# Chain Problem involving Kinetic Friction

1. Aug 20, 2012

### Tanya Sharma

1. The problem statement, all variables and given/known data

A heavy chain with a mass per unit length ρ is pulled by the constant force F along a horizontal surface consisting of a smooth section and a rough section.The chain is initially at rest on the rough surface with x=0 .If the coefficient of kinetic friction between the chain and rough surface is μ , determine the velocity of the chain when x=L .

2. Relevant equations

3. The attempt at a solution

I am applying work energy theorem . Work done by constant Force will be Force × displacement of centre of mass i.e FL but not able to find work done by friction .The friction force at an instant when chain length x lies on the rough surface should be μρxg.This force is continuously decreasing .i feel calculus is involved here but i am unable to apply it.Please help me .thanks.
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

#### Attached Files:

• ###### Chain.jpg
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2. Aug 20, 2012

### Saitama

Is the force of friction constant here?

3. Aug 20, 2012

### azizlwl

It is the sum of energy taking small mass dm across the rough surface.
μgƩxdm where dm/dx=ρ
x varies from 0 to L
From this you can find the definite integral.

Last edited: Aug 20, 2012
4. Aug 20, 2012

### voko

There is not enough information. To see that, imagine the entire chain is on a smooth surface. The work of friction is zero, so the entire work must be equal to the kinetic energy at distance L. But the velocity cannot be determined because the total mass cannot be determined.

5. Aug 20, 2012

### Tanya Sharma

The force of friction on chain is μN ,where N =(mass of chain on rough surface) ×g=ρxg.Thus frictional force at any instant when chain of length x is on rough surface is μρxg.But now as the chain is being pulled to right ,the length x decreases , hence frictional force decreases continually.Now i am not able to calculate the work done by this variable frictional force .What should be the limits of x when we consider dx displacement.Please reply...

6. Aug 20, 2012

### Tanya Sharma

i feel sufficient information is provided.here mass per unit length ρ is provided.

7. Aug 20, 2012

### Saitama

Care to check azizlwl's reply, Tanya.

8. Aug 20, 2012

### Tanya Sharma

I dont understand azizlwl's reply.Kindly explain

9. Aug 20, 2012

### Saitama

You got the force of friction right. Find the work done by force of friction for small displacement dx. Integrate it from x=0 to x=L.

10. Aug 20, 2012

### Tanya Sharma

why integrate from x=0 to x=L.This is the portion on smooth surface.

11. Aug 20, 2012

### Saitama

Yes, it is. We have to find the work done by the force of friction when x=L i.e. L length of the chain is on the smooth surface now.

12. Aug 20, 2012

### Tanya Sharma

hey Pranav...plz can u give a detailed explaination.thanks. i feel limits should be between length and length - L coz intially length (full chain)was on rough surface and finally L moves to smooth surface so length - L remains on rough surface so technically we should integrate between the limits length to lenth - L.Plz help.

13. Aug 20, 2012

### azizlwl

What is the total length of the chain?

14. Aug 20, 2012

### ehild

The force acts at the right end of the chain, and the displacement of that point can be called x. x changes from zero to L, length of the chain. If the right end is at position x, L-x long piece of chain is on the rough surface. The force of friction, as you wrote correctly is proportional to the length on the rough surface. So the total force is Ft=F-μρ(L-x)g. The work of the total force is
$$W=\int_0^L{F_tdx}=\int_0^L{\left(F-\mu \rho g (L-x)\right)dx}$$.
According to the Work-Energy Theorem the work is equal to the change of kinetic energy.

ehild

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• ###### chain.JPG
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15. Aug 20, 2012

### voko

L is not specified as the length of the chain. Nor is that apparent from the original drawing. Am I missing something?

16. Aug 20, 2012

### ehild

Ops, you are right. But it seemed plausible that L is the length of the chain. Now, if it is Lc, than the work is the integral above for L≤Lc and ∫Fdx from x=Lc to L. But one does not know the mass of the chain as you pointed out, and the problem can not be solved.

ehild

17. Aug 20, 2012

### ehild

You are right, L should be the length of the chain...

ehild

18. Aug 20, 2012

### Aimless

It seems there's some confusion amongst the previous posters on this thread, so I'm going to start from the beginning.

Two questions for Tanya:

If this problem were about a point particle of mass m, rather than a chain, would you know how to solve it?

What makes the solution for a chain different from the solution for a solid block?

My suggestion for a solution strategy for this problem is as follows. First, write down (in integral form) the equation for the change in kinetic energy of a point particle of mass m as it is pulled across both the rough and smooth portions of the table. Second, consider the chain at a given instant in time as being composed of two point particles; one on the smooth section of the surface, and one on the rough section. How does the motion of the chain affect the mass of these two imaginary point particles? Can you express that change in mass mathematically as a function of the position of the chain? If so, you should be able to write down the correct expression for the integrand of the work integral in terms of the chain's position. The last step is then to integrate.

19. Aug 20, 2012

### Tanya Sharma

I am really grateful for all the prompt responses especially echild .Yes it seems that the length of chain should be L which is not mentioned in the question, but seems plausible as mentioned in above posts.Thanks a ton...