How's it possible to push on systems with 2 parts? (Newton's third law)

[DrBanana]

An alternative title could have been "how do forces propagate" but google searches bring up things related to waves only.

Initially my problems started with a mass on a spring but I was able to boil it down to any general system comprising at least two 'parts' in succession.

Suppose you have two boxes touching each other on a frictionless surface. So if you apply a force with your hand then the boxes will move. When solving these types of problems the assumption is that, the contact force between the forces is less (not equal) than the force applied by the hand. My question is why. Suppose I apply a force F on box 1 to the right. I was under the impression that this force would propagate itself to box 2 as well (yes I know this would mean that box 2 would gain a different acceleration if it had different mass), so box 1 would be applying F on box 2, but by Newton's 3rd Law, box 2 will apply a force -F on box 1. So the net force on box 1 is F-F=0. Obviously that doesn't happen in reality, but why? Does this mean, if I apply a force on one end of some wire that only consists of a single thread of atoms, the contact force between each successive pair of atoms decreases?


By the way I realised I had this misconception because I was trying to get an understanding of applying a force to a deformable body, and why after some point it stops deforming. The usual explanation goes that the intermolecular bonds become strong enough to cancel out the applied force, but here I am left wondering (because of the thought process above) why it isn't just cancelled out immediately.

 

[kuruman]

An alternative title could have been "how do forces propagate" but google searches bring up things related to waves only.

And what's wrong with waves? Contact forces in a solid propagate at the speed of sound in the solid, which is typically several thousands of meters per second. So if you push at one end of a meter stick it takes a fraction of a millisecond for that information to get to the other end.

 

[PeroK]

First, consider box 1 and box 2 as rigid atomic components. If you apply a force $F$ to box 1, then that force must accelerate box 1 and 2. The force box 1 exerts on box 2 is less than $F$ and depends on the ratio of the masses. It is in fact:
$$F_2 = \frac{m_2}{m_1+m_2}F$$The same calculation applies when we consider the boxes as systems of atoms. The internal forces between atoms reduce from a maximum of $F$ to effectively zero at the far end of box 2.

 

[Nugatory]

So the net force on box 1 is F-F=0. Obviously that doesn't happen in reality, but why?

You haven't calculated the forces correctly, probably because you've miscalculated the acceleration.

Say both boxes have the same mass (nothing special about this case except that the math is easier)....

If we apply a force $F$ to the right-hand box, both boxes will accelerate to the left with acceleration $a=\frac{F}{2m}$ because we're applying a force $F$ to a lump of mass $2m$. Now, what is the force on each individual box to produce that acceleration? Well, we have boxes of mass $m$ accelerating at $F/2m$, meaning that the net force on each box is $F/2$. This is consistent with a force of $F/2$ being applied by the right-hand box to the left-hand box and a net force of $F-F/2$ (the force from our hand, minus the opposing force that the left-hand box applies to the right-hand box by the third law) applied to the right-hand box.

 

[DrBanana]

The same calculation applies when we consider the boxes as systems of atoms. The internal forces between atoms reduce from a maximum of F to effectively zero at the far end of box 2.

So, is there a deeper reason behind this, or is it just taken to be true (as in, do we only conclude this because assuming the opposite case would contradict calculations and reality)?

 

[Dale]

When solving these types of problems the assumption is that, the contact force between the forces is less (not equal) than the force applied by the hand. My question is why.

That is not an assumption. The assumption is that the two objects have the same acceleration. The force is derived from that assumption.

 

[DrBanana]

That is not an assumption. The assumption is that the two objects have the same acceleration. The force is derived from that assumption.

But how can we know (without experiment) that that assumption is true?

 

[Dale]

But how can we know (without experiment) that that assumption is true?

We cannot know it without experiment. In fact, it does not always happen. If you are briefly pushing a soft fluffy pillow with a rock then the accelerations may not be equal.

 

[PeroK]

But how can we know (without experiment) that that assumption is true?

I explicitly stated that the boxes be considered rigid atomic components. Under that assumption Newton's laws apply, with a single external force and an equal and opposite third-law pair between the boxes. That is a called a simplifying assumption.

We can then test this assumption and Newton's laws with a suitable experiment. Which probably would not involve springs or fluffy pillows!

 

[Ibix]

Does this mean, if I apply a force on one end of some wire that only consists of a single thread of atoms, the contact force between each successive pair of atoms decreases?

I think you can model this as a line of ideal point masses connected by springs. If you push the first mass, it must accelerate the same as your hand. But the second mass doesn't get accelerated by your hand, it gets accelerated by the first spring. So it initially doesn't accelerate because the spring is not compressed. But because it doesn't accelerate while the first mass does, the spring becomes slightly compressed and the second mass accelerates slightly. The spring is further compressed so the second mass accelerates a bit faster and so on, tending to the final state where the spring is compressed enough that the second mass accelerates at the same rate as the first.

You can make the same argument regarding the second and third masses, and the third and fourth, and so on down the line. Each mass has a lower acceleration than the one before because there are more springs between the original force and it. Eventually the whole body will settle into a compressed state. When you stop accelerating the end mass the springs expand again and the masses return to their original state, give or take any losses.

If the springs are stiff you have steel. If the springs are very weak you have jelly.

 

[DrBanana]

I think you can model this as a line of ideal point masses connected by springs. If you push the first mass, it must accelerate the same as your hand. But the second mass doesn't get accelerated by your hand, it gets accelerated by the first spring. So it initially doesn't accelerate because the spring is not compressed. But because it doesn't accelerate while the first mass does, the spring becomes slightly compressed and the second mass accelerates slightly. The spring is further compressed so the second mass accelerates a bit faster and so on, tending to the final state where the spring is compressed enough that the second mass accelerates at the same rate as the first.

You can make the same argument regarding the second and third masses, and the third and fourth, and so on down the line. Each mass has a lower acceleration than the one before because there are more springs between the original force and it. Eventually the whole body will settle into a compressed state. When you stop accelerating the end mass the springs expand again and the masses return to their original state, give or take any losses.

If the springs are stiff you have steel. If the springs are very weak you have jelly.

Your answer seems interesting but unfortunately I made this post as a precursor to understanding springs : (

 

[jbriggs444]

I explicitly stated that the boxes be considered rigid atomic components. Under that assumption Newton's laws apply, with a single external force and an equal and opposite third-law pair between the boxes. That is a called a simplifying assumption.

When I want to reconcile this simplifying assumption with the real world, I tend to think about things relaxing toward an equilibrium. This will make more sense once one has an intuition for spring-like behavior under one's belt.

As one pushes harder and harder on an object, it deflects more and more. This results in greater and greater internal stresses. Eventually, a new equilibrium is attained.

We can apply Newton's laws to the resulting equilibrium situation to determine what forces must exist for the equilibrium to hold.

The fact that an equilibrium is obtained is an empirical fact and is, in many cases, guaranteed by the second law of thermodynamics. [Not all situations -- hence turbulent flow, howling bearings, screeching tires and screeching chalk on the blackboard. We try not to teach about that stuff in the first semester].

It helps if you have worked and played outside during your formative years. Then most of this stuff is second nature.

 

[Ibix]

Your answer seems interesting but unfortunately I made this post as a precursor to understanding springs : (

Real springs, or idealised ones? These are idealised, and the force they provide is just directly proportional to the compression. They're actually standing in for the EM forces between atoms.

 

[Mister T]

So the net force on box 1 is F-F=0. Obviously that doesn't happen in reality, but why?

Because Box 1 has mass. Consider a situation where two boxes are connected with a rope; you pull on one box and because of the rope the other box moves. If you have a "massless" rope then the boxes behave differently than if you have a very heavy rope. Draw FBD's of each box as well as the rope, and compare the relative sizes of the forces acting on each of the three objects (two boxes and the rope).

 

[Dale]

Your answer seems interesting but unfortunately I made this post as a precursor to understanding springs : (

Then I think it is sufficient to know that there are a large number of common scenarios where the equal acceleration assumption is a reasonable simplifying assumption. As you learn about springs you can use that to model a bigger number of scenarios.