I think what you're aiming for here is the statement: somewhere in a black hole spacetime the Kretschmann scalar goes to infinity.

The Kretschmann scalar is the square of the Riemann tensor, or in notation R_{\alpha\beta\gamma\delta}R^{\alpha\beta\gamma\delta}.

There are no forces or coordinates or anything like that to even consider; it's a curvature invariant. ( https://en.wikipedia.org/wiki/Curvature_invariant_(general_r... )

If the Kretschmann scalar is non-infinite at a point then there is no gravitational singularity there.

If you really want to think about a black hole spacetime in a coordinate system, then in Schwarzschild coordinates for a Schwarzschild black hole, K = \frac{48M^2}{r^6} where K is the Kretschmann scalar. You can see that when r=0, K is undefined.

Other coordinate systems and black hole spacetimes are available.

However, generically, one can distinguish between a coordinate singularity and a gravitational singularity by calculating the Kretschmann scalar at a point. If the Kretschmann scalar is non-infinite then it cannot be the latter, whatever the coordinate basis is.

Continuing to think generically and without choosing coordinates:

"distance" -> it's still interesting for considering non-gravitational body forces of objects as we take the Kretschmann scalar around the objects towards infinity.

Essentially, tossing in one test particle just causes it to free-fall forever; tossing in two classical objects simultaneously (in Schwarzschild coordinates on Schwarzschild black hole spacetime)[1] means at some object- proper-time each object will collide with the other. They will still free-fall forever; there is a notional "down" if we have a gradient in the Kretschmann scalar field. However, the free-falling trajectories available narrow even after contact. For classical objects that are mutually repulsive, pressure[2] builds as they continue on converging free-falling trajectories and eventually classical forces cannot keep the two objects separated; they are forced into each other. The resulting compacted matter, whatever its nature, continues to free-fall forever.

Quantum-mechanically, who knows? Firstly we're way into the ultraviolet completion of the Standard Model, and secondly there's no reason to expect quantum mechanical effects to contribute to the stress-energy tensor in ways that do not look sufficiently classical under averaging or other tricks that let one continue using the geodesics of the black hole background metric or make only small perturbations to them. And we're ignoring quantum-mechanical effects outside the black hole, like Hawking radiation.

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[1] The simultaneity condition is a handwavy way of absorbing messy behaviour as we drop two identical charges into a black hole. See Cohen & Wald (1971), http://adsabs.harvard.edu/abs/1971JMP....12.1845C

[2] Pressure here is encoded in the stress-energy tensor T as T^{ii}, i \neq 0. Since the Einstein Field Equations say essentially G = T, we have to deal with pressure as it becomes non-negligible, because as a component of T it must be reflected in the curvature represented here as the Einstein tensor G. This is a practical example of where General Relativity is a dynamical non-linear theory. Two classical objects with identical electric charge within the horizon of a black hole makes a surprising bookkeeping mess! (compare the Schwarzschild metric with the https://en.wikipedia.org/wiki/Reissner%E2%80%93Nordstr%C3%B6... and there we are talking about what's outside the black hole and ignoring quantum mechanical effects altogether).