Timeline for Normal force, work and conservativity
Current License: CC BY-SA 3.0
20 events
| when toggle format | what | by | license | comment | |
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| S Apr 20, 2015 at 20:25 | history | suggested | Gonenc | CC BY-SA 3.0 |
texified and made minor improvements
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| Apr 20, 2015 at 20:14 | review | Suggested edits | |||
| S Apr 20, 2015 at 20:25 | |||||
| Apr 2, 2015 at 14:09 | comment | added | TheQuantumMan | (Fn-mg).dr=Km where K is the y component of the centrifugal acceleration | |
| Apr 2, 2015 at 14:04 | comment | added | Self-teaching worker | Thank you again for your answers! I meant: if $\mathbf{F}_N=k m\mathbf{g}$ for some $k$ then $\oint\mathbf{F}_Nd\mathbf{r}=k\oint m\mathbf{g}d\mathbf{r}=k\cdot 0$, but in the case that $F_N$ is not constant while going up, for ex., I cannot realise why the integral is null. I'm still searching the Web for a mathematical proof of the conservativity of the normal force, but I find nothing... | |
| Apr 2, 2015 at 13:59 | comment | added | TheQuantumMan | Yes it is non zero,but why does that make the force non conservative?Think gravity,it always has y components non zero | |
| Apr 2, 2015 at 13:57 | comment | added | Self-teaching worker | I was supposing that the $y$ component of the acceleration is non-zero when going up: $F_N>mg$ | |
| Apr 2, 2015 at 8:46 | comment | added | TheQuantumMan | That vertical normal force always opposes the gravitational force which is conservative.if you have no friction, I can not see why the vertical normal force is not conservative | |
| Apr 1, 2015 at 20:44 | comment | added | Self-teaching worker | Thank you again! I meant: constant vertical component of velocity. I understand that the seat linked to the centre of the wheel exerces a centripetal force on the passenger's side, but I am not able to prove to myself that the closed path integral of the vertical normal force acting on the seated passenger is null... | |
| Apr 1, 2015 at 8:50 | comment | added | TheQuantumMan | It never has constant speed.It is a circular motion.So the velocity always changes direction.What we forgot in the case of a ferris wheel is the force from the metal that goes to the center of the ferris.That is the centrifugal force | |
| Apr 1, 2015 at 8:29 | comment | added | Self-teaching worker | But I've supposed that, when going upwards, the wheel accelerates ($F_N>mg$ and $\mathbf{F}_N\not\propto m\mathbf{g}$) and, when going downwards, it has constant speed ($\mathbf{F}_N=- m\mathbf{g}$)... I still cannot see how the path integral can be null... :-( Thank you again! | |
| Mar 31, 2015 at 19:36 | comment | added | TheQuantumMan | 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. | |
| Mar 31, 2015 at 19:32 | comment | added | Self-teaching worker | I mean: the normal force exerced by the seat of a Ferris wheel on a person when it ascends while accelerating and then descends at constant velocity isn't orthogonal to $d\mathbf{r}$... am I wrong? Thank you a lot again!!! | |
| Mar 31, 2015 at 19:28 | comment | added | TheQuantumMan | 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 | |
| Mar 31, 2015 at 19:07 | comment | added | Self-teaching worker | Thank you so much for all the information and suggestions! I'm particularly confused by what happens, for example, in a Ferris wheel going upward while accelerating ($F_N>mg$) and then, with a reduction of the normal force, going downward with constant velocity ($F_N=mg$): how can the integral be null if the force is different on the two "side paths"? Thank you again! | |
| Mar 29, 2015 at 21:25 | comment | added | TheQuantumMan | 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. | |
| Mar 29, 2015 at 21:24 | comment | added | TheQuantumMan | 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. | |
| Mar 29, 2015 at 18:28 | comment | added | TheQuantumMan | 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 | |
| Mar 29, 2015 at 14:00 | history | edited | TheQuantumMan | CC BY-SA 3.0 |
edited body
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| Mar 29, 2015 at 12:07 | comment | added | Self-teaching worker | Thank you for the answer!!! Could you explain that $\oint\mathbf{F}_N\cdot d\mathbf{r}=0$ with a mathematical proof? If the normal force $\mathbf{F}_n$ were proportional to a rotation of the gravity, I would understand, but $\mathbf{F}_n$ is not, in general, like that: for ex. the normal force exerced by an accelerating elevator. Thank you again! | |
| Mar 29, 2015 at 10:19 | history | answered | TheQuantumMan | CC BY-SA 3.0 |