Does it make sense to compute moment of force about any axis?

This Wikipedia article gives that

If the unit vector along the axis is $e$, the moment of force about the axis is defined as $$M=e \cdot M = e \cdot (r \times F)$$

Then gives an example

which I read to be about the axis of unit vector $e_z$.

And then computes moment about axis $e_z$ as

$$M_z=e_z \cdot M = -Fx$$

So this is a moment about the axis of unit vector $e_z$. But does it make sense to have moment about axis $e_x$ (or $e_y$ for that matter) given the picture? Can one compute the moment about any axis?

• You might have to explain your difficulty a bit more. In the picture the moment about $z$ is the "natural" (or perhaps, "simplest") moment. $M_x = M_y = 0$ in the case shown. I can't quite tell what's giving you trouble. – garyp Aug 15 '16 at 17:44
• @garyp Oh wait you're right. I was mixing the "rotation about an axis" to "rotation that rotates the axis" so I was thinking that since the $e_x$ axis rotates, then it's "about axis $e_x$". But my question is still, is there a moment about any axis? Also why is $M_x=M_y=0$? – mavavilj Aug 15 '16 at 17:47
• There can be a moment about any axis, but there isn't in this case. $\vec{r}$ is oriented toward $e_x$, and $\vec{F}$ is oriented toward $-e_y$, so $\vec{r}\times\vec{F}$ is oriented directly toward $-e_z$. There is no component in the other directions. – garyp Aug 15 '16 at 17:54

Yes, it is possible to compute the moment of force (or torque) about any axis. As long as $\vec{r}$ is the vector from the axis to the point where the force is applied, $\vec{e}$ has unit norm, and $\vec{r}$ is perpendicular to $\vec{e}$, it is possible to compute the torque.
For example, the torque about $\vec{e}_y$ can be computed using the formula in your question. We see that $\vec{r} \times \vec{F}$ gives a vector that is anti-parallel to $\vec{e}_z$. Taking the dot product between this vector and $\vec{e}_y$ gives zero because they are perpendicular. $$M_y = \vec{e}_y \cdot (\vec{r} \times \vec{F}) = \vec{e}_y \cdot -r F \vec{e}_z = -r F \underbrace{\vec{e}_y \cdot \vec{e}_z}_{= 0} = 0$$ This makes sense intuitively because pushing an object along $\vec{e}_y$ would not make this object rotate around $\vec{e}_y$.