Deriving Formula for m2 in Terms of m1, a, and g in a Two Mass System

[Medgirl314]

Oh, I was overcomplicating it again! We're literally just pulling it out of the equation, I let he subscript mess me up.

m2g-m2a=m1a
m2(g-a)=m1a

Better?

 

[lightgrav]

it is still not isolated (all by itself) ... it is still multiplying that difference ...

 

[Medgirl314]

Oh, right. Shoot. But was the step correct? Can I divide both sides by m2, yielding g-a=m1a/m2 ? Or would that not be considered solving for m2, since it's not on it's own side?

 

[lightgrav]

collected it as common factor, good.
you want it on TOP, by itself ... not on bottom.
move the other factor, instead of m₂ .

 

[Medgirl314]

Oh! Duh! m2=m1a/g-a ?

 

[lightgrav]

keep the old parentheses, right? ok, done.

move on to the one with friction

 

[Medgirl314]

m2=m1a/(g-a)

Do you mean we're done here?

 

[lightgrav]

well, look at 2 special cases first:
1) what m2 will make zero acceleration?
2) what m2 will make acceleration = 9.8 m/s² ?

 

[Medgirl314]

Lightgrav, I would really love to come back to these questions, but I have so many other problems this week that I can't justify spending the time on the extra questions that they deserve. Thanks so much for helping me with the problem! I will eventually come back to these if I have a moment.