# Emulate impact forces using pneumatic piston

I am trying to test a piece of furniture based on a particular code, the code states that i can either drop a bag (with a specific shape) of 57 kg 1.2 in above the seat or I can use other methods as long as they can be shown to be equivalent. My company is interested on using vertical pneumatic cylinders instead since the test needs to be repeated 100 000 times.

My question is: If i manage to get a piston that produces the equivalent of 57kg in force and i adjust its speed to match the speed at contact for a free falling bag (0.77 m/s), would this produce equivalent effects on my test sample?

I am not sure if i am overlooking or neglecting something. Any advice will be appreciated

Thanks

• Ikea usually has these on show using air operated rams... Jul 18 '19 at 18:11
• @SolarMike yes, I noticed Ikea is performing tests based on different standards. At this point i'm still going through those standards as well. Thanks for your kind reply. Cheers! Jul 19 '19 at 15:48

It's not a question of force, its momentum m*v in a non-elastic collision that remains constant. So you have to measure the mass of your pneumatic moving rod and make sure it complies with Newton's law in a way that its momentum is equal to the momentum of the bag at the time it touches the sofa.

Edit

After OP's comment, I added detail subscripts to identify moving parts, to clarify I mean the momentum of the pneumatic punch rod should be equal to the momentum of the bag at its final velocity just before it collides with the sofa.

$$m_{rod}*v_{rod}= m_{bag}*v_{bag}$$.

This will deliver the same impact to the sofa. As you said in your comment kinetic energy will not preserve.

• Thank you for your kind response. I'm not sure the conservation of momentum is very helpful in my case since the bag and sofa will come to a rest after the collision so v2=0. The kinetic energy will be transmitted to the ground and some will be absorbed by the materials. Jul 19 '19 at 16:07
• @JCME, i guess you misunderstood my answer. i edit it now to make it clearer. i meant momentum of ram should be equal to the momentum of the bag. Jul 19 '19 at 16:28
• I see what you were trying to say now. Sorry i misunderstood your initial response. Thank you, this really helps a lot. Jul 20 '19 at 6:02