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Air friction help

  1. Feb 24, 2012 #1
    1. The problem statement, all variables and given/known data
    In formula of air friction ##f_{air}=-bv##
    v is velocity but what's b?
     
  2. jcsd
  3. Feb 24, 2012 #2
    In your formula, b would be a coefficient that includes things like drag coefficient, air density, and frontal image area. Usually though, the force is represented by v^2, not simply v.
     
  4. Feb 24, 2012 #3
    In this formula v is not squared. but if b is coefficient, how would we win Newton for friction
    coefficient =0.3 (example)
    v=3m/s
    ##f_{air}=-bv=(-0.3)(3m/s)## gives us m/s , but not newton?
     
  5. Feb 24, 2012 #4
    'b' must have units; it is not dimensionless. It's units are such that the product of b and v results in Newtons.
     
  6. Feb 24, 2012 #5

    gneill

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    Replace all the variables that you know the units for with their units (so for example v → m/s, f → N). Solve the resulting expression for the "unknown" variable. In this case solve for b. You should have b on the left and its units on the right.
     
  7. Feb 24, 2012 #6
    In this case that would be ##b=m/t→kg/s * v → m/s = kgm/s^2=N##, so mass times time times velocity gives us air friction but why is that b negative then?
     
  8. Feb 24, 2012 #7

    gneill

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    The negative sign designates the resulting force direction (it opposes the forward velocity v). It is not directly associated with the constant b.

    ## f = bv → N = b (m s^{-1}) → b = N s m^{-1} → b = \left(\frac{kg\;m}{s^2}\right) \frac{s}{m} → \frac{kg}{s}##

    That's the fundamental units of b. But for convenience you can just leave it as Nsm-1, that is, newton seconds per meter. It is then easy to see that you will be getting newtons when you multiply b by a velocity.
     
  9. Feb 25, 2012 #8
    If that would be mass divided by time , then we would have
    ##f=\frac{mv}{t}##, but that's the force formula, even friction is a kind of force, would that be friction's formula as well?
     
  10. Feb 25, 2012 #9
     
    Last edited by a moderator: May 5, 2017
  11. Feb 25, 2012 #10
    What's that b's unit, and what is it equal to?
     
  12. Feb 25, 2012 #11
    My point was that b is the product of air density, frontal area, drag coefficient divided by 2g. The above is generally written as -bv^2. The force due to air drag is a function of velocity squared, not velocity to the first power as it was initially written.
     
  13. Feb 25, 2012 #12

    gneill

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    Don't mix up units and variables. Just because the units of a constant happen to boil down to kg/s, it doesn't mean that it will be equal to the mass of the projectile divided by time!

    Units on constants are a book keeping device that keeps equations balanced and their units in agreement throughout manipulation and calculation. It also lets you convert constants for use in other unit systems.
     
  14. Feb 25, 2012 #13
    By that I understand that none answered my question, what is b , just simply and what is it equal to?
     
  15. Feb 25, 2012 #14

    gneill

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    ## b = -\frac{f_{air}}{v} ## :smile:

    In a given problem you'll either be provided with a value for b or means of calculating it.
     
  16. Feb 25, 2012 #15

    SammyS

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    It looks like there have been some very good posts here, but you don't like the answers.

    Maybr you're looking for something like the following:
    [itex]\displaystyle b=-\frac{f_{\text{air}}}{v}[/itex]​

    The quantity b, depends very much upon the shape and size of the object that is traveling through the air. It also depends upon the air density. Nobody can give a simple number for this if that's what you're asking for.

    (Opps. I see gneill beat me to it, but our answers are somewhat similar !)
     
    Last edited: Feb 25, 2012
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