Q) A very flexible uniform chain of mass M and length L is suspended vertically so that its lower end just touches the surface of a table.When the upper end of the chain is released, it falls with each link coming to rest the instant it strikes the table.Find the force exerted by the chain on the table at the moment when y part of chain has already rested on table.
1mass 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)
1yah 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 ...
1MAG 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'
24λ=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]
1i 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
1ans 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)
1i hd already given the ans.............this is nt the ans in any way.......hw cm u gt tht
1according 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.
1no...
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
1Honeys answer is correct....
however has someone got a better method ?
