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Mechanical problems converted into electrical circuits.

  1. Nov 4, 2013 #1
    Hi,
    1. The problem statement, all variables and given/known data
    Attached (Q1,Q2) are two mechanical systems I am asked to convert into analogous electrical circuits. In Q2, it was stated that B4 is the friction between each of the three masses and the surface.

    2. Relevant equations



    3. The attempt at a solution
    Attached (Q1 - Converted, Q2 - Converted) are my conversions, which I am not sure are correct. I'd sincerely appreciate some feedback on these two conversions.
     

    Attached Files:

  2. jcsd
  3. Nov 4, 2013 #2

    gneill

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    Staff: Mentor

    In your first analog, shouldn't the force due to gravity be working in the same direction for both masses? How did you end up with the inductor and resistor in parallel between the two nodes? Surely the mechanical versions are in series in the "light inextensible string" that winds through the pulleys.

    In the second analog I don't see where B4 is defined in the mechanical version, but I presume it's meant to represent friction between the masses and the "ground" surface. Should they be the same value for all three masses?
     
  4. Nov 4, 2013 #3
    I referred to B4 in my statement above. Is my analog correct then?
    Please see new attachment for the corrected analog of Q1. Is it indeed correct now?
     

    Attached Files:

  5. Nov 4, 2013 #4

    gneill

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    It looks fine to me, provided that "down" is defined to be positive in your mechanical system.
     
  6. Nov 4, 2013 #5
    Thanks :). By the way, would it be correct to say that in Q2, in the steady state (i.e. capacitors are open, inductors are short), V1 would be equal to f/(3B4) whereas V2 and V3 would be zero?
     
  7. Nov 4, 2013 #6

    gneill

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    Nope. The shorted inductors tie all three nodes together, so they must all have the same potential.

    If you consider the "original", the dampers between the masses will cause any initial oscillations to die away, but the whole ensemble will continue to move at some constant velocity limited by friction.
     
  8. Nov 4, 2013 #7
    I see. So is V1=V2=V3=f/(3B4)?
     
  9. Nov 4, 2013 #8

    gneill

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    You should be able to answer that by redrawing the circuit with the reactive components suppressed as you've stated. What does the steady state circuit look like?
     
  10. Nov 4, 2013 #9
    Please see attachment. Each branch gets f/3 so Vi should be equal to f/3B4. Unless I am mistaken. Am I?
     

    Attached Files:

  11. Nov 4, 2013 #10

    gneill

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    You are not mistaken :smile:
     
  12. Nov 4, 2013 #11
    In analog 1, are there only 2 velocity nodes/points? My guess is that there are only two, viz. V1 and V2, since the other potential nodes/points are in parallel. Am I right?
     
  13. Nov 4, 2013 #12

    gneill

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    You could define a node at the junction of the resistor and inductor in order to determine the velocity of the spring end. But this would not be an "essential node", as it occurs in the middle of a branch.
     
  14. Nov 5, 2013 #13
    Hi,
    In your first reply to my question yesterday, you wrote "surely the mechanical versions are in series in the 'light inextensible string' that winds through the pulleys." However, when you take a look at the analog (attachment '2') of the system shown in the recently added attachment '1', the masses are drawn in parallel whereas here too a light inextensible string winds through the pulleys and connects the two masses. Why the seeming discrepancy?
    Furthermore, why are the forces due to gravity now in opposite directions?
     

    Attached Files:

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    Last edited: Nov 5, 2013
  15. Nov 5, 2013 #14

    gneill

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    For masses, what is being modeled is their inertial mass. This is always referenced to the rest frame, so one leg of the analogous capacitor is always tied to ground. This is true even for masses suspended in the air by ropes.

    In this new system, note that the string ties the masses together so their movements (positions, velocities, accelerations) are linked. While the ground "leg" of both capacitors are automatically tied together simply because all masses are tied to ground, in this case the "free" legs are tied together too by the inextensible string. That's why they end up in parallel.

    Ah. I think I have an apology to deliver :blushing: Upon reflection I realize that I made an error before when I suggested that the currents ("Forces") should be in the same direction. Of course they should be opposite as you had them. If the pulley system is redrawn, stretched out horizontally and the acting forces reoriented accordingly, the correct directions for the motions and forces becomes obvious:

    attachment.php?attachmentid=63672&stc=1&d=1383653816.gif

    Here Fg is pulling "left" on M1 and "right" on M2. Opposite directions. I should have spotted that, sorry.
     

    Attached Files:

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