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

[lightgrav]

4 issues:
1) I didn't see any m_s in the problem ...
2) you've mixed cause (m₂g) onto the same side of the equation as effect (m₁a)
3) doesn't m₂ have inertia?
4) I thought you were supposed to solve for a (isolate it on the left)

 

[Medgirl314]

1) Sorry about that, I didn't hit "preview" first. I meant m2.
2) I didn't see a m2a in my equation, could you please point it out?
3) We haven't discussed inertia yet.
4) No, we're solving for m2. Sorry for the confusion!

Thanks!

 

[lightgrav]

Oops! Sorry, I thought you were writing Newton #2 .
solving for m₂ means that you need to isolate m₂ ... get both of them together first.
But it looks like you started with the correct Newton#2.

Inertia is from Newton #1: an object will maintain constant velocity due to its inertia (a property of its mass).
This rests on:
* mass is a numerically additive property, so it is a quantity ... Σm = total mass
* relative location x is a measurable property, so displacement is an additive quantity
* the product mx is an additive quantity (called the mass moment) ... Σ(mx) = m_total x_average

I fixrd my typo , but you might need to refresh screen

 

[Medgirl314]

I'm sorry, we are that far. I just had to look back. I just don't understand what you mean by "Doesn't m1 have inertia"? I think I meant that we haven't discussed inertia in a situation similar to this one.

 

[lightgrav]

1) I didn't see any m_s in the problem ...
3) doesn't m₂ have inertia?

I wrote line 3 because there should have been a term "m₂ a"
which I didn't recognize had been typo'ed into "m_s a", and then split by division.

Oops! Sorry, I thought you were writing Newton #2 .
But it looks like you started with the correct Newton#2.

So, how are you going to isolate m₂?

 

[Medgirl314]

Ah, okay. I thought you meant we needed an entire other equation to bring into this equation.

Would the following be a good place to start?

m₁+m₂=(m₂-m₁)a

 

[lightgrav]

After the diagram (displacement and velocity vectors, deduce acceleration vector, and draw Force vectors)
you usually write Newton's 2nd Law.
What is causing the Force?
How much mass is accelerating?

 

[Medgirl314]

The equation in post #41 wasn't actually where I started. I did all the steps you explained, but then I tried to rearrange Newtons 2nd Law to get the above equation. Sorry for the confusion!

 

[lightgrav]

Didn't the post 36 equation come from writing Newton's 2nd Law?
post 41 is missing gravity, and one of the masses seems to be negative.

 

[Medgirl314]

I haven't seen a new equation since post 36 besides the one in 41. Oops, I didn't notice that I neglected gravity. Can we start over from F=ma ? The next step, after writing F=ma, would be to determine the total mass, correct?

 

[lightgrav]

yes, ∑F = ∑m a ... generally, both sides are sums (recall helicopter tension).

 

[Medgirl314]

Okay, so would the mass be m₁+m₂ ?

 

[lightgrav]

yes. what's the Force?

 

[Medgirl314]

Is it m₂g?

 

[lightgrav]

yes. So Newton#2 is ...

 

[Medgirl314]

m₂g=(m₁+m₂)a ?

 

[lightgrav]

yes. now, expand the right-hand side, so that you can move the m₂ a over to the left side.

 

[Medgirl314]

Sorry, I got tied up with other subjects.

m$_{}2$g=m$_{}1$a+m$_{}2$a


I think I may have used the wrong button.

 

[lightgrav]

so that you can move the m₂ a over to the left side.

that is to isolate your desired unknown m₂ as a ratio

 

[Medgirl314]

Right, but I needed to expand first, so I just wanted to be sure that I didn't make an error.

m2g-m2a=m1a+m2a-m2a

How does that look so far?

 

[lightgrav]

now isolate the desired unknown on the left (common factor) ... done.

 

[Medgirl314]

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!

 

[lightgrav]

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)

 

[Medgirl314]

Right,thanks!

To expand we multiply, so do we divide by m2 here?

 

[lightgrav]

algebra!
c(d + e) = cd + ce ... it goes both ways ... (left to right is expanding , right to left is collecting a common factor)

 

[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.