Spring damper system without a mass

1. Feb 7, 2013

hihiip201

Hi guys:

Imagine a spring and and damper in parallel, connected to the ground on the right, and connected by a node on the left.

I never quite understand why it is said that the forces between the two must be zero.

According to my professor and the textbook, it is because the node has no mass, so the forces between them has to be zero.

But picturing a real life example where masses are being modeled as node by being relatively small compare to the damp coefficient as well as the stiffness of springs.

How does it tell us that the forces exerted on this mass on both side have to be zero? I mean surely F= ma and if m goes to zero then Fnet = 0. But isn't that under the assumption that a is finite?

Also, if a small mass is a node, how should i go about trying to visualizing this? Im trying to think of the "wires" connecting mechanical elements as a whole bunch of connected small masses but that's not very helpful.

to me F is the cause, and a is the effect, when two springs, or a damper and spring connected together in parallel or series I don't think " how forces must be equal". instead I think of "How great the acceleartion have to be"

thank you

Last edited: Feb 7, 2013
2. Feb 7, 2013

Simon Bridge

There is always a mass.
Springs are usually connected in series to the damper - otherwise, what is the point?
The book and the professor are talking about an idealized model of a spring+damper, not real life. In this model the spring and all the connectors and everything are idealized massless. The effect of any real masses in the real system are included explicitly as a separate component. Just like a circuit diagram shows zero resistance wires connecting components when we know real wires have some resistance.

3. Feb 7, 2013

hihiip201

They are connected in parallel for the sake of throwing us students off?lol
Also I believe a mechanical buffer is essentially a damper and a spring in parallel.

I understand that there are always real masses, but what I'm trying to understand it WHY the model is a good approximation of a very small mass.

is it because the net force on a tiny mass is negligible when its acceleration is finite?

Last edited: Feb 7, 2013
4. Feb 7, 2013

Simon Bridge

The model should always include a mass as a separate component. It is usually labelled "m".

A massless model won't obey the laws of motion.
For instance, the slightest unbalanced force produces infinite acceleration.
I think I need to see the specific example.

5. Feb 7, 2013

Khashishi

This situation is analogous to a circuit with no resistance. The voltage across the wire is 0, because there is no resistance. If the voltage were not 0, then there would be an infinite current that would flow, which would generate a charge imbalance which would set the voltage back to 0.

In the case of the spring, if the force on the spring were not 0, then the spring would have infinite acceleration, and it would instantly jump to an equilibrium position where the force is again 0.

These are idealized situations, so this won't happen in real life.

6. Feb 7, 2013

hihiip201

When you say the velocity of the spring, which point on the spring are you referring to?

Also, I understand that these are idealized model, but I just think that in some extreme cases of the real life example, it will approach/resemble the ideal model approximations.

7. Feb 7, 2013

AlephZero

That is fairly irrelevant IMO. A model without mass obeys a first order differetinal equation, not a second order one. It's a bit like saying you can't solve the quadratic equation $ax^2 + bx + c = 0$ if $a = 0$. Of course you can solve it - but not using the standard method for quadratics!

No. With the usual analogy between electrical and mechanical equations, it is equivalent to a circuit with no inductance. People work with circults like that all the time! Electrical resistance is analogous to the damper. Bboth dissipate energy, and they are the only parts of the systems which do that.

Exactly. If the inertia forces are neglibible compared with the other forces, the model will behave better numerically if you leave them out completely. Ignoring the fact that the acceleration might be "infinite" is a cleaner solution than trying to include an acceleration that is very large but only exists for a very short time.

Last edited: Feb 7, 2013
8. Feb 7, 2013

Simon Bridge

Depends what you are trying to model.

9. Feb 8, 2013

hihiip201

Do we need to make the assumption that a is finite (or not very big) before saying the net force/inertia force is small/0 ?