IS IT POSSIBLE TO SOLVE FOR m2 from a(m1+m2) = (m2g-m1g)?

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SUMMARY

The discussion centers on solving the equation a(m1+m2) = (m2g - m1g) for the variable m2, where g represents gravity. The user successfully isolates m2, arriving at the formula m2 = (m1a + m1g) / (g - a). This solution is particularly relevant for problems involving Atwood machines, where one mass is known and the acceleration is provided. The conversation emphasizes the importance of having known values for acceleration (a) and mass (m1) to effectively solve for m2.

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Homework Statement



a=(m2g-m1g)
------------- divided by IS IT POSSIBLE TO SOLVE FOR m2 =?
(m1+m2) where g = gravity


Homework Equations



a(m1+m2) = (m2g-m1g)
where g = gravity

The Attempt at a Solution



a(m1+m2) = (m2g-m1g)

IS IT POSSIBLE TO SOLVE FOR m2 =? where g = gravity

THANKS!
 
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tigerwoods99 said:

Homework Statement



a=(m2g-m1g)
------------- divided by IS IT POSSIBLE TO SOLVE FOR m2 =?
(m1+m2) where g = gravity


Homework Equations



a(m1+m2) = (m2g-m1g)
where g = gravity

The Attempt at a Solution



a(m1+m2) = (m2g-m1g)

IS IT POSSIBLE TO SOLVE FOR m2 =? where g = gravity

THANKS!

You can certainly isolate it on one side of the equation to get m2 = ___________

Whether you can "solve" for it depends on if you have "a" and "m1" as knowns.
 
lol..

<< most of solution deleted by berkeman >>

This however looks completely ridiculous, what is it you are trying to solve??
 
Unto said:
lol..

<< most of solution deleted by berkeman >>

This however looks completely ridiculous, what is it you are trying to solve??

Please remember not to do the OP's homework for them. We can offer tutorial help, but do not work their equations for them. Thanks.
 
Thanks! i just solved it myself lol. i got m2 = (-m1a-m1g)/(a-g) but i guess i will multyply by -1/-1 to make it (m1a+m1g)/g-a

and I am trying to solve an atwood machine problem. For example, 1 mass is given and the other is not given, and you know the acceleration so m2 = blabla really helps!
 

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