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

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The discussion revolves around deriving a formula for mass m2 in terms of mass m1, acceleration a, and gravitational acceleration g in a two-mass system. Participants clarify the need to correctly apply Newton's second law (F=ma) and the significance of forces acting on the masses. There is confusion regarding the initial equation presented, with suggestions to isolate variables and properly account for forces and mass interactions. The importance of understanding the system's dynamics, including the roles of tension and gravitational forces, is emphasized for accurate derivation. The conversation highlights the necessity of visual aids, like diagrams, to facilitate comprehension of the problem.
  • #51
m2g=(m1+m2)a ?
 
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  • #52
yes. now, expand the right-hand side, so that you can move the m2 a over to the left side.
 
  • #53
Sorry, I got tied up with other subjects.

m_{}2g=m_{}1a+m_{}2a


I think I may have used the wrong button.
 
  • #54
lightgrav said:
so that you can move the m2 a over to the left side.

that is to isolate your desired unknown m2 as a ratio
 
  • #55
Right, but I needed to expand first, so I just wanted to be sure that I didn't make an error. :smile:

m2g-m2a=m1a+m2a-m2a

How does that look so far?
 
  • #56
now isolate the desired unknown on the left (common factor) ... done.
 
  • #57
Ah, THIS is where I kept getting stuck last time! Okay, here's what we have:

m2g-m2a=m1a

This is where I haven't but a vague idea of what to do. Can I divide by ag?

Thanks!
 
  • #58
m2 on the left is a common factor in 2 subtracted terms
... "de-expand" those two, with m2 outside the parenthesis (with things that are subtracted)
 
  • #59
Right,thanks!

To expand we multiply, so do we divide by m2 here?
 
  • #60
algebra!
c(d + e) = cd + ce ... it goes both ways ... (left to right is expanding , right to left is collecting a common factor)
 
  • #61
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?
 
  • #62
it is still not isolated (all by itself) ... it is still multiplying that difference ...
 
  • #63
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?
 
  • #64
collected it as common factor, good.
you want it on TOP, by itself ... not on bottom.
move the other factor, instead of m2 .
 
  • #65
Oh! Duh! m2=m1a/g-a ?
 
  • #66
keep the old parentheses, right? ok, done.

move on to the one with friction
 
  • #67
m2=m1a/(g-a)

Do you mean we're done here?
 
  • #68
well, look at 2 special cases first:
1) what m2 will make zero acceleration?
2) what m2 will make acceleration = 9.8 m/s² ?
 
  • #69
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! :smile: I will eventually come back to these if I have a moment.
 

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