The $n$-th moment of mass of a point mass $m$ with respect to a point located at $\boldsymbol r$ is $$ \mu_n(m) = m\boldsymbol r^n $$ So the 0th moment is the total mass $m$, the 1st moment is $m\boldsymbol r$, the 2nd is $I=mr^2$, the 3rd is $mr^2\boldsymbol r$, the 4th is $mr^4$, etc.
The 1st moment of a vector quantity $\boldsymbol F$, such as force, with respect to a point located at $\boldsymbol r$, is $$ \mu_1(\boldsymbol F) = \boldsymbol r\times\boldsymbol F $$ And I assume that its 0th moment is just the total force $\boldsymbol F$. So my question is: what would be the 2nd, 3rd, etc. moments look like? I haven't been able to find even a passing reference about it. The simplest explanation would be that they are all zero because $$ \mu_2(\boldsymbol F) = \boldsymbol r\times\boldsymbol r\times\boldsymbol F = \boldsymbol 0 $$ and so on. But I don't know whether this is indeed the correct expression for the $n$-th moment of a vector quantity. It may very well be that $$ \mu_2(\boldsymbol F) = r^2\boldsymbol F \neq \boldsymbol 0 $$ $$ \mu_3(\boldsymbol F) = r^2\boldsymbol r\times\boldsymbol F \neq \boldsymbol 0 $$ etc, and I'd be none the wiser. So is there a more general expression for the moment of a vector (tensor?) quantity?