Newton's Third Law: Unpacking the Mystery

[Kurokari]

I'm kinda confused with Newton's Third Law.

Let me present my problem.

Say I kicked the floor with 50N, the floor pushes me back with 50N, there is a reaction pair.

But say if I kicked the air with 50N, why aren't I feeling the 50N? By right it should feel the same right?

 

[DennisN]

Hi! Newton's Third Law states:
"the forces of two bodies on each other are always equal and are directed in opposite directions".

Simple version:
First, imagine that the air you kick in is entirely empty (it has no molecules), that is, imagine you kick in perfect empty vacuum. Then there is only one body kicking, so there can't be a force kicking back at you.

Air version:
If you do it in air, there actually is a small force kicking back at you, as the molecules in your hand hit the air molecules. Think of drag (air resistance) (even though drag is more complicated than this). The size of the force depends roughly on the number of molecules you hit.

 

[meldraft]

And also, because of your body posture, your feet exercise some extra force on the ground, which in turn pushes back against them (there can be other ways, but I think this is the simplest one)

 

[Doc Al]

Say I kicked the floor with 50N, the floor pushes me back with 50N, there is a reaction pair.

OK.

But say if I kicked the air with 50N, why aren't I feeling the 50N? By right it should feel the same right?

The problem is that you probably cannot kick the air with 50 N of force. You will certainly have an easier time exerting a force against the floor compared with what you could exert against the air. The force you can exert against some object depends not only on you but on the object. But as Kurokari says, whatever force you manage to exert against the air, it will exert an equal and opposite force back on you.

 

[Kurokari]

Air version:
If you do it in air, there actually is a small force kicking back at you, as the molecules in your hand hit the air molecules. Think of drag (air resistance) (even though drag is more complicated than this). The size of the force depends roughly on the number of molecules you hit.[/QUOTE]

So I am correct in saying that because the air molecules in the air are so few, only a certain portion of my hand is hit back by the air molecules(others are just like you said in your first empty version), and the force is dispersed and canceled out by the arrangement of the body's molecule, thus my body is not able to detect it?

As opposed to the floor, where all the atoms are so compactly arrangement, it can be said most of my hand would experience the reaction force?

 

[Ken G]

So I am correct in saying that because the air molecules in the air are so few, only a certain portion of my hand is hit back by the air molecules(others are just like you said in your first empty version), and the force is dispersed and canceled out by the arrangement of the body's molecule, thus my body is not able to detect it?

Sort of, but rather than saying you can't feel the force, go with Doc Al's version-- you can't create the force. It takes two to tango, so if you don't have enough air molecules, you just can't generate the force you have in mnd. Here's another thing you can do-- take your right hand, and push strongly on your left hand, but do it so that your left hand does not push back at all on your right hand. Probably right now your hands are moving quickly to the left, and you are finding that you just can't do it. Forces are a relationship between two things, not something that one thing does to something else.

 

[Kurokari]

Sort of, but rather than saying you can't feel the force, go with Doc Al's version-- you can't create the force. It takes two to tango, so if you don't have enough air molecules, you just can't generate the force you have in mnd. Here's another thing you can do-- take your right hand, and push strongly on your left hand, but do it so that your left hand does not push back at all on your right hand. Probably right now your hands are moving quickly to the left, and you are finding that you just can't do it. Forces are a relationship between two things, not something that one thing does to something else.

I see, looks like I need to re-correct my whole concept of force.

But why is it that we can't generate enough force due to the lack of air molecules?

Also for the right hand pushing the left, say there is an acceleration of a, means there is an resultant force, does resultant force have a reaction pair?(Which my current understanding says no)

All of these sounds pretty ambiguous to me.

 

[Ken G]

But why is it that we can't generate enough force due to the lack of air molecules?

For the same reason you can't generate much force from your right hand on your left hand if your left hand offers no resistance-- the mass of your hand (or the air molecules) is relatively small, so a good force would need to produce a good acceleration. You can't move your hand (or foot) fast enough to generate the required acceleration. So to get a large force between your hands (or a large force between your foot and the air), you need some kind of resistance other than just inertia to partially balance the force you are providing. That's why you can kick the ground with great force, but not the air. In the case of kicking the ground, the force you provide is balanced by internal forces that set up in the ground to resist you, but in the case of the air, all you have is the inertia of the air, and that cannot support a large force unless the acceleration can be very large (i.e., if you can really move your foot fast, like sticking it out of a car window at 70 mph).

Also for the right hand pushing the left, say there is an acceleration of a, means there is an resultant force, does resultant force have a reaction pair?(Which my current understanding says no) .

All forces have reaction forces, that's Newton's third law (note we mean real forces here, not fictitious or frame-dependent apparent forces, that's a whole other issue). If your left hand has mass M, and acceleration A, then the force it can support is a force MA, and that is also the force it will push back on your right hand with (assuming you are not also exerting any force with the muscles of your left arm of course). Note the forces you would be exerting with the muscles of your right arm is 2MA, because your right hand is also accelerating with A and has mass M, so its net force is MA, so that's 2MA from your arm and -MA from the reaction force of your left hand. The 2MA makes perfect sense if you think about the combined system of both hands, which have mass 2M and acceleration A.