There is nothing wrong here.  Let $X$ be irreducible of dimension $n$.  Any sheaf of dimension less than $n$ is torsion, since it is annihilated by a function vanishing on the support of the sheaf.  Sheaves which are pure of dimension $n$ are torsion free, since if they were not torsion free they would have a torsion subsheaf supported on a proper subvariety.

A pure sheaf with irreducible support is torsion-free when view as a sheaf on its support.