Many times, people often claim nothing "weird" occurs at the event
horizon of a black hole.  They point to the case of an astronaut in
freefall falling through and experiencing no ill effect at the crossing,
which occurs in finite proper time.  But what of an astronaut NOT in
freefall?  In a related problem, people often ask something to the
effect of, "Why can't you lower a string through the event horizon and
then pull it back out again."  Often the answer given is muddled,
incorrect or a brush off.  In fact, the static force or weight of a
static or suspended object diverges as one approaches the event horizon.

Two references are:

Gravitation, by Misner, Thorne and Wheeler;        aka MTW
Essential Relativity, Wolfgang Rindler;                aka Rindler

Fact #1.  The radial proper distance to the event horizon (r=2M) from
   any Schwarzchild "r" is finite.  See MTW, eq 31.9, pg 824.

Fact #2.  The LOCAL static acceleration or "weight" of a stationary
   observer diverges or goes to infinity as "r" approaches 2M.
   The per unit weight of an object (ie, local proper acceleration)
   in conventional units is:

          (GM/r^2) (1 - 2GM/(c^2 r))^(-1/2)

   See Rindler, eq 8.71, pg 148.

[note: An easy way to remember the above is Newton's formula devided by
the redshift factor.]

By now the cogs in the brain of the astute physicist or mathematician
should be turning and he or she should begin to realize that the WEIGHT
DIVERGES OVER A FINITE DISTANCE! And this diverging is occurring
OUTSIDE the event horizon.  So the STATIC tidal forces, defined to be:
d(weight)/d(proper distance), or the weight gradient, also diverges as
one approaches the event horizon.

So now the budding general relativist is thinking to himself, "I'm so
confused. How can this be? I thought there are finite tidal forces at
the event horizon?"

Well let's look at this situation from a free-falling observer.  Since
the event horizon represents a stationary light-like surface, by
definition any infaller falls thru the horizon locally at the speed of
light.  So he is locally Lorentz contracted to zero thickness with
respect to a stationary observer at the horizon (for the purists, use
limits and pretend you are a small distance epsilon outside the horizon,
and then let epsilon -> 0.) Also at the event horizon the angle of
escape for light (and thereby anything else as well) is zero degrees
with respect to the vertical (in 3 space coordinates),  all freefalling
trajectories are vertical at the event horizon.  Since the infaller is
of zero thickness in the radial direction (ie, FLAT) his d(proper
distance) in the radial direction is ZERO.  To get the tidal force on
the faller we multiply the static tidal force times the faller's radial
extent.  We get:

        infinity * zero

which is an indeterminate form.  If you work this out in detail you
will see that the above product remains finite for a free falling
object. So the tidal forces on a freefaller remain finite, although
they vary depending on his 4-velocity.  Since the faller is in
freefall, his static weight is of course zero.

  Now an astronaut who is able to stick his arm thru the event
horizon, but remain himself outside the horizon, cannot be in
freefall. Therefore he will experience inifinite tidal forces across
the length of his arm, and it will of course be torn off.  The same
thing with a string.