Air resistance formula

Discussion in 'Physics & Math' started by kingwinner, Oct 6, 2006.

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  1. kingwinner Registered Senior Member

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    I encountered a question involving air resistance and it begins by giving a formula:
    "The force of air resistance is roughly proportional to the square of an object's speed and is directed opposite to the velocity: F(air) = -cv^2, where c is a constant that depends largely on the shape of the falling body..........."

    For F(air) = -cv^2 , is there a FLAW in the structure of the formula itself? Is there any valid reason to introduce the negative sign?

    I used the given formula to find the acceleration of a 80kg skydiver falling at 25m/s (with c=0.5 Ns^2/m^2) to be -13.7m/s^2 which is crazy...
    -cv^2 -mg = ma
    Substituing the given values gives a=-13.7m/s^2

    c>0 and v^2 >0, so the formula automatically negatives the air resistance and this means air resistance always point downward...which is wrong

    If the formula is trying to say air resistance is directed opposite to velocity, this is nonsense as well because (-30m/s)^2 and (30m/s)^2 gives the same thing and the negative sign in front of cv^2 does nothing to show that they have opposite directions.

    What should be the correct way of stating this formula?
    | F(air) | = cv^2 ??
    How about in vector form?
     
    Last edited: Oct 6, 2006
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  3. geodesic "The truth shall make ye fret" Registered Senior Member

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    I'll give you the explanation that most often crops up when explaining seemingly arbitrary signs/ indices etc. in physics - it's for historical reasons (probably). Basically, use your common sense when doing problems like this and you should be fine, if you're still not sure, just draw a picture (also lets examiners know what you're thinking)
     
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  5. kingwinner Registered Senior Member

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    Should I use the magniutde of air resistance | F(air) | = cv^2 to do my calculations and adjust the sign manually?
     
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  7. 1100f Banned Registered Senior Member

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    The minus sign here means that the drag force is opposite to the velocity of the object, what you should do is write:

    F = -(cv^2)e<sub>v</sub>, where e<sub>v</sub> is a unit vector in the direction of the velocity.
    Since e<sub>v</sub> = v/v , you can also write:
    F = -cv.v

    PS boldface fonts refer to vectors, and the same without the boldface refers to the magnitude.
     
  8. James R Just this guy, you know? Staff Member

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    kingwinner:

    In vector form, the equation for the force is:

    ma = mg - cvv

    where the bold letters are vectors.

    In converting this to a one-dimensional problem, you need to define a directional axis to call the positive direction.

    For example, suppose you choose upwards as the positive direction, and you're talking about a falling object. Then:

    g is negative (since gravity acts downwards)
    v is negative (as the object's velocity is downwards)

    So, the equation is:

    ma = m(-g) - cv(-v)

    where g means |g| and v is |v|.

    So:

    ma = -mg + cv^2

    The acceleration turns out to be negative (downwards) in this case.

    The choice of which direction to call positive is totally arbitrary, so let's look at the opposite case:

    For example, suppose you choose downwards as the positive direction, and you're talking about a falling object. Then:

    g is positive (since gravity acts downwards)
    v is positive (as the object's velocity is downwards)

    So, the equation is:

    ma = m(g) - cv(v)

    where again g means |g| and v is |v|.

    So:

    ma = mg - cv^2

    In this case the acceleration turns out to be positive, which is again downwards.

    So, the physical outcome is the same. You just need to be careful to define one direction as positive and stick to it.
     
  9. hosaf Registered Member

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    Hi all. I try to calculate air throw distance for Axial Fan.

    Do anybody know anything about it ?
     
  10. prometheus viva voce! Moderator

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    Haven't we already had this discussion here?
     
  11. prometheus viva voce! Moderator

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    In fact, since this is an old thread I will close it. Please continue the discussion in the thread linked to in my above post.
     
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