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

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...

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.

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, 200916 yr by Gareth56

I get it to work... your issue is with dropping the -ve sign on the right hand side.

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.

[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.

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.

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.

My full workings:

 

F_T - m_Eg = -m_Ea

F_T - m_Cg = +m_Ca

 

 

F_T - m_Eg -F_T + m_Cg = -m_Ea -m_Ca

 

- m_Eg + m_Cg = -m_Ea -m_Ca

 

(- m_E + m_C)g = (-m_E -m_C)a

 

( m_C - m_E)g = -(m_E +m_C)a

 

-( m_C - m_E)g = (m_E +m_C)a

 

(m_E - m_C)g = (m_E +m_C)a

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.

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.