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Homework Help: Deriving an equation for m1

  1. Jan 14, 2014 #1
    1. The problem statement, all variables and given/known data

    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.)

    2. Relevant equations


    3. The attempt at a solution

    I know I need to isolate m[itex]_{}1[/itex] on one side of the equation. I tried adding m[itex]_{}2[/itex]g to both sides and got the following:
    m[itex]_{}1[/itex]g=m[itex]_{}1[/itex]+m[itex]_{}2[/itex]+m[itex]_{}2[/itex]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[itex]_{}1[/itex] 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! :smile:
  2. jcsd
  3. Jan 14, 2014 #2


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    You don't have an equation in Section 2.
  4. Jan 14, 2014 #3


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    Shown where??
    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?

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

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

    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]

  5. Jan 14, 2014 #4
    Could you elaborate, please? Did I do the LaTex wrong, or was the algebra wrong?

  6. Jan 14, 2014 #5
    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. :smile:
  7. Jan 14, 2014 #6
    SteamKing, I think I see what you mean with the equation. When I typed it I must have done something wrong with LaTex. Thanks!
  8. Jan 15, 2014 #7
    ImageUploadedByPhysics Forums1389768314.670859.jpg

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

    Thanks again!
  9. Jan 15, 2014 #8


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    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.
  10. Jan 15, 2014 #9
    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. ImageUploadedByPhysics Forums1389770348.340184.jpg
  11. Jan 15, 2014 #10


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    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'.
    Last edited: Jan 15, 2014
  12. Jan 27, 2014 #11
    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[itex]_{}2[/itex]g-m[itex]_{}1[/itex]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.
  13. Jan 27, 2014 #12


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    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?
  14. Jan 27, 2014 #13
    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!
  15. Jan 27, 2014 #14


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    Fletcher's Trolley?
    Which mass is negative? total mass means Σm ! be sure to distinguish mass from weight!
  16. Jan 27, 2014 #15
    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.
  17. Jan 27, 2014 #16


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    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
  18. Jan 27, 2014 #17
    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?
  19. Jan 27, 2014 #18


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    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")
  20. Jan 27, 2014 #19
    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?
  21. Jan 27, 2014 #20


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    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)
  22. Jan 27, 2014 #21
    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?

  23. Jan 27, 2014 #22


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    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.
  24. Jan 27, 2014 #23
    What do you mean by track support and passive recipient mass? We haven't gotten there quite yet. :)
  25. Jan 27, 2014 #24


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    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.
  26. Jan 27, 2014 #25
    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?

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