What is the formula for the force of friction as a vector? The formula I've grown up with is $F_{fr} = \mu F_{N}$, where

*

*$\mu$ is the coefficient of friction between the object and the surface

*$F_{N}$ is the normal force of the surface acting upon the object

*$F_{fr}$ is the force of friction between the object and the surface

Treating them strictly as vectors, this formula above seems impossible to take literally, since it would imply that $F_{fr}$ and $F_{N}$ are parallel.  But in fact if you draw a free-body diagram, they ought to be orthogonal.
EDIT:
OK, so if this formula above is only for computing magnitude, then treating these forces as vectors, it seems correct to write
$\|F_{fr}\| = \mu\|F_{N}\|$ (where $\|\cdot\|$ represents the norm function)
and if we pair this with the understanding that the force of friction has the opposite direction as that of the accelerating force, this would seem to give a complete definition/formula for computing $F_{fr}$.
Is this the only way to think the force of friction?  Or can it be defined more concisely using some vector equation instead?
 A: Here's how I model the situation.
Typically, there are two cases in the frame of the surface:

*

*If there is no sliding between surfaces, with the system in static equilibrium,
then we have static friction and "$\vec F_f$  is whatever it needs to be to satisfy Newton's Law":

$$\vec 0 \stackrel{eqbm}{=} { \require{cancel} m {\cancelto{\vec 0}{\!\!\vec a}}}=\vec F_{net} \equiv \vec F_{f} + \sum_{i\neq f} \vec F_i.$$

Thus, in this case, $$\vec F_{f} \stackrel{eqbm}{=} - \sum_{i\neq f} \vec F_i.$$
In particular, it need not be as large as $\mu_s N$...it could be less (even zero) if Newton's Law requires it.
(We certainly don't want $F_f=\mu_s F_N$ if it were to violate the static equilibrium condition.)

In general, what is true about static friction is that:
whatever its magnitude is, it's no greater than $\mu_s F_N$: that is,
$F_{f,s} \not\gt \mu_s N$.

(Soapbox: I think writing $F_{f,s} \leq \mu_s F_N$ invites students to mistakenly "choose" $\mu_s F_N$ since that symbol suggests "it could be $\mu_sF_N$". My "not greater than" symbol says "what it can't be"... look elsewhere [Newton's Laws] for what it is.)



*If there is sliding between surfaces (so  $v_{rel}>0$), then
the kinetic friction opposes the sliding:
$$\vec F_f = -\hat v_{rel} (\mu_k F_N),$$
where $F_N$ is the magnitude of the normal force.
A: 
Treating them strictly as vectors, this formula above seems impossible
to take literally, since it would imply that $F_{fr}$ and $F_{N}$ are
parallel.  But in fact if you draw a free-body diagram, they ought to
be orthogonal.

The fact that the forces are orthogonal does not mean the friction force and normal force vectors are related by the friction equation by
$$\vec f_{f}= \mu\vec F_N$$
The equation
$$f_{f}=\mu F_N$$
Relates the magnitude of the friction force to the magnitude of the normal force. It is not a vector equation. The friction force vector always acts parallel to and in opposition to some applied force vector. That would be a vector equation.

It seems that $F_{fr} = -\mu\frac{\|F_{N}\|}{\|F_{a}\|}F_{a}$ does the
job.

I'm not quite sure why you using the ratio of the magnitude of the normal force to applied force, but I don't believe it "does the job" for a few reasons.
First of all you have to consider the two types of friction: Static and kinetic.
Let's start with static friction.
In the case of static friction, the normal force is only applicable when the applied force equals the maximum possible static friction force which is $\mu_{s}F_N$. Up to that point the static friction force is a variable that always matches the applied force and is independent of the normal force so that the object involved is in equilibrium. That is
$$\vec F_{fr}=-\vec F_a$$
when
$$F_{a}\lt\mu_{s}F_N$$
Now regarding kinetic friction.
When the maximum possible static friction force is reached, friction transitions to kinetic friction (generally lower than static). The magnitude of the kinetic friction is
$$F_{fr}=\mu_{k}F_N$$
The kinetic friction opposes the applied force, but unlike static friction, it is generally considered independent of magnitude of the applied force and the velocity.
Consequently, I don't see how your equation applies to kinetic friction since the kinetic friction force $F_{fr}$ is independent of $F_a$.
Hope this helps.
A: It seems that $\mathbf{F_{fr}} = -\mu\frac{\|\mathbf{F_{N}}\|}{\|\mathbf{F_{a}}\|}\mathbf{F_{a}}$ does the job,
where $\mathbf{F_{a}}$ is the accelerating force acting in an opposite direction to $\mathbf{F_{fr}}$.
A: Your question is clear and the answer is also clear: no, there is no vector equation.
The only equation that makes sense is on the moduli, as you have written.
The forces are indeed orthogonal, as you pointed out. The friction force can only be parallel to the contact surface.
And whatever the direction of the total force $F$ applied on the object, only the normal component $F_N$ enters your equation, and being normal to the surface, this force is indeed orthogonal to the friction force.
Of course the friction force need not be orthogonal to the total force. On a slope, weight is vertical, but the is friction not horizontal. Friction is along the slope, proportional to the normal component of the weight, orthogonal to the slope.
