A Rope


This one is another easily confusible problem (maybe not for some of you).
A long flexible inextensible rope of uniform linear mass density 5 kg/m is being pulled on a rough floor, with a horizontal force F, in such a way that its lower part is at rest and upper part moves with a constant speed of 2 m/s.

What should be the magnitude of this force F ??

-Shivanker

30 comments:

  1. shaanApril 27, 2011 at 2:53 PM

    hi
    hope ans is 5[2]^2

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  2. ShivankerApril 27, 2011 at 2:54 PM

    nope it ain't..

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  3. AnonymousApril 27, 2011 at 4:03 PM

    confused !

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  4. AnonymousApril 27, 2011 at 4:50 PM

    a very good question from a very good source. So, no need of me telling the answer as I already know it.

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  5. Pulkit MendirattaApril 27, 2011 at 4:54 PM

    10

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  6. mayankApril 27, 2011 at 5:54 PM

    yes, 10

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  7. AnonymousApril 27, 2011 at 6:53 PM

    shivanker and sambhav, I think that now judge should not be used. As Ankush pointed out that fault, I always confirm the answer from the page source and actually I see the answer and then try. I think now most of the students do this, so it would be better if you add answer on the judge after half day OR stop using it.
    thank you,

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  8. AnujApril 27, 2011 at 6:58 PM

    no. we'll keep the judge.
    that's another thing you should learn: trying a problem even though solutions/answers are accessible.

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  9. shaanApril 27, 2011 at 10:21 PM

    i used
    F=dp/dt= mdv/dt+ vdm/dt

    now, dv/dt=0,
    so F= v.dm/dt= vd[k.vt]/dt.....k = linear mass density

    F=kv^2= 5[2]^2

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  10. shaanApril 27, 2011 at 10:24 PM

    pls help me

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  11. Sambhav GuptaApril 27, 2011 at 11:13 PM

    yes shaan even i did the same at first...but as a habit i try both newton's laws approach and work energy approach wherever possible, and they yielded different results here

    for the given situation, F is probably not the only force acting, but certainly F is the only force doing work, so we know which approach is correct

    (i might not be correct at this explanation, but surely answer is not 20)

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  12. shaanApril 27, 2011 at 11:26 PM

    the only other force[in horiizontal plane] i can think of is friction..

    but then, it'll change the net force, as well as it'll dissipate some heat....

    why do you say- "certainly F is the only force doing work"..
    what kind of other F u talkin abt??

    though i believe , friction ke baare mein naa sochein toh better hai.

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  13. Sambhav GuptaApril 27, 2011 at 11:31 PM

    there is no relative motion between rope and ground...(in crude words, rassi jameen se ghis nhi rhi), so, though friction is actingm, it aint doing any work

    by F, i meant the given force F, the one which we need to calculate, it is not the only force as friction might also be acting, but F is the only force doing work, friction isn't

    no work by friction=>no heat

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  14. shaanApril 27, 2011 at 11:44 PM

    makes perfect sense,
    thanks..

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  15. AnonymousApril 28, 2011 at 1:17 AM

    F=dp/dt
    dp = dm * v
    F= 5*2 dt /dt=10

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  16. Sambhav GuptaApril 28, 2011 at 1:24 AM

    @Anonymous
    F=vdm/dt, we put v=2, ok, what next?
    how did you put dm/dt=5dt/dt?

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  17. Rohan SharmaApril 28, 2011 at 1:26 AM

    This comment has been removed by the author.

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  18. AnonymousApril 28, 2011 at 1:26 AM

    dm = dx * v * 5 kg/m
    dx= v dt

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  19. AnonymousApril 28, 2011 at 1:27 AM

    sorry
    dm = dx * 5 kg/m
    dx= v dt

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  20. Sambhav GuptaApril 28, 2011 at 1:29 AM

    that's just what shaan was doing, there already has been a discussion on that, and that gives answer 20 not 10

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  21. AnonymousApril 28, 2011 at 1:34 AM

    sorry again

    dx= .5 * v dt

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  22. AnonymousApril 28, 2011 at 1:35 AM

    tell me if i am correct

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  23. ShivankerApril 28, 2011 at 9:36 AM

    anonymous is perfectly correct..

    in time dt, only v.dt/2 part of the rope gains velocity v and not v.dt part

    hence F = v.dm/dt = v*5*v/2 = 10

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  24. Anonymous {ii}April 28, 2011 at 9:59 AM

    pls also tell how?

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  25. mayankApril 28, 2011 at 9:59 AM

    yes, we have to deal with the COM of the system so, though speed is 2m/s but, COM would be moving with only 1m/s. hence dm/dt=1*5=5
    so, force=vdm/dt=10

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  26. mayankApril 28, 2011 at 10:01 AM

    to get a feel of this concept, use coordinate plane and analyse the of rate of change of COM. You will find it to 1m/s

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  27. Anonymous {ii}April 28, 2011 at 10:02 AM

    kya baat kar raha hai, mayank??

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  28. mayankApril 28, 2011 at 10:07 AM

    ok let initially rope is on cartesian plan and one end be at origin and other at 10units. COM at 5 units. Now after 4sec, whole rope would be moving with 2m/s.But, COM will be at 15.So, net displacement is 10units in 4sec. dx/dt=2.5

    Hope it helps

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  29. ShivankerApril 28, 2011 at 11:55 AM

    no idea about what mayank has to say.. but here's what i meant..

    firstly, we're applying newton's law only on the curled above part of the rope.. so F=v.dm/dt.. ok..
    suppose now, that the free end being dragged by the force moves a length l units.. and we know, the part lying on the ground just below this dragged part is surely at rest.. then by constant string constraint.. dont we get that l/2 units of the rope is now curled above and moving with a velocity v, while the other l/2 is still lying on the ground??
    now u may read my previous comment again..

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  30. shivamApril 28, 2011 at 1:06 PM

    strange !!!!! situation is same as that of the figure of previous post when k=0 and please tell the solution of resonance ques because i tried it a lot and getting answer 252

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