Confused about direction of air resistance (linear and quadratic)

In summary, the conversation discusses the direction of resistive forces in different scenarios, with the main focus being on linear and quadratic resistance. It is determined that resistive forces always oppose motion, regardless of the direction in which the force is applied. The equations of motion are used to determine the direction of the resistive force, with the velocity vector and unit vector being key factors. It is also mentioned that the resistive force is independent of velocity.
  • #1
unscientific
1,734
13

Homework Statement



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!
 
Physics news on Phys.org
  • #2
hi unscientific! :smile:
unscientific said:
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.

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?

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:
 
  • #3
tiny-tim said:
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:

Hello, thanks for the reply! So which direction does ##\hat{\mathbf{v}}## point? Is it pointing in the direction that the body is moving?
 
  • #4
yes, the "hat" means that it's the vector of unit length in the same direction :smile:

(##\hat{\mathbf{v}}## = v/|v|)
 
  • #5
tiny-tim said:
yes, the "hat" means that it's the vector of unit length in the same direction :smile:

(##\hat{\mathbf{v}}## = v/|v|)

However, by using that reasoning I got all my directions wrong when i take upwards as positive...
 
  • #6
unscientific said:
However, by using that reasoning I got all my directions wrong when i take upwards as positive...

i don't see how :confused:

|v| is always positive,

so ##\hat{\mathbf{v}}## must always be in the same direction as v :smile:
 
  • #7
tiny-tim said:
i don't see how :confused:

|v| is always positive,

so ##\hat{\mathbf{v}}## must always be in the same direction as v :smile:

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...
 
  • #8
unscientific said:
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...
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.
 
  • #9
unscientific said:
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.

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:
 
  • #10
HallsofIvy said:
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.

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
 
  • #11
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.
 
  • #12
HallsofIvy said:
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.

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...
 
  • #13
bump??
 

1. What is the difference between linear and quadratic air resistance?

Linear air resistance refers to the force exerted by air on an object moving through it, which is directly proportional to the object's velocity. Quadratic air resistance, on the other hand, takes into account the object's surface area and shape, making the force exerted by air on the object non-linear.

2. How does air resistance affect the motion of an object?

Air resistance acts in the opposite direction to an object's motion, slowing it down. As an object moves faster, the force of air resistance increases, eventually reaching a point where it is equal to the object's weight, resulting in a constant velocity known as terminal velocity.

3. How can we calculate air resistance?

The formula for air resistance depends on the specific situation and is often quite complex. In general, it involves taking into account the object's velocity, surface area, and shape, as well as the properties of the air it is moving through, such as density and viscosity.

4. How does air resistance affect different objects?

Air resistance affects different objects differently, depending on their mass, surface area, and shape. Generally, larger and more aerodynamic objects will experience less air resistance compared to smaller and less streamlined objects. However, the density and viscosity of the air also play a role.

5. How can we minimize the effects of air resistance?

The effects of air resistance can be minimized by reducing an object's surface area and making it more aerodynamic. This can be achieved through streamlining the shape, adding a pointed nose, and using materials that reduce drag, such as smooth and polished surfaces. Additionally, reducing an object's speed can also decrease the force of air resistance acting on it.

Similar threads

  • Introductory Physics Homework Help
Replies
13
Views
1K
  • Mechanics
Replies
3
Views
854
  • Introductory Physics Homework Help
Replies
15
Views
346
  • Calculus and Beyond Homework Help
Replies
1
Views
1K
  • Advanced Physics Homework Help
Replies
10
Views
1K
  • Introductory Physics Homework Help
Replies
3
Views
236
  • Introductory Physics Homework Help
2
Replies
39
Views
2K
  • Introductory Physics Homework Help
Replies
18
Views
958
  • Introductory Physics Homework Help
Replies
5
Views
807
  • Introductory Physics Homework Help
Replies
8
Views
1K
Back
Top