0
$\begingroup$

I have searched very much on line, both in this site and elsewhere, but found no proof of whether the normal force is conservative or is not, in general.

Clearly, if the force is orthogonal to the velocity, $\mathbf{F}\cdot d\mathbf{r}=\mathbf{F}\cdot\mathbf{v}dt=0$, and the force does no work, as explained for example here, so it is conservative in that sense in that case.

But what happens if the normal force does work, as it can happen to do? And how can what happens be mathematically proved? I heartily thank you for any answer!

Improve this question
$\endgroup$
2
  • $\begingroup$ It would be interesting to know the reason of the downvote: have I stated anything incorrect? Thank you very much! $\endgroup$
    – Self-teaching worker
    Commented Apr 17, 2015 at 17:06
  • 1
    $\begingroup$ down votes rarely get a comment. don't worry about it $\endgroup$
    – aaaaa says reinstate Monica
    Commented Apr 20, 2015 at 19:42

1 Answer 1

Reset to default
2
$\begingroup$

$\vec F \cdot \mathrm{d} \vec r$ does not determine if a force is conservative or not. All normal forces (conservative or not) produce work equal to $\vec F \cdot \mathrm{d} \vec r$ but what determines if they are conservative is the integral of $\vec F \cdot \mathrm{d} \vec r$ in a closed loop. If that is equal to zero i.e. if $\oint \vec F \cdot \mathrm{d} \vec r = 0$ then it is conservative because no energy is lost in that loop. It is like gravity. You throw something upwards and no energy is lost(in a vacuum). So the integral I mentioned is zero.

Improve this answer
$\endgroup$
16
  • 1
    $\begingroup$ I did not fully understand your comment man.I can tell you this:if a force does not cause mechanical energy to be transformed into thermal energy,then it is conservative and the integral is equal to zero.To intuitively think about it,if a particle gets into a loop,the motions up and right produce positive work(in this example) and the motions down and to the left produce negative work that is equal(in absolute) to the positive work.So at the end of the loop,no overall work is done $\endgroup$
    – TheQuantumMan
    Commented Mar 29, 2015 at 18:28
  • 1
    $\begingroup$ Well,for a force to be conservative,test results must show you that no energy is lost in a loop.So,you know from intuition that friction is not conservative,but in order to know if the electric force is conservative,then you must test it and see that no energy is lost.If it is conservative,then you define the integral over the closed loop to be zero.I think there might also be a derivation of this,although i do not remember at the moment how to derive this.If there actually is a derivation,then you must check the very basic maxwell equations and try to use them to derive it. $\endgroup$
    – TheQuantumMan
    Commented Mar 29, 2015 at 21:24
  • 1
    $\begingroup$ If you find a way to do so,then some calculus may be required.In that case,you should just google the mathematics that you need in order to help yourself. $\endgroup$
    – TheQuantumMan
    Commented Mar 29, 2015 at 21:25
  • 1
    $\begingroup$ The integral is equal to zero because the force is ALWAYS normal to dr,so it never produces work,no matter how the MAGNITUDE of the force changes.It might change in magnitude but not in direction,it is always normal $\endgroup$
    – TheQuantumMan
    Commented Mar 31, 2015 at 19:28
  • 1
    $\begingroup$ No you are not wrong.It indeed produces work,but that force is proportional to the gravitational one(only),and thus is conservative.So in a full loop of the ferris wheel,the net work produced will be zero.It produces two equal with opposite signs works when going up and going down. $\endgroup$
    – TheQuantumMan
    Commented Mar 31, 2015 at 19:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.