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

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Homework Help Overview

The discussion revolves around the equation a(m1+m2) = (m2g-m1g), where participants are exploring the possibility of isolating m2 in the context of an Atwood machine problem. The variables include acceleration (a), mass m1, mass m2, and gravitational acceleration (g).

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • Participants are questioning whether it is feasible to solve for m2 given the equation and the known variables. Some suggest that isolating m2 is possible under certain conditions, while others express skepticism about the clarity of the problem setup.

Discussion Status

There is a mix of attempts to clarify the problem and explore the equation. One participant has claimed to find a solution for m2, while others have emphasized the importance of not providing direct answers to the original poster's question.

Contextual Notes

Participants note that the problem involves an Atwood machine scenario, where one mass is known and the other is not, along with a specified acceleration. There is an emphasis on maintaining the integrity of the homework process without directly solving the equations for the original poster.

tigerwoods99
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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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