# Dashpot has constant velocity. Why?

1. Oct 17, 2014

### yosimba2000

I was watching this video for maxwell models help:

The metal weight is being pulled down by force of gravity, which means there must be some net acceleration for it to be pulled down according to F = MA.

My question is: if acceleration is a change in velocity, then how come the dashpot moves in a constant velocity? This is heard in the video at time around 54 seconds in.

Thanks!

2. Oct 17, 2014

### Simon Bridge

The weight (in the vid) is not moving with a constant velocity.
i.e. it is swinging - so it's velocity is constantly changing.

Note: an object may fall at a constant velocity if it is acted on by another force equal in magnitude to the weight but pointing up. To fall from rest like this requires some initial acceleration - as is the case for viscous drag or some sort of special elasticity.

3. Oct 17, 2014

### yosimba2000

but won't an equal but opposite force prevent movement of the object? If gravity pulls MA on the weight down, and I the pull same force in the opposite direction, won't the weight just not move at all?

and I think you might've confused the dashpot (item in question) with the weight. the dashpot is the syringe at the top left in the video. if the weight is falling due to gravity, then there should be a net acceleration down (at least, that's what I think), right? so if the weight is falling with a net acceleration, and the weight is hooked to the dashpot, then shouldn't the dashpot also change with the same acceleration as the weight is?

4. Oct 18, 2014

### Simon Bridge

No. An unbalanced force is needed for acceleration - movement is possible without any forces at all: this is from Newton's 1st Law of motion.

... In the video, the weight does, indeed, start at rest and gain speed, this is, by definition, acceleration. So what is the problem?

The motion of the weight influences the motion of the dashpot plunger through the string. The plunger is easily seen to be initially at rest and later has some non-zero speed. Again: there must have been some acceleration at least at the start of the motion.

Therefore, at the start of the motion, the pull of the weight is greater than the pull back from the fluid in the dashpot (etc).
At some point the pull-back force came to be equal to the weight, and a constant velocity was obtained ... presumably the pull-back force was some function of velocity.

You can work out the details by drawing a free body diagram for the plunger at the moment the weight is released.

 It strikes me that you may be thinking that everything falls in gravity with the same acceleration. Is that correct?

5. Oct 18, 2014

### Staff: Mentor

In analyzing a Maxwell model (analog) such as this, inertia is considered negligible. The point of this experiment is just to illustrate how a viscoelastic Maxwell model works. So, in the experiment, ma is very small compared to the dashpot force and the spring force (which are equal). In the actual viscoelastic Maxwell Model, stress takes the place of the force and strain takes the place of the displacement.

Chet

6. Oct 18, 2014

### yosimba2000

yes, I do remember learning that all objects (disregarding air friction and all that) in freefall and relatively close to the Earth's surface will have an acceleration at 9.81m/s^2.

but I think I see what you guys are saying now. If an object is already moving with an initial velocity with no forces applied, and then during the movement we apply two equal but opposite forces, those forces will cancel and the object will continue to move with that same initial velocity.

7. Oct 18, 2014

### Staff: Mentor

This experiment doesn't really capture the essence of what a viscoelastic Maxwell model is all about. A more appropriate representation would be to replace weight W = mg with a time-dependent force at the end of the string F(t). Let δ1 be the displacement of the end of the syringe piston, and let δ2 be the displacement at the leading edge of the string. Then,

$$F=\eta \frac{d \delta_1}{dt}$$
and
$$F=k(\delta_2-\delta_1)$$

If we take the time derivative of the second equation with respect to time and then combine the two equations properly to eliminate δ1, we obtain:

$$\frac{dF}{dt}+\frac{\eta}{k}F=\eta \frac{d \delta_2}{dt}$$

This equation enables one to determine the the time dependent force variation, given the time dependent variation of the displacement.

Chet

8. Oct 18, 2014

### Simon Bridge

@yosimba2000
That's right - though this assumes that gravity is the only force acting on the falling object.

... that is correct.
You should also look at the situation for a skydiver - initially accelerating until a terminal velocity is reached.

To see the effect in detail for this situation, you really need to do the math.

I suspect that your question is really besides the visco-thingy maxwell whatsit part, but it does not hurt to talk about that too.
At what level are you trying to understand this thing?

@Chestermiller
There kind-of is though isn't there - the mass is swinging so the tension in the string immediately above the mass varies with time?

9. Oct 18, 2014

### Staff: Mentor

Yes, but that's not what the experiment was designed to illustrate. The experiment was supposed to provide a simple picture of the essence of a Maxwell model, but, in my judgement, failed to accomplish this. This judgement is based on my extensive experience with viscoslasticity, which happened to be my thesis area.

Chet

10. Oct 18, 2014

### Simon Bridge

Oh I see what you mean ... so here is your chance for internet glory: make a better video! :D

11. Oct 19, 2014

### yosimba2000

I'm just trying to understand it with intuition. I'm taking a biomechanics class and I was just wondering why the dashpot didn't accelerate. I never thought that there would be another force pulling oppositely on the dashpot to cancel gravitational force. But, where is the other force on the dashpot coming from? One is from gravity pulling on the weight which pulls on the string, and that in turn pulls on the dashpot. The second one... I have no idea, but my intuition tells me it has to do something with the restricted airflow caused by the needle.

12. Oct 20, 2014

### Simon Bridge

... there you go.
Get a syringe like that one and try to pull the plunger out really quickly - isn't that harder than when you pull slowly?
(The effect is easier to see when pushing the plunger in.)

It's hard t get a lot of stuff through a small opening quickly.