Impact force of Rigid objects and no displacement

[normal_force]

I was wondering something, I dropped a box onto my glass desk but it didn't bounce back, I also hit a hammer against a rigid wall, the matter smacked into the wall, it has KE. It never bounced back, I didn't see any deformation or rebound in either situation, I wasn't hitting too hard nor dropped the box to hard.

However, the KE of the hammer will be turned into force, now the wall didn't move nor did the hammer cause deformation nor penetration, it didn't rebound so how do I calculate the impact force?
KE is converted to force but no displacement ore rebound, it must exert a force but how do I calculate it?

If I fire a bullet at tank armor, it absorbs the impact, you won't feel it on the other side(the impact force)...right?

How would I calculate impact force on non-penetrating projectiles? the wall must have absorbed all the force and the hammer didn't rebound...but how can we calculate the force?

 

[beamie564]

Try to use the concept of momentum change and impulse to your situation.

Edit: This needs to be posted in the correct section.

 

[normal_force]

Try to use the concept of momentum change and impulse to your situation.

Edit: This needs to be posted in the correct section.

I say, the KE converts to force, thus force is applied for a time. How do I know the impact time? what about material absorption?
Is their a situation where a bullet with 500Joules(assuming impulse lasts long enough, like .451 Seconds) that hits a target, it doesn't penetrate but the bullet exerts 500 Newtons and displaces the target by 1 meter?

 

[beamie564]

Is their a situation where a bullet with 500Joules(assuming impulse lasts long enough, like .451 Seconds) that hits a target, it doesn't penetrate but the bullet exerts 500 Newtons and displaces the target by 1 meter?

This situation does seem to be possible (theoretically at least), you can also find out the initial velocity and mass of the bullet from the given data. But whether it can be actually done, I'm uncertain. You would have to take into account the elasticity/rigidity of the target.

 

[normal_force]

This situation does seem to be possible (theoretically at least), you can also find out the initial velocity and mass of the bullet from the given data. But whether it can be actually done, I'm uncertain. You would have to take into account the elasticity/rigidity of the target.

Okay, so like a bullet of 500Joules hitting a target, it absorbs the impact, let's assume its made of that material capable of absorbing that impact, so it doesn't rebound it exerts a force on the block, its SHOULD turn 500Joules, it will push that 70kg target with 500 Newtons of force, it will displace it by one meter, transferring energy.

the time is ∆t=.529150 seconds...so doing all the equation I don't want to write here...i got...
d=1 meter
Vf=3.779644m/s
Object of 70kg's...KE: 500 Joules

 

[beamie564]

Okay, so like a bullet of 500Joules hitting a target, it absorbs the impact, let's assume its made of that material capable of absorbing that impact, so it doesn't rebound it exerts a force on the block, its SHOULD turn 500Joules, it will push that 70kg target with 500 Newtons of force, it will displace it by one meter, transferring energy.

the time is ∆t=.529150 seconds...so doing all the equation I don't want to write here...i got...
d=1 meter
Vf=3.779644m/s
Object of 70kg's...KE: 500 Joules

I'm sorry, I don't quite understand what you've done here. You want to find the force on the bullet (required to stop it) right? Then use the impulse equation delta(p)*delta(t)= F
You don't need to involve the concept of kinetic energy at all ( you can, but that'll be unnecessary).

 

[normal_force]

I'm sorry, I don't quite understand what you've done here. You want to find the force on the bullet (required to stop it) right? Then use the impulse equation delta(p)*delta(t)= F
You don't need to involve the concept of kinetic energy at all ( you can, but that'll be unnecessary).

not the force to stop the bullet but the force a bullet exerts on target...but I've gotten my answer.

 

[beamie564]

not the force to stop the bullet but the force a bullet exerts on target...but I've gotten my answer.

That's the same magnitude in the opposite direction.

 

[normal_force]

That's the same magnitude in the opposite direction.

Indeed, Newtons 3rd law, which with the KE turned into force, the force may be able to do work on material.

 

[Drakkith]

Indeed, Newtons 3rd law, which with the KE turned into force, the force may be able to do work on material.

Force and kinetic energy aren't directly related to each other in this manner. You don't even need to know the kinetic energy of an object to figure out the force applied during a collision. Indeed, it doesn't even help you in many cases since it isn't a conserved quantity (total energy is, but not kinetic energy). Also note that neither KE nor force are turned into one another. That's perhaps a bit nitpicky with the terminology on my part, but these things tend to be important when trying to understand physics.

Okay, so like a bullet of 500Joules hitting a target, it absorbs the impact, let's assume its made of that material capable of absorbing that impact, so it doesn't rebound it exerts a force on the block, its SHOULD turn 500Joules, it will push that 70kg target with 500 Newtons of force, it will displace it by one meter, transferring energy.

As I explained in your other thread, this isn't the correct way to find the force exerted on the target or bullet. Not to mention the fact that your calculations here also violate conservation of momentum just like the one in your other thread. If the answer you've gotten is based on this method, then it is wrong and will not help you in any way.

 

[normal_force]

Force and kinetic energy aren't directly related to each other in this manner. You don't even need to know the kinetic energy of an object to figure out the force applied during a collision. Indeed, it doesn't even help you in many cases since it isn't a conserved quantity (total energy is, but not kinetic energy). Also note that neither KE nor force are turned into one another. That's perhaps a bit nitpicky with the terminology on my part, but these things tend to be important when trying to understand physics.
As I explained in your other thread, this isn't the correct way to find the force exerted on the target or bullet. Not to mention the fact that your calculations here also violate conservation of momentum just like the one in your other thread. If the answer you've gotten is based on this method, then it is wrong and will not help you in any way.

Well, KE will exert a force that much is true but I was wondering for what situation a object could exert 500N's without penetrating

 

[beamie564]

Well, KE will exert a force that much is true

What we're saying is that this sentence doesn't make much sense. KE doesn't exert force, they are different concepts. You're trying to make orange juice from apples, in a manner of saying.
The idea of KE is not helpful in this case. It's not even conserved. The KE of the bullet gets converted to other forms of energy such as heat and sound, NOT to a force.

 

[normal_force]

What we're saying is that this sentence doesn't make much sense. KE doesn't exert force, they are different concepts. You're trying to make orange juice from apples, in a manner of saying.
The idea of KE is not helpful in this case. It's not even conserved. The KE of the bullet gets converted to other forms of energy such as heat and sound, NOT to a force.

I know that, the conservation of energy...and No, KE and force are not apples and oranges...but they aren't similar->object with KE when slowing down is exerting a force, thus work.

 

[Drakkith]

I know that, the conservation of energy...and No, KE and force are not apples and oranges...but they aren't similar->object with KE when slowing down is exerting a force, thus work.

What you've said here is true, but it is not very useful in understanding force, KE, and work. If you want to gain a better understanding of this whole situation you're better off just forgetting about the kinetic energy of the projectile for the moment and looking at it in terms of momentum. Since momentum is conserved, it places constraints on what can happen to the projectile and target.

Well, KE will exert a force that much is true but I was wondering for what situation a object could exert 500N's without penetrating

If we're talking about bullets, then it's not possible. A bullet exerts far more than 500 Newtons of force because the collision time is a tiny fraction of a second. You literally cannot change this without drastically changing your setup.

 

[normal_force]

What you've said here is true, but it is not very useful in understanding force, KE, and work. If you want to gain a better understanding of this whole situation you're better off just forgetting about the kinetic energy of the projectile for the moment and looking at it in terms of momentum. Since momentum is conserved, it places constraints on what can happen to the projectile and target.
If we're talking about bullets, then it's not possible. A bullet exerts far more than 500 Newtons of force because the collision time is a tiny fraction of a second. You literally cannot change this without drastically changing your setup.

That is very true, this by impact it would be greater then 500N's...

The original question is being lost but...

How does a hammer work then by energy conservation laws? I used to know all of this but now I am confused and I just can't seem to understand anything anymore

The hammer converts KE to TME, like how would that work?

 

[jbriggs444]

That is very true, this by impact it would be greater then 500N's...

The original question is being lost but...

How does a hammer work then by energy conservation laws? I used to know all of this but now I am confused and I just can't seem to understand anything anymore

The hammer converts KE to TME, like how would that work?

The initial impact of hammer with nail is inelastic. However, all of the kinetic energy is in the moving hammer and as long as the nail is much less massive than the hammer, the resulting reduction in total kinetic energy is small.

The subsequent driving of the nail into the wood is a classic example of work = force times distance. The kinetic energy of the hammer head is slowly transferred to the nail over the distance that the nail is driven into the wood. This energy transfer is given by the force of the hammer head on the nail times the distance moved by the nail. Of course, this transferred energy is lost almost immediately by the nail as the frictional force of wood on nail does very nearly equal and opposite work on the nail.

The delta between the force of hammer on nail head and the force of wood on nail shaft is just enough to bring the nail to a halt, deeper in the wood than it started.

 

[normal_force]

The initial impact of hammer with nail is inelastic. However, all of the kinetic energy is in the moving hammer and as long as the nail is much less massive than the hammer, the resulting reduction in total kinetic energy is small.

The subsequent driving of the nail into the wood is a classic example of work = force times distance. The kinetic energy of the hammer head is slowly transferred to the nail over the distance that the nail is driven into the wood. This energy transfer is given by the force of the hammer head on the nail times the distance moved by the nail. Of course, this transferred energy is lost almost immediately by the nail as the frictional force of wood on nail does very nearly equal and opposite work on the nail.

The delta between the force of hammer on nail head and the force of wood on nail shaft is just enough to bring the nail to a halt, deeper in the wood than it started.

So, YES...the KE of the hammer is turned into MECHANICAL ENERGY...THUS EXERTING A FORCE, THAT IS ONLY WAY IT CAN DO WORK, that is the only way to displace a nail...because if NOT that is a violation of Newtons 2nd Law

My point being, if a bullet deforms a little but doesn't penetrate...so it presses against the block, like when I apply my hand...Im asking, if the bullet managed to stay on the surface of the object, like if I were to apply my hand, wouldn't the force allow it do work?

 

[jbriggs444]

My point being, if a bullet deforms a little but doesn't penetrate...so it presses against the block, like when I apply my hand...Im asking, if the bullet managed to stay on the surface of the object, like if I were to apply my hand, wouldn't the force allow it do work?

No.

 

[normal_force]

No.

That makes no sense, that is a violation.

 

[jbriggs444]

That makes no sense, that is a violation.

No. It is not.

Work is computed as the product of force multiplied by distance moved [in the direction of the force] by the object to which the force is applied. If the surface does not move, no work is done on it.

 

[normal_force]

No. It is not.

Work is computed as the product of force multiplied by distance moved by the object to which the force is applied. If the surface does not move, no work is done on it.

yes, but if we calculate IMPULSE or impact force truly, then we can use dynamics, as we can gather a NET force

 

[jbriggs444]

yes, but if we calculate IMPULSE or impact force truly, then we can use dynamics, as we can gather a NET force

That is just word salad.

 

[normal_force]

That is just word salad.

ugh, okay, so we know somethings absorb force, like when I hold a hammer really tight, it barely rebounds...

okay, just...think of it like how a hammer pounds in a nail or a hand pushes a box...

 

[jbriggs444]

ugh, okay, so we know somethings absorb force, like when I hold a hammer really tight, it barely rebounds...

Right. Because your hand is absorbing the recoil -- both momentum and energy.

okay, just...think of it like how a hammer pounds in a nail or a hand pushes a box...

When a hammer pounds in a nail, the nail moves. When a hand pushes a box across the floor, the box moves. When a bullet strikes a surface without making a mark, the surface does not move. No movement = no work done.

That does not mean that there is no force. It typically means that there will be a LOT of force. But it does mean that the force does no work -- transfers no energy to the surface.

 

[normal_force]

Right. Because your hand is absorbing the recoil -- both momentum and energy.When a hammer pounds in a nail, the nail moves. When a hand pushes a box across the floor, the box moves. When a bullet strikes a surface without making a mark, the surface does not move. No movement = no work done.

That does not mean that there is no force. It typically means that there will be a LOT of force. But it does mean that the force does no work -- transfers no energy to the surface.

exactly

Im asking, if a bullet deformed a little bit but managed to stick...but it falls down after it has used up all of its energy, ifs slowing down its using its energy, thus exerting force W=Fs

 

[jbriggs444]

exactly

Im asking, if a bullet deformed a little bit but managed to stick...but it falls down after it has used up all of its energy, ifs slowing down its using its energy, thus exerting force W=Fs

In a collision with an unmoving surface, s=0 and W=0. The magnitude of F is irrelevant and cannot be computed in this manner. The energy is all used up in deformation of the bullet.

 

[normal_force]

In a collision with an unmoving surface, s=0 and W=0. The magnitude of F is irrelevant and cannot be computed in this manner. The energy is all used up in deformation of the bullet.

we have a movable block, like in the original question states...a bullet that barely deforms that pushes onto the block, the block is on a frictionless surface...so positive F net

 

[jbriggs444]

Post #1 described no such thing. There are ways to analyze such a situation. But a good starting point would be with momentum, not energy.

 

[normal_force]

Post #1 described no such thing. There are ways to analyze such a situation. But a good starting point would be with momentum, not energy.

This can solved by work-energy theorem, which is a lot better...but could calculate the force applied by momentum but its much better via work-energy theorem.

 

[jbriggs444]

This can solved by work-energy theorem, which is a lot better...but could calculate the force applied by momentum but its much better via work-energy theorem.

For the work-energy theorem to apply, you have to know how much energy is transferred and over what distance the force transferring that energy acts. Without the momentum analysis you do not know the former. Without additional details on the collision, you cannot know the latter.

 

[normal_force]

For the work-energy theorem to apply, you have to know how much energy is transferred and over what distance the force transferring that energy acts. Without the momentum analysis you do not know the former. Without additional details on the collision, you cannot know the latter.

My POINT, the round let's say which deforms slightly has like 700Joules, it doesn't penetrate the 70kg box which is place on a frictionless surface...if little deformation, no rebound...in theory, could it push the 70kg box by one foot, the bullet doesn't penetrate the box...just like how a hammer doesn't penetrate a nail..couldn't it move it?

 

[DTM]

When considering impact forces, you really cannot assume perfectly rigid bodies and assume nothing is deforming. No matter how stiff the hammer or concrete wall or glass desk, they do have finite stiffness and they will deflect. The amount of time that they deflecting while contacting will dictate the time of collision and that time can be use to calculate the impulse and from the change in momentum you can calculate the contact force. However determining that impact time is very difficult and is very much dependent on the materials and shapes of the impacting bodies. Sorry no simple answers for this one.

 

[normal_force]

When considering impact forces, you really cannot assume perfectly rigid bodies and assume nothing is deforming. No matter how stiff the hammer or concrete wall or glass desk, they do have finite stiffness and they will deflect. The amount of time that they deflecting while contacting will dictate the time of collision and that time can be use to calculate the impulse and from the change in momentum you can calculate the contact force. However determining that impact time is very difficult and is very much dependent on the materials and shapes of the impacting bodies. Sorry no simple answers for this one.

I know that, not looking for simple answers...simple response...now true...nothing is perfectly rigid...but we can use work-energy and use dynamics to solve it...but i need to find out something that really takes a simple response.

 

[jbriggs444]

I know that, not looking for simple answers...simple response...now true...nothing is perfectly rigid...but we can use work-energy and use dynamics to solve it...but i need to find out something that really takes a simple response.

You keep saying this and it remains untrue. "work-energy" cannot give you the answer you seek.

 

[normal_force]

You keep saying this and it remains untrue. "work-energy" cannot give you the answer you seek.

How so? It would only make sense, from the KE thus turned in TMEi + Wext =TMEf, thus I can find impact or Favg multiplied by time and get it there

I would ask you, how would YOU solve it.

 

[Drakkith]

I would ask you, how would YOU solve it.

You can't solve it under the conditions you've specified because they are unphysical. The answer, if we attempted to use real physical laws to find it, would be meaningless. A bullet cannot strike a target and transfer all of its kinetic energy to that target unless the collision is totally inelastic. But then the bullet and the target MUST deform and most of the KE will be lost as heat and as work done during the deformation process.

There is no idealized scenario which will change this and you can't handwave any of this away. If you use perfectly rigid objects then you can't have an inelastic collision and your bullet WILL rebound.

Nothing else is physically possible.

 

[normal_force]

You can't solve it under the conditions you've specified because they are unphysical. The answer, if we attempted to use real physical laws to find it, would be meaningless. A bullet cannot strike a target and transfer all of its kinetic energy to that target unless the collision is totally inelastic. But then the bullet and the target MUST deform and most of the KE will be lost as heat and as work done during the deformation process.

There is no idealized scenario which will change this and you can't handwave any of this away. If you use perfectly rigid objects then you can't have an inelastic collision and your bullet WILL rebound.

Nothing else is physically possible.

Okay, that is true...but I've changed the situation if you have noticed...the bullet DOES deform but not by a whole lot but it doesn't penetrate its target, it may deform it a little...but WE are saying...like how a hammer and nail work...

Like, how a baseball will do work on a baseball glove...displacing it...

Like this right here...



at 0:48 you can just skip to

and ignore the political stuff obviously...

 

[Drakkith]

Okay, that is true...but I've changed the situation if you have noticed...the bullet DOES deform but not by a whole lot but it doesn't penetrate its target, it may deform it a little...

Okay. So now I think you'd need to know how the bullet responds to stress (stiffness? elasticity?), its velocity, and the mass of both the bullet and target in order to find out how long the collision takes and what the average force is.

 

[normal_force]

Okay. So now I think you'd need to know how the bullet responds to stress (stiffness? elasticity?), its velocity, and the mass of both the bullet and target in order to find out how long the collision takes and what the average force is.

Now we are getting some where, we can apply those, I just need the equation. I also assume you saw the video?

 

[Drakkith]

Now we are getting some where, we can apply those, I just need the equation.

Wish I could help you. I don't know how to figure out and apply the elasticity/stiffness/whatever to a bullet impact.

I also assume you saw the video?

I did. What about it?

 

[A.T.]

I just need the equation.

You need numerical simulations for such deformation problems.

 

[jbriggs444]

at 0:48 you can just skip to

The money shot is at 0.53. The bullet is deformed.

 

[normal_force]

Wish I could help you. I don't know how to figure out and apply the elasticity/stiffness/whatever to a bullet impact.
I did. What about it?

hmph, there is an equation...but...also, that video shows energy transfer and such...

however, I was speaking with other people another website and they were talking about this, they said some interesting stuff that can help to this.

 

[normal_force]

You need numerical simulations for such deformation problems.

which would makes sense but I am trying to find that equation...

 

[normal_force]

The money shot is at 0.53. The bullet is deformed.

like I've corrected...the bullet is deformed, true...but energy is transferred...

 

[jbriggs444]

The energy that is transferred is a function of the force applied and the distance over which that force is applied. The distance over which the force is applied is not available from the givens of the problem (mass of bullet, mass of bullet cap, muzzle velocity of bullet). So it is difficult to use work-energy to determine the energy or velocity of the bullet+cap.

An analysis based on momentum allows you to determine the velocity of the bullet+cap immediately. That, in turn, allows you to calculate its energy.

 

[normal_force]

The energy that is transferred is a function of the force applied and the distance over which that force is applied. The distance over which the force is applied is not available from the givens of the problem (mass of bullet, mass of bullet cap, muzzle velocity of bullet). So it is difficult to use work-energy to determine the energy or velocity of the bullet+cap.

An analysis based on momentum allows you to determine the velocity of the bullet+cap immediately. That, in turn, allows you to calculate its energy.

The energy that is transferred is a function of the force applied and the distance over which that force is applied. The distance over which the force is applied is not available from the givens of the problem (mass of bullet, mass of bullet cap, muzzle velocity of bullet). So it is difficult to use work-energy to determine the energy or velocity of the bullet+cap.

An analysis based on momentum allows you to determine the velocity of the bullet+cap immediately. That, in turn, allows you to calculate its energy.

Indeed but it isn't very hard to calculate...

 

[jbriggs444]

Indeed but it isn't very hard to calculate...

Are you participating in this thread in order to learn something or in order to promote a private theory?

Edit: This thread is going nowhere and a good bit of the volume is a result of my postings. I'm out.

 

[normal_force]

Are you participating in this thread in order to learn something or in order to promote a private theory?

Im learning something, that is what I am trying to do.

 

[Drakkith]

Indeed but it isn't very hard to calculate...

It may help to start with a very simple problem. If a 10 kg ball moving at 10 m/s has a perfectly inelastic collision with a stationary 100 kg ball, do you know how to find the velocity of the two balls after the collision?