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Classical Mechanics-Inclined plane tricky problem

  1. Dec 8, 2012 #1
    1. The problem statement, all variables and given/known data

    An inclined plane is given , inclined by the angle β.A small object with a given mass m is placed at the top.What distance and in what time will it go if the friction coefficient goes by the law mu=bx, where b is a constant and x is the distance that the object has traveled.

    2. Relevant equations
    All the equations from classical mechanics.


    3. The attempt at a solution
    I've scorched my brain for this one , but I can't find the solution , it's been a week now since I got the problem and i think it's time i ask for some help.

    Thanks!
     
  2. jcsd
  3. Dec 8, 2012 #2
    Hello!
    What have you tried to do so far?
    Usually, when friction is dependent on velocity/distance, it means you will need to solve a differential equation.
    Where exactly did you get stuck trying to solve this?
     
  4. Dec 8, 2012 #3
    I'm simply stuck in the math.I have no ideea how to start solving it , i haven't studied differential equations yet.(i'm in the 11'th grade)My teacher gave me this problem so i think there is a simpler way to solve it involving only the knowledge that i have so far.In math , this is my first year of calculus studying limits and strings of numbers.
    I need an idea to start with.Boy am I stuck with this one!
     
  5. Dec 8, 2012 #4
    The best way to start is going through the basics.
    Draw the issue at hand, chart all the forces, and think about this:
    You know at what speed it started moving. You know at what speed it stopped moving.
    What can you do with that knowledge?
    Be sure to choose and easy coordinate system so the problem will be simpler to understand with the formulas.
     
  6. Dec 8, 2012 #5
    I think I solved the distance problem , so writing Newton's law for the moment when the object stops moving.
    F=ma , where a=0 so F=0.(1)
    But also F=mgsinβ-MU*N=mgsinβ-b*dmgcosβ (2)

    From (1)(2) we have that mgsinβ=bdmgcosβ , where the x from MU=bx becomes the maximum distance covered by the object ,the one that i need and N is equal to the component of G that is mgcosβ.
    From here we get that d=sinβ/bcosβ.

    Is it correct?

    for the second point , finding the time , now this i think is a bit harder, i coud use it from the motion law t=√2d/a , but finding a could be a problem.
     
  7. Dec 8, 2012 #6
    a little update for the time , we know that F=ma so a=F/m but F=mg(sinβ-mucosβ) and t=√2x/a

    so t=√2x/(g(sinβ-bxcosβ)) , is it correct like this???basically I found a formula that lets you calculate the time to any given distance x.But my question is , is this what the problem asks?
     
  8. Dec 8, 2012 #7
    Regarding the distance, that's correct. You should try to check it yourself and see if it makes sense. What are the units of b? What should d's units be?
    You can also try and test your answer by going through a different way. If you have the distance traveled and the angle, you can use the work-energy theorem to see if it works out with what you found.
    The time part, however, isn't right, since when it stops a=0, and you will have t=infinity, which probably isn't the answer you are looking for. Honestly, I can only see a way to find t which you cannot use, so i'll try and think about a simpler way to do it.
     
  9. Dec 9, 2012 #8
    Does anyone have any idea how to solve the time problem?


    Ty.
     
  10. Dec 9, 2012 #9

    haruspex

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    No, that's when it stops accelerating. It will have some residual KE to burn off still.
    Can you write an expression for the energy expended overcoming friction in travelling a distance x down the plane?
     
  11. Dec 10, 2012 #10
    tu,but i talked to my teacher and he told me to ignore that part,the ine with the KE.my only trouble now
     
  12. Dec 10, 2012 #11
    is with the time now.
     
  13. Dec 10, 2012 #12

    haruspex

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    That's rather surprising. I can figure out the stopping distance from conservation of energy, but I see no way to get the stopping time without using differential equations. (It's basically simple harmonic motion.)
     
  14. Dec 10, 2012 #13
    Same here, I tried to ask others, no one seems to find a way without using differential equations.
     
  15. Dec 10, 2012 #14

    haruspex

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    squareroot, despite what you teacher side, I would recommend trying to get the right answer for the distance. It's not hard.
     
  16. Dec 11, 2012 #15
    i got a big hint from my techear, he said that there is a very simple way of solving this and avoid differential equations:), by finding a simillarity between this problem and mehanical oscillations...he said that if you observ that simillarity the prblem solves very easy...
     
  17. Dec 11, 2012 #16

    haruspex

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    I did mention that it was basically SHM, but here there is a kind of forcing term. To take the standard SHM formulae and assume they apply would be distinctly unrigorous, but if your teacher has sanctioned it...
    So, what can you write down for the period, treating this as SHM?
     
  18. Dec 11, 2012 #17
    the period is T=2pisqrt(m/k)
     
  19. Dec 11, 2012 #18

    haruspex

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    Of course, but what is k in this context? Without the differential equation, it's not obvious. When you have determined the period, T, what do you think the stopping time will be?
     
  20. Dec 11, 2012 #19
    well i won t be studying differential equation 3 years from now so this is a bit hard for me.
    Does k=MU?
    As far as what the stopping time is , i have no idea.
     
  21. Dec 11, 2012 #20
    A, as for the time. does 2t=T? because T is the time of a complete oscillation and the body will have to go all the way down and back up so that we can say that he completed a full oscillation.
     
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