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Please help.regarding drag force and Newton

  1. Mar 17, 2007 #1
    Hello, please help me I have a question regarding drag force and newton's law of motion.
    This is quotes from online pdf files from University of Toronto Scarborough.

    "the terminal velocity is F/b , where b is drag constant

    Let’s solve the equation of motion and see how
    this is reflected in the solution. Newton's law
    reads

    m (dv(t)/dt) = F - b v(t)

    which we have written entirely in terms of the
    velocity v(t) and its first derivative.

    How to solve this? Well, if F were zero, we
    would have dv/dt=–(b/m)v, which has its solution
    some constant times exp(-bt/m). By inspection,
    we find that we can account for nonzero F by
    simply adding a constant, F/b. That is,

    v(t) = F/b + (constant)exp(-bt/m)
    --------------------------------------

    Now what i dont understand is the last paragraph, and where the last equation came from. Can anyone explain more clearly? Please help :smile: Thank you very much.
     
  2. jcsd
  3. Mar 17, 2007 #2

    radou

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  4. Mar 17, 2007 #3
    Thx for the reply.
    Hmm, i am still in high school, but already learned differential and integration. Do i need to learn first about something i don't learn in high school?
    If so what is the topic title? Or maybe u can teach me if its possible?
    thx
     
  5. Mar 17, 2007 #4

    George Jones

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    Have you learnt the method of substitution for integration?
     
  6. Mar 17, 2007 #5
    i dont think so. Is there any websites that teach that?
    thx
     
  7. Mar 17, 2007 #6

    George Jones

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    Another question: do you know how to find

    [tex]\int \frac{1}{x} dx?[/tex]
     
  8. Mar 17, 2007 #7

    arildno

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    It seems you already know the solution to the simplest type of diff.equations, that is the one of the type: du/dt=ku, having u(t)=Ae^(kt) as its general solution.

    Now, ask yourself:
    What would be the next simplest type of a solution to a diff.eq?
    At least my intuition says this would be a function of the form:
    [tex]U(t)=A+Be^{kt}[/tex], where A,B and k are constants.

    Let us see what sort of differential equation such a function satisfies!
    1. We have that [itex]\frac{dU}{dt}=kBe^{kt}[/itex]
    2. We can convert the right-hand side of 1. into an expression involving U itself as follows:
    [tex]U=A+Be^{kt}\to{B}e^{kt}=U-A[/tex]
    3. Thus, we may insert 2. into 1, and gain:
    [tex]\frac{dU}{dt}=k(U-A)=C+kU, C=-kA[/tex]
    4. Note that our equation 3 is a DIFFERENTIAL equation of the general form:
    [tex]\frac{dU}{dt}=A+BU[/tex]
    But conversely, this ought to entail that we now can SOLVE such diff.eqs, gaining solutions similar to the function we studied in 1.-3.!
    We'll see that this is indeed the case:
    We may rewrite 4. as:
    [tex]\frac{dU}{dt}=B(\frac{A}{B}+U)[/tex]
    Now, let us define a new function: [tex]Y(t)=\frac{A}{B}+U[/tex]
    Now, note that by differentiation, we have the identity: [itex]\frac{dY}{dt}=\frac{dU}{dt}[/itex]

    Thus, reconverting our diff.eq in terms of Y, rather than U, we get:
    [tex]\frac{dY}{dt}=BY\to{Y}(t)=Ce^{Bt}\to{U}(t)=Y(t)-\frac{A}{B}=Ce^{Bt}-\frac{A}{B}[/tex]
    and we see that this is a solution of the form described in 1.-3.
     
  9. Mar 17, 2007 #8

    yes, ln x + c
     
  10. Mar 17, 2007 #9
    "It seems you already know the solution to the simplest type of diff.equations, that is the one of the type: du/dt=ku, having u(t)=Ae^(kt) as its general solution."

    Arildno, No i haven't known that :frown:
     
  11. Mar 17, 2007 #10

    George Jones

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    Your equation is

    m dv/dt = F - b v.

    Proceeding very informally, treat dv as one symbol, and dt as another. First, divide both sides of the above, by the right side of the above, and then multiply both sides dt.

    What do you get?
     
  12. Mar 17, 2007 #11

    AlephZero

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    If you haven't yet done differentiation and integration of exponentials, logs, trig function etc, yet you are getting a little bit ahead of yourself with this equation.

    Not that any of those topics are "hard", but there is a far amount of detail to remember, so you have to start somewhere and take it one step at a time. Most differential equations in physics, biology, etc involve exponentials and trig functions, so it makes sense for your courses to leave the topic of "differential equations" till after you have done integration and differentiation of those functions.
     
  13. Mar 17, 2007 #12
    (m dv)/(F - bv)=dt

    Btw, is there any program downloads from the internet specially made to write mathematical equations that i can easily put on microsoft word or other applications?
     
  14. Mar 17, 2007 #13
    There is LaTeX, the engine this forum uses to format mathematics. You can't put equations into Word with it, but you can generate documents separately.
     
  15. Mar 18, 2007 #14

    George Jones

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    Let v_0 be the speed of the particle at time t_0, and v be the speed of the particle at time t. Tkake this into account by integrating: (over v) the right side of the above equation from v = v_0 to v = v; the left side from t = t_0 to t = t.

    [tex]\int_{v_0}^v \frac{m}{F - bv} dv = \int_{t_0}^{t} dt[/tex]

    Can you do these integrations? Remember, m, F, and b are constants.
     
  16. Mar 19, 2007 #15
    i havent learnt that kind of integration. :frown: is there any website that can teach me that?
     
  17. Mar 20, 2007 #16

    George Jones

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    Let's look at

    [tex]\int \frac{m}{F - bv} dv.[/tex]

    The denominator is a bit messy, thus set u = F - bv, so u is function v. Find du/dv and rearrange this to find dv. In the the integral, substitute u for the denominator, and also substitute for dv.

    What does the integral look like now?
     
  18. Mar 20, 2007 #17
    is F constant? I come up with -(m/b) ln u +c
     
  19. Mar 20, 2007 #18

    George Jones

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    For consistency, all the terms in

    m dv/dt = F - b v

    must have the same units. Since the right side has the units of force, both F and bv must have units of force. F is a constant force, like the force of gravity on a particle near the Earth's surface.

    Good.

    Now change the u back to F - bv and evaluate the resulting expression at the endpoints of the integration interval.
     
  20. Mar 21, 2007 #19
    I haven't learnt about definite integral or integration intervals. I'll come back after i learned it. Thx a lot!
     
  21. Mar 21, 2007 #20

    George Jones

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    Suppose

    [tex]\int f \left( x \right) dx = F \left( x \right) + c.[/tex]

    Then,

    [tex]\int_a^b f \left( x \right) dx = F \left( b \right) - F \left( a \right).[/tex]

    Geometrically, this is the area bounded by the curve f(x), the vertical lines x = a and x = b, and the x-axis.

    Example: [itex] f \left(x \right) = x^2.[/itex]

    [tex]\int_2^5 x^2 dx = \frac{5^3}{3} - \frac{2^3}{3} = 39.[/tex]
     
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