0
$\begingroup$

In Goldstein's classical mechanics (3rd ed.) we read:

"The independence of W12 on the particular path implies that the work done around such a closed circuit is zero,i.e. $$\oint \textbf{F}.d\textbf{s}$$ Physically it is clear that a system cannot be conservative if friction or other dissipation forces are present, because $F . d\textbf{s}$ due to friction is always positive and the integral cannot vanish."

My question is: why should the work due to friction be "always positive"? Shouldn't it be nonzero instead?

Also, $F . d\mathbf{s}$ is a typo and should be $\mathbf{F} . d\textbf{s}$ (please let me know if I'm wrong)

$\endgroup$
  • $\begingroup$ In Goldstein $\textbf F$ is defined as the external force not the frictional force. $\endgroup$ – Farcher Jun 24 '18 at 7:36
2
$\begingroup$

Perhaps I misunderstand the context of Goldstein's writing, but work due to friction should be negative:

Friction always acts antiparallel to the displacement/velocity. So, when computing work from friction, drag, etc, you find that

$$ W = \oint \mathbf{F} \cdot d\mathbf{r} = \oint (F\cos\theta) dr, $$

where $\theta$ is the angle between the friction $\mathbf{F}$ and $d\mathbf{r}$. Because friction acts antiparallel, $\theta = \pi$ and $\cos\theta = -1$ always. Then,

$$ W = - \oint Fdr, $$

which is always negative because $F$ and $dr$ are vector magnitudes, and thus always positive. This is why friction is dissipative, it steals energy from the system in the form of heat and deformation. Even in the case of a line integral as presented here, each component/leg should be negative thus creating a total negative work.

Of course it makes sense that the friction force is nonconservative -- the work expelled certainly depends on the path. If you have ever moved furniture into a new apartment, of course you push it the shortest possible path, for this minimizes the energy you need. If you push it around aimlessly you will expend more energy than needed.

$\endgroup$
  • $\begingroup$ Think of two blocks on top of one another with a force applied to the bottom block to accelerate it. The work done on the top block due to the frictional force to make the top block stay in contact with the bottom block is positive? $\endgroup$ – Farcher Jun 24 '18 at 6:40
  • $\begingroup$ @Farcher The example you mentioned is the reason I think "nonzero" is more appropriate. The work on the bottom block will be negative due to friction and the work on the top block is positive. But it seems unlikely to me that Goldstein would contain such an obvious mistake! $\endgroup$ – Ali Jun 24 '18 at 7:19

Your Answer

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

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