air resistance, dimensional analysis confusion
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! 
Re: air resistance, dimensional analysis confusion
Quote:

Re: air resistance, dimensional analysis confusion
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! 
Re: 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. 
Re: air resistance, dimensional analysis confusion
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

Re: air resistance, dimensional analysis confusion
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.

Re: air resistance, dimensional analysis confusion
Quote:

Re: air resistance, dimensional analysis confusion
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.

All times are GMT 5. The time now is 01:01 AM. 
Powered by vBulletin Copyright ©2000  2014, Jelsoft Enterprises Ltd.
© 2014 Physics Forums