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Posted: 4/19/2002 12:22:49 PM EDT
[#1]
Seems my Drill Instructor mentioned the bolt spring on the M16A2 has 40 pounds of tension on it. What is a "spring co-efficient"?
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Posted: 4/19/2002 12:36:24 PM EDT
[#2]
He means "spring constant", or "force constant", "k", I think.
Also, I'm not sure why you've chosen "pressure" units for the other problem. Why not force? |
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Posted: 4/19/2002 12:38:16 PM EDT
[#3]
Thanks for the reply...I think it should help a bit.
In short, the spring co-efficient essentially means the stiffness of a spring (measured in pounds per inch or N/m). [edited for grammar] |
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Posted: 4/19/2002 12:46:50 PM EDT
[#4]
"k" (lower case) is the "spring constant", not coefficient.
"k" is the "strength" of the spring. The greater the spring constant the more force is needed to compress or stretch the spring a given distance. F=-kx |
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Posted: 4/19/2002 12:50:42 PM EDT
[#5]
Yep, Hooke's Law I:
There's a direct relationship between the force applied, and the displacement, so linear: y=mx+b, in the case of the spring: F=kx+(Y intercept is zero, right?) |
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Posted: 4/19/2002 1:52:24 PM EDT
[#6]
Please, everybody submit their resumes, NASA has openings at the Kenndey Space Center [:P]
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Posted: 4/19/2002 2:14:11 PM EDT
[#7]
Quoted: Yep, Hooke's Law I: There's a direct relationship between the force applied, and the displacement, so linear: y=mx+b, in the case of the spring: F=kx+(Y intercept is zero, right?) Yes, No force is applied when x=0. |
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Posted: 4/19/2002 6:22:17 PM EDT
[#8]
Quoted: Please, everybody submit their resumes, NASA has openings at the Kenndey Space Center [:P] You might want to try out at The Second City if you haven't already. I recognize your potential. [:O)] |
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Posted: 4/19/2002 6:31:29 PM EDT
[#9]
We just found the spring constant in physics by hanging any known weight from the spring and measuring it's displacement. That easily gives the spring constant. The force on the charging handle should simply be the necessary spring force at that displacement for full retraction of the charging handle. I can't imagine that the friction of the carrier to the reciever would be to high. I believe that it requires the most force at full retraction, more than is required to cock the hammer.
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Posted: 4/20/2002 9:11:43 AM EDT
[#10]
Quoted: We just found the spring constant in physics by hanging any known weight from the spring and measuring it's displacement. That easily gives the spring constant. Yep, you use the acceleration of gavity constant (a) with a known mass (m) to calulate the force (F) then you measure the delta or distance the spring "stretched" (x). Now you have all the variables to calculate "k". F=ma F=kx |
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