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Registered Member #101
Joined: Thu Feb 09 2006, 08:12PM
Location: Wisconsin
Posts: 41
Ok, so i got a homework assignment today. I need to find the friction force in a syringe(rubber against plastic). The plunger is pushed with the end of the syring sealed off, the internal pressure pushes back on the plunger. The initial volume is 6cc, then, when let go, it moves back to (an average of) 5.24cc. I calculated the change in volume and converted it to in^3, that is .046in^3
The plunger diameter is 1/2"; the change in volume is 0.046in^3 (6cc to 5.24cc); atmospheric pressure is 14.7psi; inital volume is 0.366in^3 (6cc). apparently that is all I ineed to know. The only friction formula that i knon is F=uN (u=coefficient of friction, N is force normal) how do find force normal in a cylinder? i think the coef. of fric. of rubber is between .5 and .85
its probably something really simple im looking over
Registered Member #193
Joined: Fri Feb 17 2006, 07:04AM
Location: sheffield
Posts: 1022
The gas in the syringe has been compressed to a smaller volume so its pressure will have risen. That pressure is pushing the piston out.Friction is holding it back so the friction force must balance the product of the overpressure and the area.
Registered Member #56
Joined: Thu Feb 09 2006, 05:02AM
Location: Southern Califorina, USA
Posts: 2445
I haven't been taught anything even remotely like that, but here is how I would do it...
force exerted by piston=force exerted by friction force on the piston = pressure* area, so by boyle's law we know P1*V1=P2V2 or 14.7*6=P2*5.24 P2=16.83psia, so deltaP=16.83-14.7=2.13psi. The area is A=3.14*(D/2)^2 A=0.196 in^2, so force (lbs) = 2.13psi*.196in^2=0.418lbs=1.85N
From there I am going in the dark, but I am assuming that you want to solve for F, so F=coefficient of friction of the syringe and N the force exerted on the plunger (in whatever unit you professor wants the F value in, lbs/N/whatever)
edit, where did you get 3lbs?
And looking at some articles on the internet, it looks like you can say that the answer is the .418lbs that I calculated (or whatever you got) as teh answer Normally it is equal to the weight of the oject? I don't know
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Engineering Problem
Normally it is equal to the weight of the oject? I don't know
