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Velocity equivalent of a given Force

  1. Oct 19, 2012 #1
    Hi everyone,

    First off I do realise that I can't directly convert force to velocity.

    My scenario is that I have a machine part and I have found that it begins to crack when a point load of approx 2800N is exerted onto its tip. What I am trying to calculate/approximate is if I dropped the machine part (approx mass = 1kg) what velocity would it need to have when it hit the floor for the force/deceleration to equate to 2800N. I realise that dropping it on different materials will change the result but again I am only looking for an approximation.

    Any help would be greatly appreciated.

  2. jcsd
  3. Oct 19, 2012 #2

    Simon Bridge

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  4. Oct 19, 2012 #3
    thanks for that, do you have any reccomendations on how to calculate this time? I am struggling. I could calculate the stiffness of the part and it could be thought of as an extremely solid spring. I believe this could help calculate the time frame but I am not sure.
  5. Oct 19, 2012 #4


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    Some background might help with this because, as you say, there is no direct equivalence between the two. Is this in the context of some claim for liability or are you trying to devise a simple strength test?

    As Simon Bridges states, the time frame is very relevant. In simple terms, this will relate to the construction of the floor as much as to the construction of your piece. If your flooring was what they have in kids play areas, the stress would be significantly less than for a concrete floor but, also, the amount by which the piece can deform before breaking is also relevant. When it hits the floor, it will have a certain amount of momentun (mass times velocity) and this can be dissipated by applying a Force for a certain length of Time (the term is Impulse). So there is no easy answer as the stopping time is not know.
    Perhaps we could come up with a suitable 'specification' that would be more meaningful if we knew more details. You may find it adequate to use a 'standard floor construction' and base a spec on that.
  6. Oct 19, 2012 #5
    thanks sophie, I work with diesel injectors and I noticed cracks appearing on parts at around 2800N so I was trying to approximate the minimum velocity that would cause cracks if an injector was dropped or hit of something while being carried.

    As I am just looking for an approximation I was going to make a couple of assumptions. One being that when the when the nozzle hits the ground there is no bounce, therefore the initial velocity is the same as the change in velocity. The parts are very strong but brittle so there will be minimum deformation as well. My main problem now is working out the time-frame of the impact.
  7. Oct 19, 2012 #6


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    Do you suspect that someone is being clumsy? Your post does read as if you are doing some sleuthing.
    I don't think there's any hope of an answer in your terms because the impact time is quite unknown without some experimenting. I really don't think there's a figure that could be suggested to be near enough to 'ball park'. If you could specify a floor construction (i.e. the hardest, thickest floor in the workshop, you could see if a drop from chest height would cause a new injector to break.
    The actual context of all this is vital to know if you want some useful information out of all this but I'm sure that you can get an answer of some sort if you are prepared to be flexible (more flexible than those injectors!)
  8. Oct 19, 2012 #7
    In crack analysis it is normal to use energy considerations, not force.

    You can easily calculate the energy available from a given height of drop or velocity arrest.

    This is the basis of the Izod and Charpy tests for materials.

    You should also look up the principles of fracture mechanics.

    In essence the energy to open up a crack is the energy to create the new surfaces. This is calculable from the material properties.

    go well
    Last edited: Oct 19, 2012
  9. Oct 20, 2012 #8


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    I absolutely agree that Energy is what counts, in principle, but in this case, it is hard to assess just how much of the energy of the falling piece actually goes into producing the damage without knowing a lot more about the situation. So much so that you'd have to do a large range of sacrificial tests on a number of injectors (£££$$$!) to calibrate your system.
    It appears that those tests (Izod and Charpy) involve standard sizes and shapes of samples and a specific impact. Very useful, of course, when used in the context of design and choice of materials. However, that doesn't instantly help in working out the equivalent stress on a random shaped piece - which is also the moving element in a drop test.

    To come up with a solution to this specific problem, it is necessary to know what is actually required. Is it all about a concern about a particular incident or is it a question about what protection would be needed in packaging - or what? I am sure there is not a simple answer, in the terms of the OP.
  10. Oct 20, 2012 #9
    My brief post was only an introduction.

    The Charpy and Izod tests are material tests, not component tests.
    My reference was meant to point the way, not be a comprehensive manual.

    Tests, specific to this application, would need to be developed, something I have done in the past and have some experience of in failure analysis.
    Last edited: Oct 20, 2012
  11. Oct 20, 2012 #10


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    . . . . . . . Which is why we need to know much more about the context of all this. It could just be an ad hoc query about a particular incident or the way into a major quality control programme.
  12. Oct 20, 2012 #11
    If you're trying to figure out a real-world problem, why not just drop the part on a scale and see what height it takes to produce 2900N? Drop it again at a different height and record that force. From those two points, you can approximate a graph and your velocity.
  13. Oct 20, 2012 #12


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    Unfortunately, the force produced by that method will depend as much upon the dynamics of the scales you are using and how long they take to register as it will on the speed of the object.
    As was said earlier, the force needed to stop the object depends on the time it is applied for. Change in Momentum equals force times time.
    Any answer you get by your method could be seriously misleading and would be quite invalid.
  14. Oct 21, 2012 #13
    Unfortunately, we all don't have high-speed cameras. I'm quite familiar with work-energy and impulse-momentum:rolleyes: btw

    If I had 4 hours to figure this out, I would do precisely what I prescribed above and get a line. Today's digital scales don't self-oscillate like spring scales Sure it wouldn't be perfect but it's a way to work towards an answer with minimal equipment while you're whining about variables for your equations.
  15. Oct 21, 2012 #14


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    They still have the equivalent to a spring constant. Something moves / deforms as the load is applied. If you don't know that value, or the mass, then you still don't know how far it will move or, hence, what energy is involved.
    E = kx2/2

    If you were able to know that plus the effective 'k' of the floor and the injector as it falls at a precise angle then you may be able to find some correspondence but this is certainly not trivial.

    But we still don't know what the OP actually wants to know.
  16. Oct 22, 2012 #15
    This problem can be solved using finite element structural analysis, including the dynamics, the material properties of the floor and the machine part, the detailed geometry of the machine part, and the effects of various initial orientations of the machine part. Finite element structural codes also includes failure criteria based on state of stress or energy. This is how mechanical engineers design parts and products such that they do not fail in typical use.
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