Laws of motion

mass of the chain = M
length of the chain = L
hence, mass of the y part of the chain = M/L*y
now, force(F) = mass*acceleration
so, F = (M/L*y)*9.8 N [as here the chain is under free fall, therefore acceleration=gravitational acceleration]
therefore, force applied by y part of the chain on the table = (M/L*y)*9.8 N (Ans)

yah d answer is absolutely ryt...shounak...
d acceleration of d chaim during d free fall has to b considered dats it abt it..!
n guys try d questions urself n den enter into ny discussion on dis site..apply ur brains 1st ...

MAG tht soln is nt full..........itz only half part
i cldn't get the ques thtz y i hd posted it here........'after applying my brain'

IS THE ANSWER.......... Mg(2L-y)/2

λ=M/L
mass of y part kept on table =λ.y=My/L

=>Weight on table= (My/L).g

Now we know that thrust force exerted by chain=λ.v²

v=√2g(L-y)
=>Thrust force=(M/L)(2g(L-y))

=> NET force=Myg/L +2Mg -2Mgy/L
=> NET force=2Mg-Mgy/L[1][1]

i am student of class x. pls discuss elaborately from the step "Now we know that thrust force exerted by chain=λ.v2" of the problem. how do u derive the equation : v=√2g(L-y)
thanx

ans is nt like tht
Force exerted by the chain on the table
F=F₁+F₂
where F₁=weight of the chain already on the table
=(My/L).g
and F₂=rate of change of momentum of chain at the instant it strikes the table
To calculate F₂,consider a small element dy of the chain at a height y above the table.
mass of the element=M dy/L
velocity of the element on striking the table
v=√2gy
dp=(M dy/L)√2gy
As dy/dt=v=√2gy
so,F₂=M/L(2gy)
Putting in F,we get
F=3(M y g/L)

i hd already given the ans.............this is nt the ans in any way.......hw cm u gt tht

according to your method the chain applies F₁+F₂=3Mgy/L on the ground. the ground also gives 3Mgy/L on the chain, thus changing its momentum. so change in momentum=3Mgy/L. But according to you dp/dt=2Mgy/L.

A contradiction.

no...
Mgy/l is the wt of chain already on table
so when chain falls on table the force exerted by table is 2Mgy/l
nd change in momentum,when chain falls on the table is 2Mgy/l

Honeys answer is correct....
however has someone got a better method ?