Hi. Consider the basic eq for a falling body with air resistance dv/dt=g-kv/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 anaysis--the 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!
good point, I had assumed k to be unit-less, but its units are kg/sec (http://oregonstate.edu/instruct/mth252h/Bogley/w02/resist.html) so F=kv would have the same dimension as F=ma. thanks!
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.
generally it is given as proportional to v or v^2--the quadratic relationship is usually for larger objects. Most introductory material on diff eq use v. thanks
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.
Introductory material uses v not because it is correct, but because it makes the differential equation a lot easier. Even for small objects, air resistance tends to have a v^{2} proportionality - the relatively low viscosity of air, and high velocity objects falling through air attain make the v^{2} relationship correct for nearly all objects in air. A linear proportionality (implying viscous-dominated drag rather than inertial) tends to happen more commonly in other fluids, especially highly viscous ones (for example, dropping a marble through corn syrup).
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.