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Confused about direction of air resistance (linear and quadratic)

  1. Jan 3, 2013 #1
    1. The problem statement, all variables and given/known data

    Here I have drawn for the following scenarios:

    Linear Resistance FR = -k[itex]\vec{v}[/itex]

    a)Free-fall starting from rest, downwards as positive.

    2lsy91k.png

    Now it gets trickier....


    b)Free-fall starting from rest, upwards as positive.

    My reasoning is as such:
    Since the resistive force always points opposite to the velocity vector, in this case the velocity is pointing downwards since the object is moving downwards:
    v = -vi. Hence F = -bv = bv i
    The force points upwards.
    2m4ubdk.png

    Quadratic Resistance

    F = -bv2 i or bv2 i since it is v2 is not a vector...How do i determine its direction using vector methods?

    a) Free-fall starting from rest, downwards as positive.

    (Common sense) Opposes motion:

    hwkaid.png


    b) Free-fall starting from rest, upwards as positive.

    The object is moving downwards, hence velocity is negative. To oppose velocity, resistive force points downwards.

    ohtiqt.png


    However, I got all my directions wrong when taking the upward as positive! I don't know what is wrong with my reasoning!
     
  2. jcsd
  3. Jan 3, 2013 #2

    tiny-tim

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    hi unscientific! :smile:
    F = -b|v|##\hat{\mathbf{v}}##

    F = -b|v|2##\hat{\mathbf{v}}##

    where ##\hat{\mathbf{v}}## is the unit vector in the direction of v :wink:
     
  4. Jan 4, 2013 #3
    Hello, thanks for the reply! So which direction does ##\hat{\mathbf{v}}## point? Is it pointing in the direction that the body is moving?
     
  5. Jan 4, 2013 #4

    tiny-tim

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    yes, the "hat" means that it's the vector of unit length in the same direction :smile:

    (##\hat{\mathbf{v}}## = v/|v|)
     
  6. Jan 5, 2013 #5
    However, by using that reasoning I got all my directions wrong when i take upwards as positive...
     
  7. Jan 5, 2013 #6

    tiny-tim

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    i don't see how :confused:

    |v| is always positive,

    so ##\hat{\mathbf{v}}## must always be in the same direction as v :smile:
     
  8. Jan 5, 2013 #7
    Linear Resistance
    For b) the force should be pointing downwards instead, changing direction when the positive direction is changed.

    Quadratic Resistance
    For d) the force should be pointing upwards (irregardless of assigning positive direction.


    I've tried googling for the answer but to no avail...
     
  9. Jan 5, 2013 #8

    HallsofIvy

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    Not only "unscientific" but nonsense! No matter whether air resistance is proportional to the velocity or square of the velocity or any other function of the velocity, resistance to motion is always directed opposite to the motion. If an object is falling downward through air, its velocity vector is pointing downward so the resistance force is pointing upward. Again, that has nothing to do with whether it is linear or quadratic.
     
  10. Jan 5, 2013 #9

    tiny-tim

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    sorry, i don't understand this at all :confused:

    if you change the coordinate system so that you replace v by -v,

    then ##\hat{\mathbf{v}}## = v/|v| is replaced by -v/|-v| = ##-\hat{\mathbf{v}}## …

    always in the same direction! :wink:
     
  11. Jan 6, 2013 #10
    Ok reading my lecture notes again carefully I have come up with a new understanding:

    The free-body diagram does not dictate the EOM, but rather the EOM leads to the free-body diagram:

    Linear Resistance

    Suppose there is a force F (in this case, weight) that accelerates the object and a resistive force that is proportional to speed. If F is chosen to be in the positive direction (downwards), here is the equation of motion:

    a3l0ls.png

    The resulting free-body diagram is as such:

    k3xgeu.png

    b) Now, taking F in the negative direction (upwards as positive) , now F is negative. However, the resistive force will still be -bv [itex]\tilde{i}[/itex] as it always opposes motion; x regardless of the direction of x (upwards or downwards).

    21435za.png


    Quadratic Resistance

    Now taking F in the positive direction, the EOM is given:

    2rolums.png




    b)Now taking F in the negative direction, the EOM is given as this. But I have no idea why... I remember my prof mentioning something about it being independent of velocity or something...



    2uf8rir.png
     
  12. Jan 6, 2013 #11

    HallsofIvy

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    What do you mean by "taking F in the negative direction". Generally an external force, F, is given, you can't just "take" a direction. You can, of course, set up your coordinate system so that F is pointing in one or the other direction, but then all quantities will change sign- so you would have a "-" on the acceleration term also.

    "I remember my prof mentioning something about it being independent of velocity or something..."
    About what being independent of velocity? Certainly not the acceleration since you have a "[itex]\dot{x}[/itex]", which is "velocity", in the equation.
     
  13. Jan 7, 2013 #12
    Alright here's what's written in my notes:

    "Although the linear case does not occur so often in reality, there are some slightly tricky issues to do with signs and directions. A typical equation of motion for this case is

    ma = F - bv

    where b is a constant. Remember that the choice of +ive direction for x determines the sign of F and other quantities. So if we choose F in the +ive x direction, F > 0 and vlimit = F/b.

    If however we choose F in the -ive x direction then F < 0 but the resistive force will still be given by -bv (why?). The terminal velocity can then be written as vlimit = -|F|/b to emphasize the fact that the object is moving along the negative x-axis.

    Consider now an object of mass m moving under constant gravity in the vertical plane subject to a linear resistance, initially at rest at height h. The coordinate y is taken positive upwards and the equation of motion is

    ma = -mg -bv

    [It is tempting here to write the RHS of the above as -mg + bv, but this is wrong since the resistive force is given by -av regardless of the choice of direction of coordinate axis.]


    "



    But they didn't explain the case for quadratic resistance when upwards is positive...
     
  14. Jan 11, 2013 #13
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