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A particle undergoing a force with friction

  1. Mar 25, 2008 #1
    What I've attempted to do here is to model the motion of a 2D particle with some force F(t) being applied. I was told in my physics class that the force of friction is proportional to velocity squared or Force-of-friction==k*v[tex]^{2}[/tex].

    I assumed that the friction force would be pushing the opposite direction as F(t).

    I then broke the acceleration, friction, and the applied force into horizontal and vertical components and used newtons second law to equate them leading to the pair of equations below the diagram. These can then be transformed into a pair of differential equations which I'm having a heck of a time trying to solve.

    If anyone could tell me whether or not I'm modeling this situation correctly or if I've gone horribly wrong somewhere that'd be great, I would also appreciate any help I could get solving these differential equations in general for Vx(t) and Vy(t) (if they are, in fact, correct).
     

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  2. jcsd
  3. Mar 25, 2008 #2
    Welcome to PF,

    Why do you model the frictional force to be going in the opposite direction as F(t)...?
    The normal thing would be to have the frictional force point in the opposite direction of the motion of the particle, that is

    [tex]
    \underline{K}= - k |v| \underline{v}
    [/tex]

    I didn't check the rest of your derivation. I want to make sure first that this somewhat strange frictional force is really what you want to do:smile:
     
    Last edited: Mar 25, 2008
  4. Mar 25, 2008 #3
    Okay, I thought that assumption was off. Could you give me any ideas as to how I would go about fixing my current model? My goal is to be able to find out what the motion of a particle undergoing this kind of friction while a force is being applied is like.

    EDIT: Okay I've started figuring stuff out, Since the frictional forces angle will depend on the angle of the velocity and the horizontal and vertical components of velocity are Vx and Vy respectively, we know the angle must be Tan[tex]^{-1}[/tex](Vy/Vx) This should allow me to make a different set of correct equations based on the principle above.
     
    Last edited: Mar 25, 2008
  5. Mar 25, 2008 #4
    You can proceed as before.

    It might be easier not to use angels as you did.

    Let [itex]F(t)=(F_x(t),F_y(t))[/itex]. The position of the particle at time t is (x(t),y(t)). The frictional force is [itex]K(t)=-k |\underline{v}| \underline v[/itex]. (What I wrote above was cubic in v, sorry) Then you have

    [tex]
    M \ddot x(t) = F_x(t) - k \sqrt{\dot x^2(t)+\dot y^2(t)}\dot x(t)
    [/tex]

    And similar for y(t). Solving these equation is of course a different matter. What kind of F(t) are you thinking of..?
     
  6. Mar 25, 2008 #5
    Okay, so I've found the angle which the frictional force will always have (Tan [tex]^{-1}[/tex](Vy/Vx)) in order to separate it into components and derive a system of differential equations. I'm still applying a force F[t] at angle theta to the particle of mass M and have separated that force into components and accounted for it. The resulting system of differential equations is pretty scary, as usual any help solving, or pointers, or pointing out of mistakes I have made would be appreciated.
     

    Attached Files:

  7. Mar 25, 2008 #6
    Ok that's equivalent to what I wrote.

    You should check some trigonometric identies to get rid of the [itex]\cos \arctan(\dots)[/itex] and [itex]\sin \arctan(\dots)[/itex].

    I continue with my expression from above (it's the same as yours)

    [tex]M \ddot x(t) = F_x(t) - k \sqrt{\dot x^2(t)+\dot y^2(t)}\dot x(t) \quad (1)[/tex]
    [tex]M \ddot y(t) = F_y(t) - k \sqrt{\dot x^2(t)+\dot y^2(t)}\dot y(t)\quad (2)[/tex]

    First try to solve for [itex]\dot y(t)[/itex]:
    [tex] F_x(t)-M \ddot x(t) = k \sqrt{\dot x^2(t)+\dot y^2(t)}\dot x(t)[/tex]
    [tex]F_y(t) -M \ddot y(t) = k \sqrt{\dot x^2(t)+\dot y^2(t)}\dot y(t)[/tex]

    Divide the second by this first..
    [tex] \frac{F_y(t) -M \ddot y(t) }{F_x(t)-M \ddot x(t)} = \frac{\dot y(t)}{\dot x(t)}[/tex]

    So [tex] \dot x(t)\frac{F_y(t) -M \ddot y(t) }{F_x(t)-M \ddot x(t)} = \dot y(t)[/tex]

    For [itex]\ddot y(t)[/itex] you can insert (2). Then it takes some algebra to solve for [itex]\dot y(t)[/itex]. This result you plug into (1) which then only involes [itex]\dot x(t),\,\ddot x(t)[/itex]. If you're lucky you don't end up with something being identically true. ( I didn't try :smile:) Then we can think about how to solve this resulting (uncoupled) ODE.

    -Pere
     
    Last edited: Mar 25, 2008
  8. Mar 25, 2008 #7
    Hooray, Glad I managed to make a functioning set of equations, btw I didn't happen to know that particular trig indentity I'm only in grade 11 (I.e. no uni trig, yet) and the grade 12 math courses didn't cover that one. Also I plugged the general form of the differential equations (simplified like you told me) into mathematica to see if it could solve it explicitly, to no avail. I'm guessing with a coupled set of equations this complex numerical methods are the way to go. Though if an explicit solution is possible in any way I'd absolutely love to get my hands on it.

    EDIT:
    Okay I've solved for Y'[t], answer is attached
     

    Attached Files:

    Last edited: Mar 25, 2008
  9. Mar 25, 2008 #8
    Well, whether or not an explicit analytic solution can be found certainly depends on what you choose for F(t) ... For F(t)=0 (hehe) the particle would move on a straight line and eventually stop, so maybe we think about this special case first. With the motion being one-dimensional we can just as well consider the equation

    [tex]
    M \ddot x(t)=-k\frac{\dot x^3(t)}{|\dot x(t)|} = -k\dot x^2(t)
    [/tex]

    where the last is true if [itex]\dot x(t)[/itex]is positive which we can assume.

    Do you know how to solve this? It gives an idea of what to expect since it is the easiest possible special case of your problem :smile:
     
  10. Mar 25, 2008 #9
    Alrighty, I managed a solution to that simpler differential equation there (X'[0]==V0,X[0]==X0), I was just wondering if there was any way to tell for sure whether or not there is a possible explicit answer, or how I could go about making some sort of infinite converging series for the answer to X[t] and Y[t].

    EDIT: Removed incorrect answer, see post below.
     
    Last edited: Mar 25, 2008
  11. Mar 25, 2008 #10
    Aha, ok. So if you sub this into (1) the [itex]\dot x^2(t)[/itex] cancels and the rest is a cmplete square [itex](F_x(t)-M\ddot x(t))^2[/itex] so we get:
    [tex]
    M \ddot x(t) = F_x(t) - k |F_x(t)-M\ddot x(t)|\dot x(t)
    [/tex]

    Interesting ... doesn't seem right though, maybe my suggestion of how to uncouple the equations was nonsense, we'll see tomorrow :smile:

    EDIT: you missed some constant term in your solutiom x(0)=x_0 does not hold...
     
    Last edited: Mar 25, 2008
  12. Mar 25, 2008 #11
    Ah, you're right, so I have, correction appended.

    Well, I sure hope that's right cause a fairly simple and nice integral equation results for the solution of X[t] with the boundary values X[1]==X0 and X'[1]==V0, solution attached. Ill try and find a similar equation for Y[t] and test if they're right, and hopefully then I'll have a general answer.

    EDIT: Boy I'm on a roll with these screw ups, the following answers are void due to what's said in the post below.
     

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    Last edited: Mar 25, 2008
  13. Mar 25, 2008 #12
    When you solved for [itex]\dot y(t)[/itex] you forgot an k somewhere .. if you cautiously keep track of everything my suggested method yields 0=0 which doesn't help much:smile:

    But I have another suggestion...

    Take (1), solve it for [itex]\dot y(t)[/itex] (easy), differentiate with respect to t to get [itex]\ddot y(t)[/itex], then plug these two results into (2) ....giving an ODE for [itex]x(t)[/itex]
     
  14. Mar 25, 2008 #13
    Ah if only the resulting ode wasn't an impossible trainwreck. I really doubt there's a general analytical expression for X[t] now. Attached is the resulting 3rd order ODE.
     

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  15. Mar 26, 2008 #14
    Yes this looks "better". As I wrote twice before, for general F(t) there will be no analytic solution (I'd guess), for the special case F(t)=0 there is. So all depends on your choice of this force. Even for numerical methods to apply you must of course specify F(t):smile:
     
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