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Value of drag forces in air?

  1. Jun 2, 2010 #1
    How much force does air resistance have on a freefalling object on Earth? No equations, please.
     
  2. jcsd
  3. Jun 2, 2010 #2
    It depends of the object's shape,size and the velocity it's traveling.
     
  4. Jun 2, 2010 #3
    Once the object reaches its terminal velocity (a constant rate of descent with no further acceleration) the drag force is exactly equal to its weight. Before that the drag force starts at zero and gradually increases to that value of the weight. Without maths there isn't much more to say.
     
  5. Jun 2, 2010 #4

    PhanthomJay

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    If the object is free falling vertically downward, if that's what you mean, the force of air resistance on that object is less than or equal to the objects weight (it is equal to the weight at terminal velocity, but less than its weight prior to that point , varying non-linearly from zero when it is first dropped, to its weight when (and if) it reaches terminal velocity).
     
  6. Jun 2, 2010 #5

    russ_watters

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    Slight definitions problem here, so I'll give two answers....

    1. By definition, an object in freefall is not experiencing any drag force at all. This is the case at the instant an object is dropped in the atmosphere.
    2. An object falling at terminal velocity experiences a drag force equal to its weight.

    PhanthomJay explained what happens in between.
     
  7. Jun 2, 2010 #6

    PhanthomJay

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    Point well taken.
     
  8. Jun 3, 2010 #7
    From Wiki: "in nontechnical usage, falling through an atmosphere without a deployed parachute or lifting device is also referred to as free fall."

    The request for "no equations" hints at a non-technical interpretation.
     
  9. Jun 3, 2010 #8

    PhanthomJay

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    Point well taken (I like to agree with everybody).:wink:
     
  10. Jun 3, 2010 #9
    I already understand air resistance. What I wanted, though, was the amount of drag force applied to a freefalling object at sea level (assuming a pressure of 10013 mb)
     
  11. Jun 3, 2010 #10
    As a suggesttion:

    assume your body is smth like a sphere, assume its velocity, prescribe some value for the viscosity and calculate the Reynolds number. Then go here and choose an estimate for the drag coefficient http://en.wikipedia.org/wiki/Drag_coefficient. The force is Cd*density*velocity^2*cross section
     
  12. Jun 3, 2010 #11
    Given that 'free falling' also includes experiencing air drag by the definition given by Stonebridge, then something that weighs 100 pounds will experience a 100 pound force due to air resistance at terminal velicity.
     
  13. Jun 3, 2010 #12

    diazona

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    The point is, it varies. It depends on the shape and size of the object and how fast it's moving (relative to the air, of course). Anything more we could say about it is best expressed as an equation.
     
  14. Jun 4, 2010 #13
    Would you say this is true if your object is not accelerating?
     
  15. Jun 5, 2010 #14
    The best way for you would be to check the equation. I havent learned this in school but last time I checked Wikipedia the equation seemed quite easy.
    That way you could create your own conditions, such as the shape of the body, density of air which you can get from your prefered air pressure, the speed of the body at an instant and so on. Thats the only way you can get it because you asked for a number...
     
  16. Jun 5, 2010 #15

    diazona

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    You mean if it's falling at terminal velocity? In that case, no - the air resistance would be simply equal to the weight. But as I understand the OP's question, it wasn't asking about that special case.
     
  17. Jun 5, 2010 #16

    diazona

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    I'd agree with that... the equation is

    force = 1/2 × (density of air) × (cross-sectional area of object) × (drag coefficient) × (speed²)

    which is written in mathematical notation as
    [tex]F = \frac{1}{2}\rho A C v^2[/tex]
    It is reasonably simple.
     
  18. Jun 5, 2010 #17
    I was writing the same stuff several posts above. However, it is NOT reasonably simple, because you have to know the drag coefficient. To get it you have to solve the full Navier-Stokes equations. So far nobody managed to do it in general way for a body of general shape, espesially for the turbulent regime. Also, as you might hear nobody proved that the solution exists in fact and standard stuff like uniqness, etc.
     
  19. Jun 5, 2010 #18

    diazona

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    Oh yeah, I missed that, sorry.

    But the equation is reasonably simple - all it involves is multiplying some numbers together. If one of the parameters in it happens to be hard to calculate, that doesn't make the equation complicated. In practice you usually look up the drag coefficient or experimentally measure it.
     
  20. Jun 5, 2010 #19
    Thanks to you both. These are by far the best answers. Oh and Diazona, the first time I looked at the coefficient of drag equation, I didn't know what the heck it meant by "reference area."
     
  21. Jun 5, 2010 #20
    Indeeed. This equation is just a definition for the drag coefficient. Really. nothing more. It is used in engeering to estimate drag of real bodies like lets say wings if you measured or calculated the drag coefficient for something more simple like a sphere or another "benchmark" wing.

    I am starting to think that most of the replies here that drag force is equal to the gravitation force when the body no longer accelerates is what the author of the topic needs. Looks like he is asking a question from mechanics, not more general one from hydrodynamics.
     
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