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Math Help - [SOLVED] Transposition of Formula

  1. #1
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    Smile [SOLVED] Transposition of Formula

    Hi everyone.

    I was hoping somebody could please give me some guidance on transposing a formula using force gravity u sin cos and tan. The u is not really a u, it stands for the coeffficient of friction.

    The formula reads; F=umgCostheta - mgCosthetaTan theta

    Looking at transposing it to read; (F=mgCostheta(u - Tantheta)

    any help appreciated

    Thanks

    David
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  2. #2
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    Quote Originally Posted by David Green View Post
    Hi everyone.

    I was hoping somebody could please give me some guidance on transposing a formula using force gravity u sin cos and tan. The u is not really a u, it stands for the coeffficient of friction.

    The formula reads; F=umgCostheta - mgCosthetaTan theta

    Looking at transposing it to read; (F=mgCostheta(u - Tantheta)

    any help appreciated

    Thanks

    David
    F = \mu \textcolor{red}{mg \cos{t}} - \textcolor{red}{mg \cos{t}} \tan{t}

    simple algebraic factoring ... just pull out (un-distribute) the common factor, \textcolor{red}{mg \cos{t}} , from both terms

    F = \textcolor{red}{mg \cos{t}}(\mu - \tan{t})

    you're done.
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    Quote Originally Posted by skeeter View Post
    F = \mu \textcolor{red}{mg \cos{t}} - \textcolor{red}{mg \cos{t}} \tan{t}

    simple algebraic factoring ... just pull out (un-distribute) the common factor, \textcolor{red}{mg \cos{t}} , from both terms

    F = \textcolor{red}{mg \cos{t}}(\mu - \tan{t})

    you're done.
    Hi Skeeter,

    Although the working looks clear what you have completed, I still have a little confusion, please advise;

    F = umgcos t - mgcos t Tan t
    F - mgcos t = umg cos t - mgcos t Tan t (subtract like terms)
    F - mgcos t = u Tan t

    F - mgcos t = uTan t (multiply both sides)

    F = mgcos t u Tan t (where am I getting it wrong)?
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  4. #4
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    Quote Originally Posted by David Green View Post
    Hi Skeeter,

    Although the working looks clear what you have completed, I still have a little confusion, please advise;

    F = umgcos t - mgcos t Tan t
    F - mgcos t = umg cos t - mgcos t Tan t (subtract like terms)
    F - mgcos t = u Tan t

    F - mgcos t = uTan t (multiply both sides)

    F = mgcos t u Tan t (where am I getting it wrong)?
    mg\cos(t)\tan(t) and mg\cos(t) are not like terms.

    mg\cos(t)\tan(t) - mg\cos(t) \neq \tan{t}

    that's like saying 5(7)(11) - 5(7) = 11 , which is not true.
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    Hi Skeeter,

    Here's my next attempt?

    F = umgcos(t) - mgcos(t) Tan(t) The subject (F = mgcos(t) (u - Tan (t))

    F - Tan(t) = umgcos(t)-mgcos(t)

    F - Tan(t) - mgcos(t)=umgcos(t)

    Fu-Tan(t) - mgcos(t) = mgcos(t)
    - mgcos(t) -mgcos(t)

    Fu-Tan(t)=mgcos(t)
    -mgcos(t)

    F - mgcos(t)(u-Tan(t))=mgcos(t)

    F - mgcos(t)=mgcos(t) (u - Tan(t))

    F - mgcos(t) = mgcos(t) (u - Tan(t))
    -mgcos(t) -mgcos(t)

    F = mgcos(t) (u - Tan(t))

    The cos (t) at small angles is more or less equal to 1, so by division of the right hand side, mgcos (t) = 1. This is how I have interpreted the last part.

    Please advise.
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  6. #6
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    Quote Originally Posted by David Green View Post
    Hi Skeeter,

    Here's my next attempt?

    F = umgcos(t) - mgcos(t) Tan(t) The subject (F = mgcos(t) (u - Tan (t))

    F - Tan(t) = umgcos(t)-mgcos(t) ... this step is incorrect, and all that follows is also incorrect. you cannot separate tan(t) from mgcos(t)tan(t) by subtraction.
    go back and look at the correct algebraic method I showed you earlier.

    why are you making this simple exercise in factoring more difficult than it needs to be?
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    Quote Originally Posted by skeeter View Post
    go back and look at the correct algebraic method I showed you earlier.

    why are you making this simple exercise in factoring more difficult than it needs to be?
    I am not deliberatley doing so,it's just i am trying to understand how to get from the beginning to the conclusion with all necessary steps involved.

    If I am now reading what you are saying correctly, are you saying;

    F = umgcos(t)-mgcos(t)Tan(t) original info

    F = 0.7x700xgxcos(t) -700xgcos(t)Tan(t) =

    F = 700xgxcos(t) (u - Tan(t))

    and if this is so, how did (u - Tan(t)) this occur?
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  8. #8
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    distribute M over the difference (u - T)

    M(u - T) = ?
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    Quote Originally Posted by skeeter View Post
    distribute M over the difference (u - T)

    M(u - T) = ?
    Sorry Skeeter

    At the moment it does not make sense to me?

    The accelerating force F = umgcos(t), this i understand is the force acting vertically downwards, then the mgcos(t)Tan(t) is subtracted, from what i see there are no like terms, umg is not the same as mg.

    If I use data as in your example, what i will then end up with is F = m(u-Tan(t))

    I still don't understand how "u" moved inside the bracket to become a minus from Tan (t)?
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  10. #10
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    you need to have a face-to-face session with your teacher.
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    Quote Originally Posted by skeeter View Post
    you need to have a face-to-face session with your teacher.
    Hi Skeeter,

    My college is close until week after next so can't ask teacher just yet, however looking at algebraic fractions;

    3x + 6 = 3(x + 2)
    6x + 12 6(x + 2) The factor x + 2 is common and by cancelling = 1/2

    So can i use this method to solve my problem, and if so how am I doing here;

    F = umgcos(t) - mgcos(t)Tan(t)
    F = u x u mgcos(t) - mgcos(t)(u - Tan(t))
    F = mgcos(t) - mgcos(t) (u - Tan(t)
    F = mgcos(t) - mgcos(t) (u - Tan(t)
    mgcos(t) mgcos(t)

    cancel the like terms on the right hand side leaves

    F = mgcos(t) ( u - Tan(t)
    mgcos(t)
    At this point I know that anything divided by itself = 1, so would i be right to think that I could now say;

    F = mgcos(t) ( u - Tan(t) from the above or am I still incorrect?
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  12. #12
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    Quote Originally Posted by skeeter View Post
    you need to have a face-to-face session with your teacher.
    Hi Skeeter,

    Thanks for your help on this topic, however, it would have been alot clearer for me if you had said;

    ab + ac = a(b + c), then F = umgcost(t) - mgcos(t)Tan(t), now simply change the terms;

    a = mgcos(t)
    b = u
    c = Tan (t), so

    F = mgcos(t)*u + mgcos(t)x-Tan(t)=mgcos(t)(u - Tan(t)
    a b a c a (b + c)

    Had I not found this identity in a maths book I would have been stumped?

    Now it has been simplied and i understand it now
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