TITLE: Self-tightening knot
QUESTION [10 upvotes]: Is there a way, for some finite $L>1$, to tie two length $L$ pieces of rope together, such that any finite force is not enough to pull them apart?
The type of rope I have in mind is something like cylindrical with radius 1, unbreakable, unstretchable, perfectly flexible, non-self-intersecting and has length $L$, but I am open to other models. 
Suppose that rope 1 has ends $A$ and $B$, rope 2 has ends $C$ and $D$. Tie $B$ and $C$ together. Pull $A$ and $D$. Is there a knot that holds for every coefficient of friction $e>0$ and every force $F>0$ applied to the two ends?

REPLY [3 votes]: The answer to your question is obviousely "NO", since you want $L$ to be fixed.
So let me consider the following question instead:

Given the coefficient of friction $e>0$, is there a knot that holds any force? 

I would bet that the answer to this question is "YES".
Obviously we should have $L\to\infty$ as $e\to 0$. 
A right way to proceed would be to take the Ashley's big book of knots suggested by Matt and look for a knot which admits a sequence of iterations of some kind.
Even if you made right guess for iterated knot, actual proof that it holds any force might be difficult.
Now let me explain why I would bet for "YES". 
Assume in addition you are allowed to make any metal ring with zero coefficient of friction  with the rope. Then this 

(source: psu.edu) 
would solve your problem.
The metal ring is marked by black, the green and red ropes alternate;
i.e., they go under and over the black ring in turn.
(The width of the ring should be about 1.