3
$\begingroup$

I have an assignment, where I’m required to derive the formula for air resistance for a falling object.

$$ F = \frac{1}{2}CAdv^2 $$

where $C$ is drag coefficient, $A$ is the cross-sectional area, $d$ is the density of the air and $v$ is the speed, through dimensional analysis.

In the process, I’m required to explain why the air resistance on the object is independent of the mass, $m$, and the position, $x$, of the object. However, I can’t really explain why that is? How do I clearly explain and prove, that those things do not have an effect on the air resistance?

Edit
Now I have a pretty good understanding of why air resistance is independent of mass, since it's actually quite straight forward. However, something that is not so straight forward, is explaining why air resistance is also independent of acceleration, $a$. How would you convince someone, that this is the case, because I don't see a clear way of doing that?

Improve this question
$\endgroup$
  • 2
    $\begingroup$ Essentially a dupe of physics.stackexchange.com/q/197106/25301 $\endgroup$ – Kyle Kanos Dec 7 '18 at 23:50
  • $\begingroup$ Carl, you should use $\rho$ for density, not $d$. $\endgroup$ – David White Dec 8 '18 at 1:01
  • $\begingroup$ It's weird something depends on acceleration. Anyways, think this way: speeding from $1m/s$ to $2m/s$ in $1s$ is the same acceleration as in $80m/s$ to $81m/s$ in the same interval. Our experience tells us that the second case has more resistance. That's because it depends on $v$, not $a$. $\endgroup$ – FGSUZ Dec 10 '18 at 11:51
  • $\begingroup$ Drag (say as measured in wind tunnels) is a time-average for steady states. There could be transient effects related to acceleration. $\endgroup$ – Bert Barrois Dec 10 '18 at 14:51
1
$\begingroup$

Drag or air resistance doesn't depend on acceleration, only on speed. Because that resistance is a result of particles being in the way that now have to move out of the way.

Compare with you running through a crowd of people at prime-time Christmas shopping at the mall. If you run slow, it is semi-tough to get through. If you run fast, it is tougher to get through. How quickly you change from slow to fast speed, isn't relevant - only for how long time you are at each individual speed-level.

Air resistance can be thought of in the same way. A bunch of particles are in the way. They have to be moved away. The "heavier" they are, the tougher (density). The more of them you need to move, the tougher (size or frontal/perpendicular area). And the faster they must be moved, the tougher (speed). The constant $C$ encompasses all other constant influences.

And that is it. The particles have no other dependence on the object that is being resisted - they do not attract as gravitational or magnetic forces do or hold up anything by balancing out the weight like normal forces do. There is no other interactivity with the bulk. The bulk doesn't matter. The object could be hollow and have the same air resistance; a bathing ball and a bowling ball of equal shapes have the same air resistances.

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.