The problem of just considering engine force and rolling friction $$\sum F=ma=F_{engine}+\mu_{rolling}mgv$$ is that when you solve for velocity like $$\dot v=\frac Fm-\mu gv$$ with initial condition $$v(0)=0$$ and get $$v(t)=\frac{F}{\mu mg}\left(1-e^{-\frac{t}{\tau}}\right)$$ and time constant, $$\tau=\frac{1}{g\mu}$$, you find out that with a steel-on-steel rolling coefficient of $\mu=0.02$ and a gravitational acceleration $g=9.81m/s^2$, in four time constant's time, or $4\tau=0.7848$ secs, $v(t)$ will settle to an equilibrium velocity.

But realistically a locomotive can only go from 0 to 60mph in 60 secs. Is it possible to model such a locomotive with my equations? If not what other factors besides rolling friction and the engine should I consider?

  • $\begingroup$ Your solution is wrong. Look at my answer to your previous question $\endgroup$ – Aaron Stevens Jul 16 at 16:02
  • $\begingroup$ Is $v$ here velocity? If so, you have issues with dimensions in your expression. You are summing a force ([$kgms^{-2}$] ) with something that is of units [$kgms^{-2}*ms^{-1}$]. Edit unless you forgot to add dimensions for $\mu$ $\endgroup$ – Vangi Jul 16 at 16:12
  • $\begingroup$ @AaronStevens, fixed it $\endgroup$ – El Lago Jul 16 at 16:59
  • $\begingroup$ @Vangi, if you are referring to my first equation, $\mu mgv$ has the units $\frac{s}{m}\times kg \times \frac{m}{s^2} \times \frac{m}{s}$. $\endgroup$ – El Lago Jul 16 at 17:01
  • $\begingroup$ Is the coefficient you report what you think it is? You report it without units, which is usually what it is. But in this case it can't be. $\endgroup$ – Aaron Stevens Jul 16 at 21:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.