If you had a string of very large tensile strength and lowered it into a black hole from a safe distance, would its tension where you are holding it ever surpass its mass per unit length times the speed of light squared? Because if it did, then every inch of string you let out would do more work (let's say it was pulling on a pulley, making it turn, which does useful work for you) than the mass of that inch of string times the speed of light squared, and thus every inch of string you let out from that point on would presumably decrease the mass of the black hole because you could use the energy of the work it does for you to generate another inch of string and have some left over. And if you let the string out to the edge of the universe and let the cosmological constant pull on it instead, then it would function as a perpetual motion machine, one that doesn't even break the laws of thermodynamics because from your perspective the string at the other end would have a negative mass per unit length, and it would, at some finite distance, pass through 0 in switching from positive mass per unit length to negative. The alternative is that no matter how much string you let out, the tension never reaches the string's mass per unit length times the speed of light squared. But that is also very strange because that would mean you could have an infinite supply of string and lower it all the way into the black hole. And what if you have a weight at the end of the string? That would mean that you could surpass a tension of its mass per unit length times the speed of light squared, but as you let more string out, the tension would go DOWN to asymptotically approach its mass per unit length times the speed of light squared, which would mean the highest tension would not be achieved where you are holding it but somewhere in its middle, that there would be a region on the string where for any point in this region, the differential higher up than it, which is holding everything below it, is under less tension than the differential right below it, and that's very counterintuitive.