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Dynamics of a particle

  1. Mar 25, 2012 #1
    1. The problem statement, all variables and given/known data
    Untitled.png


    2. Relevant equations

    I converted the given data to SI units, because of my native calculation system.
    I used an index for particles and named rope force BC: [tex]T_{BC}[/tex]

    3. The attempt at a solution

    Kinematic expression for particle B:
    [tex]a=constant=1,22 \frac{m}{s^{2}}[/tex]
    [tex]v=a.t[/tex]
    [tex]x(2)=a.\frac{t^{2}}{2}=2,44 m[/tex]

    Dynamic expression for particle B:
    [tex]m_{B}.a_{B}=44,5+P-T_{BC} => 4,54 kg.1,22\frac{m}{s^{2}}= 44,5+P-T_{BC}[/tex]

    Finding unknown rope force BC form dynamic expression of particle C:
    Acceleration of particle C is the same as particle B
    [tex]-5,54=44,5-T_{BC}[/tex]
    [tex]T_{BC}=-50,.04N[/tex]

    Apply this in equation for particle B gives: [tex]5,54=44,5+P-50,04 => P=11,08N[/tex]

    (Correct) solution has to be: 7,37 N

    What did I do wrong?
     
  2. jcsd
  3. Mar 25, 2012 #2

    gneill

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    Staff: Mentor

    Notice that your solution doesn't include any effects from body A. That should make you suspicious :wink:

    If pulley D accelerates downwards (as body A accelerates upwards), then this will affect the tension you find. If the pulley is accelerating downward then the whole "subsystem" consisting of the pulley with body B and body C will share the same net acceleration. One effect of this will be to reduce the effective gravitational acceleration acting on B and C. If we call the downward acceleration of the pulley A1, then the effective gravitational acceleration operating on B and C will be g - A1.
     
  4. Mar 28, 2012 #3
    I reviewed the problem and came to follow set of equations:

    Subsystem with particle B and C (and the pully) has an acceleration [tex]a_{1}[/tex]
    Newton second law:
    [tex]+9,08.a_{1}=+P+89-T_{AD}[/tex]

    Particle A move with same acceleration upward, Newton second law:
    [tex]-9,08.a_{1}=+89-T_{AD}[/tex]

    Substituting the equations gives me: [tex]2.T_{AD}=178+P[/tex]

    (This gives the correct answers)

    Now I do have to find a relation between the acceleration of the subsystem and particle B.
    I know that the effective acceleration of particle B is:[tex]1,22 m/s^{2}[/tex].
    But I don't understand why [tex]a_{1}[/tex] equals [tex]9,81 m/s^{2}-1,22m/s^{2}[/tex]



    Could you explain this to me?
     
  5. Mar 28, 2012 #4

    gneill

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    Staff: Mentor

    I don't think I can explain it, because I don't think that it's true if I'm correctly understanding that "9,81" is g and "1,22" is the net acceleration of block B :smile:

    I believe that the acceleration a1 of the subsystem as a whole (and hence block A also) should be closer to 0.4 m/s2.

    What sort of "relation between the acceleration of the subsystem and particle B" are you looking for and why? Didn't already solve for force P?
     
  6. Mar 28, 2012 #5
    Here are my usefull equation to solve problem:
    [tex]2.T_{AD}=178+P[/tex]
    [tex]9,08.a_{1}=P+89-T_{AD}[/tex]
    The acceleration of particle B consist of 2 components:
    [tex]a_{1}=a_{pully}[/tex]
    [tex]a_{Beffectif}=1,22 m/s^{2}=a_{1}+a_{particle B}[/tex]
    [tex]m_{particle B}.a_{particle B}=P+44,5-T_{BC}[/tex]

    Particle C will accelerat upwards because of force P and the subsystem will move down.

    With this configuration I still cannot solve it..

    By the way the solution you proposed is indeed correct.

    grtz
     
  7. Mar 31, 2012 #6
    Oke, I think I got the correct appoarch with the relevant assumptions according to the kinematics. It was actualy a bit more tricky than just putting in some variabeles.

    I will write the solution-equation in the assumption the system is moving:

    - particle A: [tex]-m_{A}.a_{A}=F_{g,A}-T_{AD}[/tex]
    - subsystem (containing pully D en particle B and C): [tex] +(m_{B}+m_{C}).a_{A}-T_{AD}+P[/tex]

    This lead to: [tex]178-2.T_{AD}+P=0[/tex]
    [tex]18,14.a_{A}=P[/tex]

    Now the tricky part I first looked over: the given kinematic expression of particle B is abolute. This means that: [tex]1,44m/s^{2}=a_{B/D}+a_{D}[/tex]

    For particle C:[tex]a_{C}=a_{D}-a_{C/D}[/tex]
    [tex]a_{C/D}=a_{D}-a_{C}[/tex]

    We know that: [tex]a_{C/D}=a_{B/D}[/tex]
    The dynamics of particle C: [tex]a_{C}.m_{C}=+44,5-T_{BC}[/tex]
    [tex]T_{BC}=44,5+m_{A}.(a_{D}-a_{C/D})[/tex]
    So now I can write rope force BC as function of acceleration:
    [tex]T_{BC}=44,5+m_{A}.(a_{D}-a_{A/D})[/tex]


    The dynamics of particel B: [tex]m_{B}.a_{B}=+P+44,5-T_{BC}=6,5N[/tex]
    [tex]m_{B}.a_{B}=+18,14.a_{A}+44,5-T_{BC}=6,5N[/tex]
    [tex]m_{B}.a_{B}=+18,14.(1,44-a_{B/D})+44,5-T_{BC}=6,5N[/tex]


    Putting this all together get's me: [tex]a_{D}=0,479 m/s^{2}[/tex]

    Still a bit incorrect, but more important to me is the method. So my final question are the assumptions I used correct?

    grtz
     
  8. Mar 31, 2012 #7

    gneill

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    Staff: Mentor

    Your assumptions and approach look okay to me.

    Personally, I like to leave everything in symbolic form for as long as possible in order to avoid having to "carry around" a lot of numbers through a derivation and to avoid accumulating rounding errors.
     
  9. Apr 1, 2012 #8
    Indeed, but I wanted to make it clear that I got the good approach. So with a couple of number it's more easily.

    Offcourse the errors occured at my final solution is due to the fact of rounding solution at the steps I took.

    thank for all the help!
     
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