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Blazing goIITian

Joined:	13 Jan 2007
Post:	1289

4 Nov 2007 19:02:50 IST

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471

linear momentum.............

None

it is easy 1..

q a uniform chain of mas M & length L is held vertically in such a way that its lower end just touches the horizontal floor. the chain is released from rest in this posn. any portion that strikes the floor comes to rest. assuming that the chain doesn't form heap on the floor, calculate the force exerted by it on the floor when a length x has reached the floor.

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Comments (4)

'',' Off to a nu start!!

Blazing goIITian

Joined: 13 Jan 2007

Posts: 1289

5 Nov 2007 19:09:50 IST

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common solve this......

Omi Tipnis

Hot goIITian

Joined: 2 Jun 2007

Posts: 164

5 Nov 2007 19:36:39 IST

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see i cant rite the whole answer its too lng n complicated.....

consider a vry small part of it..... n nw find change in momentum.......

initial one is before this small part falls.... n final is aftr this small part falls....

k nw change in momentum is equal to impulse.....

integrate find the force..... nw add the normal reaction exerted by the part already present lying on the table this gives u ur answer....

try it urself...... its a vry good question.... if u dun gt it.... dun wrry i ll try n put the whole soln by scanning it k.......

'',' Off to a nu start!!

Blazing goIITian

Joined: 13 Jan 2007

Posts: 1289

6 Nov 2007 13:34:35 IST

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answer is

3mgx/L
is it complicated???

plz write..

sowjanya gudipati

Cool goIITian

Joined: 13 Oct 2007

Posts: 75

6 Nov 2007 20:10:39 IST

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hey i am using the normal reaction offered by the floor on the chain which is equal to the force exerted by the chain on the floor

N=N_wt+N_thrust

N_wt=(m/l)xg

N_thrust=V_relative(dm/dt)

V_relative=u-v = v

dm/dt=(dm/dy)(dy/dt)=m/l(v)

N_thrust=v²(m/l)

v=(2gx)^1/2

N_thrust=2gx(m/l)

N=(m/l)gx+2gx(m/l)=3(m/l)gx

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