A question about opposite and equal reactions

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In summary, Newton's third law of motion is not accurate. Momentum was conserved, the window acclerated in the direction of the rock, and the rock decelerated.
  • #1
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A question about "opposite and equal reactions"

Newton's third law of motion says "For every action there is an opposite and equal reaction". So when I throw a rock at a brick wall (action), that rock bounces off that wall (opposite and equal reaction). But if I jump up and while in the air and my shoes are not touching the ground when I throw a rock at a glass window (action) that rock smashes through that window (reaction but not opposite since that rock goes in the same direction I threw it). It appears Newton's third law of motion is not accurate. Can someone clarify this third law of motion? Did I disprove it or is Newton somehow correct? Please explain.
 
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  • #2


Momentum was conserved, the window acclerated in the direction of the rock, and the rock decelerated. The accelerations correspond to the equal and opposing forces, divided by the mass of the rock and the affected part of the window.
 
  • #3


But what is the opposite and equal reaction when that rock smashed through that window? That rock didn't deaccelerate very much; that window didn't have enough resilency to totally deaccelerate that rock to force it backwards. So that rock still carried with it most of its forward thrown momentum knocking glass fragments forward. It doesn't look like there was an "opposite reaction" and/or an "equal reaction".
 
  • #4


Newtype said:
Newton's third law of motion says "For every action there is an opposite and equal reaction". So when I throw a rock at a brick wall (action), that rock bounces off that wall (opposite and equal reaction).
You throwing a rock and the rock hitting a wall are not "action/reaction" pairs.

Newton's third law describes how bodies exert forces on each other when they interact. When you are throwing the rock, your hand exerts a force on the rock; Newton's third law says that the rock will exert an equal and opposite force on your hand.

When the rock hits something, such as a wall or window, that's a totally different interaction. The rock exerts a force on the wall, and thus the wall exerts an equal and opposite force on the rock.
 
  • #5


But that window didn't exert an equal and opposite force onto that rock like the force that wall exerted onto that rock. Please provide an example to explain what you're saying. Here's an example to explain what I'm saying: A 10kg rock thrown at 1 meter per second northwise will bounce off that wall (wall supposedly exerts 10kg of 1m/s southwise force), but a 10kg rock thrown at 1 meter per second northwise will smash through that window; that window will not exert an opposite and equal 10kg at 1 m/s southwise force to repel that rock since that window doesn't have the resilency to do so, that window will be smashed to pieces northwise instead.
 
  • #6


Newtype said:
But that window didn't exert an equal and opposite force onto that rock like the force that wall exerted onto that rock. Please provide an example to explain what you're saying. Here's an example to explain what I'm saying: A 10kg rock thrown at 1 meter per second northwise will bounce off that wall (wall supposedly exerts 10kg of 1m/s southwise force), but a 10kg rock thrown at 1 meter per second northwise will smash through that window; that window will not exert an opposite and equal 10kg at 1 m/s southwise force to repel that rock since that window doesn't have the resilency to do so, that window will be smashed to pieces northwise instead.
When the rock hits the wall, the wall and rock exert equal and opposite forces on each other. When the rock hits the window, the window and rock exert equal and opposite forces on each other. And if the rock hit a butterfly, the butterfly and rock would exert equal and opposite forces on each other. Of course, since these are all different interactions the forces created will be different. But in every case, Newton's third law will apply.

Clearly the force exerted by the window was not enough to slow the rock down much, nevermind send it bouncing backwards. But it was clearly enough force to break the window.

Note that momentum and force are different concepts. A 10 kg rock moving at 1 m/s has a certain amount of momentum, not force.
 
  • #7


Newtype, the window DOES exert an equal and opposite force, but ONLY to it's own limit to withstand the impacting force of the rock.

Thus, to balance things off, the impacting force continues through the window, yet exiting with a reduced force EXACTLY as less as the window inhibited it.

Hopes that makes any sense...
 
  • #8


Please provide an example like I did. Northwise moving glass fragments are not opposite a northwise moving rock. Force = mass multiplied by acceleration. Acceleration is the change of velocity, no change of velocity, then no acceleration (same velocity). Momentum = mass multiplied by velocity. So force and momentum are pretty much the same.
 
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  • #9


pallidin said:
Newtype, the window DOES exert an equal and opposite force, but ONLY to it's own limit to withstand the impacting force of the rock.

Thus, to balance things off, the impacting force continues through the window, yet exiting with a reduced force EXACTLY as less as the window inhibited it.

Hopes that makes any sense...

Yes, from the rock's frame the window represents a certain amount of resistance to its forward motion. That resistance constitutes a force exerted on the rock: it is acting to change the rock's inertia. The rock is inert at this point, discounting gravity, and the window, therefore, represents a force capable of accelerating it: changing its state of motion or rest.

From the window's frame the rock is a force which is capable of changing the window's state of motion or rest. The window is inert, in its own frame, and the rock will exert a force to change the state of motion or rest of the window.

The rock will experience an impulse in its encounter with the window: I = ft, or I = mat, and the value of that impulse will be exactly the same as the impulse experienced by the window. In this case the magnitude of the impulse will be capped by the window's material strength. In the case of an immovable wall, the impulse received by both will have a higher value, but will still be the same for both masses.

Newton III does not mean that both masses which are in relative motion can automatically match the full possible force of the other. It simply means that, whichever wins, informally speaking, they will both have experienced the same amount of acceleration from the other.
 
  • #10


Newtype said:
Please provide an example like I did. Northwise moving glass fragments are not opposite a northwise moving rock. Force = mass multiplied by acceleration. Acceleration is the change of velocity, no change of velocity, then no acceleration (same velocity). Momentum = mass multiplied by velocity. So force and momentum are pretty much the same.
Force is the rate of change of momentum. The change in momentum of the glass fragments is equal and opposite to the change in momentum of the rock.

Here's an example (with made up numbers) that might help. A 10 kg rock moves north at 1 m/s (let's ignore gravity) and thus has a momentum of 10 kg-m/s north. It encounters a window pane that is able to exert a force of 10 N on it for 0.1 seconds toward the south. (In the course of exerting that force, the window pane shatters, but so what?) According to Newton's 3rd law, both window and rock exert equal and opposite forces on each other, thus the rock exerts a force of 10 N on the window for 0.1 seconds toward the north.

What happens to the rock? Since the window has exerted an impulse (FΔt) on the rock equal to 10 N * 0.1 sec = 1 N-s south, the rock's new momentum is 10 - 1 = 9 kg-m/s north. So the rock is still moving north, only now its speed is only 0.9 m/s.

What happens to the window? If the window is attached to a window pane, etc, more forces act on it than just the rock. For simplicity, let's ignore all that and just pretend that the window pane was suspended in mid-air when the rock hit it. The window shatters from the 10 N force in the north direction, and glass bits go flying. Since the rock exerted an impulse of 1 N-s north on the glass, we know that the new momentum of all the glass pieces (you have to add them up) will total 1 kg-m/s north.

Note that the total momentum of "rock + window" hasn't changed. It started out as 10 kg-m/s north (all rock), and it ended up as 9 (rock) + 1 (window fragments) = 10 kg-m/s north.

Make sense?
 
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  • #11


I'm guessing the issue is the difference between the wall and window case. The forces in the window case are much less than in the wall case. Still, the smaller forces between rock and window are equal and opposing, and the larger forces between rock and wall are equal and opposing.
 
  • #12


Jeff Reid said:
I'm guessing the issue is the difference between the wall and window case. The forces in the window case are much less than in the wall case. Still, the smaller forces between rock and window are equal and opposing, and the larger forces between rock and wall are equal and opposing.

I think the problem is the wording of Newton III, which is tricky and requires explication:

To every action there is always opposed an equal reaction: or, the mutual actions of two bodies upon each other are always equal, and directed to contrary parts.

It's quite possible to read that and suppose he's making the extravagant claim that when, say, the rock hits a butterfly the rock should, by this law, always bounce off the butterfly in the opposite direction from, but at the same velocity with which it hit the butterfly.

To avoid that misreading, explication and context are required, which Newton provided:

"Whatever draws or presses another is as much drawn or pressed by that other. If you press a stone with your finger the finger is also pressed by the stone. If a horse draws a stone tied to a rope, the horse (if I may say so) will be equally drawn back toward the stone; for the distended rope, by the same endeavor to unbend or relax itself, will draw the horse as much toward the stone as it does the stone toward the horse, and will obstruct the progress of the one as much as it advances that of the other. If a body impinge upon another, and by its force change the motion of the other, that body also (because of the equality of the mutual pressure) will undergo an equal change, in its own motion, toward the contrary part. The changes made by these actions are equal, not in the velocities but in the motions of bodies; that is to say, if the bodies are not hindered by any other impediments. For, because the motions are equally changed, the changes of the velocities made toward contrary parts are inversely proportional to the bodies. This law also takes place in attractions and will be proved in the next scholium."

-Law III
Mathematical Principles of Natural Philosophy
 
  • #13


I have a tendency to resort to frames fo reference to resolve these kinds of ideas. Look at it from the rock's point fo view; it is moving forward with a certain amount of momentum (the "force" to which you refer). From the rock's point of view, it is not moving at all. The rock hits the wall, and experiences a tremendous jolt. The wall experiences exactly the same jolt. But the wall, being much more massive, doesn't move much, so the rock moves alot.

When hitting a window, the rock experiences a much lesser jolt. This slight jolt is not enough to impart much acceleration to the rock, but for the tiny shards of glass it is sufficient force to accelerate them quite a bit.

In each case, the same amount of force is applied to the rock and the obsticle, and both accelerate in the direction opposite one another. And the difference in acceleration is determined by the difference in the masses of the two objects involved in the colision.

Does that help any?
 
  • #14


I find that, as a general rule of thumb, if you ever think one of Newtons laws is wrong, don't tell anyone.
 
  • #15


Ok, here goes, with an example (your example in fact):

The rock hits the wall and bounces off (here we're assuming an infinitely hard rock and wall). The energy is conserved. The force on the wall from the rock is the same as the force on the rock from the wall and in oppsite directions at the point and exact moment of impact

Now you throw your infinitely hard rock through a window. At the point and exact moment of impact the force on the rock is equal and opposite to the force on the glass. Because this force is not large enough to stop the rock, the window breaks and momentum (motion) is given to all the glass shards (they will all move in different directions and at different velocities depending on their mass) and the same amount of momentum (motion) will be removed from the rock, so it will slow down slightly and may change direction or begin rotating. The rock will eventually hit something else or more things and come to a complete stop when it has transferred all its momentum to those things. At every moment and exact point of impact there will be a force on the rock from the object and a force from the object on the rock. This force will be equal and will act in opposite directions. The question if the rock will stop or not is only related to how much of its momentum it will give to the objects that it hits.

If this is not clear for now, don't worry, it'll become clear as you advance in your studies. Unlike archosaur (no offence archosaur and you're right in 99.9% of the cases (rule of thumb, like you say)) I think that it's ok to question everything, even Newton's laws, but keep an open mind and allow people with experience to guide you to a better understanding of these and other laws and accept them when you find out how to use them better.
 
  • #16


redargon said:
no offence archosaur

None taken! I'm just having fun.


Great example, redargon!
 
  • #17


Archosaur said:
I find that, as a general rule of thumb, if you ever think one of Newtons laws is wrong, don't tell anyone.
Yes, they are liable to knock you UP!

(Our British friends may not get that joke.)
 
  • #18


Redargon, How is there a "force on the rock is equal and opposite to the force on the glass"? Where does this force come from?
 
  • #20


You seem to have it in your head that just because something breaks that that in any way negates reactive forces. Take a wooden board and punch through it martial arts style. Yes you broke the board, but I bet your fist really hurts.
 
  • #21


Archosaur said:
I find that, as a general rule of thumb, if you ever think one of Newtons laws is wrong, don't tell anyone.

LOL!

I hope I'm not derailing the thread, but Newton probably didn't mean for his third law simply to mean that [itex]F_{12} = - F_{21}[/itex] as we commonly summarize it today. Elsewhere in Principia he applies it a bit more powerfully, anticipating D'Alembert's principle.
 
  • #22


Newtype said:
How is there a "force on the rock is equal and opposite to the force on the glass"? Where does this force come from?
maverick_starstrider said:
The glass.
No: the rock hitting the glass. The force is due to the interaction of (collision between) the pair of objects. If the rock is thrown from the left and the glass is on the right, the rock exerts a force on the glass from left to right and the glass exerts a force on the rock from right to left. Equal and opposite.
 
  • #23


If rock hits glass -> glass hits rock.
 
  • #24


Yes, the glass is made up of particles and since the glass is stationary it will put forth a resistive physical force. But how do you determine the opposite and equal force that the glass is putting forth against the rock's physical force? To say that the glass is putting forth an equal and opposite force against the rock's force is like saying that an eye of yours can put forth an equal and opposite force against a spear thrown at it. Are you going to tell me that it's impossible for a spear that's thrown to destroy a human eye because that eye will put forth a force equal and opposite to the spear's thrown force? Are you going to tell me that on September 11, 2001 airplanes could not have crashed into the World Trade Center buildings because the buildings' "opposite and equal" reaction forces canceled the airplanes' impact forces and that the airplanes "simply bounced off the World Trade Center buildings"? Really, why are you so adamant to continue putting forth that "for every action there is an opposite and equal reaction"? What's really going on?
 
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  • #25


Newtype said:
To say that the glass is putting forth and equal and opposite force against the rock's force is like saying that an eye of yours can put forth an equal and opposite force against a spear thrown at it.
Well, that's true! When the spear exerts a force on your eye, your eye exerts an equal and opposite force on the spear. So what?
Are you going to tell me that it's impossible for a spear that's thrown to destroy a human eye because that eye will put forth a force equal and opposite to the spear's thrown force?
:confused: What does exerting equal and opposite forces on each other--which always happens when two objects interact--have to do with the effects of such forces. If I smack an egg over my head, the forces they exert on each other are equal--but the egg is destroyed while my head is just fine, albeit messy. So what? Same thing with the spear and a human eye. The eye will not survive the experience unscathed, but the spear will.
 
  • #26


Newtype said:
To say that the glass is putting forth and equal and opposite force against the rock's force is like saying that an eye of yours can put forth an equal and opposite force against a spear thrown at it.
The eye does in fact put forth an equal and opposite force against a spear thrown at it.
Newtype said:
Are you going to tell me that it's impossible for a spear that's thrown to destroy a human eye because that eye will put forth a force equal and opposite to the spear's thrown force?
A given force will deform an eye much more than a spear. So the fact that the eye is destroyed and the spear is not does not in any way contradict the fact that the forces on each are equal.

EDIT: Doc Al beat me to it!
 
  • #27


DaleSpam, how are those forces "equal"?
 
  • #28


Newtype said:
DaleSpam, how are those forces "equal"?
Again, you are mistaking the effect of a force on a particular body with the fact that the forces exerted are equal. The same force acting on different objects can easily have different effects.
 
  • #29


In your wonderfully morbid spear-eye example, the forces felt by both objects are equal, but an eye cannot handle as much force as the metal tip of a spear can, so it breaks.
 
  • #30


Alright, Object X is thrown at material Y. If material Y experiences Z amount of force it will shatter. So if object X is chucked with a force GREATER THAN Z when it comes in contact with a material Y, X will exert a force Z (since it is all Y can muster) on Y (the result of this on Y will be that it shatters), Y will also exert an equal and opposite force Z on X (which, if the total collision/shattering took t seconds then this will result in object X loosing Z*t momentum), this may also dent X or what have you. However, if the force with which is much greater than Z it will still have a non zero momentum after its shattering collision with Y and will keep on going. It's like static friction, the force of static friction equals [itex]f_s \leq \mu_s N [/itex]. Notice the less than equal to.
 
  • #31


It looks like everybody did a pretty good job of getting at this, but I thought I would just chip in my two cents with an alternative sort of answer.

First let's consider the throwing of the rock. Maybe you are familiar with the notion of a "free body diagram". Basically it just consists of a picture of the object in question, and arrows representing all the forces acting on the body. Now, if you draw a FBD of the rock when it's being thrown, this picture consists in the rock, and only the rock. In the picture, you will see a force arrow which represents the hand pushing on the rock. (You will also see an arrow pointing downwards representing gravity. Let's just ignore that for now.) Now at the same instant, if you draw a FBD of the hand (and only the hand), then you see an arrow of the same length, pointing in the opposite direction. This represents the rock pushing back on the hand. That's all that Newton's third law means.

If you do the same thing for the window and the rock at the initial point of impact, again, you will have a similar situation if you draw instantaneous pictures of the window or the rock. The window breaks because it is unable to sustain this force, which it feels in terms of a sudden, nonuniform increase in load. After the window shatters, the rock is no longer exerting a force on the window at all, so it is correct to say that there is not an equal and opposite reaction for a long time. The rock does not stop because the work done by the window is insufficient to fully dissipate the rock's kinetic energy.

Question: since the window is trying to push backwards against the flight of the rock, do you see bits of the window flying outwards, in the opposite direction? The answer is yes, I believe. However, the distribution of forces in the shattered glass is much more complicated, and it is not straightforward to predict the trajectories of the shards. Elastic and nonelastic deformation are complicated problems.
 
  • #32


So the more momentum the rock has, the more force it will take to make that momentum change direction, because if the rock doesn't change direction, it will break the glass apart. The amount of force the glass can exert on the rock is an intrinsic property of the glass-rock interaction, so a piece of wood with the same momentum won't feel the same change in momentum, or force, because it's a different interaction.
Am I right?
 
  • #33


JanClaesen said:
So the more momentum the rock has, the more force it will take to make that momentum change direction, because if the rock doesn't change direction, it will break the glass apart. The amount of force the glass can exert on the rock is an intrinsic property of the glass-rock interaction, so a piece of wood with the same momentum won't feel the same change in momentum, or force, because it's a different interaction.
Am I right?
No one?

Is the law of Archimedes also a reaction force?
If it is: (m is the mass of the object I drop in a cup of water)
mg = pVg (p is mass density), so the mass of the object I drop in the water (pV) must be the same as the mass of the water, what's wrong here?
And what about Hooke's law?
 
  • #34


JanClaesen said:
So the more momentum the rock has, the more force it will take to make that momentum change direction, because if the rock doesn't change direction, it will break the glass apart. The amount of force the glass can exert on the rock is an intrinsic property of the glass-rock interaction, so a piece of wood with the same momentum won't feel the same change in momentum, or force, because it's a different interaction.
Am I right?
Not bad. I'll nitpick a bit: Force is not just change in momentum, it's the rate of change of momentum--change in momentum per unit time. I can throw a rock and a pillow against a wall with the same momentum. Assuming they come to rest, they both experience the same change in momentum, but the pillow-wall interaction will take longer and produce less force.

JanClaesen said:
Is the law of Archimedes also a reaction force?
Not sure what you mean by that. (Certainly if the water pushes on the object, the object pushes back on the water with an equal and opposite force.)
If it is: (m is the mass of the object I drop in a cup of water)
mg = pVg (p is mass density), so the mass of the object I drop in the water (pV) must be the same as the mass of the water, what's wrong here?
Archimedes's principle says that the buoyant force (exerted by the water on the object) will equal the weight of the displaced water. Only if the object has the same density as water will its mass equal the mass of the displaced water.
And what about Hooke's law?
What about it?
 
  • #35


On a second thought, those questions about the Archimedes force were rather stupid. :)
So about Hooke's law ;-), is that 'Hooke force' a reaction force on the force I pull with?
 
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