I've only recently been introduced to the concept of static vs dynamic friction and I was looking to check my understanding by determining the max angle a wheeled body can climb. In the diagram below the car is moving up the hill.

enter image description here

So by resolving parallel to the hill, the max angle a car could climb at a constant velocity is when $$ N \mu_{s} = N\mu_{k} + mgsin\theta $$ Because $ N = mgcos\theta $ , the equation can be re-arranged to give $$ tan\theta = \mu_{s} - \mu_{k} $$

So is the tan of the max angle simply the difference between the two coefficients? This confused me because a car is capable of climbing a much steeper hill than a train however the difference between rubber and asphalt coefficients is not much greater than the difference between steel on steel and would only give a angle difference of 1 or 2 degrees, so I am wondering if I have made a mistake or a wrong assumption?

  • $\begingroup$ What's $\mu_k$? There's normally no $N \mu_k$ force in a scenario like this for a car (or it is negligible). A train, on the other hand, probably lacks the power/engine torque, to climb steep hills. Also, remember, that friction is distributed somewhat evenly between all wheels, but only some are driving the vehicle. In the case of a train, just the locomotive is powered. $\endgroup$ – LLlAMnYP Jun 1 '16 at 12:11
  • $\begingroup$ I am confused why $ N \mu_{k} $ is negligible as according to en.wikibooks.org/wiki/Physics_Study_Guide/… the coefficient of kinetic friction on asphalt on rubber is 0.67, surely that is too large to be made negligible? $\endgroup$ – OneOneFour Jun 1 '16 at 12:24
  • $\begingroup$ @oneonefour i don't think a wheeled vehicle can climb up a hill without any engine force, what is $N_{\mu (s)}$ if it is force of static friction then your equations are wrong $\endgroup$ – user5954246 Jun 1 '16 at 12:24
  • $\begingroup$ @DeNiSkA , yeah I was a bit confused on the difference between the engine force and the traction force, are they not the same thing? $\endgroup$ – OneOneFour Jun 1 '16 at 12:26
  • $\begingroup$ @OneOneFour by engine force i mean any force which is making the vehicle go up. And traction force and engine force are not the same. $\endgroup$ – user5954246 Jun 1 '16 at 12:31

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.