Calculate Distance for 5ft-lbs Impact w/ 1.18lb Sphere

[mrb112103]

Hello,

I am probably going to suffer some embarassment, once I see the answer to my question. Anwyay, I need to calculate the distance needed to create a 5ft-lbs. impact with a 1.18lb. sphere being dropped vertically onto a surface.

How would I go about calucating the vertical distance needed?

This is for an impact test from UL, where the mass and impact force are defined, but the distance is not.

Please help a newbie. :uhh:

 

[denverdoc]

(well I abhor those english units, but I reckon the principles are the same). what i like even less are words like "impact." since its in ft-lb, I assume they are looking for energy.

there are a few ways to go about this, but usually the most direct is to equate potntial and kinetic energies

wt*h=1/2Mv^2=impact energy

 

[mrb112103]

Still unclear...

So if the Mass of the sphere (M) = 1.18 lbs = .535kg & the impact force = 5 ft. lbs. = 6.8 Joules, then is the velocity = to acceleration by gravity ~ 9.8m/s^2?

So it's .5(.535kg * (9.8 m/s^2)) = 6.8 Joules ??

 

[Dick]

If your you want your final kinetic energy to be E, then use E=mgh. Or like denverdoc said, E=F*d (force times distance). Gravitational force is 1.18 lbs, you want 5 ft-lbs of impact energy, so d=E/F=(5/1.18) ft.