# Kinematics problem

1. Sep 5, 2008

### naggy

A chain, total length h meters long, is sitting on a table. C meters hang down from the table. The chain is let go from rest at time t=0 . The table is frictionless. Find the equation of motion, x(t)

Hint: Second order differential equations that can be reduced to cosh if x is defined posative downward.

I need a jump start I guess. No masses are given or anything....

I can write a DE if there is air resistance.

a + kv = g ?

Last edited: Sep 5, 2008
2. Sep 6, 2008

### naggy

I should add the part of the chain that is on the table is straight, not curled up.

MY ATTEMPT

The part on the table is of length h-c and the part hanging down from the table is c. I'm just can´t think of what to use to find the differential equation. Maybe there is air drag?

If I would write a force equation then it would be mg = ma because the table is frictionless?

3. Sep 6, 2008

### Staff: Mentor

That does make a difference--the entire chain moves at a single speed.
Forget air drag (that just adds complexity).
Yes, something along those lines: Apply Newton's 2nd law.

4. Sep 6, 2008

### naggy

Applying Newtons second law is sort of the problem. I can´t see how I'm supposed to treat the chain as one mass.

Can I think if it as two blocks attached together with a string? There's no friction on the table so the force on the chain on the table( tension in the string) will just be equal to mcg, where mc is the mass of the part of the chain hanging from the table.

What other forces are there?

I then have mcg - T =mca

but mc changes as more part of the chain drags from the table right?

Last edited: Sep 6, 2008
5. Sep 6, 2008

### Staff: Mentor

All you care about are forces parallel to the direction of motion, so that's the only force you need. (The weight of the chain on the table is balanced by the normal force.)
Forget about tension--that's an internal force and thus cancels out. (You want the net force on the chain.)
Of course. Write it as a function of x, where x is the amount of chain hanging off the table.

6. Sep 6, 2008

### naggy

what about mx/h where m/h is mass per unit length?

$$\frac{mxg}{h}$$ = ma

that leads to x'' -xg/h = 0 a second order DE

Last edited: Sep 6, 2008
7. Sep 6, 2008

### Staff: Mentor

Good, but correct that right hand side.

8. Sep 6, 2008

### naggy

already did I think it's just ma not ma/h

9. Sep 6, 2008

### Staff: Mentor

Good. Now all you need to do is solve the DE.

10. Sep 12, 2008

### naggy

There's a sequel to this problem. Now there is friction on the table

Could I write the frictional force as umgx/(h-c)?

u=frictional coefficient

11. Sep 12, 2008

### naggy

No wait. It must be

$$\frac{umg(h-c-x)}{(h-c)}$$

12. Sep 12, 2008

### naggy

this should be correct

13. Sep 13, 2008

### Staff: Mentor

That latest version looks good to me.