
#1
Jan1104, 03:51 PM

P: 2,292

This problem is bugging me, and I know that those far more knowlegable than I can help lift me out of my ignorance:
I have a gear on a spindle which has a 10:1 ratio with respect to the spindle(armature) When I spin the armature at 1 mph, the outside diameter of the gear spins at 10 mph, but with a 10 times reduction in force. The gear and armature has "teeth" I now connect a second gear, such to where the first gear turns the armature of the second, identical gear. The second gears' outer rotation is now 100 times that of the initial input. 100 times the speed, 100 times less force. If I continue this ganged arrangement, a point in time will occur when the continually reduced force is not sufficient to even turn the next gear. What happens to this force? Is it dissipated? 



#2
Jan1104, 04:56 PM

Sci Advisor
PF Gold
P: 478

I'd like to help you out, but I don't quite follow. Are you saying that you have a driven gear (pinion) that begins a gear train and that each successive gear in the train gives a 10x reduction in torque? With such a gear train, you will quickly run into a problem with the existence of a small enough gear: say the pinion has 100 teeth, then the next gear has 10 teeth, then the next gear has only 1 tooth and we're in trouble.
Or maybe I'm misunderstanding and you're talking about some kind of planetary gear arrangement where you have a sun gear with, say 100 teeth, and a number of planet gears around it, each with 10 teeth. If that's what you're talking about, then the gear ratios will not multiply; each planet will have the same 10:1 reduction. Note that I am not talking about your typical epicyclic gear train here, where you also have an external ring gear as the output gear; the axes for these planets is fixed in space (at least I'm guessing that might be what you mean). Anyway, if you have a gear train that will not move, the likely culprit is friction. (your input force will be eqilibrated by static friction in the train) Since friction (and inertia, for that matter) reflects back through a gear train with a factor of the gear ratio squared, it will actually be easier to drive the big gear than the small gear. I hope somehting in there is intelligible/helpful. 



#3
Jan1104, 06:46 PM

Mentor
P: 22,008





#4
Jan1904, 02:05 PM

HW Helper
P: 2,327

Gear problemEventually, the last gear in the chain will be "heavier" than the reduced force supplied. This "weight" will most likely be do to the friction between the teeth of two gears in contact. The gears will eventually stop, and remain stopped, because the static frictional force of the teeth exceeds the provided force that has been reduced. 


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