Determining force to crush an object using a pendulum

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The discussion centers on calculating the force exerted by a pendulum impacting an object, specifically using a 4.25-pound pendulum swinging from a height of one foot to crush a part by .025 inches. The agreed formula for force is F = (2*m*g*h)/s, leading to an initial calculation of over 4000 pounds of impact force, which seems implausible given the pendulum's weight. A correction was noted, adjusting the formula to F = (mgh)/s, resulting in a force of 2040 pounds, still higher than expected. Participants express skepticism about the accuracy of these calculations, particularly regarding the omission of time in the analysis. The conversation highlights the complexities of translating theoretical physics into practical applications.
RobStoll
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Afternoon all,

It has been WAY too many years since I have had to apply my schooling in physics to the real world. There are 3 of us arguing over how to solve the following. Let me lay this out:

We have a 4.25 pounds (.1321 slugs or 1.9278 kg) pendulum that is 1 foot (.3048m) long
We are starting it 90° from impact and releasing it to impact the face of our part. We are trying to determine the amount of force that was applied to crush the part .025 inches (.002083 feet or .000635m)

The formula they both eventually agreed on was:
Force = (2*m*g*h)/s and was developed from:

m = mass
h = drop height
s = crush distance
PE = Potential Energt
KE = Kinetic Energy
v = velocity
a = acceleration
g = gravity constant

PE = KE or mgh = 1/2mv^2
v^2 = 2gh
They say we can get the acceleration (decel in this case) by taking the v^2 at the moment of impact from the previous line and dviding by the crush distance so
a=v^2/s

From there they applied the good old fashioned:
F=ma so
F = m * v^2/s or m2gh/s

This results in the following numbers then:
F = (.1321slugs*2*32.17ft/sec^2*1foot)/.002083feet
F = 4080 pounds.

We all agree that it seems unlike that a 4.25 pound weight is impacting the part with 4000+ pounds of impact force when being swung on a 1 foot pendulum.

Is this the correct approach therefore the correct answer? I seem to think not. I think we are missing the element of time in this. I approached it from the Energy to Momentum To force which has a time component.

Any help would be greatly appreciated.

Thank you,
Rob Stoll
Design Release Engineer
Jiffy-tite Co. Inc.
 
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2 things:

1. I accidently left in the 2 so the final formula should have been F=(mgh)/s (therefore 2040 pounds)

2. We determined that this is likely to be the impact force. It is much higher then our original instincts, but it appears to be correct. I still wouldn't mind you thoughts.

Rob Stoll
 
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