Sign of work done by friction

Question

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.
  $$oint textbf{F}.dtextbf{s}$$
  Physically it is clear that a system cannot be conservative if friction or other dissipation forces are present, because $F . dtextbf{s}$ 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, $F . dmathbf{s}$ is a typo and should be $mathbf{F} . dtextbf{s}$ (please let me know if I'm wrong)

zhutchens1 · Answer

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

$$ W = oint mathbf{F} cdot dmathbf{r} = oint (Fcostheta) dr, $$

where $theta$ is the angle between the friction $mathbf{F}$ and $dmathbf{r}$. Because friction acts antiparallel, $theta = pi$ and $costheta = -1$ 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.

Jon · Answer

Indeed, it's an errata. You can find a list of them here.

Sign of work done by friction

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