Newton's Third Law box push

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TheShapeOfTime

Newton's third law states something along the lines of "For every action, there is an equal and opposite reaction." The problem I have is that I don't understand how it stays true for something as simple as pushing a box. If a push the box (action), how can it be giving an equal opposite reaction if I'm moving it? Why wouldn't it just stay still? I know I'm missing something here...
 
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Gokul43201

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The action and reaction act on different bodies.

You push the box with a force F. There is some frictional acting on the box along the floor. The box's motion is a function of these forces and its mass.

The box exerts a reaction force on you, which is also equal to F. There is some frictional force acting on the soles of your shoes, due to the floor. Your movement (which one would expect to match the box's) is a function of these forces and your mass.
 
TheShapeOfTime said:
Newton's third law states something along the lines of "For every action, there is an equal and opposite reaction." The problem I have is that I don't understand how it stays true for something as simple as pushing a box. If a push the box (action), how can it be giving an equal opposite reaction if I'm moving it? Why wouldn't it just stay still? I know I'm missing something here...

Maybe to clarify Gokul's explanation a little. To analyze the state of motion of the box you look at the forces acting only on the box. If they are unbalanced, the box will accelerate in the direction of the greater force. The box's reaction force doesn't act on the box, it acts on you, so it has no influence on the box's state of motion.
 
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TheShapeOfTime said:
If a push the box (action), how can it be giving an equal opposite reaction if I'm moving it? Why wouldn't it just stay still? I know I'm missing something here...
I think the simultaneity principle will help clear this up.

The word 'action' and 'reaction' can seem to imply that action occurs before reaction. But the concept in physics teaches that it's simultaneous.

Let us test to see which one is true by using logic.

Ether there is a simultaneous action and reaction or there isn't in your example with hand meets box.

1.) If there isn't a simultaneous action and reaction, we say:

The hand will exert a force, then the box will exert an equal and opposite force.

2.) If there is a simultaneous action and reaction, we say:

The hand will exert a force, while the box exerts an equal and opposite force.

In number one, a force from the hand is asserted to be exerted upon the box. Then after that force is exerted, the box exerts an equal and opposite force upon the hand.

The problem with this assertion is that force only occurs when an object changes velocity (accelerates). No object has a force exerted upon it that is in uniform motion or rest. Two objects must interact for force to exist: the outside force and the object it exerts a force upon. The hand and the box must meet for force to exist.

The hand and box must hit each other at the same time for a collision to occur. One object can't hit the other while the other is not hitting it also. Since they both hit each other at the same time, they both change velocity at the same time. If they are not touching, they can't change velocity. So they only change velocity while they touch. Since they change velocity at the same time, force is exerted at the same time. Equal and opposite force occurs simultaneous. The box exerts and equal and opposite force!

Therefore the second statement must be true. Action and reaction is simultaneous.

Furthermore.

Statement one makes a statement that fuses Newtons first and third law. But there is a distinguishement between the two. We see this distinguishment with our senses when we see the difference between constant motion and acceleration motion. An object may only be in two types of motion: constant or accelerating.
 
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lol yes true , but there is a limit, over the limit (in ur case the friction of box to the ground) the extra froce was used for moving the box,

in other case, if u tries to punch a wall, the harder u punch , the more you hurt, why? the more force u apply in , the more force the wall apply return, becasue wall will not move or break, no extra force is existed.(except maybe heat and sound but nogeotiable)

alternatively if punch a wall that can easily break, ur hand would not hurt that much why? the force that the wall apply to u is less than the force u apply to wall, extra force was used to break the wall into parts and fly off, plus sound and heat.
 
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TheShapeOfTime said:
Newton's third law states something along the lines of "For every action, there is an equal and opposite reaction." The problem I have is that I don't understand how it stays true for something as simple as pushing a box. If a push the box (action), how can it be giving an equal opposite reaction if I'm moving it? Why wouldn't it just stay still? I know I'm missing something here...
You can think of the principle as: forces always come in pairs - equal and opposite pairs. This is perhaps the only universal fundamental principle of all physics. It applies to Classical physics, special and general relativity, quantum physics and the Standard Model. It applies in all frames of reference. It is the underlying basis for the conservation of momentum.

Moving a box on a non-frictionless surface is complicated. If the box moves in uniform motion, the force moving the box is just equal to the force required to balance dynamic friction. But this is further complicated because you are pushing back on the floor to exert a force on the box. Because it is so complicated, it is not very useful or clear in illustrating Newton's third law.

Instead, think of a collision of two bodies M and m in space. A body of mass M applies a force F to mass m to give it acceleration dv/dt = F/m. But m (by its inertia) resists force F by applying an equal and opposite force -F on M, causing M to acclerate in the opposite direction with acceleration -F/M. The two forces operate for exactly the same time.

[tex]F_m\triangle t = m\frac{\triangle v_m}{\triangle t}\triangle t = m\triangle v_m = \triangle p_m
[/tex]

[tex]F_M\triangle t = M\frac{\triangle v_M}{\triangle t}\triangle t = M\triangle v_M = \triangle p_M[/tex]

Since:
[tex] F_m\triangle t = -F_M\triangle t [/tex]

Then:
[tex]\triangle p_m + \triangle p_M = 0
[/tex]
 
expscv said:
lol yes true , but there is a limit, over the limit (in ur case the friction of box to the ground) the extra froce was used for moving the box,

in other case, if u tries to punch a wall, the harder u punch , the more you hurt, why? the more force u apply in , the more force the wall apply return, becasue wall will not move or break, no extra force is existed.(except maybe heat and sound but nogeotiable)

alternatively if punch a wall that can easily break, ur hand would not hurt that much why? the force that the wall apply to u is less than the force u apply to wall, extra force was used to break the wall into parts and fly off, plus sound and heat.
This is a wrong concept. Whether the wall moves or not only decides how long you are in contact with the wall, and how long your hand can hit it. Also, even when the wall breaks, you still get hurt, if you punched hard. The same thing is that when you get a soccer ball very hard, even if it zooms across half the field, your foot feels the pain from the impact, because the ball applied a reaction force on your foot.

Force and motion are two separate things that are related, but one doesn't imply the other. Whether it moves or not doesn't affect the magnitude of the reaction force due to the force you exert on it.
A simplified version of Newton's third law states that when A applies a force on B, then B applies an equal force in the opposite direction on A. It's the same as saying that when I push you, you also push me. Or think of tug-a-war(spelling?)...when team A pulls the rope towards themselves, they also feel a force forward from the rope. However, it doesn't mean that they're moving.
 

russ_watters

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To expand just a little, breaking something involves energy - a force, a distance, and a time. Walls are strong, so the force is high, but they are also brittle, so the distance and time are small. The result is a relatively low energy. http://www2.latech.edu/~jordan/courses/me215/ManualF03/04CharpyImpact.htm [Broken] is a Charpy impact/fracture energy energy tester that is used to measure this very phenomena. Its essentially a pendulum with an axe on the end.

One interesting thing you can demonstrate is that while it takes more force to break iron than aluminum, it takes less energy.
 
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Doc Al

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expscv said:
alternatively if punch a wall that can easily break, ur hand would not hurt that much why? the force that the wall apply to u is less than the force u apply to wall, extra force was used to break the wall into parts and fly off, plus sound and heat.
This is incorrect. To add to comments already made: Whatever force you exert on the wall always equals the force the wall exerts on you. It doesn't matter how "hard" you punch or what happens to the wall. That's a direct consequence of Newton's 3rd law.
 

Alkatran

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I like to picture the wall coming out and smashing in to my fists when I punch it.

If you stood still with your first out in front of you with a wall hurtling towards you, it's the same basic idea as you punching a wall.
 

Andrew Mason

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Doc Al said:
This is incorrect. To add to comments already made: Whatever force you exert on the wall always equals the force the wall exerts on you. It doesn't matter how "hard" you punch or what happens to the wall. That's a direct consequence of Newton's 3rd law.
What matters is the time over which the force (deceleration of the hand) operates. The hand doesn't hurt as much when you punch through the wall because the hand does not slow down as rapidly so the force is not as great on the hand (as when you punch the wall and it doesn't break). The force on the hand, however, lasts longer because your hand has to push matter forward and that takes more time. So the momentum imparted to the wall is the same.

Perhaps a karate chop to a brick provides a better example. It is always more painful when you don't break the brick. As any karate student.

Andrew Mason
 
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wow thx all for correcting me in such wrong concept over a basic principle.

force is force

motion is motion

energy is energy

=)
 
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TheShapeOfTime

I will absorb what you have all said and get back to you with any questions. Thanks for all your replies!
 
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TheShapeOfTime

Hey,

I still have a few questions, but I understand where the reaction "goes" now.

When a box is moving with uniform motion, the forces are balanced. This means that the force moving the box forward is balanced by friction just so that the box keeps from accelerating? This would imply that my hand is still pushing the box after it's initial accerleration or period of unbalanced forces. When my hand leaves the box, the forces become unbalanced (friction overcomes the now zero force pushing the box?), and the box would slow down.

What I'm really confused about is the fact that you (omin) said there are no forces acted upon something in uniform motion. If I'm understanding correcting, this wouldn't make what I said above incorrect. But wouldn't there always need to be a force of friction acting upon an object? Wouldn't there need to be something balancing that friction? If so, how can that be considered "no force". Maybe I just misunderstood you.

The problem with this assertion is that force only occurs when an object changes velocity (accelerates). No object has a force exerted upon it that is in uniform motion or rest. Two objects must interact for force to exist: the outside force and the object it exerts a force upon. The hand and the box must meet for force to exist.
 
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Omin probably meant to say "no net force/no unbalanced force" rather than "no force at all".
 
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TheShapeOfTime

kuenmao said:
Omin probably meant to say "no net force/no unbalanced force" rather than "no force at all".
Oh, ok. Makes sense then. :)
 

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