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

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Homework Help Overview

The discussion revolves around the formula for air friction, specifically the term 'b' in the equation ##f_{air}=-bv##. Participants are exploring the meaning of 'b' and its implications in the context of air resistance and drag forces.

Discussion Character

  • Conceptual clarification, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the nature of 'b', questioning its units and how it relates to the force of air friction. Some suggest that 'b' is a coefficient that incorporates factors like drag coefficient and air density, while others express confusion about how to derive its value from the equation.

Discussion Status

The discussion is active, with participants providing insights into the units of 'b' and its dependence on various physical parameters. There is a lack of consensus on a simple definition for 'b', and multiple interpretations of its role in the equation are being explored.

Contextual Notes

Some participants note that the formula is typically represented with velocity squared, leading to further questioning of the initial setup and assumptions regarding the relationship between force and velocity in the context of air friction.

-Physician
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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
 
Last edited by a moderator:
  • #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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