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

  • Thread starter tigerwoods99
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In summary, the conversation discusses solving for m2 in an Atwood machine problem where one mass is given and the other is unknown. The solution provided is m2 = (-m1a-m1g)/(a-g), but the user plans to multiply by -1/-1 to simplify it to (m1a+m1g)/g-a.
  • #1
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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  • #2
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.
 
  • #3
lol..

<< most of solution deleted by berkeman >>

This however looks completely ridiculous, what is it you are trying to solve??
 
  • #4
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.
 
  • #5
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!
 

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

Can this equation be solved for m2?

Yes, the equation can be solved for m2. However, it is important to note that the solution may not always be unique.

Is this equation valid for all values of m1 and m2?

Yes, this equation is valid for all values of m1 and m2. The equation is based on the principles of Newton's Second Law of Motion and is applicable to all objects, regardless of their mass.

What is the significance of the g in this equation?

The g in this equation represents the acceleration due to gravity. It is a constant value that is approximately equal to 9.8 meters per second squared on Earth. This value is used to calculate the weight of an object, which is necessary for solving for m2 in this equation.

Can this equation be used to solve for m1 instead of m2?

Yes, this equation can be rearranged to solve for m1 as well. Simply switch the positions of m1 and m2 in the equation and follow the same steps to solve for m1.

Is this equation limited to a specific type of problem?

No, this equation can be used to solve for m2 in a variety of problems involving forces and masses. It is commonly used in physics and engineering to calculate the unknown mass of an object when given other values such as acceleration, force, and weight.

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