IIT JEE , AIEEE Forum - Physics , Mechanics - How much is the force created due to the motion of the falling chain ?

adisahay

Q. A uniform chain of length 'L' and weight 'W' is hanging vertically from ends A & B which are close together. At a given instant the end B is released. Find the tension at A when B has fallen a distance 'x' (x<L).

Here I have understood almost everything that the total tension at A is due to 2 forces i.e. one due to the weight of the new length of the chain below A (L/2+x/2) and the other due to the rate of change of momentum of the falling chain. My doubt is that in the 2nd force, whether a small mass 'dm' falls through a distance x or x/2. In my book it's given that it falls through x/2 but I think that it falls through x distance.

Anyways, the answer given in the book is T = W/2[1 + 3x/L] whereas I'm getting

T = W/2[1 + 5x/L]. How much are you getting ?

adisahay

C'mon yaar! Is there no one to try it ? Please, I want an answer as quick as possible.

rohitjhaji

look dude. what ur book says is correct. it does go down by x/2 as here we consider the mption of com and not the tip

adisahay

Please explain it in detail. I'm totally confused after thinking on it the whole night yesterday. I've even checked it practically! It should be x.

elessar_iitkgp

The answer given in the book seems to be incorrect. My approach is a bit different from urs.
The left portion of the string is a variable mass system in which mass is entering.
By Merchersky's equation,
m(dv/dt) = F_ext + u_rel (dm/dt) = F_ext + u_rel (dm/dx)(dx/dt)
As the left portion of the chain is in equilibrium,
0 = {T-(m/L)[(L/2)+(x/2)]}j +(-

2gxj)(m/L)(

2gx)
T = (mg/L)[(L/2) + (5x/2)]
= (W/2)[1+(5x/L)]

nick

hi elessar
pls tell me what is
Merchersky's equation,

elessar_iitkgp

m(dv/dt) = F_ext + u_rel (dm/dt)

This is Merchersky's equation for a variable mass system
u_rel= Relative velocity of the mass being gained/ejected w.r.t the system

krishna.gopal

Hi adisahay. I woul like to know what is dm in your book.
Let chain be ACDB (in that order) Such that AC = L/2 (C is lowest point to begin with)
AD = L+x/2 (D is lowest point when B has fallen x units)
Now fall for each element is different.
Fall for a point P in AC = 0
Fall for a point P in CD (CP =y) = 2y which will vary from zero for a P close to C to x for P close to D
And it x for P lying on DB.

For section AC as a whole center of mass falls by 0
For section CD as a whole center of mass falls by x/2
For section DB as a whole center of mass falls by x

adisahay

Sir, 'dm', as given in the book is a small mass which I think is the mass of the point D mentioned by you. This point D, before falling, should be at height 'x' distance above(but it's x/2 in the book).

This mass after falling throgh the distance x is stopped suddenly when it reaches the lowest point. Let the mass/unit length be 't', then dm = t*dx, where dx is the length of the segment dm. Now, the mass dm after falling through x distance acquires a velocity of

2gx. Suppose it is stopped in dt seconds, then the rate of change of momentum of this mass will be = the additional force due to this motion = dm*v/dt = t*dx*

2gx/dt =t*(dx/dt)*

2gx = t*(

2gx)² = 2gxt. Since the book has taken x/2 instead of x, the force comes out to be 'gxt'. THIS IS THE POINT OF MY CONFUSION.

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