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. $$\oint \textbf{F}.d\textbf{s}$$ 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)

Answer 1

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.

Answer 2

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

Answer 3

Of Course, I also have the same question. Technically the work done by friction is negative and what that means is the decrease in the energy of the system. But there is similar principle in thermodynamics that decrease in the energy of the system is due to the positive work done by the system. So if we consider a system in which there are frictional forces in action, then the energy of the system must decrease and that decrease in the energy is reflected like the positive work has been done by the frictional forces in that system.

Answer 4

The force of friction, a vector, opposes the actual or impending motion, and the work done by friction can be positive or negative depending on the situation.

We are addressing the mechanical work done by friction on the system $\int_{a}^{b} \vec F \cdot d \vec s$. Typically, physics mechanics assume a rigid body for which there are no "heating" effects (no change in internal energy).

Thermodynamics broadens the definition of work to be "energy transferred between a system and its surroundings, without mass transfer, due to an intensive property difference other than temperature". This broader definition of work includes mechanical work. Work, and heat, can change the internal energy of a system. The work done by the system is the negative of the work done on the system.

For friction, we are addressing the mechanical work done on the system.