables - road surface, Vi, tyre compound, mass, etc. I do however, need some equations to simulate this. So in most cases, the tyres are the limiting factor. Assuming that the car has no ABS we can assume [tex]\mu[/tex]Fk . But anyway, the main problem is calculating how fast the car will stop, based upon the stopping force the tyres can provide (a % of force from brakes, depending on friction) and the other input variables. So, the force will be in Newtons. But these Newtons are "eating away" at the KE of the car. (or the momentum? I assume we'd be dealing with KE here).
    Can anyone please derive an equation relating stopping force and deceleration of the car, for say a 2000kg car travelling at 30m/s?
    2. Relevant equations
    Not sure...


    3. The attempt at a solution
    None have come out nicely yet...
     
  2. jcsd
  3. Mar 3, 2008 #2

    CompuChip

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    Welcome to PF. I will sketch the way I would handle this problem, but I don't know how far you can take this yourself (e.g., what your level of physics is).

    You will want to use Newtons law, F = ma where the force is in principle air friction ([itex]\propto -c_V A \vec v^2[/itex], A the frontal area) + road friction ([itex]\mu_F N[/itex]).
    Once you have the acceleration, you would have to solve the differential equation
    [tex]x''(t) = a[/tex]
    with x'(0) = initial velocity and x(0) = 0.
     
  4. Mar 3, 2008 #3
    Thanks for the welcome and quick reply.
    I am not sure that we would include air friction, but as you included a nice formula for me we may include it. As for the tire/road friction, I suppose that I would just need to vary the overall coefficient of friction based upon the input variables such as road surface coefficient of friction and tire type etc.
    Is that differential eq'n a "standard" equation for calculating this type of thing? I presume that the equation is solving for time taken?
    Sorry I'm only just learning about differential equations and I'm not too familiar with them yet.

    As for my Physics background, I have taken Physics 1, 2, and AP (non-calculus based) in High school, and am now doing a Foundation Engineering year in Uni. However AP physics was more difficult than what I'm doing this year :-)
     
  5. Mar 3, 2008 #4

    CompuChip

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    That equation is just
    F = m a
    where the acceleration is the second derivative of the position x(t). If the force is also linearly dependent on mass, which is usually is, then the mass cancels out. For example, in a gravitation problem the force would be F = m g and you just get m g = m a so g = a = x''(t) (of which the solution is the - hopefully familiar - x(t) = 1/2 g t^2).
    Once you solve this equation, you get the displacement function x(t) which will give you, for any time t, the position of the car. If you want to do this numerically, I bet there are plenty of packages out there which do it. If the acceleration is a very simple function, you might even solve it by hand which will make the program faster (instead of solving a differential equation each time, you just hard-code the formula for x(t) and only need to plug in a t).

    I have to go now, when I get back I'll see if I get can you some reading material. You should be able to get quite far yourself with Physics 1, 2 and AP (though I don't know the contents of those exactly, but Newtonian mechanics should be in any physics 1 course IMO :smile:). The only problem you may run into is solving the differential equation, if you've never done that before.
     
  6. Mar 3, 2008 #5
    Thanks for clarifying that, it makes sense now.
    I may have some more questions later but for now this should keep me busy. Figuring out the programming in Visual Basic might be a bit harder though :-)
    One question relating to the friction though... and I know that you won't even get tyre engineers to decide entirely on some issues, but as for things like tire width... technically they shouldn't matter, since area does not come into Fn x Coef Fr, but surely having more tire area gives better stopping ability... at least ths is what I have found from my own experience. Sure, compound makes more difference but it just seems that area must also play a role. But maybe this is a little abstract for what we are hoping to accomplish?
     
  7. Mar 3, 2008 #6

    CompuChip

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    These are details of the problem which we usually hide in the friction coefficient. Obviously, wider tires gives more contact area, hence the coefficient of friction will be larger. If you are a racing fan the "Cv" value may mean something to you. A lot is explained on this page. Basically, once you get the friction coefficient (Crr, or Cv) from the manufacturer, you needn't worry about all those details, they're all in there.
     
  8. Mar 3, 2008 #7
    Ahh, that makes sense now, how convenient of them!
    Cheers! :cool:
     
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