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

[Medgirl314]

I think I understand the concept of how the machine moves well enough, but for some reason I'm having a hard time understanding your directions. Would the left side of the equation be m2g-m1g?

Thanks!

 

[lightgrav]

for Atwood's machine #9, but not for Fletcher's trolley (#7). the trolley's weight is supported by the table.

 

[Medgirl314]

Okay. Since we're only working on the trolley, we don't include weight, correct? But you mentioned that we don't have to subtract the masses, so is the left side a(m1+m2) ?

 

[lightgrav]

Draw the Force vectors that are applied to the objects in the system.
... both of them have mass, so both of them have weight (mg) pulling down.
The trolley's weight is perpendicular to the rope, and perpendicular to the ensuing displacement,
which makes it perpendicular to the eventual velocity and perp to the acceleration
- so that weight does not help nor hinder the acceleration (besides, the table's push upward cancels it).
But the hanging block weight is along the rope, so THAT one does cause acceleration
(sometimes only one component of the Force is along the acceleration, found by trig).

The acceleration term goes on the RIGHT side of the equation, as the effect.

 

[Medgirl314]

Could the answer possibly be m_s=m₂g-m¹a, with the right side all divided by a?

Thank you again!

 

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


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)