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Newton's laws problem: K&K 2.19

  1. Jul 28, 2011 #1
    1. The problem statement, all variables and given/known data
    A pedagogical machine is illustrated in the sketch, yatta yatta, what force F must be applied to M_1 to keep M_3 from rising or falling? No friction. Here's the http://www.slideshare.net/brigittperalta/sol-maquina-pedagogica-1546585" [Broken].

    3. The attempt at a solution
    If we examine the problem from M_1's frame, M_2 seems to be experiencing a force to the left of magnitude $F/(M_1+M_3)\cdot M_2$. The fraction is the acceleration F imparts on M_1 and M_3 (as they are horizontally constrained to be together) and by multiplying it by M_2 we find the (inertial) force on M_2.
    This should be equal to the force of gravity on M_3 so
    M_3 g=F/(M_1+M_3)\cdot M_2.
    Solving for F, we get
    F=2Mg
    where M=M_1=M_2=M_3. However, K&K says I should be getting 3Mg. Can anyone explain why F acts on all three masses?

    Thanks.
     
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Jul 28, 2011 #2
    Why wouldn't it? The force is giving an acceleration to the whole system, which consists of three masses, not two.
     
  4. Jul 28, 2011 #3
    Actually, now that I think about it, I guess it makes sense.
    If M_3 is not moving (in respect to M_1), M_2 is moving to the right, and therefore the force is acting on it as well (through tension).
     
    Last edited: Jul 28, 2011
  5. Jul 28, 2011 #4
    I'd think so as well, the pulley itself is connected to the large block which experiences the force.
     
  6. Jul 28, 2011 #5
    Well, I dunno - I'm kinda confused about the pulley. I was thinking that the pulley can't really "pull" M_2, because it just slides past the string, redirecting the force.

    What I was thinking is that if just the right force F is applied, M_3 doesn't accelerate, and due to the string-connection, M_2 doesn't accelerate either, which means M_2 moves along with M_1 as well. Thus, the force must be acting on M_2 through tension (the string is taut) or something like that.
     
  7. Jul 28, 2011 #6
    Well, the string itself is also moving along with the pulley (due to F). The sliding would be due to the tension force.

    m3 will always tend to go down, regardless of the acceleration. It's the force on m2 which will balance this tendency to go down.
     
  8. Jul 29, 2011 #7

    PeterO

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    Beautiful set-up:

    I would look at a partial system - to get my head around what is/might happen.

    If you held on to M1, keeping it still, M3 would fall, and m2 would move to the right. That is a familiar system and with equal masses the acceleration would be g/2 [half the acceleration due to gravity]. The tension in the string would be 1/2Mg, but M3 is falling.


    If M3 is not to rise or fall, the tension in the string has to be Mg, so that there is no net vertical force on M3. Since that tension also acts on M2, M2 will be accelerating right at acceleration g.

    If m2 is not to move relative to M1, then M1 must also be accelerating right with acceleration g - so the applied force has to cause all 3 masses to accelerate right with acceleration g - so the force will be 3Mg.

    Peter
     
    Last edited by a moderator: May 5, 2017
  9. Jul 29, 2011 #8

    PeterO

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    Not quite! if you apply a larger force to m1, you may cause it to accelerate at more than g [lets say 1.2g]. If m2 was then supposed to have a similar acceleration [no slipping remember or m3 will be moving up or down], the tension in the string would have to be 1.2Mg.

    But if the tension in the string was 1.2Mg, then m3 will rise, as the upward force [Tension 1.2Mg] exceeds the downward force [Mg]; the weight of m3

    But that result is not allowed to happen.

    Peter
     
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