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Determining the 'Instantaneous Force' during impact

  1. Dec 19, 2011 #1
    I have been doing some calculations on a falling object and its impact force. We use heavy weights on our belt conveyors to maintain tension on the belts. One of our calculations relates to the amount of force placed on the structure that one of these weights lands on. I have the calculations and the answers, but as a matter of curiosity, I started looking at the formulas a little closer. The force of impact is Favg. Since this force is the change in Kinetic energy divided by the distance after impact, it seems to me that this is the average force over that distance. Is there a way to graph this force during deceleration? How can the instantaneous force (for lack of a better term) be found? Is there a formula that I am overlooking? Not having to work very often with the dynamics of the structures I engineer, my physics in this area is very rusty, and being a pretty new engineer, I like to dig a little deeper into the things I am doing. Can anybody help? Thanks
     
  2. jcsd
  3. Dec 19, 2011 #2
    Your logic is correct since work done = gain in KE = Force x distance
    When a weight lands on the conveyor belt there is slipping between the weight and the belt.
    This friction force is what acts on the weight and speeds it up to the speed of the conveyor belt. When the weight is up to speed there is no resultant force on the weight. So the foirce on the weight starts at some 'high' value and decreases to zero.
    You must use the average force to calculate the gain in energy.
    If the force decreases uniformely then the maximum force will be 2 x the average force.
    If there was no friction (ice covered conveyor !!!!) I think you could say that the weight would not be accelerated up to the speed of the conveyor belt.
    This is not a particularly rigorous analysis but I hope it goes some way to explain why an average force is needed in any calculation.
     
  4. Dec 19, 2011 #3
    That makes sense for using the average force. Since the force does change between the initial energy and the final energy with respect to the distance traveled it would have to be the average over that distance. I don't know if the force changes uniformely over that distance though. I am using a crushable honeycomb product to 'catch' the takeup weight, and so I can probably find out from them what the force curve looks like. Thank you
     
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