Mechanics problem involving a Differential Equation

AI Thread Summary
The discussion focuses on solving a mechanics problem involving a differential equation related to forces acting on an object. The user is trying to express the net force equation in terms of velocity, v(x), but is struggling with integrating the equation due to the presence of v². Participants suggest that the user consider the applicability of drag laws and provide insights on rewriting the equation for easier manipulation. They also discuss the importance of differentiating and integrating the equations correctly to progress towards a solution. Overall, the conversation emphasizes the need for clarity in handling differential equations in mechanics.
darrenhb
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Homework Statement



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)

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!
 
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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
 
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
 
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'"
 
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