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
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)?
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.
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?
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)?
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?
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