# How is it possible to define work for friction in several dimension?

I have been taught that, given a force field F, the work done by the force over a certain curve $\gamma$ is defined as the line integral of said field along $\gamma$.

But this makes sense only if force can be written as a function of position, as it is the case with gravity, or a spring. Unlike these, friction does not depend on position only: the same body might go through a point in space at two different times and experience a different friction (the magnitude wouldn't change, but direction and sense might).

So how can it make sense to talk about work done by friction if you can't define a force-field for it in the first place?

• If you know it's motion and a relationship between motion and friction you can use that to determine positions and therefore friction at a position. This makes sense only if you can resolve force into a function of position, which is hard to rule out without knowing what information we have/are able to collect. – JMac Apr 19 '17 at 17:44
• @JMac Does this rule out the possibility of calculating work in certain cases? Could you please expand a little more on that? – Nicol Apr 20 '17 at 18:52
• This just adds a time dependence to the equations. While it might become a more difficult problem to solve, I don't see the issue you're having. – StephenG Apr 23 '17 at 12:25
• it does not matter if you are able to find a closed expression for your force or not, if the force changes each time you take the same path or not, the work will be performed regardless of your capacity to calculate it, at each position dW=F.dr, wheter you know F or not – user126422 Apr 23 '17 at 15:59
• @StephenG Do you mean that I should take the line integral of a field $F:(x,y,z,t)\in \mathbb{R^4}\rightarrow \mathbb{R^3}$? – Nicol Apr 23 '17 at 20:57

Wonderful question. You are absolutely correct, we cannot define a force field for friction like we do for gravity. But the formula for work

$$W=\int_\gamma \vec{F}\cdot\vec{dr}$$

still holds. We just need to be a bit more careful about how we write $\vec{F}$. Naturally (like for gravity), we want to write down $\vec{F}$ solely as a function of position. That is, $$\vec{F}=F_x(x,y,z)\,\hat{i}+F_y(x,y,z)\,\hat{j}+F_z(x,y,z)\,\hat{k}.$$

For example, gravity has a constant force field, $\vec{F} = -g\,\hat{k}$. Then, to solve a problem, we would parametrize a curve $\gamma = (x(t),y(t),z(t))$, replace all the x's, y's, and z's in the equation for $\vec{F}$ with these new expressions (in terms of $t$) and then do the line integral.

But we didn't have to write $\vec{F}$ solely in terms of $x$, $y$, and $z$. In fact, the form I wrote above may not particularly useful - we didn't even use it directly! We only used it as a tool to get $\vec{F}$ at each point along the curve, and by extension to get $\vec{F}$ at each time $t$. But if we already know either of these things, then we don't have to go through those gymnastics.

For example, how does the story for kinetic friction go? Well, it always oppose the motion, or is in the direction opposite the velocity. Furthermore, it has a constant size dependent on the mass of the object ($F_{f} = \mu_k N$). So, we know $\vec{F}$!

$$\vec{F} = - F_f \,\vec{\gamma'(t)}$$

Here I wrote the vector sign for emphasis. From here hopefully you can see that, given a curve, we can find the work due to friction. Let me know if you're still having trouble with the details.

I have been taught that, given a force field F, the work done by the force over a certain curve $\gamma$ is defined as the line integral of said field along $\gamma$.

Friction is a force, but is not derived from a force field (in any useful sense). So your statement, while true, is inapplicable. (We were not "given a force field" in this case!) So we fall back on the more general definition that work is net energy expenditure, which in turn equals the time-integral of power applied to the system: $$W = \int \mathbf{F}(t) \cdot \mathbf{v}(t) \, dt$$ where v is velocity and t is time.

I have been taught that, given a force field F, the work done by the force over a certain curve $\gamma$ is defined as the line integral of said field along $\gamma$.

All well and good but perhaps that statement is not the only way to evaluate the work done given that it may not be practical to evaluate what the force field.

The force field for the two examples that you have given, gravity and a spring, are easy to evaluate because they are static.
For your example of gravity that force field might have to be dynamic in that the gravitational attraction at a point might depend on any number of moving masses.
In such a case you could say that there is a force field at an instant of time and then that force field changes at the next instant of time.
Do you then really need to evaluate the whole of the force field at every instant of time to evaluate the work done?
Is it not easier to imagine that at an instant of time there is a force field but only work out what the force is at the particle's position?

Friction is a little different in that the direction of the frictional force depends on the velocity of the particle which is experiencing this frictional force.
In some ways this can make life a little easier in that all one needs to know is the magnitude of the frictional force as its direction will be in the opposite direction to that of the velocity of the particle.
In such a case where the direction of the frictional force is only determined by the velocity of the particle you could set up a scalar force field which gives the magnitude of the frictional force as a function of position, the speed of the particle and time.

Again although you could imagine that there is a scalar force field, you are only concerned about what happens at a certain position.

So throw a ball vertically up in the air and allow it to fall down again.
The force field due to gravity is not a problem as it is static.
The force field due to friction is very much dynamic but your only concern at a given instant of time would be the magnitude of the frictional force which would depend on the speed of the ball.
Coming down the speed of the ball at a particular height might well be different so the force field at that instant will be different from that when the ball was moving up but all you need to do is work out the magnitude of the frictional force at that point.

For friction with some medium:

From Rayleigh dissipation function you can take the friction as function of velocity, and define it as a gradient (special one) of some scalar field.

$$\vec{f}=\vec{f}(\vec{v})=\vec\nabla_v \:(\mathcal{F})$$ where $\vec\nabla_v$ is defined as follow $$\vec\nabla_v =\dfrac{\partial}{\partial v_x}\hat x + \dfrac{\partial}{\partial v_y}\hat y +\dfrac{\partial}{\partial v_z}\hat z$$ In this case you have to know the object's velocity instead the position to know the force at one moment.

For "normal" friction:

In fact you can define the friction due to movement over a surface (over the xy plane for example)

$$\vec f = \begin{cases} f_x\hat x +f_y \hat y& \text{over the surface} \\ 0 & \text{otherwise} \end{cases}$$