Gareth56 Posted August 26, 2009 Posted August 26, 2009 In my physics book it has the following 2 equations:- F(subT) - m(subE)g = -m(subE)a F(subT) - m(subC)g = +m(subC)a To get rid of F(subT) it says to subtract the first equation from the second. here's where I get confused, I was to told (years ago) do this change the signs of the bottom equation and then add both equations but when I try that I don't get their answer of:- (m(subE) - m(subC))g = (m(subE) + m(subC))a --------------(3) Could anyone describe the process, as I don't get what's on the right hand side of equation (3) Thanks
Klaynos Posted August 27, 2009 Posted August 27, 2009 What answer do you get? The way I used to do it, is put one above the other and write a (-) sign out the front and work through it the same way you would a normal subtraction list...
D H Posted August 27, 2009 Posted August 27, 2009 Suppose you have two equalities, [math]\aligned a&=b\\c&=d\endaligned[/math] Step-by-step, here is why [math]c-a=d-b[/math] 1. Adding the same value to both sides of an equality does not change the equality. If [math]c=d[/math] then [math]c+x=d+x[/math] for all x. So, add [math]-a[/math] to both sides of [math]c=d[/math]: [math]c-a=d-a[/math] 2. Equality is transitive and reflexive. Since [math]a=b[/math], then [math]d-a=d-b[/math]. Thus [math]c-a=d-b[/math] In short, you can subtract one equality from another.
Gareth56 Posted August 27, 2009 Author Posted August 27, 2009 (edited) What answer do you get? The way I used to do it, is put one above the other and write a (-) sign out the front and work through it the same way you would a normal subtraction list... As I said I changed the signs of the bottom line then added so I got F(subT) - m(subE)g = -m(subE)a -F(subT) + m(subC)g = -m(subC)a and got:- (m(subE) - m(subC))g = (m(subE) - m(subC))a which is wrong but when it comes to dealing with letters instead of numbers my head explodes in confusion. I mean how do you subtract -m(subE)g from +m(subC)???? I need numbers!!!!! I'm sorry D H but your post was too advanced for my simple brain Edited August 27, 2009 by Gareth56
Klaynos Posted August 27, 2009 Posted August 27, 2009 I get it to work... your issue is with dropping the -ve sign on the right hand side.
Gareth56 Posted August 27, 2009 Author Posted August 27, 2009 My issue is that I haven't a clue how the answer in the book was arrived at as they've assumed a certain level of knowledge on the subject and left a couple of steps out.
swansont Posted August 27, 2009 Posted August 27, 2009 [math]F_T - m_eg = -m_ea[/math] [math]-F_T + m_Cg = -m_Ca[/math] ————————————— [math](m_C - m_e)g = -(m_e + m_C)a[/math] multiply by -1 [math](m_e - m_C)g = (m_e + m_C)a[/math] LaTex makes it easier to look at, IMO Learning to manipulate the symbols and do algebra is a powerful tool, and a skill that needs to be mastered. Failure to do so will result extra work in a lot of problems, when you aren't able to cancel terms, which leads to a whole host of potential errors.
Gareth56 Posted August 27, 2009 Author Posted August 27, 2009 many thanks Swansont. Can I just ask how do you arrive at [math](m_C - m_e)g [/math] from adding [math]- m_eg [/math] & [math] + m_Cg[/math] Sorry about the LaTex this is something in addition to basic maths manipulation that I need to master.
insane_alien Posted August 27, 2009 Posted August 27, 2009 g is a common factor. you can take common factors outside the brackets. it means that if you wish to expand the equation you multiply the terms inside by g which returns you to where you started.
Klaynos Posted August 27, 2009 Posted August 27, 2009 My full workings: FT - mEg = -mEa FT - mCg = +mCa FT - mEg -FT + mCg = -mEa -mCa - mEg + mCg = -mEa -mCa (- mE + mC)g = (-mE -mC)a ( mC - mE)g = -(mE +mC)a -( mC - mE)g = (mE +mC)a (mE - mC)g = (mE +mC)a
Gareth56 Posted August 27, 2009 Author Posted August 27, 2009 Many thanks for your time and trouble. It's 99.99% clear now so if you could indulge a dimwit for one more time and explain the last two steps i.e. from -( mC - mE)g = (mE +mC)a to (mE - mC)g = (mE +mC)a I would be eternally grateful. I understand that in Step 4 you multiplied both sides by -1 but not too sure how you got rid of the minus sign outside the left hand side bracket in Step 5 Just popped to my local library to get Engineering Mathematics by Stroud which I understand is a good book for this sort of thing.
insane_alien Posted August 27, 2009 Posted August 27, 2009 gareth. you can take the - sign as being -1. as he is multiplying the inside by -1 he must divide the outside by -1. -1/-1 is 1 which dosn't have to be shown as a multiplication by 1 is trivial.
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