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  1. Mr M
    01-06-2012
    yes
  2. Mr M
    31-05-2012
    Yes but it will be about 6 p.m. when I post it as I have a meeting after school and I'm teaching all day.
  3. Mr M
    15-05-2012
    Haven't got time to help right now but if you don't know about things like contact forces you should read my OCR M1 powerpoints - I will send them if you give an email address.
  4. 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. Which part are you stuck on?
  6. 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?
  7. Sorry for the late response.

    We know  x=y , sub this in  x^2 -2xy+4y^2 =12

    So we get,  x - 2x^2 + 4x^2 = 12 \implies 3x^2 = 12 \implies x^2 = 4 \implies x = \pm 2

    As  x= y , hence the coordinates are  (2,2) \ \ and \ \ (-2, -2)

    Do my answers agree with your book's answers?
  8. When tangent is parallel to the x-axis then its gradient is zero.
  9. Can a denominator be equal to zero?
    Denominator being equal to zero mean  \dfrac{x}0 , is this possilbe?

    Give me the exact question.
  10.  p^3 = 3py -2x

    Sub in x=-10 and y=7, you will get,  p^3 - 21p -20 =0

    Now find the one root by inspection, we can see that p=-1, is a root of the expression. Hence ((p+1) is a factor of the expression,  p^3 - 21p -20 =0 . So to find the other factors divide,   p^3 - 21p -20 =0 by (p+1) using long division.

    So in this way you will find the other 2 roots.

    So now sub them in  x= p^3 \ \ and \ \ y = p^2 to find the possible coordinates.

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  • Last Activity 24-03-2013
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