Dynamics of a particle

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Homework Statement


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Homework 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]

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?
 

Answers and Replies

  • #2
gneill
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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.
 
  • #3
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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?
 
  • #4
gneill
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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?

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?
 
  • #5
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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
 
  • #6
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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
 
  • #7
gneill
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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.
 
  • #8
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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.

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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