Conversation Between Smiley Face :) and raheem94

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  1. Smiley Face :)
    10-06-2012
    Doesn't matter just figured it out
  2. Smiley Face :)
    26-05-2012
    hi raheem, you free atm, i am still stuck on the M1 q, Its the post 2 down from this one...
  3. Smiley Face :)
    23-05-2012
    Hey ,
    Would you be able to help me out on a couple of q's for M1 please?
    http://www.xtremepapers.com/papers/O...hs/2005%20Jun/
    It's M1 and its questions 5 part 1 and c
    and questions 6 part 1 - a.b and c
    Thanks in advance... BTW not entering my working as it is shocking as I genuinely had very litttle idea as to what the question was asking...
  4. raheem94
    14-05-2012
    It will be best that you post it on the forum.

    I have never done a question of that type, so i can't help you on it.
  5. Smiley Face :)
    14-05-2012
    Q7 part 2
    a and b
    thanks
  6. raheem94
    14-05-2012
    Which part are you stuck on?
  7. Smiley Face :)
    14-05-2012
    Hi could I have some help on Q7 on this paper pls??
    I am struggling with the wording of it..
    http://www.ocr.org.uk/download/pp_11...gce_472801.pdf
  8. Smiley Face :)
    14-05-2012
    Um, I kind of understand the top 9N bok it was the bottome 3 N force as on the markscheme it said that the coewfficient of frcition for it will not change and it also mentioned something about traction force, which I have never heard off :/
  9. raheem94
    14-05-2012
    I haven't given much attention to your working, though i will try to explain it.

    Now the lower particle(3N) will exert a vertical normal reaction on the upper particle(9N).

    The diagram will look like this:




    Now resolve the forces,
     \displaystyle ( \uparrow ) \ \ \ R + 5 cos60^o = 9 \implies R = 6.5N

     \displaystyle ( \rightarrow ) 5sin60^o = \mu R \implies \boxed{ \mu = \frac{5\sqrt3}{13} }

    NB: I forgot to show friction on the diagram.

    Does it makes sense?
  10. Smiley Face :)
    14-05-2012
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