Chain Problem involving Kinetic Friction

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 Thumbnails

 

Pranav-Arora

Is the force of friction constant here?

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.

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.

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...

Tanya Sharma

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

Pranav-Arora

Care to check azizlwl's reply, Tanya.

Tanya Sharma

I dont understand azizlwl's reply.Kindly explain

Pranav-Arora

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.

Tanya Sharma

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

Pranav-Arora

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.

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.

azizlwl

What is the total length of the chain?
I made assumption length=L.

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 F_t=F-μρ(L-x)g. The work of the total force is
[tex]W=\int_0^L{F_tdx}=\int_0^L{\left(F-\mu \rho g (L-x)\right)dx}[/tex].
According to the Work-Energy Theorem the work is equal to the change of kinetic energy.

ehild

Attached Thumbnails

 

voko

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

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

ehild

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

ehild