# Homework Help: Second Law of Motion Question

1. Oct 21, 2012

### Vroc

1. The problem statement, all variables and given/known data
A heavy metal chain of mass 0.6 kg hangs over the edge of a table. When half the chain is hanging over the edge, what is the magnitude of the external net force that causes the chain to accelerate? What's the acceleration of the chain when half of it is over the edge?

2. Relevant equations
Fnet = ma
Fnet = mg (Object in free fall)

3. The attempt at a solution
I have an idea of what to do here but I'm not sure if I'm correct or not.
Fnet= mg
0.3 (half over the edge) times 9.8 (gravity) = 2.94N. I can only guess that 2.94 is the external net force. I think I might have to factor in the other 0.3 in some way but I'm not completely sure.

2. Oct 21, 2012

### lendav_rott

So F1 pulls the chain off the table - 0.3g N - ok
But the other part of the chain is preventing that from happening , well to the best of its ability which ought to be friction.
We have to assume that the other part of the chain isn t parallel to the side of the table and right on the edge or it would fall off immediately.

Tricky question, right now, all I can think of is friction that is stopping the loose end of the chain moving.

3. Oct 21, 2012

### Vroc

Damn, I guess I'll have to wait a little bit longer for someone to reply that knows what they're doing.

4. Oct 21, 2012

### mishek

Your first equation (Fnet = ma) could be useful. What are the forces acting on a chain?

5. Oct 21, 2012

### Vroc

The forces acting on a chain are Fg, Fn, Ft and Ff, why? How can Fnet= ma help?

6. Oct 21, 2012

### mishek

Could you draw a fbd for the half of the chain that lies on a table?

7. Oct 21, 2012

### Vroc

Okay I drew it. FBD of hanging object only has Fg and Ft

The one on the table has FN, Fg and Ft. ?

8. Oct 21, 2012

### mishek

Exactly. Could you now tell what is Fnet?

9. Oct 21, 2012

### Vroc

Tension? But I don't know how to calculate it.

10. Oct 21, 2012

### mishek

Check the fbd I attached. Does it help?

#### Attached Files:

• ###### fbd.JPG
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11. Oct 21, 2012

### Vroc

A little but I just calculated each of those and subtracted. It comes to zero.

Can you please tell me how you got those formulas?

12. Oct 21, 2012

### mishek

Well, I "cut" the chain on half. I know that only force on the hanging chain is m/2*g. That force is also "transferred" on the part which lies on the table. After that, you make the equations.

But, without fbd you can guess that mg of the hanging chain is pulling him down and the Ff which comes from lying half of the chain is trying to hold him in place.

#### Attached Files:

• ###### fbd2.JPG
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16.3 KB
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13. Oct 21, 2012

### Vroc

What does N in your equation stand for? Just normal force?

Also, If the magnitude of the external net force is 0, That makes the acceleration also 0 correct because Fnet= Ma
0= 0.3 x a
0 divided by 0.3 = 0.

That makes them both zero.

14. Oct 21, 2012

### mishek

Let's try like this: you have a half chain lying on the table. The friction force is mass of that chain (half of it which is on the table) multiplied by friction factor and gravity.

The other force that is acting on it is the force that is "pulling him down". That is the mass of half the chain that is hanging multiplied by gravity.

Those two forces make Fnet, ant they differ in direction: Fnet= 0.5*m*g-0.5*m*g*u, but i don't see gravity factor anywhere...

15. Oct 21, 2012

### Vroc

What the does "u" stand for in your equation?

16. Oct 22, 2012

### mishek

Friction coefficient. But I don't see you have it here...

17. Oct 22, 2012

### szimmy

My question is why are you guys trying to incorporate friction? It says in the question the chain is accelerating, which means there is a net force. So the net force of the object when it is half over is simply m/2 * g. The object is not in equilibrium, so there is nothing that definitely says there is friction. You're trying to find acceleration, the whole chain is accelerating, not only the falling half, so the mass of the system is the entire .6. So .6a = .3(9.8) so a = (.3 * 9.8)/.6 and a = 9.8 / 2 and a = 4.9m/s^2

Also your external net force is correct at 2.94N