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air resistance, dimensional analysis confusion

 
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Dec6-12, 01:46 PM   #1
 

air resistance, dimensional analysis confusion


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!
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Dec6-12, 01:58 PM   #2
 
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Quote by pjordan View Post
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!
Does k have units?
Dec6-12, 02:12 PM   #3
 
good point, I had assumed k to be unit-less, 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!
Dec6-12, 03:01 PM   #4
cjl
 

air resistance, dimensional analysis confusion


Another problem: your proportionality is wrong. Air resistance follows a v2 proportionality, so in reality, it should be:

dv/dt = g - kv2/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.
Dec6-12, 03:21 PM   #5
 
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
Dec6-12, 03:55 PM   #6
K^2
 
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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.
Dec6-12, 06:22 PM   #7
cjl
 
Quote by pjordan View Post
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
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 v2 proportionality - the relatively low viscosity of air, and high velocity objects falling through air attain make the v2 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).
Dec6-12, 06:35 PM   #8
K^2
 
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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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