I was trying to find the distance a machine system I designed would potentially go. The machine works like this: A mass $M$ drops down and through a system of pulleys, pulls a wheel wound with the cable which pushes the system forward. Given a speed of wind $w$, coefficient of friction $\mu$, mass of entire system $m$, and the force of air resistance $F_a=c(w+v)^2$, and a maximum drop height $h$ before the cable fully unwinds from the wheel and let's go of it, what is the maximum distance $s$ the system will travel?Problem

$_{ \text {(pardon my mspaint)} }$

My Work: I first split this up into two problems: Find the kinetic energy of the velocity after the cable has been detached based on work, then set that kinetic energy equal to the work done by air resistance and friction.

$W_1 = \int (\sum F_x)\cdot v \ \mathrm dt$

$W_1 = \int v(-c(w+v)^2+g(M -\mu m)) \ \mathrm dt$

$W_1 = \int (-cw^2v-2cwv^2-cv^3+gv(M -\mu m)) \ \mathrm dt$

But since $v = {\ \mathrm ds \over \ \mathrm dt}$, things get really weird, as:

$W_1 = -c \int (w^2{\ \mathrm ds \over \ \mathrm dt} + 2w({\ \mathrm ds \over \ \mathrm dt})^2 + ({\ \mathrm ds \over \ \mathrm dt})^3) \ \mathrm dt + g \int {\ \mathrm ds \over \ \mathrm dt} (M -\mu m) \ \mathrm dt$

$W_1 = -c \int w^2 \ \mathrm ds + \int 2w({\ \mathrm ds \over \ \mathrm dt} + ({\ \mathrm ds \over \ \mathrm dt})^2) \ \mathrm ds + g \int (M -\mu m) \ \mathrm ds $

$W_1 = -c \int (2w{\ \mathrm ds \over \ \mathrm dt} + ({\ \mathrm ds \over \ \mathrm dt})^2) \ \mathrm ds + gs(M -\mu m) - csw^2$

And I could be wrong, but that looks like a really invalid integral, so how might one solve this problem? I'm thinking there should be a way to do it with Diff. EQ, but am yet to figure that out.

  • $\begingroup$ I think I'm thinking of air resistance correctly, treating the velocity of the air in the reference frame of the machine (thus $v_{total} = v_{machine} + w$ in the $F_a = cv^2$ equation) $\endgroup$ Nov 5, 2014 at 8:21
  • $\begingroup$ Ok, can you stop editing this, unless it's to add something important? I'm trying to read it here :) $\endgroup$ Nov 5, 2014 at 8:30
  • $\begingroup$ Yeah, I'm finished. Just tiding up and fixing some parts. Never was good at proofreading... Sorry about that. $\endgroup$ Nov 5, 2014 at 8:34
  • $\begingroup$ What is $g(M-\mu m)$? Doesn't make much sense to me... $\endgroup$ Nov 5, 2014 at 8:35
  • $\begingroup$ g is the earth gravitation constant, should have addressed that. But then that's the same thing as $T-F_f = Mg- \mu m g$ $\endgroup$ Nov 5, 2014 at 8:37

1 Answer 1


It looks good to me, that is, not invalid at all.

In my opinion, what you have to do now, is (I prefer $\dot{s}$ instead of $\frac{ds}{dt}$):

$$E_k = W_1 \Rightarrow \frac{m {\dot{s}}^2}{2} = -c \int \left(2w{\dot s} + {\dot s}^2 \right) \mathrm ds + gs(M -\mu m) + - csw^2$$

Taking $\frac{d}{ds}$ on both sides:

$$m \dot{s} \frac{\mathrm d \dot{s}}{\mathrm ds} = -c \left(2w{\dot s} + {\dot s}^2 \right) \ \mathrm + g(M -\mu m) + - cw^2$$

This will be a differential equation of the form $y\ y\prime = ay+by^2+c$, where $y$ is your $\dot s(s)$. I'm not sure right now whether you can solve it explicitly, but if not I'd try to simulate it in Scilab.

  • 1
    $\begingroup$ I'm afraid you don't know very much what you're doing at this point, do you? The D.E. that must be solved is the one I wrote above. I don't think anything can change that. If you need help understanding what it means to solve that, just ask. $\endgroup$ Nov 5, 2014 at 9:59
  • $\begingroup$ Sorry, I was only looking at the integral part, so I assumed your where saying: −c∫(2ws˙+s˙2)ds = −c(2ws˙+s˙2). My mistake for not looking at the entire equation... Thanks for the help. $\endgroup$ Nov 5, 2014 at 10:06
  • $\begingroup$ Anytime! Feel free to ask if you have further problems. $\endgroup$ Nov 5, 2014 at 11:35

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