Air Friction Help: What Is 'b' in ##f_{air}=-bv##?

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In the air friction formula f_{air} = -bv, 'b' is a coefficient that incorporates factors such as drag coefficient, air density, and frontal area, and it has units of N·s/m, which ensures the product of b and velocity results in force measured in Newtons. The negative sign indicates that the force opposes the direction of velocity, not that 'b' itself is negative. The formula typically uses velocity squared (v^2) for air resistance, but the discussion focuses on the linear version. The value of 'b' can vary based on the object's shape and size, and it is not a fixed number. Understanding 'b' requires context from specific problems, as it is derived from the conditions of the scenario.
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Homework Statement


In formula of air friction ##f_{air}=-bv##
v is velocity but what's b?
 
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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.
 
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?
 
'b' must have units; it is not dimensionless. It's units are such that the product of b and v results in Newtons.
 
-Physician said:
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?

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.
 
gneill said:
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.

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?
 
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.
 
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?
 
LawrenceC said:
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. [PLAIN]http://www.vvio.info/jpg1[/QUOTE]

In your formula, b would be a coefficient that includes things like drag coefficient
 
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  • #10
shiliangtu said:
In your formula, b would be a coefficient that includes things like drag coefficient

What's that b's unit, and what is it equal to?
 
  • #11
shiliangtu said:
In your formula, b would be a coefficient that includes things like drag coefficient

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.
 
  • #12
-Physician said:
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?

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 let's you convert constants for use in other unit systems.
 
  • #13
By that I understand that none answered my question, what is b , just simply and what is it equal to?
 
  • #14
-Physician said:
By that I understand that none answered my question, what is b , just simply and what is it equal to?

## 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.
 
  • #15
-Physician said:
By that I understand that none answered my question, what is b , just simply and what is it equal to?
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:
\displaystyle b=-\frac{f_{\text{air}}}{v}​

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 !)
 
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