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

[Medgirl314]

Homework Statement

1. Suppose two masses are arranged as shown and released from rest. There is no friction.
Suppose we have a certain mass m1
and we want to cause a certain acceleration a. How big
should m2be? (In other words, derive a formula for m2in terms of m1, a, and g.)

Homework Equations

m$_{}1$-m$_{}2$+m$_{}2$g(a)

The Attempt at a Solution

I know I need to isolate m$_{}1$ on one side of the equation. I tried adding m$_{}2$g to both sides and got the following:
m$_{}1$g=m$_{}1$+m$_{}2$+m$_{}2$g (a)
If I divided by g on the left side, the equation would go unchanged and I'd have my answer. Except for that little pesky m$_{}1$ in the parentheses. I can't seem to figure out how to get it out. Could someone please help me along here?

Thanks in advance!

 

[SteamKing]

You don't have an equation in Section 2.

 

[HallsofIvy]

Homework Statement

1. Suppose two masses are arranged as shown and released from rest.

Shown where??

There is no friction.
Suppose we have a certain mass m1
and we want to cause a certain acceleration a. How big
should m2be? (In other words, derive a formula for m2in terms of m1, a, and g.)

Why should the mass of one object affect the mass of the other? Perhaps that was shown in a figure that you do not have here?

Homework Equations

m$_{}1$-m$_{}2$+m$_{}2$g(a)

That is not an equation unless one of those "+"s was supposed to be "=".

The Attempt at a Solution

I know I need to isolate m$_{}1$ on one side of the equation. I tried adding m$_{}2$g to both sides and got the following:
m$_{}1$g=m$_{}1$+m$_{}2$+m$_{}2$g (a)

Are there parentheses missing on the right side of the equation? If you meant
$m_1g= (m_1+ m_2+ m_2)g(a)$
Then you could divide both sides by g to eliminate g.

If I divided by g on the left side, the equation would go unchanged and I'd have my answer. Except for that little pesky m$_{}1$ in the parentheses. I can't seem to figure out how to get it out. Could someone please help me along here?

If it really is [itex]m_1g= (m_1+ m_2+ m_2)g(a)[/oitex] then the two "[itex]m_1g[itex]" terms will cancel out and you cannot solve for [itex]m_1[/tex]

Thanks in advance!

 

[Medgirl314]

Could you elaborate, please? Did I do the LaTex wrong, or was the algebra wrong?

Thanks!

 

[Medgirl314]

Oops, sorry Halls of Ivy, my page didn't refresh.

I completely neglected the diagram when I posted this! I'll go to mobile and attach it and then perhaps this will be easier. I can't read the last LaTex equation, is that on my end or yours?

Thanks so much.

 

[Medgirl314]

SteamKing, I think I see what you mean with the equation. When I typed it I must have done something wrong with LaTex. Thanks!

 

[Medgirl314]

Here we are. Can you see the diagram and the question easily? The problem I'm on right now is the first.

Thanks again!

 

[Dick]

View attachment 65628

Here we are. Can you see the diagram and the question easily? The problem I'm on right now is the first.

Thanks again!

Can't see it easily, but I can see it. So how did you derive an equation relating all of the variables? What you've stated before doesn't look correct.

 

[Medgirl314]

Sorry about the pic, maybe I can figure out how to take it better next time. The equation I used is the root of the problem. I clicked on a video and watched my teacher use the equation that I will attach. I forgot that that video was part of a series, and he said that that equation was for one specific situation. I will rewatch the video and get back to this thread tomorrow, thank you all for pointing out that the equation was wrong for this situation.

 

[Dick]

Sorry about the pic, maybe I can figure out how to take it better next time. The equation I used is the root of the problem. I clicked on a video and watched my teacher use the equation that I will attach. I forgot that that video was part of a series, and he said that that equation was for one specific situation. I will rewatch the video and get back to this thread tomorrow, thank you all for pointing out that the equation was wrong for this situation. View attachment 65629

Yes, it's wrong. Atwood's machine has both masses hanging. You have one mass on a horizontal frictionless surface. There are ways to derive these on your own. Use the notion of 'tension'.

 

[Medgirl314]

So sorry for the delay! I had masses of items thrust upon my to-do list at once.

This is my first time deriving an equation like this without having numbers to start with, so please excuse any faulty logic that ensues.

So far I've come up with m$_{}2$g-m$_{}1$g(a) ,

but this doesn't seem right because I didn't even touch it algebraically, I just looked at the diagram and tried to see what may work. I'm slightly stuck on the first step, going from F=ma.

Edit: Sorry about the faulty LaTex. I'm still figuring it out.

 

[lightgrav]

the pic in post #7 is called "Fletcher's Trolley". a hanging mass propels itself and the trolley.
the pic in post #9 is called "Atwood's machine" either way:
Easy way: treat the pulley as a "bend" in the coordinate along the string
left side: How much external Force is there (total), along the string?
right side: how much mass is there (total) that needs to be accelerated?

 

[Medgirl314]

Okay, so we start with F=ma.

I'm pretty detailed with my work, so I'm not really going to understand how to work this unless I find the total mass first, if possible. Would it be m2-m1?

Thank you!

 

[lightgrav]

Fletcher's Trolley?
Which mass is negative? total mass means Σm ! be sure to distinguish mass from weight!

 

[Medgirl314]

Yes, starting from the diagram in #7. If m2 is going down, and down is positive, then m1 is negative, correct? I still can't really tell if my statement of total mass= m2-m1 is correct.

 

[lightgrav]

mass is a _scalar_ , it has no direction, and is always positive.

weight is a Force, a vector that is applied to any mass (downward) when it is immersed in gravity (a field).
velocity is a vector that describes how fast something is moving.

be careful: ΣF = (Σm) a

 

[Medgirl314]

What does the symbol mean in the context of your equation? I'm not familiar with it quite yet. So would the total mass simply be the two masses added together?

 

[lightgrav]

Σ is capital Greek Sigma ... usually read it as "Sum __", it means "add them all".
for vectors, you worry about direction ... for scalars you don't.

the trolley has just been released, with zero speed, so nothing is "going down" yet.
(its velocity might become faster, if it is "pulled" <= that's the Force, called "weight")

 

[Medgirl314]

Oh, I see it now! The reason my teacher made one of the numbers negative is because he was calculating weight, mG. Thanks! I'm still confused on the total mass. Would it be the two masses added together?

 

[lightgrav]

yes, in Atwood, one object is pulled "rightward", while the other object is pulled "leftward" .
left is opposite right ... left = - right.
(didn't you practice adding and subtracting mass, maybe day 1 or 2?)

caution: gravity field (9.8 N/kg) has symbol g , lower-case! (G is reserved for how planets cause g)
so left side (cause) = right side (effect)

 

[Medgirl314]

Nope, we dove into more complex ideas. :)

Ah, thanks for the warning! Maybe I'll bold next time I want to emphasize the variable.

So does this mean that the left side of the equation is m1+m2 ? I don't know how to bring acceleration or gravity into this. Are we actually calculating the weight on the left side instead of the mass?

Thanks!

 

[lightgrav]

we usually write "cause => effect" ... ΣF => (Σm) a ;
the sum of all Forces cause the whole thing's mass to accelerate.
so yes, the Forces (weights, track support perhaps cancelling) are on the left,
and the (passive recipient) mass is influenced by whatever the total Force is.

 

[Medgirl314]

What do you mean by track support and passive recipient mass? We haven't gotten there quite yet. :)

 

[lightgrav]

does the trolley fall thru the track? why not? Isn't it pulled down by gravity? it has mass, right? F=mg ...

subject predicate object ... there are active subjects (Earth distorts space here, making gravity g)
and there are passive objects (cart gets pulled by that gravity) ; pull (Force) is the predicate (verb)

so add all the Forces, and have that Sum accelerate all the mass.

 

[Medgirl314]

Ah, got it. I'm not entirely sure which part is the "trolley". Are we using that to describe the two masses that are being pulled? By 'have that sum accelerate all the mass", do you mean multiply it by the acceleration?

Thanks!

 

[lightgrav]

trolley is the cart on wheels that rolls on the table (or track). but ALL the mass needs to have the same velocity,
because the 2 parts are connected by a cord (rope) that does not stretch.

"set the Force Sum equal to" the effect that it produces ... (Σm) a

 

[Medgirl314]

I don't see a literal trolley in this diagram, just the masses connected by a rope pivoting over a ball. But I think you're saying that we do multiply the masses by the acceleration? So we have m1+m2(a), correct? Is that the total external force on the top of the string?
Is there a way we can do this algebraically? Or do we need to do it by looking at the diagram and just thinking about the forces that apply?

Thanks!

 

[lightgrav]

in post #7's pic, block m1 slides along a frictionless table toward the pulley
m1 can be set on wheels, to approximate frictionless motion, if you want to actually do this experiment.

Atwood's machine acceleration is fully solved on post #9's pic (but not the rope Tension)
this approach looks at the unbalanced Forces pulling leftward (ccw) or rightward (clockwise) ... not up and down.

 

[Medgirl314]

Oh, so it wasn't a literal trolley. Okay, makes sense.

I'm somewhat confused by the last sentence. Is there a way to derive this particular equation algebraically, or do we need to make inferences based on the picture?

 

[lightgrav]

Do you see in post #9's pic, how the Forces are drawn?
the weight on the right tries to turns the system rightward (clockwise),
the weight on the left tries to turn the system leftward (ccw):
the total rightward Force is : m_right g - m_left g ... that is what Sum Forces means
... which the pic writes as "net Force" , which means all F's added up, sign-wise.

Using that Atwood as a pattern, can you do it for Fletcher in post #7?

There IS only one equation to derive acceleration ... Newton #2: ΣF = (Σm) a
(some situations might be tedious, to extract vector components or rotations but)
you always just add the forces on the left, add the masses on the right.

 

[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!