# Homework Help: [help please] Difficult Dynamics Challenge

1. Aug 2, 2012

### HeZ

Hello all,
Semester has just started and i am dong a 4th year dynamics unit, so i am reviewing the previous years unit which was a pre-requisite for the unit i am currently doing. I have come accross a rather difficult dynamics problem that was given as a challenge question at the beginning of last years course - one which i never managed to solve. It is really doing my head in and I am hoping someone here can help me with it?

It is as follows:

Two masses connected by a linear spring, constant k. A force is applied to mass m2.
There IS friction.
What is the minimum constant force F that will cause mass m1 to move?

I am assuming that the spring can be considered massless, as i have never been given a problem in which this was not the case, and a spring mass was not given.
And I forgot to indicate in the image that gravity acts downwards (perpendiculat to the surface), not that it matters.

I tried many solutions, all of which I was told were wrong. Below is one of them, I will type more of them up later when i have some spare time if necessary.

Chris

2. Aug 3, 2012

### andrien

Re: [help plz] Difficult Dynamics Challenge

that is good.

3. Aug 3, 2012

### Studiot

Re: [help plz] Difficult Dynamics Challenge

Some things to consider.

1) Why do you say the "force across the spring must be zero" What is in equilibrium to justify this?

2) You have not offered a coefficient of friction. Is it the same for m1 & m2

3) have you considered what happens if

m2 > m1

m2< m1

m2 = m1

4. Aug 3, 2012

### AlephZero

Re: [help plz] Difficult Dynamics Challenge

Your solution is OK if you increase the force F very slowly from zero, so you can ignore the inertia of the masses. (Obviosuly m2 has to move to stretch the spring, before the spring applies enough force to move m1).

But if you suddenly apply a constant force F, m2 will accelerate and then slow down as the spring stretches. You will be able to move m1 with a smaller force than your solution.

You could find F by by solving the equations of motion, but an easier way is to think about how much work you need to do to the system to get it to the state where m1 just starts to move.

5. Aug 3, 2012

### HeZ

Hey guys thanks for the replies.

@Studiot:
1) If the spring is mass less then the net force must be zero. Energy may be stored in the spring however as the mass is zero the internal components have no momentum and therefore no internal forces. The force at one end is equal and opposite at the other end.

2) Coefficient of friction is got given or required. It is sufficient to answer in terms of the frictional forces.

3) The problem is the same regardless of which mass is heavier I believe, since we are answering in terms of m1, m2 and the forces. I believe if memory serves me correct the tutor said to just make them equal if we like, for simplicity. Since that will not change the problem.

@AlphaZero:

Ohh thanks that's a nicer way of looking at it. Since it says a constant force I guess it would would imply a step change.
So if i were to solve it from a work point of view, I would need to determine when enough energy has been put into the system (stored in the spring + dissipated due to friction on m2) for the force on the spring connected at m1 to be greater than the friction force of m1 (force at spring-m1 connection: Ff1 = kx + F).

Does that look like a good starting point for this method?

Again, thanks for the replies
Kind Regards
Chris

6. Aug 4, 2012

### Studiot

I thought you were supposed to analyse and prove things?
That was why I suggested breaking the problem down to m1 = < > m2

When Aleph Zero says a gradually increasing force he does not mean literally starting from zero and slowly increasing it in a single experiment. We do not normally consider how the force F arises, just that the situation has been going on for a long enough time for things to have settled down.

Consider it this way. It is what happens when you do a series of (many) experiments with F constant, each experiment having F increased incrementally.

You can only apply the conditions of equilibrium to the blocks in equilibrium. Conventionally we say the block(s) are just about to move. Any greater value of F will lead to the need to apply the equations of motion instead of static equilibrium.

Now consider the case of m1 > m2 and do the (thought) experiments.

Let F1 and F2 be the friction forces required to just move m1 & m2 respectively.

Then F1 > F2

At each experiment nothing moves until the experiment where F = F2
When this is reached m2 moves.
And F = F2 is transmitted through the spring to m1

m1 does not move since F2 < F1, the force needed to move it.

The spring extends until it offers a restoring force equal to F2, given by the spring equation. At this point the system is restored to equilibrium and halts.

The next increment of F moves m2 a little further so the spring restoring force is now (F2+δF).
Again the system stops at equilibrium.

This continues for each increment of F until (F2+ƩδF) = F1, at which point m1 begins to move.

This is a start for you to continue.

7. Aug 4, 2012

### andrien

you can think both masses as your system and the spring will not play role because it will be internal.so you will have to consider only friction for the purpose of motion.

8. Aug 4, 2012

### SammyS

Staff Emeritus
It appears to me that you're proposing viewing the system as a whole. The center of mass of the system will move with a smaller external force than what's required to move m2.

Consider this:

Mass, m1 will begin to move when the spring is stretched just enough so that the spring force overcomes the static friction on m1.

9. Aug 4, 2012

### AlephZero

Yup, that's the right idea.

For the minimun value of F, the mass m2 will have zero velocity at the instant when m1 starts to move, so the system has no kinetic energy at that moment in time.

10. Aug 5, 2012

### andrien

but I think that if force is applied on m2 it is the sole one on the system and for m1 to move m2 should get a little movement too because the force on m1 in forward direction will only be when the spring will get some stretched and it will happen only because of m2 because m1 movement will compress it.

11. Aug 5, 2012

### HeZ

Thank you both Studiot and AlephZero for your replies. Two very different approaches to the problem. I plan on attempting them both as I believe it is important to have multiple ways of solving a problem available to me.

@Studiot:
Do we not need to consider the momentum of the mass m2 at this point?

@AlephZero
I think I have a solution here.

A step force is applied. Assuming that this force is sufficient to break static friction of m2 (F > Fm2-static) then m2 begins moving.

Block continues to move until the spring stops it. If we are considering the minimum force applied to move m1, then spring deflection at this point will be x1.

Where: Fm1-static = kx1

ENERGY in system whilst m2 is sliding:
0.5kx2 + 0.5m2v2 - Fm2-kineticx = Fx
ENERGY in system at instant when m2 comes to a rest:
0.5kx12 - Fm2-kineticx1 = Fx1

A little bit of algebra later:
0.5kx1 - Fm2-kinetic = F

Recall that:
kx1 = Fm1-static

Hence:
0.5Fm1-static - Fm2-kinetic = F

Thus in order to move m1 (Assuming already that F > Fm2-static):
F ≥ 0.5Fm1-static - Fm2-kinetic

Regards,
Chris

Last edited: Aug 5, 2012
12. Aug 5, 2012

### Studiot

If you have been asked this question in a mechanics course I assume you have studied the simpler rules for the application of friction?
If so, the quote 1 above worries me.

I do not see this as a dynamics problem - that would be the difficult way to approach it.

Quote 2 above - A static object has no momentum.

Quote 3 above is the standard way to approach this situation as a statics problem.

So do we need to review what happens ie what forces, including friction, act on a simple body as we slowly increase an external traction force F?

So have you considered what happens when m2 > m1 ?

Last edited: Aug 5, 2012
13. Aug 5, 2012

### HeZ

Yes, that sentence was poorly articulated, my apologies. Obviously if the mass is sliding up/down an incline then that matters a lot, what I intended for that quote to mean was the angle that gravity makes with the surface does not matter in terms of the friction force since it is sufficient to answer in terms of friction forces, not values. If a numerical solution is ever to be required then the sine of the angle gravity make with the surface will be used to calculate the force. However having said that I do understand that if the surface were at an incline to gravity then the entire nature of the problem will change. I have edited the OP to make more sense.

This is the main thing that I am confused about, I am probably way off base with this understanding, but this is how I visualize the situation:
When the applied force F is greater than the restoring force of the spring and friction, the block will have a net force to the right and begin accelerating.
The spring will stretch and the restoring force will therefore increase.
When the spring extends to the point that the net forces on the block are zero, the acceleration of the block will be zero also. However the velocity must be non zero as it has just been accelerating.
The block will then therefore continue to extend the spring, raising the restoring force to a value greater than the applied force - friction.
It will continue to stretch the spring, decelerating due to a net force to the left.
When the velocity reaches zero, it will continue to have an acceleration to the left and the spring will compress again.
Thus at the very instant that the velocity reaches zero, the spring will be at maximum deflection, providing maximum force. If this force at this instant is greater than the static friction of m1, then m1 will move.

EDIT: Apparently I cant edit the OP...

Last edited: Aug 5, 2012
14. Aug 5, 2012

### Studiot

It's not a point about articulation it's that you didn't consider gravity important.
Without gravity there would be no friction.

Yes so how about trying to answer my question what happens when m2 > m1?

And why are you still considering the system moving with any significant velocity?

This is the classic condition for static analysis.
If you do not understand this we need to go through it.

Last edited: Aug 5, 2012
15. Aug 5, 2012

### HeZ

I am pretty sure that the hint we received was that the analysis should be the same regardless of the masses.. but:
If m2 > m1 then F2static > F1static

If the applied force F is greater than F2static then m2 will begin to move.
F = F2static is transferred through the spring to m1 hence m1 will move

That's following the analysis from your previous post. However I don't know what you mean by:
ANY velocity at the point where the spring offers a restoring force equal to the applied force is going to be significant as it will cause the spring to stretch a little further, increacing the force that is transferred to the mass m1. Since the mass cannot stop instantaneously.
I'm not sure how this is a static analysis problem, I understand that just moving m1 is a static problem as we don't care about it the very instant after it begins to move, however the mass m2, which generates the slowly increasing force applied to m1 via the spring is moving.

Regards
Chris

16. Aug 6, 2012

### andrien

you need not to think about different cases for masses.As I have said in beginning that your answer is good ,it is not ok.I have written something after that which was just saying that you can not move m1 without moving m2 BECAUSE untill m2 moves there will not be elongation in the spring ,and that is what matters for m1 ,m1 movement will compress the spring and that is not the desirable part.so your answer is good.
edit:if you think a little you will get it,but it is simple enough.so just for static balance purposes it holds.

17. Aug 6, 2012

### Studiot

If this is true (and yes it is) then you should get the same result calculating either way so you should be able to prove this - it is instructive to do so.

Another tip - it is fundamental to be able to extract or draw good isolation diagrams or free body diagrams to solve mechanics problems. Knowing what to isolate is key to most solutions. Knowing also when a system is even momentarily in equilibrium is also useful as the equations of equilibrium cna be applied at that instant.

I have worked through a chain of reasoning in the attachment, starting from rest and considering situations with successively increased values for F.
I have assumed that the coefficient of friction is the same for both masses for convenience.

Firstly in (1) F is not large enough to move anything so the system remains at rest.
In this situation there is no force at all transmitted through the spring so there is only vertical equillibrium acting on mass1.
Mass2 has horizontal forces acting but note that the friction force is only large enough to oppose F . The mass has not reached limiting friction.

At limiting friction mass2 is just about to move, the remainder of the system is the same as in (1)

Once mass2 starts moving it is no longer in equilibrium.

However at this time F is not strong enough to also move mass1 so mass 1 remains in equilibrium.
However mass1 now has horizontal forces acting which we can write an equilibrium equation for the friction force and the spring force.

(so we are considering mass1 as an isolated or free body)

This gives us a spring force to substitute into the equation of motion for mass2, which is being accelerated by the difference between F and (the mass2 friction plus the spring force).
The minimum possible difference is an infinitesimal amount above zero which is the answer to the question asked.

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