# Thread: [SOLVED] Transposition of Formula

1. ## [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

2. Originally Posted by David Green
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.

3. Originally Posted by skeeter
$F = \mu \textcolor{red}{mg \cos{t}} - \textcolor{red}{mg \cos{t}} \tan{t}$

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)?

4. Originally Posted by David Green
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.

5. ## Algebraic Fractions

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.

6. Originally Posted by David Green
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?

7. Originally Posted by skeeter
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?

8. distribute M over the difference (u - T)

M(u - T) = ?

9. Originally Posted by skeeter
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)?

10. you need to have a face-to-face session with your teacher.

11. Originally Posted by skeeter
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?

12. Originally Posted by skeeter
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)