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This isn't exactly an answer to your question, because as it stands your question can't be answered, but I thought I'd post this because the answer really surprised me.

Firstly, the reason your question can't be answered is that you can never get your rope below the event horizon. From the perspective of an observer stationary with respect to the black hole anything dropped into it takes an infinite time to even reach the event horizon, let along cross it. So you could not find yourself holding one end of a rope that had its other end below the black hole - not even if you waited an infinite time.

But provided the bottom end of the rope is above the event horizon then it's perfectly reasonable to ask what force you feel holding the end of the rope, and it's also perfectly reasonable to ask what happens to this force in the limit of reaching the event horizon. So let's do this.

But the force on a rope is hard to calculate because the mass is distributed evenly along its length. To keep things simple replace the rope by some mass $m$ dangling on the end of a weightless rope. With this setup calculating the force is easy.

Suppose the mass $m$ is at a distance $r$ from the centre of a black hole of mass $M$. Twistor59's answer to the question What is the weight equation through general relativity? tells us that relative to a shell observer hovering at a distance $r$ the gravitational acceleration is:

$$ a_{shell} = \frac{GM}{r^2} \frac{1}{\sqrt{1 - \frac{r_s}{r}}} $$

where $r_s$ is the radius of the event horizon. But relative to you standing a large distance from the black hole the time of the shell observer is dilated by a factor of:

$$ t_r = \frac{1}{\sqrt{1 - \frac{r_s}{r}}} $$

And the acceleration you measure far from the black hole is $a_{shell}$ divided by this factor squared so:

$$\begin{align} a &= \frac{GM}{r^2} \frac{1}{\sqrt{1 - \frac{r_s}{r}}} \left( 1 - \frac{r_s}{r} \right) \\ &= \frac{GM}{r^2} \sqrt{1 - \frac{r_s}{r}} \end{align}$$

And the force is simply the acceleration multipled by the mass of your weight $m$:

$$\begin{align} F &= \frac{GMm}{r^2} \sqrt{1 - \frac{r_s}{r}} \\ &= F_N \sqrt{1 - \frac{r_s}{r}} \end{align}$$

where $F_N$ is the force predicted by Newtonian gravity i.e. the force you'd measure in the absence of realtivistic effects.

So the force you would feel is actually less than you'd expect from Newtonian gravity and indeed the force goes to zero as the weight approaches the event horizon. To illustrate this I've graphed the force you would feel compared to the force predicted by Newton's equation:

The force is in units of $GMm$. At distances of around four times the event horizon radius and greater the force is similar to the one calculated by Newton's equation, by as the weight approaches the event horizon the force you feel peaks around $1.4r_s$ then falls to zero at the horizon.

This isn't exactly an answer to your question, because as it stands your question can't be answered, but I thought I'd post this because the answer really surprised me.

Firstly, the reason your question can't be answered is that you can never get your rope below the event horizon. From the perspective of an observer stationary with respect to the black hole anything dropped into it takes an infinite time to even reach the event horizon, let along cross it. So you could not find yourself holding one end of a rope that had its other end below the black hole - not even if you waited an infinite time.

But provided the bottom end of the rope is above the event horizon then it's perfectly reasonable to ask what force you feel holding the end of the rope, and it's also perfectly reasonable to ask what happens to this force in the limit of reaching the event horizon. So let's do this.

But the force on a rope is hard to calculate because the mass is distributed evenly along its length. To keep things simple replace the rope by some mass $m$ dangling on the end of a weightless rope. With this setup calculating the force is easy.

Suppose the mass $m$ is at a distance $r$ from the centre of a black hole of mass $M$. Twistor59's answer to the question What is the weight equation through general relativity? tells us that relative to a shell observer hovering at a distance $r$ the gravitational acceleration is:

$$ a_{shell} = \frac{GM}{r^2} \frac{1}{\sqrt{1 - \frac{r_s}{r}}} $$

where $r_s$ is the radius of the event horizon. But relative to you standing a large distance from the black hole the time of the shell observer is dilated by a factor of:

$$ t_r = \frac{1}{\sqrt{1 - \frac{r_s}{r}}} $$

And the acceleration you measure far from the black hole is $a_{shell}$ divided by this factor squared so:

$$\begin{align} a &= \frac{GM}{r^2} \frac{1}{\sqrt{1 - \frac{r_s}{r}}} \left( 1 - \frac{r_s}{r} \right) \\ &= \frac{GM}{r^2} \sqrt{1 - \frac{r_s}{r}} \end{align}$$

And the force is simply the acceleration multipled by the mass of your weight $m$:

$$\begin{align} F &= \frac{GMm}{r^2} \sqrt{1 - \frac{r_s}{r}} \\ &= F_N \sqrt{1 - \frac{r_s}{r}} \end{align}$$

where $F_N$ is the force predicted by Newtonian gravity i.e. the force you'd measure in the absence of realtivistic effects.

So the force you would feel is actually less than you'd expect from Newtonian gravity and indeed the force goes to zero as the weight approaches the event horizon. To illustrate this I've graphed the force you would feel compared to the force predicted by Newton's equation:

The force is in units of $GMm$. At distances of around four times the event horizon radius and greater the force is similar to the one calculated by Newton's equation, by as the weight approaches the event horizon the force you feel peaks around $1.4r_s$ then falls to zero at the horizon.

This isn't exactly an answer to your question, because as it stands your question can't be answered, but I thought I'd post this because the answer really surprised me.

Firstly, the reason your question can't be answered is that you can never get your rope below the event horizon. From the perspective of an observer stationary with respect to the black hole anything dropped into it takes an infinite time to even reach the event horizon, let along cross it. So you could not find yourself holding one end of a rope that had its other end below the black hole - not even if you waited an infinite time.

But provided the bottom end of the rope is above the event horizon then it's perfectly reasonable to ask what force you feel holding the end of the rope, and it's also perfectly reasonable to ask what happens to this force in the limit of reaching the event horizon. So let's do this.

But the force on a rope is hard to calculate because the mass is distributed evenly along its length. To keep things simple replace the rope by some mass $m$ dangling on the end of a weightless rope. With this setup calculating the force is easy.

Suppose the mass $m$ is at a distance $r$ from the centre of a black hole of mass $M$. Twistor59's answer to the question What is the weight equation through general relativity? tells us that relative to a shell observer hovering at a distance $r$ the gravitational acceleration is:

$$ a_{shell} = \frac{GM}{r^2} \frac{1}{\sqrt{1 - \frac{r_s}{r}}} $$

where $r_s$ is the radius of the event horizon. But relative to you standing a large distance from the black hole the time of the shell observer is dilated by a factor of:

$$ t_r = \frac{1}{\sqrt{1 - \frac{r_s}{r}}} $$

And the acceleration you measure far from the black hole is $a_{shell}$ divided by this factor squared so:

$$\begin{align} a &= \frac{GM}{r^2} \frac{1}{\sqrt{1 - \frac{r_s}{r}}} \left( 1 - \frac{r_s}{r} \right) \\ &= \frac{GM}{r^2} \sqrt{1 - \frac{r_s}{r}} \end{align}$$

And the force is simply the acceleration multipled by the mass of your weight $m$:

$$\begin{align} F &= \frac{GMm}{r^2} \sqrt{1 - \frac{r_s}{r}} \\ &= F_N \sqrt{1 - \frac{r_s}{r}} \end{align}$$

where $F_N$ is the force predicted by Newtonian gravity i.e. the force you'd measure in the absence of realtivistic effects.

So the force you would feel is actually less than you'd expect from Newtonian gravity and indeed the force goes to zero as the weight approaches the event horizon.

This isn't exactly an answer to your question, because as it stands your question can't be answered, but I thought I'd post this because the answer really surprised me.

Firstly, the reason your question can't be answered is that you can never get your rope below the event horizon. From the perspective of an observer stationary with respect to the black hole anything dropped into it takes an infinite time to even reach the event horizon, let along cross it. So you could not find yourself holding one end of a rope that had its other end below the black hole - not even if you waited an infinite time.

But provided the bottom end of the rope is above the event horizon then it's perfectly reasonable to ask what force you feel holding the end of the rope, and it's also perfectly reasonable to ask what happens to this force in the limit of reaching the event horizon. So let's do this.

But the force on a rope is hard to calculate because the mass is distributed evenly along its length. To keep things simple replace the rope by some mass $m$ dangling on the end of a weightless rope. With this setup calculating the force is easy.

Suppose the mass $m$ is at a distance $r$ from the centre of a black hole of mass $M$. Twistor59's answer to the question What is the weight equation through general relativity? tells us that relative to a shell observer hovering at a distance $r$ the gravitational acceleration is:

$$ a_{shell} = \frac{GM}{r^2} \frac{1}{\sqrt{1 - \frac{r_s}{r}}} $$

where $r_s$ is the radius of the event horizon. But relative to you standing a large distance from the black hole the time of the shell observer is dilated by a factor of:

$$ t_r = \frac{1}{\sqrt{1 - \frac{r_s}{r}}} $$

And the acceleration you measure far from the black hole is $a_{shell}$ divided by this factor squared so:

$$\begin{align} a &= \frac{GM}{r^2} \frac{1}{\sqrt{1 - \frac{r_s}{r}}} \left( 1 - \frac{r_s}{r} \right) \\ &= \frac{GM}{r^2} \sqrt{1 - \frac{r_s}{r}} \end{align}$$

And the force is simply the acceleration multipled by the mass of your weight $m$:

$$\begin{align} F &= \frac{GMm}{r^2} \sqrt{1 - \frac{r_s}{r}} \\ &= F_N \sqrt{1 - \frac{r_s}{r}} \end{align}$$

where $F_N$ is the force predicted by Newtonian gravity i.e. the force you'd measure in the absence of realtivistic effects.

So the force you would feel is actually less than you'd expect from Newtonian gravity and indeed the force goes to zero as the weight approaches the event horizon. To illustrate this I've graphed the force you would feel compared to the force predicted by Newton's equation:

The force is in units of $GMm$. At distances of around four times the event horizon radius and greater the force is similar to the one calculated by Newton's equation, by as the weight approaches the event horizon the force you feel peaks around $1.4r_s$ then falls to zero at the horizon.

This isn't exactly an answer to your question, because as it stands your question can't be answered, but I thought I'd post this because the answer really surprised me.

Firstly, the reason your question can't be answered is that you can never get your rope below the event horizon. From the perspective of an observer stationary with respect to the black hole anything dropped into it takes an infinite time to even reach the event horizon, let along cross it. So you could not find yourself holding one end of a rope that had its other end below the black hole - not even if you waited an infinite time.

But provided the bottom end of the rope is above the event horizon then it's perfectly reasonable to ask what force you feel holding the end of the rope, and it's also perfectly reasonable to ask what happens to this force in the limit of reaching the event horizon. So let's do this.

But the force on a rope is hard to calculate because the mass is distributed evenly along its length. To keep things simple replace the rope by some mass $m$ dangling on the end of a weightless rope. With this setup calculating the force is easy and the result is surprising.

Suppose the mass $m$ is at a distance $r$ from the centre of a black hole of mass $M$. Twistor59's answer to the question What is the weight equation through general relativity? tells us that relative to a shell observer hovering at a distance $r$ the gravitational acceleration is:

$$ a_{shell} = \frac{GM}{r^2} \frac{1}{\sqrt{1 - \frac{r_s}{r}}} $$

where $r_s$ is the radius of the event horizon. But relative to you standing a large distance from the black hole the time of the shell observer is dilated by a factor of:

$$ t_r = \frac{1}{\sqrt{1 - \frac{r_s}{r}}} $$

And the acceleration you measure far from the black hole is $a_{shell}$ divided by this factor squared so:

$$\begin{align} a &= \frac{GM}{r^2} \frac{1}{\sqrt{1 - \frac{r_s}{r}}} \left( 1 - \frac{r_s}{r} \right) \\ &= \frac{GM}{r^2} \sqrt{1 - \frac{r_s}{r}} \end{align}$$

And the force is simply the acceleration multipled by the mass of your weight $m$:

$$\begin{align} F &= \frac{GMm}{r^2} \sqrt{1 - \frac{r_s}{r}} \\ &= F_N \sqrt{1 - \frac{r_s}{r}} \end{align}$$

where $F_N$ is the force predicted by Newtonian gravity i.e. the force you'd measure in the absence of realtivistic effects.

So the force you would feel is actually less than you'd expect from Newtonian gravity and indeed the force goes to zero as the weight approaches the event horizon. Isn't that surprising?

This isn't exactly an answer to your question, because as it stands your question can't be answered, but I thought I'd post this because the answer really surprised me.

Firstly, the reason your question can't be answered is that you can never get your rope below the event horizon. From the perspective of an observer stationary with respect to the black hole anything dropped into it takes an infinite time to even reach the event horizon, let along cross it. So you could not find yourself holding one end of a rope that had its other end below the black hole - not even if you waited an infinite time.

But provided the bottom end of the rope is above the event horizon then it's perfectly reasonable to ask what force you feel holding the end of the rope, and it's also perfectly reasonable to ask what happens to this force in the limit of reaching the event horizon. So let's do this.

But the force on a rope is hard to calculate because the mass is distributed evenly along its length. To keep things simple replace the rope by some mass $m$ dangling on the end of a weightless rope. With this setup calculating the force is easy.

Suppose the mass $m$ is at a distance $r$ from the centre of a black hole of mass $M$. Twistor59's answer to the question What is the weight equation through general relativity? tells us that relative to a shell observer hovering at a distance $r$ the gravitational acceleration is:

$$ a_{shell} = \frac{GM}{r^2} \frac{1}{\sqrt{1 - \frac{r_s}{r}}} $$

where $r_s$ is the radius of the event horizon. But relative to you standing a large distance from the black hole the time of the shell observer is dilated by a factor of:

$$ t_r = \frac{1}{\sqrt{1 - \frac{r_s}{r}}} $$

And the acceleration you measure far from the black hole is $a_{shell}$ divided by this factor squared so:

$$\begin{align} a &= \frac{GM}{r^2} \frac{1}{\sqrt{1 - \frac{r_s}{r}}} \left( 1 - \frac{r_s}{r} \right) \\ &= \frac{GM}{r^2} \sqrt{1 - \frac{r_s}{r}} \end{align}$$

And the force is simply the acceleration multipled by the mass of your weight $m$:

$$\begin{align} F &= \frac{GMm}{r^2} \sqrt{1 - \frac{r_s}{r}} \\ &= F_N \sqrt{1 - \frac{r_s}{r}} \end{align}$$

where $F_N$ is the force predicted by Newtonian gravity i.e. the force you'd measure in the absence of realtivistic effects.

So the force you would feel is actually less than you'd expect from Newtonian gravity and indeed the force goes to zero as the weight approaches the event horizon.