Impulse on Rubber and Metal Hammers

AI Thread Summary
The discussion focuses on the effectiveness of rubber versus metal hammers in driving nails, highlighting the elastic properties of rubber that increase contact time. While a rubber hammer requires less force for the same change in momentum, it may impart less force to the nail, potentially making it less effective for driving nails into hard surfaces. Participants debate the implications of Newton's third law and the relationship between impulse and momentum, suggesting that the rubber hammer's bounce-back could complicate its effectiveness. The conversation also touches on energy transfer, resistance forces, and the nature of impacts, emphasizing that the context of use significantly affects hammer performance. Ultimately, the consensus leans toward metal hammers being more effective for driving nails due to their ability to deliver a sharper impact.
Ballbearing101
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Homework Statement
Between a metal or wooden hammer of equal mass, which is more effective at hammering a nail into the ground.
Relevant Equations
Impulse = Ft
a=f/m
The rubber hammer is more elastic so the time of contact will increase. Ft=m(v-u)
Does this mean it is more effective as less force is needed for the same change in momentum of the hammer (and the nail) so it requires less force from you?
Or does it mean it is less effective as it imparts less force on the nail and not move it as much?
 
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Ballbearing101 said:
Homework Statement:: Between a metal or wooden hammer of equal mass, which is more effective at hammering a nail into the ground.
Relevant Equations:: Impulse = Ft
a=f/m

The rubber hammer is more elastic so the time of contact will increase. Ft=m(v-u)
Does this mean it is more effective as less force is needed for the same change in momentum of the hammer (and the nail) so it requires less force from you?
Or does it mean it is less effective as it imparts less force on the nail and not move it as much?
Rubber or wooden?
 
Ballbearing101 said:
Does this mean it is more effective as less force is needed for the same change in momentum of the hammer (and the nail) so it requires less force from you?
Or does it mean it is less effective as it imparts less force on the nail and not move it as much?
Which are you as user supplying, force or momentum?
 
berkeman said:
Rubber or wooden?
Rubber
 
Ballbearing101 said:
Rubber
Thanks for the clarification.

So what is your intuition? And given that intuition, what do you think the quantitative difference is that makes it so much more ineffective to hammer nails with a rubber hammer? :smile:
 
My intuition went to bed at 11.
 
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Ballbearing101 said:
My intuition went to bed at 11.
If it is awake now, please try to answer my question in post #3.
 
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The important points to be kept in mind:

1. Newton’s third Law will be followed. Nail will push on the hammer with same force.

2. The deformation caused in the rubber hammer would try to come at its origins shape.
 
Try to think which ball bounces back higher: rubber/tennis ball or metal ball (when dropped from terrace).
 
  • #10
Adesh said:
The important points to be kept in mind:
1. Newton’s third Law will be followed. Nail will push on the hammer with same force.
2. The deformation caused in the rubber hammer would try to come at its origins shape.
Adesh said:
Try to think which ball bounces back higher: rubber/tennis ball or metal ball (when dropped from terrace).
No, as @Ballbearing101 surmised in post #1 it is time of contact that is key. But that has to be coupled with an appreciation of what is the same for both hammers.
 
  • #11
It turns out that the physics of hammering might not be so simple after all.

I am thinking that since the rubber hammer will bounce back more it will transfer more momentum to the nail, so seems to be more effective in that way, but of course the bouncing back will make harder the life and cause arm pain to the guy that hammers with it. Where am i wrong @haruspex cause we know i am wrong otherwise people would use rubber hammer from the middle ages...
 
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  • #12
Delta2 said:
I am thinking that since the rubber hammer will bounce back more it will transfer more momentum to the nail, so seems to be more effective in that way,
I too think that, because the deformation that will be caused in the rubber hammer by the reaction force of the nail would try to come at its original shape and hence would apply force on the nail again.
 
  • #13
Adesh said:
I too think that, because the deformation that will be caused in the rubber hammer by the reaction force of the nail would try to come at its original shape and hence would apply force on the nail again.
Not sure about that, i was thinking in terms of the momentum-impulse theorem, since the momentum of the rubber hammer is say MV before it hits the nail, after the hit it will bounce back with velocity -v, so the momentum impulse theorem says that $$-Mv-MV=J$$ where J is the impulse of the force from the nail to the hammer. By Newton's 3rd law the impulse of the force from the hammer to the nail will be ##-J=M(v+V)##. The metal hammer will not bounce back rather it will continue with some smaller velocity v' towards the same direction so the impulse ##-J'=M(V-v')## (we assume before the hit metal hammer has the same momentum MV as the rubber hammer) which obviously is smaller than ##-J=M(v+V)##.
 
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  • #14
Delta2 said:
Not sure about that, i was thinking in terms of the momentum-impulse theorem, since the momentum of the rubber hammer is say MV before it hits the nail, after the hit it will bounce back with velocity -v, so the momentum impulse theorem says that $$-Mv-MV=J$$ where J is the impulse of the force from the nail to the hammer. By Newton's 3rd law the impulse of the force from the hammer to the nail will be ##-J=M(v+V)##. The metal hammer will not bounce back rather it will continue with some smaller velocity v' towards the same direction so the impulse ##-J'=M(V-v')## (we assume before the hit metal hammer has the same momentum MV as the rubber hammer) which obviously is smaller than ##-J=M(v+V)##.
As I wrote in post #10, this is not what is important. Please do not distract @Ballbearing101.
 
  • #15
haruspex said:
As I wrote in post #10, this is not what is important. Please do not distract @Ballbearing101.
You are not saying that it is wrong, just that it isn't important. I wonder what is important then but ok i ll make a private conversation with you as to not reveal too much or distract the OP.
 
  • #16
This is what evie and irene Curie must of felt like
 
  • #17
@haruspex, your first post
What would be the difference? I assume your applying the force as your wielding the hammer and the nail is providing the change in momentum as its stopping the hammer.
 
  • #18
Post #10 by @haruspex is a good hint. What thing (or things) are the same for both hammers? What thing (or things) are different about the impacts that ensue?

@haruspex has mentioned time of impact. How does that differ for the two hammers? What is the effect of that difference?
 
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  • #19
The time of impact will increase for the rubber hammer, so for the same change in momentum the force decreases. So if the force needed to change the hammers momentum is less, +considering Newtons 3rd law, therefore, the force on the nail is less.
But if the change in momentum remains the same, (m(v-u)) doesn't that mean the nail will be displaced the same distance regardless of the hammer's property.
 
  • #20
Ballbearing101 said:
if the change in momentum remains the same, (m(v-u)) doesn't that mean the nail will be displaced the same distance regardless of the hammer's property.
You are not considering what opposes the advance of the nail. We are driving it into a fixed block. To make progress we have to overcome the resistive force.
 
  • #21
How does that affect it if we're considering a closed system.
 
  • #22
Ballbearing101 said:
How does that affect it if we're considering a closed system.
The nail is being hammered into the ground, and the question relates to the relative movement of the two, so it is not a closed system.
 
  • #23
I think this is a very dubious pedagogical example. For one thing soil is often not a Newtonian fluid (nor for that matter are the fascia in wood with a nail).
Maybe one should be trying to crack a walnut instead? Just a comment..
 
  • #24
hutchphd said:
soil is often not a Newtonian fluid
Why does that make it unsuitable for such a question? I would model it in terms of static and kinetic friction, increasing somewhat as the nail is driven in.
 
  • #25
I don't understand why the discussion revolves around impulse, force and momentum. From my point of view, this is all about energy. Momentum is conserved, but kinetic energy isn't. There are two ways to look at it:
  1. Inelastic collision and coefficient of restitution
  2. Model the hammer and the ground as springs and determine the amount of elastic energy necessary to deform them (either elastic or plastic deformation). Whatever energy is used to deform one, will not be used to deform the other.
 
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  • #26
jack action said:
From my point of view, this is all about energy.
Not so; see post #24. If the peak force does not overcome the resistance of the ground the nail makes no progress. The ground will deform and spring back.
 
  • #27
haruspex said:
Not so; see post #24. If the peak force does not overcome the resistance of the ground the nail makes no progress. The ground will deform and spring back.
Speaking as someone who has swung a hammer into a nail in a soft target (unbacked board) and a hard one (board backed with plank end-on), the difference is dramatic. The effect to which @haruspex refers applies. Nobody who has ever tried it would drive a nail with a rubber hammer. You want a sharp, well-backed impact.

In my mind's eye, I see an ideal impact of hammer on spike being driven into the work as an inelastic collision of hammer head with spike followed by a period of deceleration as the spike plus hammer head move slightly into the work.
 
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  • #28
The simple model of a nail being driven into anything is zero movement until a certain force on the nail, then continuous movement as long as that (constant) force is maintained.

Try breaking down the problem like this:

Step 1: What is the force vs time (qualitative plot - hand sketch) of rubber and steel hammers hitting a rigid, immovable object. If the two hammers have equal mass and velocity, how does the area under the two curves compare? How does the time in contact compare? And the peak force? Show us a pair of sketches on the same axes.

Step 2: Apply the knowledge from Step 1 to a nail that needs a certain force to start and keep moving. Look at two possibilities:
1) The hammer bounces off the nail because the peak force is not enough to move the nail.
2) The hammer contacts the nail with enough force to move, then stays in contact while moving the nail. The force is F = ma, where m is the mass of the hammer, and a is the hammer acceleration while in contact. The force F is the force to move the nail.

Speaking as a person who has used heavy, light, hard, and soft hammers. Also sledges, splitting mauls, baseball bats, and axes.
 
  • #29
haruspex said:
hy does that make it unsuitable for such a question? I would model it in terms of static and kinetic friction, increasing somewhat as the nail is driven in.
I guess I am thinking mostly about splitting firewood with a maul or wedges. It is not at all clear to me that simple static and moving friction model makes any sense for that process. In particular the resistance force often seems to rapidly increase with the speed of the maul or wedge so there seems a "natural speed limit" because this nonlinearity. Perhaps soil is a simpler system...I've not spent much time pounding in the dirt.
Seems breaking rocks (or walnuts) would provide a more straightforward idea.
 
  • #30
jbriggs444 said:
I see an ideal impact of hammer on spike being driven into the work as an inelastic collision of hammer head with spike
I don't think inelasticity is in itself an advantage. What matters is the stiffness. Treating the hammer as a massive spring, a high k during compression means a large force. If completely inelastic then there is zero k under decompression, whereas elasticity might give a sufficiently large k to achieve a bit more movement of the nail.
hutchphd said:
splitting firewood with a maul or wedges. It is not at all clear to me that simple static and moving friction model makes any sense for that process. In particular the resistance force often seems to rapidly increase with the speed of the maul or wedge so there seems a "natural speed limit"
I would have thought the resistance would increase as the split progresses (up to a point) and the same would be true for driving a nail into the ground or into a block of wood. Thus, the threshold force to be exceeded increases with each blow. But to address the question in the thread it suffices to consider a single blow.
hutchphd said:
breaking rocks (or walnuts) would provide a more straightforward idea.
It might be a simpler problem to analyse, but it is not the one in the thread.
 
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  • #31
haruspex said:
I don't think inelasticity is in itself an advantage.
Agreed, not an advantage. But it is somewhat an inevitability for a heavy and hard hammer driving a light nail.

Even if the collision of hammer with nail is highly elastic (and brief), the result is to propel the lighter nail into the work. The hammer will follow, of course. The nail quickly stops in the work and another collision ensues. The process continues until the hammer is brought to a stop. So elastic or inelastic, the hammer winds up at a stop either way. The bulk of the hammer's initial kinetic energy winds up dissipated in the work either way.

A series of elastic collisions may save some of the heat energy that would otherwise result from the inelastic collision of hammer with nail, dissipating it in the work rather than in the nail/hammer interface. But for a heavy hammer and light nail, this is guaranteed to be a minor loss.

Agreed also that stiffness matters -- so that the force of static friction holding nail in the work is exceeded.
 
  • #32
This is conceptual. Its requiring an explanation no calculation. The topic is impulse, momentum and energy.
It says there are two answers as to what happens if you switch the hammer to rubber. 1. is its less effective as the time of contact increases. 2. is its more effective as the change in momentum increases.
I sort of get 1. In order for it to be less effective when time of contact increases. F must decrease, which means that delta p for both hammers must be considered to be equal. So it supplies less force to nail.
I don't get 2. Is the change in momentum increase just due to the time increase. I wouldn't use a rubber hammer so why is it more effective.
 
  • #33
Ballbearing101 said:
This is conceptual. Its requiring an explanation no calculation. The topic is impulse, momentum and energy.
It says there are two answers as to what happens if you switch the hammer to rubber. 1. is its less effective as the time of contact increases. 2. is its more effective as the change in momentum increases.
I sort of get 1. In order for it to be less effective when time of contact increases. F must decrease, which means that delta p for both hammers must be considered to be equal. So it supplies less force to nail.
I don't get 2. Is the change in momentum increase just due to the time increase. I wouldn't use a rubber hammer so why is it more effective.
The extra momentum change is because of the bounce. The hammer bounces back, so has a greater change in momentum.
But as I posted, this will generally be much less significant than the drawn out contact time.
 
  • #34
Ballbearing101 said:
The topic is impulse, momentum and energy.
It says there are two answers as to what happens if you switch the hammer to rubber. 1. is its less effective as the time of contact increases. 2. is its more effective as the change in momentum increases.
It is all about energy. I really don't like the choices you are given, even though 1. is the correct answer. I also prefer explaining the phenomena with springs, as it is easier to visualize. But once you see that, you can explain it with inelastic collision and the coefficient of restitution (post #25). I'll show you the concept with springs and let you relate it to the inelastic collision on you own.

As @haruspex told us before, the force of impact is determined by the force needed to deform the ground. Say it takes 100 N to deform the ground. If the impact force is less than 100 N, then the nail doesn't move. If it reaches 100 N, then the nail begins to move. Once it moves, the force required to break the ground further is still 100 N (let's assume it's constant for simplicity). How we reach that force is irrelevant, if it is 100 N, it will move.

Once it moves, it means that we have kinetic energy involve. This our goal: we want the nail to move. The less energy is transformed into kinetic energy for the nail, the less efficient the hammer will be. The amount of energy coming from the hammer is ##\frac{1}{2}mv^2## where ##m## is the same whether it is made of rubber or metal and ##v## is the velocity at the moment of impact, which we can assume to be the same in both cases (same input from the person holding the hammer).

The difference between the two materials is their stiffnesses. Because the hammers can be modeled as springs, they will deform under a force, any force. So before the force of impact reaches the required 100 N (if it does), the hammer will compress. At 10 N it will compress a little bit, at 50 N it will compress a little bit more and it will stop compressing at 100 N because then the nail will begin to move away from the hammer. And the force will never go over 100 N.

But compressing a spring requires energy. The amount of energy required is ##\frac{1}{2}Kx^2##, where ##K## is the spring stiffness and ##x## the spring displacement. But we also know that the force required to compress a spring for a displacement ##x## is ##Kx##. So the energy required for a known force ##F## (100 N in our case) is ##\frac{1}{2}Fx##. What this tells us is that for a given force acting on a spring, the greater the displacement, the greater the amount of energy is stored in the spring. The rubber hammer will have a greater displacement because it is less stiff than the metal hammer.

Now, the energy coming from the hammer (##\frac{1}{2}mv^2##) is split into 3 ways:
  1. First, the hammer deforms until the force reaches 100 N;
  2. Then, some energy is required to deform the ground (##= Fd## where ##d## is the displacement of the nail)
  3. Finally, whatever is left is transformed into kinetic energy to move the nail into the ground.
And when all the energy has been disposed of, the nail stops, and the elastic energy stored into the hammer comes back out. But at that point, the impact force will go from 100 N to 0 N as the spring (i.e. hammer) goes back to its original shape. So the nail will not get driven into the ground, and the stored energy will be converted to pure kinetic energy for the hammer. The amount of energy stored in the spring will determine how high the hammer will bounce back.

Of course, the result is that the momentum that the hammer goes back with, was not transferred to the nail (conservation of momentum). The greater momentum difference can be explain by the time taken for the impact (because it takes time to compress a spring). But momentum alone cannot explain why there is such a transfer and why the nail stops at one point. Only conservation of energy can.

Now try to explain this with the principle of inelastic collision (which is all about momentum and energy and doesn't need the concept of spring).
 
  • #35
jack action said:
It is all about energy.
Thought experiment: push nail a short way into ground; drop a pingpong ball onto the nail, and repeat that 1,000,000 times. How much energy have you expended in pingpong KE? How much further has the nail gone in?
 
  • #36
haruspex said:
Thought experiment: push nail a short way into ground; drop a pingpong ball onto the nail, and repeat that 1,000,000 times. How much energy have you expended in pingpong KE? How much further has the nail gone in?
The OP's question is about the effectiveness of a material for the hammer versus another one. That is all about energy. Less effective means that some of the input energy was not used into driving the nail.
 
  • #37
jack action said:
Less effective means that some of the input energy was not used into driving the nail.
That doesn’t make it "all about energy". That would be true if all you needed to know to determine the distance the nail advances is the energy delivered by the hammer.
You can deliver the same energy with a rubber hammer as with a steel one. The question is why the steel hammer is more effective for the same energy.
You could equally claim it is all about momentum - after allowing for the momentum that gets wasted.

The key to understanding why you need the hammer to be made of a stiff material is thinking about static friction and the force threshold that has to be exceeded.
 
  • #38
haruspex said:
You can deliver the same energy with a rubber hammer as with a steel one.
You can deliver the same [ouput] energy with a rubber hammer, but it requires more [input] energy. Hence why it is less effective. With a rubber hammer, my body does more work (energy) for the same result. The extra energy is wasted into elastically deform the rubber hammer.
haruspex said:
You could equally claim it is all about momentum - after allowing for the momentum that gets wasted.
With another type of material, the hammer could break or be permanently deformed and it wouldn't bounce back and you could even have no - or very little - wasted momentum. It would still be less effective because it has wasted energy to break or deform the material.

Why does the hammer bounces back and gets that extra momentum? Only the energy stored in the hammer while it compresses explains where this momentum comes from.
haruspex said:
The key to understanding why you need the hammer to be made of a stiff material is thinking about static friction and the force threshold that has to be exceeded.
I already explained that in post #34 and cited you as a reference, so I understand what you are saying. But this threshold force doesn't depend on the material of the hammer. It doesn't matter if the hammer is stiff or not. If it can deliver the force, the nail will move. How far will it move (work done, ##Fd##) depends on the energy transferred from the hammer to the ground. To evaluate this energy transfer, the stiffness of the hammer has an influence, as demonstrated in post #34. Heck, the stiffness of the nail is just as important too.
 
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  • #39
jack action said:
It doesn't matter if the hammer is stiff or not. If it can deliver the force, the nail will move.
If the hammer is stiff, it can deliver more force for a given impulse. If it is not, it cannot deliver as much.
 
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  • #40
jbriggs444 said:
jack action said:
It doesn't matter if the hammer is stiff or not. If it can deliver the force, the nail will move.
If the hammer is stiff, it can deliver more force for a given impulse. If it is not, it cannot deliver as much.
The only reason why a rubber hammer wouldn't deliver the same force as a steel hammer is if the rubber hammer would store all its kinetic energy while deforming (into elastic energy) before it would be able to reach the threshold force required to break the ground. But if it can reach the threshold force, it's a maximum, nothing more is needed of the hammer, except energy. This is what I meant.
 
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  • #41
jack action said:
The only reason why a rubber hammer wouldn't deliver the same force as a steel hammer is if the rubber hammer would store all its kinetic energy while deforming (into elastic energy) before it would be able to reach the threshold force required to break the ground. But if it can reach the threshold force, it's a maximum, nothing more is needed of the hammer, except energy. This is what I meant.
I am still not certain what point you are trying to make here. Even if a rubber hammer can succeed in delivering enough force during some portion of an elastic collision to cause forward acceleration by the nail this does not provide an assurance that the total amount of kinetic energy delivered to the nail during that elastic collision exceeds that from a rigid but inelastic collision.

In particular, the phrase "it's a maximum" conveys nothing to me. Exactly what thing is a maximum out of what set of things?

We can agree that if the force delivered exceeds that of static friction at some point, this does guarantee that the collision has succeeded in contributing some energy to the nail and that the nail has accordingly made some progress into the work.
 
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  • #42
Here's a video. Not sure it's all that relevant to the question at hand. But, it's fun to watch :oldsmile:

 
  • #43
If you hit a nail with a rubber mallet, all that will happen is, the rubber head will get stuck on the nail. You will drive the hammer head "around" the nail, since the rubber is soft and the nail isn't.

I have quite a few hammers, ranging from lead, rubber, plastic, rawhide, copper, brass, steel... from an ounce to 8 pounds. They all have their uses. Driving nails is not one of the rubber mallet's strong points.

I just did a little fooling around with various hammers. I am beating them on the flat surface of a vise.

My rubber mallet has a big head and the rubber is pretty hard. It does not "bounce" much, if at all. It is more of a "dead blow" tool. Another (rubber) hammer has a smaller diameter head and the rubber is softer, this one bounces a little, but not much. The steel ball-pein bounces more, a lot more.

I think the entire premise of this thread ("rubber bouncy") is off-base.
 
  • #44
jack action said:
if it can reach the threshold force
Well, that's a big if for a start. So you do accept that a stiffer hammer is better able to exceed the threshold, and if the threshold is not exceeded the nail does not advance?

Next, what happens if the threshold is exceeded, but only when the rubber is almost at maximum compression, so has nearly come to a stop? Yes, there's energy stored in the rubber, but what happens to it?
@gmax137 makes the point that we've all been assuming rubber is more elastic than steel, in the sense of returning a higher fraction of stored energy, but that may well be false. But even supposing it is true, what will happen to that energy?
The nail starts to advance, but the hammer now reaches instantaneous rest and starts to rebound. Both the small advance of the nail and the hammer's rebound reduce the compression force, making it fall back below the threshold. The remaining recoverable energy in the hammer goes into the rebound. Very little has gone into advancing the nail.
 
  • #45
gmax137 said:
If you hit a nail with a rubber mallet, all that will happen is, the rubber head will get stuck on the nail. You will drive the hammer head "around" the nail, since the rubber is soft and the nail isn't.

I have quite a few hammers, ranging from lead, rubber, plastic, rawhide, copper, brass, steel... from an ounce to 8 pounds. They all have their uses. Driving nails is not one of the rubber mallet's strong points.

I just did a little fooling around with various hammers. I am beating them on the flat surface of a vise.

My rubber mallet has a big head and the rubber is pretty hard. It does not "bounce" much, if at all. It is more of a "dead blow" tool. Another (rubber) hammer has a smaller diameter head and the rubber is softer, this one bounces a little, but not much. The steel ball-pein bounces more, a lot more.

I think the entire premise of this thread ("rubber bouncy") is off-base.
We can avoid the nail penetrating the rubber hammer by supposing a broad flat head.
But yes, we should not assume a rubber mallet bounces more than a steel one. It will depend on the nature of the rubber.

That said, I suspect whoever set the question was under the same delusion regarding relative bounce, so to appreciate the intent of the question we should compare a stiff non-bouncing hammer with a soft bouncing one.
 
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  • #46
jbriggs444 said:
In particular, the phrase "it's a maximum" conveys nothing to me. Exactly what thing is a maximum out of what set of things?
I made a mistake here. As I said in post #34, I was imagining that if the resistive force requires, say, a 100 N before the nail moves, then this would be the maximum force that will exist in the system. Even if you hit the nail "harder", the force will still be 100 N, because the resistive force does not increase when you hit "harder". I was wrong because the nail will go in faster; which means an acceleration of the nail, which means the input force must be greater than the friction force.

Still it doesn't change my analysis based on energy, where the kinetic energy of the nail is equal to the kinetic energy of the hammer less the energy required to deform the different components (the ground, the nail, the hammer). The more deformation there is, the less energy is available to move the nail.
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haruspex said:
Well, that's a big if for a start. So you do accept that a stiffer hammer is better able to exceed the threshold, and if the threshold is not exceeded the nail does not advance?
Yes.
haruspex said:
Next, what happens if the threshold is exceeded, but only when the rubber is almost at maximum compression, so has nearly come to a stop? Yes, there's energy stored in the rubber, but what happens to it?
@gmax137 makes the point that we've all been assuming rubber is more elastic than steel, in the sense of returning a higher fraction of stored energy, but that may well be false. But even supposing it is true, what will happen to that energy?
The nail starts to advance, but the hammer now reaches instantaneous rest and starts to rebound. Both the small advance of the nail and the hammer's rebound reduce the compression force, making it fall back below the threshold. The remaining recoverable energy in the hammer goes into the rebound. Very little has gone into advancing the nail.
But you are repeating exactly what I said in post #34:
jack action said:
And when all the energy has been disposed of, the nail stops, and the elastic energy stored into the hammer comes back out. But at that point, the impact force will go from 100 N to 0 N as the spring (i.e. hammer) goes back to its original shape. So the nail will not get driven into the ground, and the stored energy will be converted to pure kinetic energy for the hammer. The amount of energy stored in the spring will determine how high the hammer will bounce back.
Did anyone read it?
 
  • #47
jack action said:
But you are repeating exactly what I said in post #34:
So why do you keep saying it is "all about energy"??
And why do you write
jack action said:
. It doesn't matter if the hammer is stiff or not.
?
 
  • #48
haruspex said:
But yes, we should not assume a rubber mallet bounces more than a steel one. It will depend on the nature of the rubber.
The question of elastic versus inelastic is much ado about nothing.

If the nail is much less massive than the hammer head then the collision between the two does not waste much energy regardless. The collision energy of nail with hammer scales with the mass of the nail. It is a small fraction of the kinetic energy in the hammer head.

If the distance by which the nail advances is large compared with the deformation of nail and hammer upon impact then the loss of kinetic energy to that deformation also becomes irrelevant.

But -- if that deformation of the hammer head becomes significant relative to the advance of the nail, then that deformation, elastic or not, results in lost energy.

Stiff beats soft regardless.
 
  • #49
haruspex said:
So why do you keep saying it is "all about energy"??
Because I'm only talking about energy in my explanation.

Although, I've been waiting on your explanation for a few days now; just to see how you can explain it any other way.

I don't understand what you expect to your question in post #3 and what you will do with that answer:
haruspex said:
Which are you as user supplying, force or momentum?
(I would say kinetic energy)

Also, I'm not sure what you are trying to prove with post #20:
haruspex said:
You are not considering what opposes the advance of the nail. We are driving it into a fixed block. To make progress we have to overcome the resistive force.
After 24 posts over a period of 2 days, that is in when I chimed into bring something that would push the discussion forward a little bit.

So how do you explain that the rubber hammer is less effective than the steel one?
 
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jack action said:
Because I'm only talking about energy in my explanation.
That you are only talking about energy is not an explanation for why you are only considering energy.
jack action said:
I don't understand what you expect to your question in post #3 and what you will do with that answer:
Since we are assuming the hammers are equal mass, it doesn't matter whether you are supplying energy or momentum. I could have written either. My point was that you are not supplying force directly. But it is force that is needed to overcome static friction, so what matters in the characteristics of the hammer is how well it translates its energy/momentum into a large force, i.e. its stiffness.
jack action said:
So how do you explain that the rubber hammer is less effective than the steel one?
At the risk of repeating myself, steel wins by virtue of its stiffness.

If, as you claim, it does not matter whether the hammer is stiff, how do you explain that of two hammers with the same mass, energy and momentum, steel does better?
 
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