Gareth56 Posted August 26, 2009 Share 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 Link to comment Share on other sites More sharing options...
Klaynos Posted August 27, 2009 Share 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... Link to comment Share on other sites More sharing options...
D H Posted August 27, 2009 Share 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. Link to comment Share on other sites More sharing options...
Gareth56 Posted August 27, 2009 Author Share 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 Link to comment Share on other sites More sharing options...
Klaynos Posted August 27, 2009 Share Posted August 27, 2009 I get it to work... your issue is with dropping the -ve sign on the right hand side. Link to comment Share on other sites More sharing options...
Gareth56 Posted August 27, 2009 Author Share 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. Link to comment Share on other sites More sharing options...
swansont Posted August 27, 2009 Share 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. Link to comment Share on other sites More sharing options...
Gareth56 Posted August 27, 2009 Author Share 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. Link to comment Share on other sites More sharing options...
insane_alien Posted August 27, 2009 Share 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. Link to comment Share on other sites More sharing options...
Klaynos Posted August 27, 2009 Share 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 Link to comment Share on other sites More sharing options...
Gareth56 Posted August 27, 2009 Author Share 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. Link to comment Share on other sites More sharing options...
insane_alien Posted August 27, 2009 Share 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. Link to comment Share on other sites More sharing options...
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