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M1 Calculating the coefficient of friction Help Needed? Watch

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    The question is " A particle is sliding with acceleration 3ms-2 down a 40 degree slope. Calculate the coefficient of friction."

    I don't really know how to answer this question without been given the mass of the particle. The only thing i could do is give the coefficient of friction in algebra form and not as a number, is this the correct way of approaching this?
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    Must be without a mass.
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    (Original post by Robert Green)
    The question is " A particle is sliding with acceleration 3ms-2 down a 40 degree slope. Calculate the coefficient of friction."

    I don't really know how to answer this question without been given the mass of the particle. The only thing i could do is give the coefficient of friction in algebra form and not as a number, is this the correct way of approaching this?
    You do not need the mass

    You will get R in terms of m
    You will get friction in terms of m

    And then you will divide them and the m will cancel



    This has to be the case otherwise mu would depend on m and it doesn't, it os the same for all m
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    Maybe not then ^ this is probably why I got that M1 score :lol:
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    just find it in terms of m, they should cancel.
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    (Original post by TenOfThem)
    You do not need the mass

    You will get R in terms of m
    You will get friction in terms of m

    And then you will divide them and the m will cancel



    This has to be the case otherwise mu would depend on m and it doesn't, it os the same for all m
    Name:  Screen Shot 2013-03-13 at 18.32.33.png
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    I drew this quickly to model whats happening.

    So i got R = mgCos(40)

    And F =mgSin(40) - 3m

    Is this correct so far? And could you help in telling me what i'd do next?
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    (Original post by Robert Green)


    Is this correct so far? And could you help in telling me what i'd do next?
    yes

    so F = mu R

    therefore mu = F/R

    So do the division
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    (Original post by TenOfThem)
    yes

    so F = mu R

    therefore mu = F/R

    So do the division
    I got

    mgSin(40) - 3m / mgCos(40)

    Therefore would that cancel to

    Sin(40) - 3m / Cos(40) = mu

    ?
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    (Original post by Robert Green)
    I got

    mgSin(40) - 3m / mgCos(40)

    Therefore would that cancel to

    Sin(40) - 3m / Cos(40) = mu

    ?
    You have not cancelled the m in the 3m

    Where has g gone?

    \mu = \dfrac{mg\sin 40-3m}{mg\cos 40}


    You just have to cancel the m
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    (Original post by TenOfThem)
    You have not cancelled the m in the 3m

    Where has g gone?

    \mu = \dfrac{mg\sin 40-3m}{mg\cos 40}


    You just have to cancel the m
    So i end up with...

    mgSin(40) - 3 / gCos(40) ?
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    Don't you end up with gTan(40)-3?
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    (Original post by Robert Green)
    So i end up with...

    mgSin(40) - 3 / gCos(40) ?
    You still are not cancelling the m

    mgSin40 - 3m = m(gSin40 - 3)
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    (Original post by Fool In The Rain)
    Don't you end up with gTan(40)-3?
    no you don't
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    f= μr
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    (Original post by TenOfThem)
    no you don't
    Do you end up with gSin40-3/gCos40? Since m(gsin40-3)/m(gcos40) will cancel the m's out. Apologies in advance for my terrible M1 skills .
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    ..Just do F=ma downwards......
    mgsin40-uR=3m
    now R=mgcos40
    mgsin40-umgcos40=3m
    cancel all m's and rearrange for u.
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    (Original post by TenOfThem)
    You still are not cancelling the m

    mgSin40 - 3m = m(gSin40 - 3)
    Ahh i didn't factorise... so is the coefficient of friction 0.439 (3 sig fig)?
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    (Original post by Robert Green)
    Ahh i didn't factorise... so is the coefficient of friction 0.439 (3 sig fig)?
    yes
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    (Original post by Fool In The Rain)
    Do you end up with gSin40-3/gCos40? Since m(gsin40-3)/m(gcos40) will cancel the m's out.
    yes
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    (Original post by Robert Green)
    Ahh i didn't factorise... so is the coefficient of friction 0.439 (3 sig fig)?
    Yeah, I got that answer as well. Using gsin(40)-3/gcos(40). Let's hope it's right!
 
 
 
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