# Mechanical problems converted into electrical circuits.

1. Nov 4, 2013

### peripatein

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.

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2. Nov 4, 2013

### 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?

3. Nov 4, 2013

### peripatein

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?

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4. Nov 4, 2013

### Staff: Mentor

It looks fine to me, provided that "down" is defined to be positive in your mechanical system.

5. Nov 4, 2013

### peripatein

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?

6. Nov 4, 2013

### Staff: Mentor

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.

7. Nov 4, 2013

### peripatein

I see. So is V1=V2=V3=f/(3B4)?

8. Nov 4, 2013

### Staff: Mentor

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?

9. Nov 4, 2013

### peripatein

Please see attachment. Each branch gets f/3 so Vi should be equal to f/3B4. Unless I am mistaken. Am I?

#### Attached Files:

• ###### Q2 - Converted1.jpg
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10. Nov 4, 2013

### Staff: Mentor

You are not mistaken

11. Nov 4, 2013

### peripatein

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?

12. Nov 4, 2013

### Staff: Mentor

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.

13. Nov 5, 2013

### peripatein

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?

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• ###### 2.JPG
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Last edited: Nov 5, 2013
14. Nov 5, 2013

### Staff: Mentor

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 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:

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

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