IIT JEE , AIEEE Forum - Physics , General - rope falling down.........................force exerted.................

jaysunantony

A uniform rope of mass m per unit length hangs vertically from a support so that the lower end just touches the table top as in fig. it is release from rest.

Find the force exerted by it on the table top when a length y of the rope has already fallen.

nitin62225

ans= mgl(1-2y)+mgy² --------------l is the total length of rope.

mlg-N =U_rel (dm/dt)

and udm/dt = mv² = mgl²-mg(l-y)²

=-mgy²⁺ 2mgly

so N=mgl+mgy²-2mgly

=mgl(1-2y)+mgy²

plz rate....

jaysunantony

the answer is wrong........................

l is not given in the question.........then how can it come in the answer?

options]

2myg 3myg myg 3/2myg

Please dont guess ..................

kushalbhatia

ans is 3mgy i have done this before

mgy due to its own weight and 2mgy due to the rate of change of momentum concept .note that the velocity of a body falling freely under gravity is(2mgy)^1/2

jaysunantony

kushalbhatia,

YOU are right..............................take a thumbs up.

Can you explain it with detailed steps and sure, I will rate you.

akku

Hi jasun antony,
the force exerted by chain on the table when 'y' length has come on the table=3*wt of chain on the table.
DETAILED SOLUTION:
Suppose an element of length dy at y distance from ground (at t=0 )
Since every part of rope falls freely
vel of dy=v=(2*g*y)^.5 (v^2=u^2+2gh ; u=0 as chain is released from rest)
final momentum=0
initial momentum= -dmv=-dm(2*g*y)^.5 (downward dirn=-ve)
Dp(of chain)= pf-pi=dp=dmv=dm(2*g*y)^.5
Dp transfered to the table= -(2*g*y)^.5
where dm=mass of dy=mdy
time during which p is transfered=dt=dy/v
FORCE(due to impact)=dp/dt=mv^2=2mgy

NET FORCE ON THE TABLE=force(due to impact)+weight
=2mgy+(M)g (M=mass of y length of chain)
=2mgy+mgy
=3mgy(Answer)

sanchay_1991

here what will be ans ifwe take frictional forces in account

then how will the change be observed in the ans

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