Understanding the Impact of Static and Sudden Forces on Pain Perception

[Hepic]

I keep a book that his weight is 10kg(100 Newton,lets say). My hand will feel a force of 100 Newtons.
Now let's say that the book is up of a desk,and with some force(lets say 60 Newtons) I hit that.

Why I will feel more pain the second time,while I feel less Newtons?
It is about the time?

 

[arildno]

Why do you feel less Newtons??

When you hit the book, you start out with a huge velocity difference between your hand and the book.
That velocity difference must be reduced to 0 in a fraction of a second, i.e, a huge force couple is generated between your hand and the book.

 

[Hepic]

Ι feel less Newtons,because I hit the book with less Newtons,but I feel more pain from when I keep the book(that weight 10 kg)

 

[Drakkith]

I'm having a very difficult time understanding you. Is the following correct?

You hit a 10 kg book with your hand and feel some pain. I assume it's in the air or upright so it just flies off after you hit it.
You then hit the same book with less force, but this time it's lying on a desk. You feel a greater amount of pain.

Is that correct?

 

[Hepic]

Nope. See
1)You just keep a book that weight 10 kg,so your hand feel 100 Newtons.
2)You leave that book somewhere,and you hit this book with force of 60 Newtons.

My hand will feel more Newtons at 1,but more pain at 2.

 

[Drakkith]

Nope. See
1)You just keep a book that weight 10 kg,so your hand feel 100 Newtons.
2)You leave that book somewhere,and you hit this book with force of 60 Newtons.

My hand will feel more Newtons at 1,but more pain at 2.

You don't hit the book with 60 Newtons of force. You hit it with a certain velocity and the maximum force will depend on how quickly your hand decelerates. The force builds from zero to maximum before falling to zero again over the course of the deceleration time.

 

[sophiecentaur]

Nope. See
1)You just keep a book that weight 10 kg,so your hand feel 100 Newtons.
2)You leave that book somewhere,and you hit this book with force of 60 Newtons.

My hand will feel more Newtons at 1,but more pain at 2.

Your thought experiment is not providing you with valid results, I'm afraid - presented as it is.

It is not the 'Force' that counts in the context of collisions. People are always asking about the Force of an impact (vehicle collisions and injuries) but the damage done will depend upon the Kinetic Energy and Momentum involved rather than just the Force involved.
When you apply 100N to keep the book suspended then the force is defined and you have an equilibrium situation. If the book is suspended in space and you 'hit it' with a force that you say is 60N, is this an instantaneous force? Is it applied for 1s, 1ms or 1μs? Whatever time it's applied for, if it's only 60N then there will be less deformation / damage / pain to your hand, simply because the force is less. If you have the book on Earth and can only give it an upwards force with your hand of 60N, then its weight will actually be pushing (accelerating) your hand towards the ground.

If you hit a sharp corner of the book with your hand then the local pressure can give more pain than when you support the book on your palm - but that's another layer of complication.

 

[arildno]

Hepic:
Your hand starts out with velocity, ending with velocity 0 in time interval "dt".

thus, the AVERAGE force experienced by it is mV/(dt), which can be very huge, if dt is tiny, V "ordinary".

 

[Hepic]

So to undestand. Does not matter only the force,but how quicly you hit that.
Right?

 

[arildno]

So to undestand. Does not matter only the force,but how quicly you hit that.
Right?

Not at all!
The (average) force CANNOT be deduced as you believe it can.

You can estimate it as the ratio between the momentum change and the time interval over which the momentum change happened.

 

[sophiecentaur]

Not at all!
The (average) force CANNOT be deduced as you believe it can.

You can estimate it as the ratio between the momentum change and the time interval over which the momentum change happened.

Hepic still appears to have his (her?) original problem of misplaced intuition. It would be possible to apply a force, as in the original model (say, with a small calibrated rocket jet, applied for some known time) but that is not the case with a hand hitting the book. You could even measure the force during a real impact but it would not be constant over the time of impact and would be only one of the factors involved in determining what would be happening.

Vehicle insurance claims would be a very different thing if it were as simple as Hepic is suggesting.

 

[arildno]

Hepic still appears to have his (her?) original problem of misplaced intuition. It would be possible to apply a force, as in the original model (say, with a small calibrated rocket jet, applied for some known time) but that is not the case with a hand hitting the book. You could even measure the force during a real impact but it would not be constant over the time of impact and would be only one of the factors involved in determining what would be happening.

Vehicle insurance claims would be a very different thing if it were as simple as Hepic is suggesting.

True enough.
But understanding how the AVERAGE force can be found (and why it can be extremely huge) should dispel his idea that the magnitude of the force somehow could be calculated from the weight of the object.

 

[sophiecentaur]

I think you mean that the peak force can be huge. An average force would presumably refer to the change of momentum divided by the duration of the contact. I can only hope that Hepic will follow this in a more formal and less intuitive way. It's the only way to get a proper understanding of this topic.

 

[arildno]

The average force can be huge as well, not just the peak.
Is mV/dt a huge number, the average force was huge (and the peak force even huger)
And "change in momentum"/"time interval" IS the formal definition of average force over that time interval.

 

[ZapperZ]

So to undestand. Does not matter only the force,but how quicly you hit that.
Right?

Please look up the concept of "Impulse".

Zz.

 

[sophiecentaur]

But aren't we starting off with a 60N scenario? That certainly limits the 'Average' force - and it could even be the limit for the Maximum force.
And what is the time interval? Where do you actually say that the interaction of hand and book starts and finishes? What's the time profile of the force? My point is that the quantity Impulse is the relevant one, because it contains most of what's relevant and measurable in a situation like this.
[Edit: You beat me to the term Impulse, Z]

 

[arildno]

"My point is that the quantity Impulse is the relevant one"
--
Quibble.
Since the impulse always equals the change in momentum.
We have:
$$\vec{I}=\int_{t_{0}}^{t_{1}}\vec{F}dt=\bigtriangleup\vec{p}$$
Average force equals:
$$\vec{F}_{av}\equiv\frac{1}{\bigtriangleup{t}}\int_{t_{0}}^{t_{1}}\vec{F}dt=\frac{\bigtriangleup\vec{p}}{\bigtriangleup{t}}$$

And, this is by far the most practical way to find the average force, rather than demanding to use the instantaneous force distribution over the time interval.
We don't know that distribution, nor do we need it, in order to compute the average force.

 

[arildno]

OOPS! It seems that I accidentally in one post used "velocity" rather than "momentum".
Damn!

 

[sophiecentaur]

But does 'average force' matter? Is it of interest to anyone? It doesn't indicate the 'pain' or damage involved. It doesn't even represent anything about the Work done or energy input.

 

[arildno]

But does 'average force' matter? Is it of interest to anyone? It doesn't indicate the 'pain' or damage involved. It doesn't even represent anything about the Work done or energy input.

It sure matters.
Because OP seems hung up on measuring the forces involved in a particular situation, and does that in a totally skewed manner.

Therefore, it DOES matter to point out to him WHAT sort of force he actually might calculate in a correct manner for a collision, and also to show how he is missing orders of magnitude of those forces by thinking they somehow are comparable to the object's weight.

Beyond that, I certainly agree with you that they aren't particularly important.

 

[sophiecentaur]

My problem with even considering any 'average' force is that the same effect can be achieved with a single, short burst of high force (with a hammer), which could be many times the weight of the object or a long acting small force (using a bungee) which would involve a very small force - much less than the weight. Exactly the same 'average' force in each case, but very different maximum forces / damage / pain.
I'm not sure whether you are just playing devil's advocate, here. You may be. But it could encourage the OP to maintain that inappropriate approach. And we both agree that it is not appropriate.
Sometimes, 'Average' tells you very little about a situation.

 

[arildno]

Well, I've had enough with your snide sarcasms against me, with your nonsensical, malevolent outbursts against me personally.
You should apologize.

For example:
" which could be many times the weight of the object or a long acting small force"

Here, you are insinuating that I did NOT write:
"also to show how he is missing orders of magnitude of those forces"

the premise has been a COLLISION.
You know that, and you ALSO know that knowing a typical average force during a collision would dispel the idea of it being comparable to the object's weight.
Instead, despicably, you insinuate I don't know that bungee jumping might occur in the world, as well.

 

[sophiecentaur]

I apologise unconditionally if you have taken offence. It was not my intention so calm down, friend. I cannot actually see any "sarcasm" or personal attack in what I wrote. I am not even sure whether or not you are disagreeing with me, basically.
I possibly misunderstood your language, which was why I just spelled out the distinction between short and long term collisions. "Missing orders of magnitude" was bit unspecific and I couldn't understand why, if you meant the same as I was saying (and perhaps you were were) that you were still in favour of the notion of 'average force'. If you really think that Average force is a worthwhile idea then I have to disagree with you.

 

[arildno]

"If you really think that Average force is a worthwhile idea then I have to disagree with you. "
It sure is. As the most easily calculable lower bound for the peak force during a collision. Which is, on its own, entirely sufficient to dispel the idea that the forces involved in the described collision by OP has comparable magnitude to the object's weight.

Beyond that? Not very worthwhile at all.

 

[sophiecentaur]

"If you really think that Average force is a worthwhile idea then I have to disagree with you. "
It sure is. As the most easily calculable lower bound for the peak force during a collision. Which is entirely sufficient to dispel the idea that the forces involved in the described collision by OP has no comparable magnitude to the object's weight.

Beyond that? Not very worthwhile at all.

Only if you know about the time for the collision. That's my problem and the whole point of my contribution to this thread. Merely by appearing to think it's worth while, you could be giving the OP what I would call the wrong idea.

(No response to my apology then? Can you still be sore?)

 

[arildno]

No, if you DON'T know about the duration of collision, there does not exist any more readily available quantities of force, either. (Since, for example, the instantaneous force distribution is NOT a readily available quantity in any collision).
So, the average force IS, in all cases, THE MOST easily calculable force within the prob.
But not necessarily easy, or possible to calculate that, either. And I never said it was.

Furthermore, as for collision time interval:
If you have a rough estimate of that time interval, your calculated av. force, relative to the actual average force won't be off by more than by the error bound for the time, either.
--
PS:
If you don't know whether you are bungee jumping or just hitting a book lying on a table, your answers won't count much, anyway.

 

[sophiecentaur]

No, if you DON'T know about the duration of collision, there does not exist any more readily available quantities of force, either. (Since, for example, the instantaneous force distribution is NOT a readily available quantity in any collision).
So, the average force IS, in all cases, THE MOST easily calculable force within the prob.
But not necessarily easy, or possible to calculate that, either. And I never said it was.

Furthermore, as for collision time:
If you have a rough estimate of that time, your calculated av. force, relative to the actual average force won't be off by much more than by the error bound for the time, either.

How has that crept into the sentence? There is no available value for the force without knowledge of the time. The "error bound" for the force is way worse than the 'error bound' for the observable duration (without the help of a high speed slo-mo camera, perhaps, to look at distortions in the object). It's an irrelevance in most cases. You could at least try to make that very clear to the OP.

 

[arildno]

"The "error bound" for the force is way worse than the 'error bound' for the observable duration"

Not for the AVERAGE force, no.

The average force, with error bound is given by (since we know intial and final momentum):
$$\vec{F}_{av}=\frac{\bigtriangleup{\vec{p}}}{\bigtriangleup{t}(1 \pm\epsilon)}\approx\frac{\bigtriangleup{\vec{p}}}{\bigtriangleup{t}}(1 \mp\epsilon)$$
i.e, if the relative error in the observed duration is fairly small, the relative error in the average force is of the same magnitude.

 

[sophiecentaur]

That assumes the force is remotely likely to be constant. It would more probably follow some complicated law - even for a rectangular object, end on. You would need to know the dynamic distortions before you could have a hope.
I admit, if it was absolutely vital to know the maximum forces in a collision, and there were no alternative to this method, you could get inventive with it. But where is all this getting you? You would end up with a 'number', is all.
Could we just let this one die?

 

[arildno]

"That assumes the force is remotely likely to be constant. "
No it does not.
It is strictly correct for the average force, for whatever type of force distribution you like.
We have:
$$\vec{F}_{av}=\frac{1}{\bigtriangleup{t}(1\pm\epsilon)} \int_{t_{0}}^{t_{0}+\bigtriangleup{t}(1\pm\epsilon)} \vec{F}dt=\frac{\bigtriangleup\vec{p}}{\bigtriangleup{t}(1\pm\epsilon)}$$
No assumptions whatsoever has beein laid upon the variations of the instantaneous force here.

 

[sophiecentaur]

The average force is just a number. It has absolutely no relevance to anything. The above Maths looks right (apart from the Delta sign which doest come out on my screen) but what has that got to do with the OP, or anything? I'm not sure what you are trying to justify or prove, any more. I am merely anxious that the OP would not get the impression that his original ideas were a suitable way into 'collisions'. Could you even suggest an occasion when the average force could be of use?

 

[arildno]

Being the most eaily calculable force within an extremely complicated collision scenario makes it conceptually useful for establishing a lower bound for the magnitudes of forces involved.
Why should it be less useful than the value of the non-obtainable value of the peak force, for example?

 

[sophiecentaur]

I have similar reservations about the peak force, too. But at least the peak force gives an idea of the strength of structure needed. I'm not sure that "it will need to be at least as strong as xyz" would be a sound way to start an engineering design. I guess it would be useful in deciding whether to reject a design out of hand, though.

Have you not read the plethora of posts which ask "What was the force of collision when my car was hit by my neighbour's Ferrari?" This was a similar question and it doesn't come from any relatively well informed direction. It comes from the direction of intuition, which is why I have been banging on in this way.

 

[Dale]

I keep a book that his weight is 10kg(100 Newton,lets say). My hand will feel a force of 100 Newtons.
Now let's say that the book is up of a desk,and with some force(lets say 60 Newtons) I hit that.

Why I will feel more pain the second time,while I feel less Newtons?
It is about the time?

If you do not feel pain at 100 N then you won't feel pain at 60 N (assuming same contact area).

 

[russ_watters]

But does 'average force' matter? Is it of interest to anyone? It doesn't indicate the 'pain' or damage involved. It doesn't even represent anything about the Work done or energy input.

Yes, average force times distance traveled in decelerating during the collision equals work or energy absorbed.

But the OP just says 60N. Given the level of understanding that the OP has, I'd expect that to be maximum force. And it is what it is; it is less than 100N, so the feeling of the impact is less than the feeling of lifting the book (neither would involve pain, as Dale said). That said, I'm sure the OP is thinking of hitting the book hard enough to feel pain, which would be much more than a 60N force. There is probably a conceptual error there.

My problem with even considering any 'average' force is that the same effect can be achieved with a single, short burst of high force (with a hammer), which could be many times the weight of the object or a long acting small force (using a bungee) which would involve a very small force - much less than the weight. Exactly the same 'average' force in each case, but very different maximum forces / damage / pain.

That doesn't compute. Obviously, large force is larger than a small force; the averages are not the same. The energies or impulses might be, but impulse and energy aren't force (I know you know that).

For example, if you drop a brick from a meter onto a sidewalk, the average force it applies during the collision is vastly larger than the average force applied if you drop it on a thick carpet even though the energy absorbed is the same.