# Not sure how I am wrong

1. Jul 26, 2004

### TopCat

Here's the problem:
A perfectly flexible cable has length L. Initially, the cable is at rest, with a length x0 of it hanging vertically over the edge of a table. Neglecting friction, compute the length hanging over the edge after a time t. Assume that the sections of the cable remain straight during the motion.

The way I see it, the solution will be a formula such that after a certain time t', the solution of the formula is always L, since after t' all of the cable has fallen off of the table. Ignoring a finite L, the physics of the problem should lead to
x = |-x0 - .5gt²|, where x is the length of cable hanging over the edge. Setting x equal to L I see that t' = √[2(L-x0)/g]. Therefore I have as a solution the piecewise formula:

x = |-x0 - .5gt²| for t < √[2(L-x0)/g]
X = L for t ≥ √[2(L-x0)/g]

The textbook, on the other hand, has x = x0 Cosh[√(g/L) t]. I don't have the foggiest idea how they ended up with with solution, unless I missed something big and fundamental. A push in the right direction is greatly appreciated.

2. Jul 26, 2004

### e(ho0n3

You're assuming the acceleration of the system is g which it isn't. Draw yourself a free body diagram to see what is happening. You're going to end up solving some sort of differential equation I gather.

3. Jul 27, 2004

### HallsofIvy

Staff Emeritus
Let M be the mass of the cable. Then its density is M/L (I'm assuming it is uniform).
The force pulling the cable off the table is the weight of the portion hanging down: the mass of that portion is (M/L)x and the force is (M/L)xg. The mass used in
"m d2x/dt2= F" is the entire mass: M d2/dt2= (M/L)xg so d2x/dt2= (g/L)x (notice this is not "-" since the chain falling will increase x).

4. Jul 27, 2004

### TopCat

Ah, thanks for the replies, guys.