mperegrinefalcon
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I did a test on with some bulkheads and 8-32 threaded rod where I had a mock Av-bay suspended from a beam, and the other eyebolt tied to a 25.4 lb (11.5Kg) weight. I dropped the weight 16.5" (0.4191m) and it held, but the eyebolts pulled open. The 8-32 thread was perfectly fine though. My question is how might I go about calculating the amount of force exerted on the eyebolts by the weight being stopped. I have only taken high school physics and it has been a while, so please be patient.
I used the equation v=(2gh)^1/2 where G=-9.81 m/s^2 and h=.4191 and I got a Velocity of 2.87m/s.
How would I figure out how much time it would take to slow down?
How would I figure out the distance required to slow down?
I calculated the average velocity of 0m/s and 2.87m/s and used that to find .00885 seconds for the time it took to slow down, assuming it slowed down in 0.5"
This would equal 324.3 m/s^2, or 33.07 gees. This would then simply be 33.07x25.4=839.9 lbs of force.
Would this be correct?
Thanks for any help
I used the equation v=(2gh)^1/2 where G=-9.81 m/s^2 and h=.4191 and I got a Velocity of 2.87m/s.
How would I figure out how much time it would take to slow down?
How would I figure out the distance required to slow down?
I calculated the average velocity of 0m/s and 2.87m/s and used that to find .00885 seconds for the time it took to slow down, assuming it slowed down in 0.5"
This would equal 324.3 m/s^2, or 33.07 gees. This would then simply be 33.07x25.4=839.9 lbs of force.
Would this be correct?
Thanks for any help