7
$\begingroup$

According to Newton's third law, action force equals reaction force in terms of magnitude.

When I punch a glass, the glass punches me back. If I exert a greater force on the glass, it will break.

Suppose a glass could sustain 100N force and that my muscles can exert up to 200N force: if I went all out, I couldn't punch the glass with a 200N force because the glass would break, which means it's not able to apply a 200N force on me. I apply F = 200N and the reaction is only f = 100N.

Now suppose I punch a feather in a vacuum, can you explain this?: does it matter if someone is holding the feather for you to punch or it's free?

$\endgroup$
12
$\begingroup$

You will be punching the feather with a really small force. That doesn't mean your arm is punching lightly. Your arm can have a lot of momentum and internal tension due to internal forces that makes you punch hard, but the actual force is defined on the interaction with another object, in this case: the feather, so the force you apply to the feather is pretty small compared with the maximum force your arm would be capable of doing.

Update: In a comment Jim says:

Unless the collision is elastic, in which case the feather will experience the full force and accelerate to an enormous speed

This is incorrect (although it depends on the definition of "enormous"). Here I will show that the "enormous" speed that you can get for the feather is at most twice that of the arm. If you calculate the final speed for an elastic collision (which I'll leave as homework), you will obtain:

$$v_{feather}=\frac{2v_{arm}}{1+\mu}$$

where $\mu=\frac{m_{feather}}{m_{arm}}$

Thus, in the limit of a negligible feather mass, and in an elastic collision, the speed of the feather will be at most twice that of the arm. The speed will be still less if the collision is not elastic.

UPDATE: For those who find it difficult to get the equation.

From conservation of momentum: $m_{arm}v_{arm.ini}=m_{arm}v_{arm.final}+m_{feather}v_{feather}$

From conservation of energy: $.5m_{arm}v_{arm.ini}^2=.5m_{arm}v_{arm.final}^2+.5m_{feather}v_{feather}^2$

if you eliminate $v_{arm.final}$ you get $$v_{feather.initial}=\frac{2v_{arm}}{1+\mu}$$

$\endgroup$
  • $\begingroup$ @jonny, Where does the formula come from? Conservation of momentum? How do you know the total momentum of the feather and the arm is two times that of arm? $\endgroup$ – most venerable sir Dec 29 '14 at 17:23
  • $\begingroup$ @user11355 from both, the law of conservation of momentum and kinetic energy, I just did it again and got the same result. I'll edit my answer $\endgroup$ – Wolphram jonny Dec 29 '14 at 17:39
  • $\begingroup$ You did it again? What is it? $\endgroup$ – most venerable sir Dec 29 '14 at 17:42
  • $\begingroup$ @user11355 I meant the derivation of the equation I used, I updated the answer. $\endgroup$ – Wolphram jonny Dec 29 '14 at 17:54
  • 1
    $\begingroup$ This post doesn't answer the main question about the reaction of the glass. $\endgroup$ – user75744 Mar 31 '15 at 9:53
2
$\begingroup$

The question as it stands is not very clear. What you describe as a "200N force" is what you would measure if your fist hit a force probe. What is in fact physical is the momentum of your fist. If you imparted some momentum to your fist and set it in motion on a collision course with the feather, then Newton's laws would predict that your hand would keep moving almost unimpeded forever until it encountered a force. The force that your hand then encounters is a pulling force because it is attached to your arm, which doesn't get any longer.

$\endgroup$
2
$\begingroup$

I read that Wolphram Jonny has already made the exact same point that I wanted to make. I will still try and present the idea in an easier to understand manner.

You say your arm can punch with 200N force. OK, so if you punch in empty space, where is all this energy going? Well it might come surprising to you, but the force that your arm exerted is absorbed by your arm itself when your punch comes to its end. It is the shock/jerk you feel when your punch stops in empty air. You feel the sense of shock in your shoulder area when your punch ends up in the empty space. So in the absence of any foreign body, you are in fact punching yourself in the shoulder. It might sound funny, but thats the truth. Keep punching in empty air with full force and gradually your shoulder will start hurting. This is due to the absorbed force of all your apparently "empty" punches.

This is exactly what happens when you punch a feather (or nearly anything lesser in mass than your arm and having little elasticity). Very little of your punch force is transferring into the feather and most of it it getting absorbed by your shoulder (your metacarpals transfer the shock to the radius and ulna which transfer it to the scapula). As a general rule, heavier bodies (e.g. your punch) cannot convey their full momentum to lighter bodies (e.g. the feather) if the lighter bodies aren't elastic enough to absorb all the incoming energy. I mean to say, if you replace the feather with a very light, but strong spring, you could convey all your punch energy to the spring, because it has the ability to absorb your full force gradually by bending in. The feather can't do that.

The same principle applies when you punch a glass with 200N. Your punch/hand will get a shock of 100N (which is the force transferred to the glass, resulting in breaking the glass) while the remaining 100N will be absorbed back by your shoulder.

To understand the underlying principle of punch force absorption by your shoulder, try punching a sand-bag 10 times and then punch in the empty air 10 times. Each time, feel which part of your arm absorbs the shock. For the sandbag, it would be your hand/punch which gets to get the shock while for the empty air, you will feel a jolt in your shoulder. If you keep punching the sandbag, your shoulders will get tired, while if you keep punching in empty air with full force, your shoulders might suffer internal injury, because you are punching your own shoulder. Interesting, isn't it?

$\endgroup$
1
$\begingroup$

I agree with Wolphram jonny's answer that even if your fist has the same momentum in both cases, the force your fist exerts on the feather will be less than the force on the glass, and in each case the force your fist exerts is matched by an equal and opposite force from the object you're punching. But I thought I'd also add a simple argument for why this should be the case. Suppose we simplify things by assuming the collision between your fist and the object is an elastic one--no kinetic energy from your fist is dispersed as heat or sound waves. In this case, by combining an equation expressing conservation of momentum with an expression expressing conservation of linear kinetic energy, it's possible to derive these equations for the case of a head-on collision where object 2 is initially at rest:

$v^{\prime}_1 = \frac{m_1 - m_2}{m_1 + m_2} v_1$

$v^{\prime}_2 = \frac{2m_1}{m_1 + m_2} v_1$

Here $m_1$ is the mass of the initially moving object, in this case your fist, while $m_2$ is the mass of the object initially at rest, the feather or glass. $v_1$ is the initial velocity of your fist before the collision, $v^{\prime}_1$ is the velocity of your fist after the collision, and $v^{\prime}_2$ is the velocity of the feather or glass after the collision. So the change in momentum of your fist would be $\Delta p_1 = m_1 (v^{\prime}_1 - v_1)$, while the change in momentum of the glass or feather would be $\Delta p_2 = m_2 v^{\prime}_2$ (A little algebra will verify that $\Delta p_2 = - \Delta p_1$ as required by conservation of momentum).

Since the change in momentum is the impulse, for the period $\Delta t$ the fist and the object are actually in contact, the average force on the object $F_{avg}$ will obey $F_{avg} \Delta t = \Delta p_2$. So, this gives us:

$F_{avg} \Delta t = \frac{2m_1 m_2}{m_1 + m_2} v_1$

During the period of an elastic collision, the material of the object being punched briefly gets compressed like a spring, then expands like a compressed spring with one end free, and that (along with your fist doing the same thing) is what causes it to fly off (see this pdf for some detailed analysis of collisions with equations). So the period $\Delta t$ between when your fist starts to contact the object and when the object springs off depends on the material the object is made of, but as a further simplification let's assume we are punching two objects made out of the same type of material, so $\Delta t$ is the same in both cases, but with the two objects having different values for the mass $m_2$. In that case, you can see from the above equation that the value of $F_{avg}$ will depend a great deal on the value of $m_2$. In the limit as $m_2$ goes to zero, $F_{avg} \Delta t$ goes to zero as well. And in the limit as $m_2$ becomes very large compared to $m_1$, $\frac{2m_1 m_2}{m_1 + m_2} v_1$ becomes very close to $\frac{2m_1 m_2}{m_2} v_1 = 2 m_1 v_1$, twice the original momentum of the fist approaching it. So, you can see that if you punch different objects with your fist having the same momentum $p_1$, $F_{avg} \Delta t$ can have any value between $0$ and $2p_1$, depending on the mass of the object you're punching.

In reality a more elastic object like feather should remain in contact with your fist for a greater time $\Delta t^{\prime}$ than the time $\Delta t$ for a more rigid object like a sheet of glass. So given that the equation indicates the average force multiplied by the time interval is greater for the glass than the feather, if the average collision force is $F^{\prime}_{avg}$ for the feather and $F_{avg}$ for the glass, we have $F_{avg} \Delta t > F^{\prime}_{avg} \Delta t^{\prime}$ and $\Delta t < \Delta t^{\prime}$, which implies that $F_{avg}$ for the glass must be greater than $F^{\prime}_{avg}$ for the feather by an even larger amount than if the time intervals were equal.

Edited to add: Bert and bobie's answers appear to deny that the third law is applicable to cases where one object breaks through another one, with Bert saying "you are right, the reaction of a (window) glass is not always equal to the action", and bobie saying " If we punch it with 200N, it cannot give an equal 'reaction'. If we call 'action' the punch, the 'reaction' of the glass cannot exceed 100N." I suppose it's possible they might just be saying that if you punched a glass with enough momentum so your fist would exert 200N if it was hitting a less breakable force meter instead of a piece of glass, then the glass will exert less than 200N on your fist, without them actually claiming that your fist was exerting 200N on the glass in this case. But I have asked both of them for clarification on this point in comments (my comments on bobie's post were apparently deleted) and neither of them have been willing to discuss it. This makes me suspect they probably are just disagreeing with the point I (and others) have made that the forces between two bodies are always equal and opposite, though the value of these equal and opposite forces can vary depending on what it is you're punching.

If they are denying that the forces are equal and opposite even for examples like an object smashing through a glass, I just wanted to add that this is definitely not the mainstream physics view. In my comments to Bert I pointed to the snippets from the solutions manual to a college physics textbook here and here (and some additional snippets which can be seen googling specific phrases) which pose the following question and answer:

A brick hits a window and breaks the glass. Since the brick breaks the glass,

(a) the force on the brick is greater than the force on the glass,

(b) the force on the brick is equal to the force on the glass,

(c) the force on the brick is smaller than the force on the glass,

(d) the force on the brick is in the same direction as the force on the glass.

Solution: According to Newton's third law, the answer is (b). Why isn't answer (c) correct since the brick breaks the glass? It takes a considerably smaller force to break a glass than to break a brick. When the brick hits the glass, they exert equal but opposite forces on each other, say 150 N. It may take only 100 N to break the glass and 1000 N to break the brick. So the glass breaks and the brick remains basically undamaged

Likewise, from the same page, there's this snippet:

There is another misconception concerning the third law. The third law states that the two forces are equal no matter what. For example, an egg and a stone collide with each other, the egg breaks and the stone is intact. Since the egg breaks, we often conclude that the force by the stone on the egg is greater than the force by the egg on the stone. This is not so. The forces are always equal. The egg breaks because it is simply easier to break.

More generally, if you go to google books and perform this search for "Newton's third law" along with the word "always", you can find a large number of textbooks saying that the forces between two interacting objects are always equal and opposite, with no exceptions noted for breakage or any other classical scenario. And note that when Bert and bobie suggest the forces would be unbalanced, they never cite any physics expert making this claim, nor do they offer any mathematical derivation of this claim or any experimental evidence to believe it's true.

$\endgroup$
0
$\begingroup$

according to what i have got about the law (maybe wrong or right): the breaking of the glass is the reaction of the glass to the excess force you have exerted on it. and you fill as much as the glass can react when it is not broken and is in touch with your hand.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.