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We all know that if I punch the wall with $100\,\mathrm{N}$ force, the wall pushes me back with with $100\,\mathrm{N}$ and I get hurt. But if I punch air with $100\,\mathrm{N}$, does air punches me with $100\,\mathrm{N}$? I mean I don't get $100\,\mathrm{N}$ back. I don't get hurt. Does it violate Newton's third law? Appreciate if someone can answer. This question has been bothering me for more than 5 years.

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    $\begingroup$ How do you propose to "punch the air with 100 N"? (for that matter, how do you punch the wall with 100 N...). $\endgroup$ – Floris Oct 20 '15 at 15:14
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    $\begingroup$ @floris with much pain, on the second part at least. $\endgroup$ – Trotski94 Oct 20 '15 at 15:17
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    $\begingroup$ In the second case, it's not the air that slows you down - it's your body. That's where the force is applied. If you throw a particularly strong punch, it's possible to hurt your elbows, shoulders or some of the tendons or muscles. $\endgroup$ – Luaan Oct 20 '15 at 17:23
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    $\begingroup$ The simple answer is yes, punching air with 100N would hurt as much as punching a wall with 100N (but the nature of hurt may be different: blunt force trauma vs friction burn). Only, in order to punch air with 100N reaction force you need to have either a very large surface area for a fist or move your fist really fast. I've been hit by a board that's been blown by the wind before and it really hurts - that's the air punching me. $\endgroup$ – slebetman Oct 21 '15 at 6:05
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    $\begingroup$ In short: Newton's laws are not violated. You are not exerting the same force on the air as on the wall even though your fist is moving the same speed. The air moves around your fist. $\endgroup$ – Neo Oct 21 '15 at 16:35
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The assumption that you are making here is that with the same motion of a punch, that you are applying $\text{100 N}$ of force to both a wall and to the air. However, you should think about the most fundamental equation of Newton's laws, namely,

$F=ma$

The most important part of this in relation to what you are talking about is that the force applied, $F$, is proportional to the acceleration, $a$. When you hit the wall, your hand goes from full speed to a complete stop rather quickly. This is a fast deceleration, or a high value of $a$, so the value of the force is high.

However, when you punch through the air, the air molecules hardly slow down your hand at all. This means that your deceleration is low, or a low value of $a$, meaning that the force, $F$ is also low.

In the absence of a wall to stop your fist, what is really stopping your punch is your own body, not the air, as your arm socket will have to pull back on your arm to keep your fist from flying away. This also will slow down your arm more slowly than a wall however, since the tendons and ligaments in your arm tend to stretch, reducing the deceleration compared to the wall, and thus the force.

So, what I'm trying to say here, is that yes, the forces are always equal and opposite which is in line with Newton's laws. However, your assumption that force from hitting a wall with a punch is the same as a force from swinging your fist through the air is incorrect.

I hope this clears things up for you.

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    $\begingroup$ In the absence of 100N of force for the air to counteract the momentum of your arm, you'll notice your arm socket will have to supply the 100N force to eventually stop your arm. And that will hurt. $\endgroup$ – Señor O Oct 20 '15 at 17:11
  • $\begingroup$ This is true :), I used the $\text{100 N}$ figure to coincide with the question. An arbitrary place holder should probably be used instead. At least we have the humorous thought experiments that come with that number. $\endgroup$ – tmwilson26 Oct 20 '15 at 17:14
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    $\begingroup$ The comment about about the socket of your arm is more apropos. It's not "air friction" slowing you down, but [different] muscles, stretching of ligaments, etc and arm itself. (It would work likewise in a vacuum). These are all "pushing" against the weight of your body. If you were wearing ice-skates doing this, you might jerk a bit backwards during the initial punch, then come to a stop during the "recoil". Punch and recoil both apply forces and acceleration in opposite directions - netting a "zero" result. $\endgroup$ – Brad Oct 20 '15 at 17:52
  • $\begingroup$ Right, I understand where the comment is coming from now. I can add a line clarifying this point. $\endgroup$ – tmwilson26 Oct 20 '15 at 17:54
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    $\begingroup$ @SeñorO: It'll supply the impulse, but not necessarily the same force. $\endgroup$ – user2357112 Oct 21 '15 at 5:00
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This is the same problem as the famous question states: which is heavier, 1kg of feathers or 1kg of iron?

It requires many more feathers to get 1kg of them, so it confuses most people. In this case, it requires you to have a big wing to be able to apply a force of 100N on air.

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  • $\begingroup$ Maybe a parachute or umbrella helps too. $\endgroup$ – Paŭlo Ebermann Oct 20 '15 at 18:36
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No, Newton's third law is not violated.

According to Newton's Second Law, we have that force is the rate of change of momentum with time, i.e. $$F=\frac{\Delta p}{\Delta t}$$ where $p$ is momentum and $\Delta t$ is time elapsed.

When Punch Strikes Wall

Initial momentum of fist = $mv$
Final momentum of fist = $0$
Force applied = $\frac{mv-o}{t} = \frac{mv}{t}$

When Punch Strikes Air

Initial momentum of fist = $mv$
Final momentum of fist = $0$
Force applied = $\frac{mv-o}{T} = \frac{mv}{T}$

Hence force $F$ applied in both cases is not same.

As you cannot touch an air molecule as such and air molecules have very low or no rigidity, so change in momentum takes place very slowly and $T$ is considerably large.

But you can feel striking the wall molecules as such and the wall molecules have very high rigidity, so change in momentum takes place very quickly and $t$ is quite small.

I think the difference is now quite obvious.

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  • $\begingroup$ Am I not punching air with 100 Newton? If not then why not? Cause I am doing the same thing...punching the air as violently as punching the wall... – Tough questions just now edit $\endgroup$ – Tough questions Oct 20 '15 at 15:57
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    $\begingroup$ Your force is not the same, your $\mathbf{impulse}$ is the same. Impulse = $F \times t$ = $mv - 0$. When you say you are punching both violently, you mean you are experiencing the same impulse, not same force. $\endgroup$ – SchrodingersCat Oct 20 '15 at 16:05
  • $\begingroup$ Then how do birds fly? They use reaction of air by flapping their wings.air must give birds enough reaction. $\endgroup$ – Tough questions Oct 20 '15 at 16:18
  • $\begingroup$ Birds have a very light body and an aerodynamic shape to use air drag to its advantage. So they can use reaction of air by flapping their wings. $\mathbf{But \,\ your \, fist \,\ is \,\ much \,\ heavier}$. $\endgroup$ – SchrodingersCat Oct 20 '15 at 16:21
  • $\begingroup$ @Toughquestions You apply force to your hand while you accelerate the fist, but you are the one decelerating it as well (and if you throw violently, you can hurt yourself in the process). The air is not applying a significant force to your fist. With water you can have a resistance at normal speeds. With air, you must drive maybe 70 mph (my Mach 0.1) to feel your 100N when you stick the hand out of the passenger window, and then you do feel the air's force. $\endgroup$ – Peter A. Schneider Oct 20 '15 at 16:22
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Let's revisit "force on impact" for a moment.

I will consider a "sticky" ball of mass $m$ traveling at velocity $v$ at a stationary wall. When it hits the wall (and sticks), what is the maximum force felt by the ball / wall?

That is actually not a trivial question to answer. The ball has momentum $p=mv$, and if the impact time (time for the ball to come to a complete stop in an inelastic collision) is $\Delta t$, the average force $F_{av} = \frac{p}{\Delta t}$

This helps explain why, if you throw a punch, there is a difference (in pain) between a gloved punch vs bare knuckles. The glove causes the fist to slow down over a greater distance (longer $\Delta t$), resulting in less average force. It also spreads the force over a greater area, meaning that the local pressure will be less. Again, reducing the pain.

You can look at it in terms of acceleration as well: if you slow down over a shorter distance, the acceleration (and the force) must be larger.

This gets me to the question I asked in a comment above: "how do you throw a 100 N punch into the air?". If your fist is moving at the same speed, it will experience very little force from the air; that's not a "100 N punch". If you could somehow move your fist fast enough that it would experience a force of 100 N from the air resistance, then I still expect there to be less "pain" (from the air) since the force will be more evenly distributed (fist hitting wall = just a small contact area; fist hitting air = large contact area).

How fast would your fist have to move? If we model the fist as a sphere with 10 cm diameter, and use the air drag equation

$$F = \frac12 \rho v^2 A C_D$$

we can solve for $v$:

$$v = \sqrt{\frac{2F}{\rho A C_D}}\approx 200\; \rm{m/s}$$

If you can move your fist that fast you should give up on physics and take up professional boxing. The force needed to accelerate a fist (attached to an arm) to that speed over the range of a punch is significant. If you consider a muscular arm to have a mass of 10 kg, and the lower half (below the elbow) 5 kg, then accelerating that to 200 m/s over a 50 cm range requires an average force of 200 kN. roughly the force needed to benchpress a family of elephants. With one hand.

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So there's two situations where forces are "equal and opposite", and I see undergraduates getting confused all the time. The one time I've taught a recitation section for this stuff, I made this distinction a point of one lesson, and it seemed to help them; maybe it helps you.

Newton's third law.

Newton's third law states, in its deepest and most abstract theoretical sense, that the laws of physics are the same everywhere. It turns out that this sort of "continuous symmetry" (space-translation symmetry) always corresponds to a conserved quantity, a number that you can calculate which never changes its total value. You can then treat this quantity like a "stuff" which distributes over different parts of the system. For space-translation symmetry, this "stuff" is called momentum (more precisely, the numbers are the components of momentum in each of three perpendicular directions). The principle of conservation of momentum says that each component is a "stuff" which cannot be created or destroyed, but only redistributed: if this object has a momentum going this way, then that object must get the corresponding momentum going that way.

Newton defined any disposition to change momentum, per unit time, as a force. Therefore conservation of momentum manifests as "every force comes in a force-pair: when you push on a wall with some force, it pushes on you with the same force." This has been a ludicrously successful perspective for the sake of engineering. But it comes from the fact that when you punch a wall, the forward momentum of your punch has to be absorbed by something: so when your hand stops, the wall (and whatever it's attached to) must keep going. These are the same principle; treat them the same.

In particular, if you look at the center-of-mass frame of a system, where the momentum of that system is zero, no internal forces of the system can possibly change it. If you look at the center of mass of a rocket before it takes off, that rocket cannot change that center of mass no matter what it does with its fuel: the fuel has to go backwards really fast for the rocket to go slowly forwards, so that the total fuel-plus-rocket momentum is zero.

Force balance

Now there's another, completely different thing that also happens due to Newton's definition (a force is a disposition to change momentum during a unit of time): if the momentum of an object is not changing, then it exists in a state of force balance: all of the components of all of the forces on the object must, in each of the three directions, cancel out. This looks very similar, because for the simplest case, like sitting in your chair, the force of gravity downward on you is balanced out by an "equal and opposite force" of your chair upwards on you, to keep you still. But it's not the same thing, it doesn't have to exist: your chair could be spring-loaded, sending you up; or it could break, letting you fall. It's just that the chair happens to be working as planned, that these things balance. The law of conservation of momentum says that your force on the chair is equal-and-opposite to the chair's force on you; the principle of force balance says that gravity's force on you is also equal-and-opposite to the chair's force on you.

Conservation of momentum is somewhat easy to locally violate here: if I plop down into the chair, surely I come to rest, no? That's because the "stuff" (momentum) leaves the "system" (me plus chair) to join a much larger system (the floor, and eventually the planet). So conservation of momentum might be a very useless principle if momentum escapes the system you're studying, and force-balance might be a very useless principle if something is not keeping a constant momentum (which could be zero, but more generally any uniform motion in a straight line is constant-momentum due to the symmetry that generates its conservation). Sometimes one is useful and the other isn't, independently of their details.

Punching a wall vs. punching air.

Here's a great example. The wall is in a state of force-balance: it's not moving no matter how hard you punch it! Any momentum you pour into it with your puny little fists will tend to escape into the planet itself, so momentum is not conserved locally but only globally. But as long as your fist doesn't go through it, that force balance is also going to apply to the leading edge of your fist: however hard you punch the wall, the wall will punch you back.

You can similarly punch air, but it takes a lot longer for air's drag force to slow you down: in fact it takes so long that the main thing which slows down your fist is your arm not stretching enough. Most of that momentum does not go to moving the air forward. Still, there are similar forces, like the air drag that slows you down when you're on a bicycle: and they all obey conservation of momentum.

Well, this is much easier if you punch underwater, or so: if you hit the water with a paddle, like in a canoe or kayak, you can see the water "spinning off" your paddle: and you move the other way, conserving momentum. This is a state where force-balance is not necessarily a useful idea (each push moves both you and the water, neither momentum stays constant) but momentum conservation is a great way to look at the problem.

The role of pain and damage

The last thing to notice is that none of these things have to do much with you feeling pain; you feel pain when your body gets damaged and your nerves send those signals of "hey, stop it!" to your brain.

Damage tends to happen due to stresses, which are pressures or forces per unit area. I like to tell students "it takes a lot less force to stub your toe," because the cross-section of your toe is very small, and that perpendicular cross-sectional area matters just as much as force does. If you imagine a string that is loaded with so much weight that it is near breaking, some weight W and I put a second string next to it, you'd expect that I could now hang 2W before being similarly near breaking: W from one chain, W from another. But if you hung the string from another string, you'd just picture that they both stretch and they're both near-breaking with just W on them.

There are two effects here. The first one is that a force is a change in momentum **per unit time*: longer times make for lower forces. So if you are in a car going 60mph, you're going to sustain less damage going to 0 mph if that's done normally with your brakes over the course of 50 seconds than if you are crashing suddenly into a tree, because the forces can be a hundred times larger in the 0.5 seconds of tree-crash than in the 50 seconds of slow braking.

The second one is that a pressure is a force per unit area. This happens a lot during my preferred sport, ultimate, where it is not uncommon to dive after a flying disc that is otherwise just out of your reach. When you do, the evolution-hardened tendency (that you have to resist) is to reach out with your arms: evolution would rather go the force-reduction route by sticking out your arms to twist and break slowly, so as to minimize the forces to your chest and head. But in ultimate, it is a lot safer if you can absorb the impact on the grass with your chest and stomach and thighs, pulling your head and arms up out of the way. There is also some force-reduction -- because you slide a little, so you only need to dissipate your downward momentum immediately, not your forward momentum -- but the biggest win is the added surface area that the force dissipates over.

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  • $\begingroup$ Conservation of momentum and balance of forces are two faces of the same issue. Just because two interacting "entities" (say: solid bodies) experience the same (but opposite) forces for the same time (which is rather a truism) is it that the overall system "conserves momentum". That is true whether the body falls in earth's gravity field or whether it sits motionless in a chair so that all forces could be considered internal forces of the body earth. After all, nothing is really "touching" or "resting": everything interacts through dynamic field interactions. $\endgroup$ – Peter A. Schneider Oct 20 '15 at 16:36
  • $\begingroup$ @PeterSchneider Yes, there is a way to refer to conservation of momentum as a force balance, but no, pedagogically we must treat them very differently, because students get very confused about the difference between "the force of gravity is balanced by the force of my chair on me" (which happens because I'm not accelerating -- what I've called a "force balance") and "the force of my chair on me is equal to the force of me on my chair" (Newton's third law). $\endgroup$ – CR Drost Oct 20 '15 at 16:44

protected by Qmechanic Oct 20 '15 at 18:27

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