air resistance, dimensional analysis confusionby pjordan Tags: analysis, confusion, dimensional, resistance 

#1
Dec612, 01:46 PM

P: 3

Hi. Consider the basic eq for a falling body with air resistance
dv/dt=gkv/m I dont understand air resistance as a force, since it seems irreconcilable to the force equation F=ma. How is a force a function of velocity? I am also not sure how this equation makes sense in terms of dimensional anaysisthe right side is m/s^2, the left m/s^2+(m/s)/kg. I am apparently the only one troubled by this, as extensive googling has yeilded nothing. Thanks! 



#3
Dec612, 02:12 PM

P: 3

good point, I had assumed k to be unitless, but its units are kg/sec (http://oregonstate.edu/instruct/mth2...02/resist.html)
so F=kv would have the same dimension as F=ma. thanks! 



#4
Dec612, 03:01 PM

P: 975

air resistance, dimensional analysis confusion
Another problem: your proportionality is wrong. Air resistance follows a v^{2} proportionality, so in reality, it should be:
dv/dt = g  kv^{2}/m, in which k = ρ/2*Cd*A, where ρ is the density of the fluid, Cd is the drag coefficient (unitless), and A is the reference area. 



#5
Dec612, 03:21 PM

P: 3

generally it is given as proportional to v or v^2the quadratic relationship is usually for larger objects. Most introductory material on diff eq use v. thanks




#6
Dec612, 03:55 PM

Sci Advisor
P: 2,470

Precisely. Drag equation can be different under different conditions. Quadratic drag is more common in practical situations, but slow motion through viscous medium will often produce linear drag.




#7
Dec612, 06:22 PM

P: 975





#8
Dec612, 06:35 PM

Sci Advisor
P: 2,470

I missed the bit about it being specific to drag in air. Yes, with air, you are unlikely to see linear drag outside of Millikan Oil Drop, or similar setup.



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