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Differential Equations in simple mechanics?

  1. Mar 2, 2015 #1
    I came across a few problems in the Kleppner and Kolenkow book in which you must find the force of tension at specific lengths on a rope of mass m. They said you must use differential equations to solve these types of problems. How can you solve and use differential equations like this to get knew information? I would appreciate a technical explanation as i am only in high school.
  2. jcsd
  3. Mar 2, 2015 #2
    Is there a specific situation you have in mind? Not many high schoolers have experience with differential equations. Do you know what they are?
  4. Mar 2, 2015 #3
    I know what differential equations are but i do not know to apply them mechanical problems.
  5. Mar 2, 2015 #4
    Ok. I'm still not sure what type of problem you have in mind so I will give you a simple example of how differential equations are used straight from Newton's second law:
    Suppose that a particle at rest is subject to a constant net force ##F##. Newton's second law tells us that (as you've probably seen it written) ##F=ma##. This is actually a differential equation which I'll write as ##F=m dv/dt##. The 'unknown' function in this case is the velocity function. All we have is information about its derivative


    So the job is to find the function who's derivative is a constant. In general, that is a linear equation. In this case the function is ##v(t)=F/m *t##.

    As for modeling tension in a rope or chain you need to be more specific. Note that if the rope is in equilibrium then the tension is uniform throughout.
  6. Mar 3, 2015 #5


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    There's an introduction to Differential Equations at high-school level here:

    http://www.examsolutions.net/maths-revision/syllabuses/Index/period-1/Further-Pure/module.php [Broken]

    Towards the bottom of the list of topics.
    Last edited by a moderator: May 7, 2017
  7. Mar 3, 2015 #6
    Is F = m*a the F definition?
    Then what you would substitute instead F in
    dv/dt = F/m ???
  8. Mar 3, 2015 #7
    The sum of the forces along a particular direction. In my post I called 'F' a constant force for simplicity.
  9. Mar 3, 2015 #8
    The tension of a dangling rope (a rope in equilibrium) is simply the weight of whatever it is supporting.

    Specifically, the force of tension on that rope goes in the direction of the rope. So if you have multiple ropes attached to a single object, you can derive using trigonometry.

    I am guessing that instead of giving you initial values, it just gives you a differential equation describing the system.

    If, in fact, you mean finding the tension on parts of a partially slack rope using differential equations... well, that's hard.
  10. Mar 4, 2015 #9
    Simplicity is Ok, but what about the origin of F? Another formulae (wich?), experiment or???
  11. Mar 4, 2015 #10
    F is just the name I gave to a constant force, the origin is unimportant. Think of it as the pull of a string or something.
  12. Mar 4, 2015 #11
    That's actually the best analogy I've heard for vector forces.
  13. Mar 4, 2015 #12
    Somehow somewhere isn't a case of differential equation. In order to solve it us must define F pure math. It could not be done from F = m*a, then? If you define F as constant, where to find the value of it????
  14. Mar 4, 2015 #13
    I think we are talking past each other here. Let me illustrate the example in a totally arbitrary way using different notation. Suppose I know that a differential equation is


    where v is a constant; it does not need any physical meaning for the sake of this example, but it COULD stand for something physical... perhaps a constant velocity.

    The goal of solving this equation is to find an unknown function x(t) knowing that its derivative is a constant value. This function is, of course


    where x0 is the integration constant.

    This problem is exactly the one that I explained above except the derivative physically represented acceleration (the derivative of velocity) and was equal to a constant net force.
  15. Mar 6, 2015 #14
    I could easily agree with you latest example. Us just wanted to underscore the difference between "definition" and "equation".
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