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## Main Question or Discussion Point

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.

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.