Given a uniform chain on an incline, find the work done by friction

[Kanda ryu]

Homework Statement

A uniform chain of mass 'm' and length 'l' rests on a rough incline (inclination is angle 'Q') with its part hanging vertically. The chain (inclined) starts moving up the incline (and the vertical part moving down) provided the hanging (vertical) part equals to 'n' times the total chain length. Find the work done by friction as it completely slides off the incline.

Homework Equations

Newton's 2nd law

The Attempt at a Solution

I assumed friction coefficient to be 'u'.
Frictional force would be N u.
I took the inclined part a system to draw the free body diagram.
N = m'gCosQ where m' is the mass of chain on the incline.
Work (fric) = N u s (displacement) x Cos 180° = - m'gCosQ x s
As mass m' will decrease as chain slides up the incline. I considered a small part of chain 'x' of mass dm. As chain is uniform density,
m/l = dm/x => dm = m x /l
For a small work dW
= -dm gCosQ dx
= - (m/l) gCosQ xdx
Let the length of vertically hanging part be 'a'
a/l = n => a=nl
So length of inclined part initially is l-nl
So I integrated xdx putting limits (l-ln) to 0 and got W= mlgCosQ u / 2 but the problem is I'm unable to find a way to write 'u' in terms of given variables.
I was hoping for an idea to find 'u' or any other better method to solve this (within the scope of basic calculus and mechanics).

 

[haruspex]

small part of chain 'x'

That is an unusual choice and could lead to confusion. The standard approach would be to say the element is length dx and is at x from one end (which needs to be specified).

For a small work dW

The context is unclear. A small amount of work done on/by what and during what event? I guess you mean the work done by friction on the element as it moves its whole distance up the incline.

= -dm gCosQ dx

Two "d" terms looks very wrong. What is dx here?

Let the length of vertically hanging part be 'a'

When, at the start or at some point during the movement?

= - (m/l) gCosQ xdx

By some miracle, you have obtained the right expression. The element length dx has mass (m/l)dx and moves distance x up the incline.

putting limits (l-ln) to 0 and got W= mlgCosQ u / 2

Why does n not appear in your result?

I was hoping for an idea to find 'u'

You are given this for the initial state:

The chain (inclined) starts moving up the incline

It is not entirely clear, but I would take that to mean it started in near equilibrium.

 

[haruspex]

PS.. I assume the question expects you to take the tip of the incline over which the chain slides as being frictionless. More reasonably, it would have the same coefficient as anywhere else. The tension in the chain as it runs around that point would lead to a significant normal force there and consequently a lot more friction. This becomes quite complicated because the friction will mean the tension is not uniform, and will anyway depend on the acceleration. This makes for a much more advanced problem.

 

[Kanda ryu]

There is a frictionless pulley at the tip of incline. The dimensions of the pulley is to be neglected.(view image)

When, at the start or at some point during the movement?

At the start.

The equation I get is W(by friction) = (mlgCosQ(1-n)^2 u)/2

 

[haruspex]

There is a frictionless pulley at the tip of incline. The dimensions of the pulley is to be neglected.(view image)At the start.

The equation I get is W(by friction) = (mlgCosQ(1-n)^2 u)/2

Ok, but now use the hint I gave you at the end of post #2 to find u.

 

[Kanda ryu]

So I take the initial part of chain on incline as a system and draw it's free body diagram. The exernal forceks are normal by surface, gravitational force, friction down the incline and tension up the incline by the other part of chain. But I am finding difficulty finding friction coefficient as it seems to be cancelling each other out in the equilibrium equations. Did I mess up taking the system or making the equations?
Here are my equations;
N=MgCosQ
NCosQ + TSinQ = fSinQ + Mg
NSinQ+fCosQ=TcosQ
f+MgSinQ=T
Where N and f are normal and frictional force exerted by rough surface on inclined part of chain initially and T is tension applied by the other part of chain. As you said, I assumed the system to be in near equilibrium initially so no acceleration.

 

[haruspex]

N=MgCosQ
NCosQ + TSinQ = fSinQ + Mg
NSinQ+fCosQ=TcosQ
f+MgSinQ=T

Those four equations are all obtained by considering force balances on the sloping part of the chain. There are only two dimensions, so only two independent such equations are available. Whichever two you pick, the other two are redundant.
You need to relate the mass of that portion to the mass of the hanging portion (using n) and that to the tension. You also need to relate N to f using u.

 

[Kanda ryu]

. Whichever two you pick, the other two are redundant.

Ok so I picked the the equation balancing the forces pulling it up and down the incline.
Hence f + m(1-n)gSinQ = T
So tension here would be equal to weight of the part hanging so T = mng
f=m(1-n)gCosQ u
So on substituting these values in the above equation, I got;
u= [n-(1-n)SinQ]/(1-n)CosQ
And on substituting this value to the Work by friction equation;
Wf= mgl(1-n)(n-(1-n)SinQ)/2
At n=1,Wf=0 (from this equation) which is to be expected as entire chain is hanging now. I hope this equation is correct and would be glad if you confirm it.
Thanks for the support

 

[haruspex]

Ok so I picked the the equation balancing the forces pulling it up and down the incline.
Hence f + m(1-n)gSinQ = T
So tension here would be equal to weight of the part hanging so T = mng
f=m(1-n)gCosQ u
So on substituting these values in the above equation, I got;
u= [n-(1-n)SinQ]/(1-n)CosQ
And on substituting this value to the Work by friction equation;
Wf= mgl(1-n)(n-(1-n)SinQ)/2
At n=1,Wf=0 (from this equation) which is to be expected as entire chain is hanging now. I hope this equation is correct and would be glad if you confirm it.
Thanks for the support

Looks right.