Common Factor.

[Gareth56]

I know the following involves common factors but I just cannot see how the 1appears in the second equation could some kind soul add an explanation?

 

= mg + m v^2/r .................(1)

 

???????????

 

= mg (1 + v^2/rg) ................(2)

 

It's what happens between (1) & (2) that's bugging me. Maths was so long ago!!!

 

Thanks

[DrP]

Looks like you are dividing through by mg!

 

So - put the whole lot in brackets, then take the mg outside. Leaving the 1 on the LHS. (because 1 x mg is mg)

 

 

To check it out - take whats in the brackets and multiply by mg!

You are left with eq.1. mg x 1 = mg and mg x v^2/rg = mv^2/r

[ajb]

Are you in any doubt that it is true?

 

Just expanding (2) gives (1).

 

It is easiest to go from (1) to (2) in a few steps.

 

I) take out the common factor of m,

 

[math]m (g + \frac{v^{2}}{r})[/math]

 

II) now take out a factor of [math]g[/math], but gang on, there is no factor g in the second part! So, put one in

 

[math]m(g +\frac{v^{2}}{r} \frac{g}{g})[/math]

 

as [math]g/g[/math] = 1

 

III) Now take out the factor to get

 

[math]mg (1 +\frac{v^{2}}{g r} )[/math]

[Gareth56]

You guys are brilliant. It's so simple if you can "see" these things. 30+ years have dulled my eyesight somewhat

 

Although I'm not too sure how you can just put things in like g/g, clearly it's allowed but why?

 

Also when you say "take out the factor" do you mean take the factor outside the brackets?

 

Thanks again

Edited September 4, 2008 by Gareth56

[ajb]

Yes indeed I just mean take factor outside the bracket.

 

Provided we are dealing with things that a notion of divide ( or more generally inverses) exist then we can always have [math] a/a = a a^{-1} = a^{-1} a = 1 [/math], (identity) for "invertible elements".

 

As [math]b= 1b = b1 = 1b1[/math] etc we can insert such a factor [math]a/a[/math] where ever we like. The trick is knowing where to stick it!

[Gareth56]

I see. I shall always look to where I can stick my "invertible elements", hopefully I won't get arrested!

 

Thanks again ajb and give my regards to Manchester where I was born and educated (in a place on Oxford Road!)

[DrP]

I see. I shall always look to where I can stick my "invertible elements", hopefully I won't get arrested!

 

 

Quite - as ajb said g/g or a/a are equal to 1. So you can multiply anything by 1 anytime you like without changing the equation.

Edited September 4, 2008 by DrP
multiple post merged