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My lack of problem solving skills - Mechanics

  1. Aug 30, 2005 #1
    I am taking a senior level analytical Mechanics course, which is very math based. I am having major difficulty in bridging the gap between what a given problem is asking for and manipulating the problem to solve it. I have had all the necessary math courses and did respectable, but now I feel as though I have never been really taught on how to apply that knowledge to solve real physics problems. It is being assumed that I know exactly what to do and I don't. How do I know when to integrate or differentiate or both for any given problem? Is there a good reference out there for just this sort of dilema? I really want to understand what to do and not just skate by. The problem below is indicitive of what I am trying to solve. I am not asking anyone to solve it as it is merely a reference. Any help would be very grateful.

    Given: F(x)=Fo+ct, Fo is independent of v
    Find x dot, and x.
  2. jcsd
  3. Aug 30, 2005 #2


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    "Given: F(x)=Fo+ct, Fo is independent of v
    Find x dot, and x."

    If the problems are phrased like that, it's no wonder you have problems! There is no "x" or "x dot" in the problem. Also you have "F(x)" but Fo+ ct doesn't depend on x!

    I assume that you mean that F(t)= Fo+ ct is the force function for some object, that "x dot" is the speed function, and that "x" is its position. The problem ought to say those things!

    One very important "problem solving skill" is learning basic definitions and formulas. If you are taking a mechanics class then surely someone has mentioned F= ma! If your object has mass m, then ma= F= F0+ ct so [itex]a= \frac{F_0+ ct}{m}[/itex]. a ("x dot dot") is the derivative of x dot (that's one of the definitions you should have learned)so x dot is the anti-derivative of [itex]\frac{F_0+ ct}{m}[/itex] which is [itex]\frac{F_0t+ \frac{1}{2}t^2}{m}+ v_0[/tex]. v0 is the initial speed. Force on gives acceleration: the rate of change of speed. You have to know what the initial speed was to tell what it changed to!
    Now, speed (x dot) is the derivative of position (x) so to find x, you have to find the anti-derivative of that.
  4. Aug 30, 2005 #3
    You are right, I was going from memory and it should not have been F(x) but Fx which is really F. This makes it significantly easier. I still struggle a bit with understanding the role of integration and differentiation and deriving solutions to problems. Is there a good reference (book or link) that you know pertaining to this matter that would aid?
  5. Aug 31, 2005 #4
    Alright, I have a specific homework question with regard to attaining more skills with problem solving. I have been given this specific problem:

    b is positive constant, find force acting on particle as function of x.

    I am not sure where to even start. I know F = ma = m(xdot)*(d(xdot)/dx). I think I am getting tripped up on converting the derivative m(xdot)*(d(xdot)/dx) into something I can use on left side of my initial equation. Is this the correct first step?
  6. Aug 31, 2005 #5


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    I'm not sure what you mean, but we have that
    [tex]a = \frac{{d^2 x}}{{dt^2 }} = \ddot x = \frac{{d\dot x}}{{dt}}[/tex]

    We know that [tex]\dot x = bx^{ - 3} [/tex]

    [tex]a = \frac{{d\left( {bx^{ - 3} } \right)}}{{dt}}[/tex]

    [tex]F = ma = m\frac{{d\left( {bx^{ - 3} } \right)}}{{dt}}[/tex]
  7. Aug 31, 2005 #6
    But shouldn't this be with respect to x instead of t as you have shown as I am looking for F(x).
  8. Aug 31, 2005 #7


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    Well to get a, you have to find the derivative of v (so that x dot) with respect to time (t) and not position (x).
  9. Aug 31, 2005 #8


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    Not if you are using stardard notation! In mechanics, the "dot" is used to mean differentiation with respect to the time, t, so that [itex]\dot x[/itex] is the velocity function and [itex]\ddot x[/itex] is the acceleration function.

    If [itex]F(x)= m\ddot x[/itex], you will need to use the chain rule: [itex]\frac{dv}{dt}= \frac{dv}{dx}\frac{dx}{dt}[/itex]. But, of course, [itex]v= \frac{dx}{dt}[/itex] so [itex]\frac{dv}{dt}= v\frac{dv}{dx}[/itex]. With v= bx-3, that gives
    F(x)= m(bx-3)(-3bx-4).
  10. Sep 2, 2005 #9
    Just wanted to say thanks for the help. Your pointers helped.
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