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I How do you find the force that will be exerted by something?

  1. Dec 15, 2016 #1
    Given a certain situation, how do you figure out the force that will be exerted onto something (you can only use F=ma to find the average net force over a certain time period, and only after you know the acceleration)? For example, if you have a non accelerating object colliding with another non accelerating object, given their mass and velocity, how do you find the acceleration that the objects will experience upon collision?

    According to Newton's Third Law of Motion, momentum is always conserved, but how do you figure out the rate at which the two objects will change their velocity (the objects won't immediately reach their final velocity)?

    Can you figure out the amount of time it takes for the objects to reach their final velocity before empirical observation?

    How do you figure out what their final velocities will be?
     
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  3. Dec 15, 2016 #2

    Nugatory

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    You have to know something about the properties of the two colliding objects and the details of the collision. For elastic collisions, you can often get the right order of magnitude if you know the modulus of elasticity of the colliding objects. For inelastic collisions you may have to make some good guesses, or do experiments - there's a reason why auto safety engineers use real crash tests, complete with crash-test dummies.

    For elastic collisions, conservation of energy plus conservation of momentum will do it. For inelastic collisions, it going to depend on the details of the collision and especially how much energy is dissipated.
     
  4. Dec 18, 2016 #3
    Once you find the final velocities in the head-on elastic collision (for simplicity, although the equation would be just as easy to solve given the angle of contact) using the two conservation equations, how do you figure out the rate at which the velocities change? The change in momentum (and by extension, velocity): impulse, is equal to the force multiplied by the time that the force is exerted for. However, without knowing the time that it takes, you cannot find the acceleration or divide the impulse by time (so it is impossible to find the force).
     
  5. Dec 18, 2016 #4

    Nugatory

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    If you assume that the modulus of elasticity is constant, you can calculate it. It's the same problem as changing the speed of an object by bouncing it off an ideal spring, and there's only one combination of time and compression that will produce any given change of speed. However, that's a somewhat dubious assumption, which is why I said that this is an order of magnitude sort of calculation.
     
  6. Dec 20, 2016 #5

    Saw

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    It is a good question, that I was also asking myself, and a good answer. Based on the latter, I realize now that the problem with F = ma is that it is half dynamic (m) but half cinematic (a). Therefore it does not provide a complete rule that can enable you to predict the time that the interaction will take, by sheer analysis of the characteristics of the system; you can only do so ex post by measuring with clocks and rods. If you want to overcome this limitation, you have to resort to some trick to get rid of all cinematic terms (except time itself!), which I think is what Nugatory is suggesting. This is something akin to what you do when deriving the anugular frequency and period of simple harmonic motion: 1/ you start with F = -kx (also half cinematic....) ; 2/ you consider Newton's second law and conclude that a = -kx/m; 3/ based on an analogy with uniform circular motion, you get a cinematic equation a = ω2x. But fortunately the x cancel out and thus you get a purely dynamic formula for ω and hence T. I suppose that in a collision you would be interested in T/2.
     
  7. Dec 20, 2016 #6
    The calculation Nugatory is describing is typically very complicated, and involves taking into account the 3D stress-strain behavior of the materials comprising the objects. The deformational equations, including local inertia effects within the objects, needs to be solved using a set of partial differential equations (similar to the wave equation, except more complex). You are looking at the local deformations, stresses, strains, and accelerations on a local basis within the objects during the very short time interval in which the objects are in contact. Contact loading (with non-uniform loading in the contact region) make the solution to the problem extra complicated.
     
  8. Dec 20, 2016 #7
    What do you mean by half dynamic, half kinematic? Why is mass a dynamic (i.e. one that changes based on motion) value in a non-relativistic situation?
     
    Last edited: Dec 20, 2016
  9. Dec 20, 2016 #8

    Saw

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    First, consider by comment superseded by what Chestermiller explained. Things appear to be more complicated than I thought.

    As to my expressions, I was referring to "dynamics" versus "kinematics", meaning the study of the causes of motions versus the study of motion through the measurement of time and distance traversed. Hence by "dynamic" I did not mean anything that changes with motion, but the physical cause for a change of motion. For example, for simple harmonic motion you have formulas that predict the period of oscillation based on the physical characteristics of the system, like the k (= constant of stiffness) of a spring and its mass. I would call that a purely dynamic formula because it anticipates what is going to happen (how long the cycle will take) based on prior knowledge about what the system is, as compared with posterior analysis (through measurement of time and distance) of how the system reacted to a given stimulus.

    But I confess that I was thinking aloud and probably not making orthodox use of the words: in fact, in usual parlance, "dynamics" is said to be the study of the causes of motion but when you get down to it the main formula is F=ma, where the factor "acceleration" is more an effect of the interaction, but nobody cares about that.
     
  10. Dec 20, 2016 #9
    When I first saw your reply, I initially thought about the definitions of those two fields. However, I had not previously seen someone describe terms in an equation using them as adjectives. Thanks for clarifying.
     
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