Basics:



This question belongs to a special type of physics questions. They involve a 50-50 blend of maths and physics.

A small block (1kg) slides down the smooth circular surface of a wedge (5kg).
What should be the minimum coeff. of friction b/w the wedge and ground such that the wedge does not slip during the motion of the block??

11 comments:

  1. Not AnonymousApril 17, 2011 at 12:58 AM

    1/(2)^3/2

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  2. AnonymousApril 17, 2011 at 6:52 AM

    wedege is 3kg lr 5 kg.

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  3. Limit of AnonymityApril 17, 2011 at 5:20 PM

    i got 3sqrt(10)/40

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  4. shaanApril 17, 2011 at 10:44 PM

    acc to me

    mew= (root15)/54 if M=3[as writen in text]

    or mew=(root35)/130 if M=5[as writen in image]

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  5. AnujApril 18, 2011 at 12:22 AM

    i've corrected the text question.

    ReplyDelete
  6. ShivankerApril 18, 2011 at 2:11 PM

    m also getting 3sqrt(10)/40

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  7. AnonymousApril 18, 2011 at 9:44 PM

    1/(10+3*root(2)) or approx 0.071

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  8. AnonymousApril 19, 2011 at 5:31 PM

    0.98

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  9. ShivankerApril 20, 2011 at 9:54 AM

    Bhaiya plz add this post too to the judge..

    ReplyDelete
  10. AnujApril 20, 2011 at 12:40 PM

    i would if i was so sure of the answer. i'll try and post the answer soon.

    ReplyDelete
  11. Ankush SachdevaApril 20, 2011 at 12:45 PM

    yup same....3sqrt(10)/40

    ReplyDelete

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