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Homework Help: Forces question, a pedagogical machine

  1. Jan 22, 2010 #1
    forces question, a "pedagogical machine"

    1. The problem statement, all variables and given/known data

    All surfaces are frictionless. How much force must be applied to M1 in order to keep M3 from rising or falling?

    Picture:

    http://www.slideshare.net/brigittperalta/sol-maquina-pedagogica-1546585

    2. Relevant equations


    F = Ma

    3. The attempt at a solution

    I think my understanding of these force questions is way off.. if the surfaces are frictionless, how will M1 affect the M2 on top of it? I think in order to keep M3 from falling, you need the tension to equal to the weight of M3..

    thanks
     
  2. jcsd
  3. Jan 23, 2010 #2

    tiny-tim

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    Hi holezch! :smile:

    Your diagram shows a block of mass m1 being pulled by a horizontal force F.

    On the horizontal top of the block, a mass m2 is attached to a horizontal string which goes over a pulley to become vertical, and is attached to a mass m3 which is free to rise or fall in a vertical hole in the block.

    ok … there is no vertical acceleration, so yes, T = m3g, and yes, the block (m1) cannot affect m2.

    Hint: you need the acceleration of m2 to equal the acceleration of the block. :wink:
     
  4. Jan 23, 2010 #3
    Re: forces question, a "pedagogical machine"

    okay, would I have to pull M1 enough so that the whole vertical acceleration of the system is equal to the weight of M3? Since how M3 falls or risings is horizontal, vertical forces won't affect that

    thanks a lot!!
     
  5. Jan 23, 2010 #4

    tiny-tim

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    "the whole vertical acceleration of the system" doesn't make any sense …

    the system doesn't move "as a whole".

    and "since how M3 falls or rises is horizontal" doesn't make any sense either. :confused:

    Just do Newton's second law three times, once for each of the three masses.​
     
  6. Jan 23, 2010 #5
    Re: forces question, a "pedagogical machine"

    wouldn't it move together? if M2 sits on top of M1..

    thanks
     
  7. Jan 23, 2010 #6

    tiny-tim

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    No, because all surfaces are frictionless …

    if there's no friction, and if there was no tension from m3, there's no force, so (good ol' Newtons first law) m2 ain't going nowhere :biggrin:
     
  8. Jan 24, 2010 #7
    Re: forces question, a "pedagogical machine"

    okay, so I guess the big m1 block has to push m3 which pulls the cord and pulls m2.. so then you have to push m3 until the acceleration of m2 is equal to m1? Is that so the tension of the string is balanced out?

    thanks
     
  9. Jan 25, 2010 #8

    tiny-tim

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    (just got up :zzz: …)

    No, you're not thinking about this the right way.

    You need to think about each mass separately … what are the forces on it? … what is its acceleration?

    Concentrate on m3 (the one in the vertical hole) …

    what are the horizontal forces on m3? … what is its horizontal acceleration?

    what are the vertical forces on m3? … what is its vertical acceleration?
     
  10. Jan 25, 2010 #9
    Re: forces question, a "pedagogical machine"

    thanks tiny-tim, I drew my free body diagram focusing individually on masses.. for M3, I have that ma1 is pushing it sideways, mg is pulling it down, and so the total force is diagonal m(a1+g) and this is what is pulling the cord as well. Then the cord pulls on M2 by m(a1+g).. and then by newton's 3rd law, M2 should pull back by m(a1+g) as well.. but then that would cause an equilibrium in the pulley system for any force you apply to M1 ?? that's definitely not right...
     
  11. Jan 25, 2010 #10
    Re: forces question, a "pedagogical machine"

    my concept of tension is obviously wrong.. if I imagined a case when M2 is not moving and M3 is just pulling down, I'll get an equilibrium!

    If M3 is pulling on the cord, wouldn't that become the tension?
     
  12. Jan 25, 2010 #11

    tiny-tim

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    Sorry, but that's completely wrong. :redface:

    A cord (string, rope, etc) only experiences a force (tension) along its length.

    The length is vertical, so the force is vertical.

    (Yes, I agree m3 is also getting a horizontal acceleration, but that's from being pushed horizontally by the wall of the hole … it has nothing to do with the cord.)

    Start again, calculating the tension T only from the vertical part of F = ma for m3.
     
  13. Jan 25, 2010 #12
    Re: forces question, a "pedagogical machine"

    but if you pushed the block sideways so it starts moving horizontally, wouldn't it pull the cord sideways as well? :S
     
  14. Jan 25, 2010 #13

    tiny-tim

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    You need to consider one mass at a time …

    For m3, the cord is vertical, so the tension T3 acting on m3 is vertical.

    For m2, the cord is horizontal, so the tension T2 acting on m2 is horizontal.

    (and obviously in this case T2 = T3)
     
  15. Jan 25, 2010 #14
    Re: forces question, a "pedagogical machine"

    thanks, but I meant for M3, if M1 pushes M3 horizontally, the cord will start getting pulled horizontally? :S
     
  16. Jan 25, 2010 #15

    tiny-tim

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    If you mean that the cord will start moving horizontally, the answer is no, not necessarily.

    If you mean that the cord will start getting a horizontal tension, that has nothing do to with the force of m1 on m3.

    Why did you ask this?

    Which of the three masses were you considering the forces on, and in which direction? :confused:

    (btw, i'm going to bed now :zzz: …)
     
  17. Jan 25, 2010 #16
    Re: forces question, a "pedagogical machine"

    I was trying to focus on m3 and then m2.. :( Could you please show me a worked out solution so I know what to do for next time? I'd really appreciate it.. thanks
     
  18. Jan 26, 2010 #17

    tiny-tim

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    erm :redface:noooo

    You have three unknowns: a, the acceleration of m1, b, the acceleration of m2 and m3, and T, the tension.

    Do three separate F = ma equations for forces and acceleration:

    i] for m1 in the horizontal direction

    ii] for m2 in the horizontal direction

    i] for m3 in the vertical direction​

    That will give you three equations for three unknowns, which you can then solve.
     
  19. Jan 26, 2010 #18
    Re: forces question, a "pedagogical machine"

    okay, if we want m2 to have the same acceleration as m1, don't we only need to push m1 so that the total force is equal to mg/2 (if they had equal masses) ? since that's how m2 will be moving.. thanks
     
    Last edited: Jan 26, 2010
  20. Jan 26, 2010 #19
    Re: forces question, a "pedagogical machine"

    acceleration of m3 = mg/2
    T = mg/2
    acceleration of m2 = mg/2
    acceleration of m1 = m1a1 (we can choose this force)
     
  21. Jan 26, 2010 #20
    Re: forces question, a "pedagogical machine"

    I still need to produce a force to counter balance the mg from m3.. I don't know what kind of force would do that :(
     
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