# Hanging Cable (Projectile Motion?)

1. Dec 4, 2007

### TopCat

(Solved) Hanging Cable (Projectile Motion?)

1. The problem statement, all variables and given/known data
A perfectly flexible cable has length L. Initially, the cable is at rest, with a length $$x_0$$ 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.

3. The attempt at a solution
I figured that I could take a reference point, that being the very tip of the cable hanging over the edge. Since friction is neglected the only force on the cable is gravity. If I take the table edge to be x=0 then the position of the tip of the cable over the edge will also give the length of cable behind it hanging. So then it should simply be $$x = x_0 + \frac{1}{2}gt^2$$ but the book lists the answer as $$x = x_0 cosh (\sqrt\frac{g}{L} t)$$. I have no idea where to even start. Please help.

Last edited: Dec 5, 2007
2. Dec 5, 2007

### Shooting Star

Assume the linear density, that is, mass per unit length is d. The weight of the hanging portion is dxg, which is moving the whole mass of the chain. The accn of the whole chain is the same at time t. So,

dxg = Md^x/dt^2. Now, solve.

3. Dec 5, 2007

### TopCat

Outstanding! Thanks for the help. I've gotten the solution though I have one last question.

When describing the force on the portion of cable on the table why is the mass M used d*L and not d*(L - x)?

4. Dec 5, 2007

### Shooting Star

I've never taken into account the force on the mass on the table separately, because the whole mass, the mass on the table and the hanging portion both, is moved with the same accn by the weight of the hanging portion only.

So, where you have to write mass*accn, it's the whole mass M=d*L.

5. Dec 5, 2007

### TopCat

Okay, thank you very much. It all makes sense now.