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Kleppner - Pedagogical Machine

  1. Mar 19, 2014 #1

    jbunniii

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    Edit: I think I may have found my problem. See my second post below.

    1. The problem statement, all variables and given/known data
    I'm trying to solve this problem from Kleppner and Kolenkow:
    qXhH0.png
    But I seem to be missing something because my answer doesn't match the clue. I have worked out the entire problem but I want to focus on the physical content so I'll just show my force equations below. I think the mistake is with one of them because I've checked my algebra in the rest of the problem several times.

    2. Relevant equations
    ##F=ma##


    3. The attempt at a solution
    I chose coordinates ##x## pointing to the right and ##y## pointing upward, stationary relative to all of the masses.

    Mass 2 - vertical forces
    We have an upward normal force of ##N_2## and a downward weight of ##M_2 g##. This mass does not accelerate vertically, so ##N_2 = M_2 g##.

    Mass 2 - horizontal forces
    The only horizontal force on ##M_2## is the tension in the rope. I call it ##T##. It is pointing to the right and satisfies ##T = M_2 \ddot x_2##.

    Mass 3 - vertical forces
    This mass is acted upon by the same tension ##T## in the upward direction, and weight ##M_3 g## in the downward direction. Thus, ##T - M_3 g = M_3 \ddot y_3##.

    Mass 3 - horizontal forces
    I assume mass 3 is touching the walls of the slot in which it is situated, so there are normal forces from each wall. I denote these by ##N_3## pointing rightward and ##N_4## pointing leftward. Thus, ##N_3 - N_4 = M_3 \ddot x_3##.

    Mass 1 - vertical forces
    Mass 1 is acted upon by downward weight ##M_1 g##, and upward by the normal force exerted by the floor, ##N_1##. It is also acted upon by the downward normal force ##N_2## due to the weight of mass 2. Thus, ##N_1 - N_2 - M_1 g = 0##, where the right hand side is zero because mass 1 is not moving vertically.

    Mass 2 - horizontal forces
    Mass 1 is acted upon horizontally by the applied force ##F## in the rightward direction. It is also acted upon by the same normal forces as mass 3, but in the opposite directions, so ##N_3## points leftward and ##N_4## points rightward. Thus, ##F + N_4 - N_3 = M_1 \ddot x_1##.

    Constraints
    Note that mass 3 is constrained horizontally so that it moves in lockstep with mass 1, so ##\ddot x_3 = \ddot x_1##.

    Also, the rope connecting masses 2 and 3 causes them to move in lockstep. The horizontal motion of mass 2 relative to mass 1 must match the vertical motion of mass 3. Thus, ##\ddot x_2 - \ddot x_1 = -\ddot y_3##.

    Concerns
    As I mentioned, if I work out the problem to completion, my answer does not match that of the clue. Moreover, I think the error is conceptual, not algebraic. For suppose I combine the Mass 3 Horizontal equation with the Mass 2 Horizontal equation:
    $$N_3 - N_4 = M_3 \ddot x_3$$
    $$F + N_4 - N_3 = M_1 \ddot x_1$$
    Add these together to obtain
    $$F = M_1 \ddot x_1 + M_3 \ddot x_3$$
    As noted above, we have the constraint ##\ddot x_1 = \ddot x_3##, so this reduces to
    $$F = (M_1 + M_3) \ddot x_1$$
    This seems wrong, because the total mass in the system is ##M_1 + M_2 + M_3## and surely ##F## must push all three of these in order to accelerate mass 1. So I think this should be ##F = (M_1 + M_2 + M_3) \ddot x_1##. But I can't see the where my error is.
     
    Last edited: Mar 19, 2014
  2. jcsd
  3. Mar 19, 2014 #2

    jbunniii

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    Oh, wait a minute. I didn't take into account the forces on the pulley, which is attached to ##M_1##. Thus ##M_1## has an additional horizontal force, ##T##, acting leftward, and an additional vertical force, ##T##, acting downward. The fact that I didn't have "equal and opposite" forces somewhere in the system to balance the ##T## forces I had already identified should have raised a red flag. I'll rework this and post a revised set of equations, might have to wait for tonight because work calls. :tongue:
     
  4. Mar 19, 2014 #3

    AlephZero

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    What you are doing will get there in the end, but the "trick" with this sort of problem is to spot questions that have easy answers, for example

    From the kinematics of the machine, if M3 does not move vertically, how does M1 move relative to M2?
    If M3 does not move vertically, what is the tension T in the string?
    Knowing the tension T, what is the acceleration of M2 relative to the ground?
     
  5. Mar 19, 2014 #4
    It's not necessary to consider the movement of M1 in the problem. When I first did a version of this problem, I used only M2 and M3 ;).
     
  6. Mar 19, 2014 #5

    AlephZero

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    Well, there are lots of ways of making mistakes. And sometimes you get lucky and the mistakes cancel out :smile:

    I can't see how you can ignore the movement of M1 and get the right answer here. (But let's not spoil jbunniii's fun by giving too many clues).
     
  7. Mar 19, 2014 #6

    jbunniii

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    I don't trust my physical intuition (yet) so I'm happy to write everything out in gory detail. :smile: Even then I failed to notice two forces.

    Also, I prefer to plug in ##\ddot y_3 = 0## at the end, because the general equations might be insightful. In particular, I want to check whether there is any value of ##F## (maybe negative) which would wreck any of the assumptions I've made. For example, any answer that indicated a negative value of ##T## would be physically meaningless. The previous exercise about the painter on a scaffold has made me paranoid about such things. :tongue:

    By the way, it occurs to me that there is an important unstated assumption. The condition ##\ddot y_3 = 0## does not ensure that mass 3 is not moving vertically, only that it is not accelerating. If it already had a nonzero vertical velocity, I believe there would be no way to stop it by applying a constant force.* Maybe this assumption is implicit in "keep ##M_3## from falling or rising."


    *OK, maybe apply some glue to the bottom of ##M_3## and let it hit the bottom of the slot...
     
  8. Mar 19, 2014 #7
    You can't exactly ignore the movement of M1. However, you don't need to draw an FBD for it or anything (if you use the strategy I'm thinking of). I remember that the first time I solved this problem, I did it in that way- largely ignoring M1, and it all worked out fine. The second time I did this problem (at a physics program, and I had forgotten the problem) I did pay attention to M1 but got it wrong for the same reason as jbunniii xD. I thought about it for a bit though, and like him realized where I went wrong (and then subsequently remembered I had done essentially the same problem while learning from Kleppner).
     
  9. Mar 19, 2014 #8
    Yeah, I assumed that while doing this problem.
     
  10. Mar 19, 2014 #9

    AlephZero

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    ##y_3 = \text{constant} ## implies ##\ddot y_3 = 0##. But as you said ##\ddot y_3 = 0## does not necessarily imply ##y_3 = \text{constant} ##.

    The question is not physically realistic, because the machine can't have constant horizontal acceleration for ever in real life. Adding some more complications (e.g. extra forces that are removed at time ##t = 0##) could make it a bit more realistic, but wouldn't make modeling its behavior, a "better" question, IMO.
     
  11. Mar 19, 2014 #10

    jbunniii

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    The question doesn't need to be more realistic, but a statement that "initially the objects are all at rest" would make it more correct. If ##\dot y_3 \neq 0## at ##t = 0## then there's no solution. But I recognize that being a pedantic math guy isn't going to be helpful when doing physics problems... :biggrin:
     
  12. Mar 19, 2014 #11

    AlephZero

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    That's not so much about mathematical pedantry, as deciding how smooth the relevant functions have to be to model the physics. (But most engineers don't worry about that too much, until it turns round and bites them).
     
  13. Mar 19, 2014 #12

    jbunniii

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    OK, after adding the missing forces, the answer came out right, and I also got the expected ##F = (M_1 + M_2 + M_3) \ddot x_1## when ##\ddot y_3 = 0##. The general equations with ##\ddot y_3## not necessarily equal to zero were pretty heinous and unenlightening.

    The next problem asks for ##\ddot x_1## when ##F = 0##. My expression doesn't exactly match the solution, but this time I'm pretty sure it's just an algebra error from all the tedious symbol-pushing. Will look at it again in the morning.
     
  14. Mar 20, 2014 #13

    AlephZero

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    The neat way to do that is probably to use conservation of energy and momentum (I say "probably" because I haven't actually worked through it). You only need two parameters to represent the motion of the system (e.g. the positions of masses 1 and 2) so two equations should be enough, and you don't need to bother about the internal forces.
     
  15. Mar 20, 2014 #14

    jbunniii

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    Any technique that requires fewer equations and unknowns makes me happy. The probability that I make an algebra error is on the order of ##1 - (P\{\text{no sign error}\})^{n}##, where ##n## is the number of opportunities for a sign error. Looking forward to Chapter 4 (Momentum) and Chapter 5 (Energy).
     
    Last edited: Mar 20, 2014
  16. Jun 28, 2014 #15
    Dosent this mean that Mass 2 is moving to the right , and mass 3 is moving upwards , whichis impossible ?
    working the same problem and i'm stuck here
     
  17. Jun 28, 2014 #16
    help?
     
  18. Jun 28, 2014 #17
    got it nevermind, you have to assume that the acceleration is decreasing in order to get the correct result
     
  19. Jun 28, 2014 #18

    BvU

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    Acceleration decreasing ?????
     
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