# Homework Help: Mechanics problem involving a Differential Equation

1. Sep 23, 2011

### darrenhb

1. The problem statement, all variables and given/known data

Fnet = Ffriction + Fairresistance = μmg + cv2 = ma

Essentially I just need to get this in a v(x) form. Prof said that we need to go about it by doing this:

F = m$\ddot{x}$ = m $\dot{x}$ * d$\dot{x}$/dx

and then integrating to get v(x)

3. The attempt at a solution

so...

μmg + cv2 = m $\dot{x}$ * d$\dot{x}$/dx

And this is where I just can't figure it out. I'm terrible with DEs.

I thought of moving the 1/dx to the other side and then integrating both sides, but I get stuck, especially since v2 can be written as $\dot{x}$2, right? And that messes with my sleep-deprived brain a little, lol. Any nudge in the right direction would be greatly appreciated!

2. Sep 23, 2011

### danielakkerma

Hi!
If I recall correctly, the v^2 proportionality law(of drag) applies to high speeds, only; Are you sure that you need to have v^2 there? Stoke's law, in this case, I suppose, is a better fit(i.e k*v);
How did you arrive at the following relation:?${\frac{d}{dx}\dot{x}} = 0$, I can't see its use here...
We'll work on it,
Daniel

3. Sep 23, 2011

### darrenhb

The question unfortunately doesn't give much of a hint as to how fast this might be traveling, so I'm not sure I could cancel out the v2. I also can't see where I typed d$\dot{x}$/dx = 0. :S

4. Sep 23, 2011

### danielakkerma

It's not so much that you typed it, it's what it is... :)
Meaning:
$\frac{d}{dx}{dx}{dt} = \frac{d}{dt}\cdot\frac{dx}{dx} = 0$ Meaning that it can't be used...
The first thing to do, in my view would be to rewrite it, such that:
$\Large mx''(t) = mk(x'(t))^2 + C$ Where C is a constant carrying your constant friction force.
Then one would lean towards differentiating again(to get rid of the C) with respect to t:
$\large mx'''(t) = 2mkx'(t)x''(t)$ and dividing by m, and substituting z(t) = x'(t), and for simplicity, taking k=1:
$\large z'' = zz', \frac{d^2z}{dt^2} = z\frac{dz}{dt}$.
Multiplying by dt:
$\large zdz = (\frac{d^2z}{dt^2})dt$.
Now one must integrate, and continue the process further.
Try it,
Hope that works for you,
Daniel
P.S
Here's some "cheating"(:D), just to verify your steps:
http://www.wolframalpha.com/input/?i=-x''+==+x*x'"

Last edited by a moderator: Apr 26, 2017