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Duration of Contact 
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#1
Jun406, 02:02 PM

P: 13

What is the best way to estimate or measure the duration of a contact in a collision?
I have a stick applying a 1.5 N force onto a stationary plastic box. Assuming that the plastic is incompressible, how long will the time have elapsed before the stick with 1.5 N no longer has an effect on the box? I might be leaving out some important information on materials, etc.. Please make the necessary assumptions. Thanks, Mike 


#2
Jun406, 02:43 PM

HW Helper
P: 7,049

The issue is that it's the compression or deformation that determines contact time. The stress/strain or force/compression ratio are part of the factors it takes to determine how long it takes for the two objects in a collision to undergo acceleration until they reach their new velocities.



#3
Jun706, 05:58 PM

P: 20

Hi,
It's a bit fuzzy what you mean by a 1.5N force on a stationary box in the context of a collision. If it's a 1.5N force, that means it will have an effect on your box as long as the force between them is 1.5N. When the stick stops having an effect, the force has gone to 0, but that doesn't sound like a collision to me. That sounds like someone pushing a box with a stick. I think you're talking more about whacking a box with a stick right? If so, this is not exactly a simple problem, but here is the simplist way to go about it that I can think of. Since we are assuming the stick is non compressable we should treat the box as a simple spring obeying Hooke's law. F = kx (force = spring constont times displacement, with minus sign) This means that as you push on the box, it will push back. If you can somehow measure how much force makes the box compress how much, you can come up with a value for k. you can then consider a 'mass on a spring' system and take half the peroid to be your answer. Period = 2*Pi*sqrt[m/k] where m is actually the reduced mass of the Stick/box system m = (mass of stick)*(mass of box)/[(mass of stick) + (mass of box)] half the period is your answer unless you whack your box so hard that it doesn't obey Hooke's law anymore. This explanation is really what we call a first order approximation, but I bet it will get you in the ballpark. I hope that helps. If you want help with any particular part of what I wrote, let us know. 


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