How Do Forces Cause Motion?

[david smith]

Dear all

When I first learned about forces and motion I was taught F=ma in terms of apply a force to a mass and the acceleration of that mass = x. I was quite clear about how something moved ie apply a force to a mass - it moves or tends to move.

But then I came onto the D' Alembert variation of F-ma=0 ie everything in equilibrium.
Very useful it is too.

In terms of F= ma so ma = F and F-F = 0. So the I thought well how can things move then? I was told by many that the insertion of inertia into the equation so that F-ma = 0 is just a mathematical trick and useful for resolving forces and moments in a mechanical system.

However to me at least inertia is real and easily proved. You can't apply a force unless an equivalent force pushes back.

So I decided that motion is caused by the transfer of kinetic energy of momentum in the direction of the force applied. In terms of conservation of momentum, so that even though there is equal and opposite inertial force there is not equal and opposite momentum and the motion carries on in the direction of the force.

But I can't quite resolve that concept! Since the force applied to the mass does work on the mass as it accelerates therefore (in my mind) the inertial force must do equal and opposite (negative) work to the applied force. As Ek = 1/2mv^2 and work = fd theta and this becomes W = 1/2mv^2theta then Ek of applied force should equal negative Ek of inertial force.

So how do things move??

Can anyone explain? have I got my concepts mixed up?


Thanks Dave Smith

 

[Andy Resnick]

A rock climber can ascend narrow chimneys using a standard method. The climber can even rest in place partially up the chimney. At rest, the climber is in equilibrium. Does that mean when at rest, there are no forces?

 

[Mapes]

Hi Dave, welcome to PF. One serious problem is that your terms "kinetic energy of momentum" and "Ek of applied force" are very difficult to understand. Energy, momentum, and force are all very different, and you probably would not say something like "length of temperature" or "height of voltage."

To use your example, when A accelerates B, it does work on B and increases B's energy. The energy of A is decreased, so we could say that B does an equal amount of work on A, except opposite in sign. (I find the term "negative work" confusing.) So total energy remains constant. I'm not sure why you think you can set the two values (one positive, one negative) equal to each other ("Ek of applied force should equal negative Ek of inertial force.")

Edit: Oh, I think I see what you meant. Is it that the energy change in B is, say, 5 J; the energy change in A is -5 J; and 5 = -(-5)?

 

[david smith]

A rock climber can ascend narrow chimneys using a standard method. The climber can even rest in place partially up the chimney. At rest, the climber is in equilibrium. Does that mean when at rest, there are no forces?

Andy

Yes there are forces at work and they are in equilibrium

Dave

 

[david smith]

Hi Dave, welcome to PF. One serious problem is that your terms "kinetic energy of momentum" and "Ek of applied force" are very difficult to understand. Energy, momentum, and force are all very different, and you probably would not say something like "length of temperature" or "height of voltage."

To use your example, when A accelerates B, it does work on B and increases B's energy. The energy of A is decreased, so we could say that B does an equal amount of work on A, except opposite in sign. (I find the term "negative work" confusing.) So total energy remains constant. I'm not sure why you think you can set the two values (one positive, one negative) equal to each other ("Ek of applied force should equal negative Ek of inertial force.")

Mapes

Its quite possible I am getting confused here but, I can't imagine a way length an temperature are linked however energy and momentum and force: Kinetic energy = 1/2 mv^2 and momentum = mv so kinetic energy requires momentum. The applied force does not exist on its own it requires a mass + an acceleration so acceleration = d/t^2 or a change in momentum so force = mass * change in momentum. Therefore a force = change of momentum = change in Kinetic energy. And so, in my mind, I can (perhaps loosely) say the energy of momentum and the Ek of force.

So how do you explain motion in terms of change from a state of rest or constant velocity.

To use your example, when A accelerates B, it does work on B and increases B's energy. The energy of A is decreased, so we could say that B does an equal amount of work on A, except opposite in sign.

I get that, so are you saying the forces are equal and it is the exchange of energy that causes motion?


I don't require a definition as I think I know all the definitions. Don't get me wrong I can and do work with all the equations of Newtonian mechanics (even if the above doesn't sound like it). I just suspend my disbelief and do the maths.

Thanks for your help, Dave Smith

 

[Andy Resnick]

David,

Let me back up: there's a couple of places you say something like "the origin of an applied force is the acceleration of some object". Is that what you mean?

 

[david smith]

David,

Let me back up: there's a couple of places you say something like "the origin of an applied force is the acceleration of some object". Is that what you mean?

Andy

Yes both, if there is a force there must be an acceleration (or tendency to acelerate) and you can't apply a force unless you have another force to resist it ie 2 accelerations in opposite directions. EG you can't jump in the air if you don't have the ground to push off.

Cheers Dave

 

[Andy Resnick]

Ah. Now I understand your confusion.

The formula F = ma (or F = dp/dt) is, for various reasons, very subtle. First, it's important to realize that we never measure forces. There's no such thing as a 'force-o-meter'. This is not the case for acceleration (or velocity, or position)- we have rulers and stopwatches to measure these things.

There may be objections to this- surely a spring measures force? Or a strain gauge? Or a pressure gauge? None of those things really do, they rely on constitutive (material properties) relationships which relate a force (dynamic quantity) to a displacement (kinematic quantity).

F=ma conceals this important point- it sets equal a *kinematic* quantity (acceleration), with a *dynamic* quantity (force). We do not ultimately know what causes a force. We have lots of models, to be sure, in as much detail and complexity as you can stomach. But in the end, we still have no force-o-meter.

If you like, you may think of 'mass' as being a constant of proportionality between an applied force (whatever the origin of said force) and a resultant motion- but that has limits as well (massless particles). Consequently, it may be helpful to spend time temporarily expunging 'force' from your mental concepts and instead think in terms of momentum and energy.

 

[russ_watters]

I didn't read the entire thread, but I'm not seeing in the first half of the OP anything to be misunderstood. f=ma means f=ma. Both forces are represented in the equation: the left side is the force applied to the object and the right side is the equal and opposite force applied back.

Re-arranging the equation to f-ma=0 doesn't change anything. 0 isn't the acceleration, it is the sum of the forces.

But then you repeat your question without actually connecting it to the math you just showed, so it isn't clear what you don't like about this.

Anyway, though, by throwing energy in there in the second half of the post, you are just confusing yourself. If you ever do sort out the math of it, you'll find it just reduces back to f=ma anyway. So just drop that entire thing and just get your arms around the idea that f=ma is a true and complete statement of how a force results in acceleration.

 

[russ_watters]

Hmm, maybe this will help. In order to apply a force to an object and not have it move, you need two force pairs on it, not one. If you have an object bolted to the ground and you push on it, you and the object exchange a force and the object and the bolt exchange a force. The forces sum to zero, with two positive and two negative.

If you have only one force pair, one of those forces manifests itself as ma in order to maintain the balance. If you start with F1=F2 and know that there is no corresponding force pair to stop the acceleration, then F2 must be a force due to inertia and acceleration. F2=ma.

Physicists often treat gravity as equivalent to acceleration, so the same principle applies to the unmoving climber on the side of the mountan. The force pair between the climber and mountain is F1=F2 and F2 still equals ma.

 

[Mapes]

I enjoyed reading Andy's answer. As supplement, Greenwood says this in Principles of Dynamics:

"The concept of force as a fundamental quantity in the study of mechanics has been criticized by various scientists and philosophers of science from shortly after Newton's enunciation of the laws of motion until the present time. Briefly, the idea of a force, and a field force in particular [Greenwood defines field forces as those involved with action at a distance], was considered to be an intellectual construction which has no real existence. It is merely another name for the product of mass and acceleration which occurs in the mathematics of solving a problem. Furthermore, the idea of force as a cause of motion should be discarded since the assumed cause and effect relationships cannot be proved." (boldface mine)

 

[rcgldr]

Equal and opposite forces when an object is accelerated.

Reaction forces are the result of acceleration times inertia (inertia equals mass in the case of translational acceleration). For angular movement, you have reaction torque = angular acceleration times angular inertia. Reaction forces or torques don't stop acceleration, they are the response to acceleration. Without acceleration, reaction forces or torques don't exist.

When calculating accelerations, reaction forces are not included.

Say a pushing force is applied to a block on a frictionless surface, the block accelerates according to the equation, A = F/M. The block has inertia, and resists acceleration with an equal and opposite "reaction" force that exactly equals the force that is producing the acceleration (F = MA) but this determines the rate of acceleration, it doesn't cancel the acceleration.

There may be objections to this- surely a spring measures force?

Close enough for me. If the object involved has a known stress / strain curve, then the change in length of the object under tension or compression could measured be used to calculate force. Tension could be measured by the vibration rate, for example the strings on a piano or guitar, higher tension => higher pitch.

 

[Cyrus]

The D'almebert "force" is not a real force, you treat ma as if it were a force - but its not..

 

[Cyrus]

Ah. Now I understand your confusion.

The formula F = ma (or F = dp/dt) is, for various reasons, very subtle. First, it's important to realize that we never measure forces. There's no such thing as a 'force-o-meter'. This is not the case for acceleration (or velocity, or position)- we have rulers and stopwatches to measure these things.

There may be objections to this- surely a spring measures force? Or a strain gauge? Or a pressure gauge? None of those things really do, they rely on constitutive (material properties) relationships which relate a force (dynamic quantity) to a displacement (kinematic quantity).

F=ma conceals this important point- it sets equal a *kinematic* quantity (acceleration), with a *dynamic* quantity (force). We do not ultimately know what causes a force. We have lots of models, to be sure, in as much detail and complexity as you can stomach. But in the end, we still have no force-o-meter.

If you like, you may think of 'mass' as being a constant of proportionality between an applied force (whatever the origin of said force) and a resultant motion- but that has limits as well (massless particles). Consequently, it may be helpful to spend time temporarily expunging 'force' from your mental concepts and instead think in terms of momentum and energy.

I enjoyed reading this post, nice.

 

[Andy Resnick]

<snip>

Close enough for me. If the object involved has a known stress / strain curve, then the change in length of the object under tension or compression could measured be used to calculate force. Tension could be measured by the vibration rate, for example the strings on a piano or guitar, higher tension => higher pitch.

Yes, but do you see what you did? You switched from talking about 'force' (Newton) to STRESS (Cauchy).

IMHO, Cauchy's first law is much more relevant that Newton's second law, for this very reason. We can indeed measure stress- pressure and strain gauges. This is not measuring a force! Applying a Newtonian force to a pressure gauge will give different results, depending on the geometry of the pressure gauge.

 

[david smith]

All, Russ, Jeff, Mapes, Andy, Cyrus, thakyou for you time and effort and your input.

Aha! this is more like it, these are the types of answers I've been looking for.

Let me explain, I work as a Podiatrist and deal with pathological problems of gait and posture in terms of biomechanics. Therefore I am dealing with kinetics and kinematics -forces and movement.

In my degree we were always taught that 'forces and moments in a mechanical system , whether stationary or in motion,are always in equilibrium' IE D'Alemberts version of Newtons law F=ma => F-ma=0.

(Russ

Re-arranging the equation to f-ma=0 doesn't change anything. 0 isn't the acceleration, it is the sum of the forces.

- I know zero is in terms of balance of forces and moments and not acceleration.)

So my quandry has always been if the forces and moments are equal on each side of the equation how can there be motion?

The usual answer are:

1A: Just accept it (My reply: Can't) Russ wrote-

So just drop that entire thing and just get your arms around the idea that f=ma is a true and complete statement of how a force results in acceleration.

Maybe I should Russ but isn't it a complete statement of how we explain acceleration in terms of force?

Russ

If you have only one force pair, one of those forces manifests itself as ma in order to maintain the balance. If you start with F1=F2 and know that there is no corresponding force pair to stop the acceleration, then F2 must be a force due to inertia and acceleration. F2=ma. Physicists often treat gravity as equivalent to acceleration, so the same principle applies to the unmoving climber on the side of the mountan. The force pair between the climber and mountain is F1=F2 and F2 still equals ma.

confused! - so are you agreeing with me then, ie that forces are always in equilibrium and do not cause motion??


2A: Inertia is an imaginary force, Cyrus wrote-

The D'almebert "force" is not a real force, you treat ma as if it were a force - but its not..

(My reply: seems real enough to me -EG try punching a wrecking ball hanging by its chain and see how the inertial force breaks your hand)

Jeff wrote

Reaction forces are the result of acceleration times inertia (inertia equals mass in the case of translational acceleration). For angular movement, you have reaction torque = angular acceleration times angular inertia. Reaction forces or torques don't stop acceleration, they are the response to acceleration. Without acceleration, reaction forces or torques don't exist.

So there is inertia and there are reaction FORCES

Jeff

When calculating accelerations, reaction forces are not included.

Oh! so there aren't any reaction forces, which is it? Is it valid (even if useful) to ignore a parameter just for convenience?

Jeff

Say a pushing force is applied to a block on a frictionless surface, the block accelerates according to the equation, A = F/M. The block has inertia, and resists acceleration with an equal and opposite "reaction" force that exactly equals the force that is producing the acceleration (F = MA) but this determines the rate of acceleration, it doesn't cancel the acceleration.

Exactly! so therefore if the forces can't cause the accleration, what does? (by your argument one ignores the inertial forces when calculating the acceleration, so then there is the inertia of the negative acceleration of the mass applying a force and the force of inertia of the positive acceleration of the mass resisting the applied force. Ignore the inertial forces, which is both sides of the equation ma = ma, and you get acceleration. How?


3A: Don't mix force theory with energetics theory. (My reply: Why?, isn't momentum and energy transfer directlly related to force? You can't transfer kinetic energy without applying a force and potential energy only becomes kinetic with the application of a force.)

Rus wrote

Anyway, though, by throwing energy in there in the second half of the post, you are just confusing yourself.

I'm definintely confused.!

Andy wrote

If you like, you may think of 'mass' as being a constant of proportionality between an applied force (whatever the origin of said force) and a resultant motion- but that has limits as well (massless particles). Consequently, it may be helpful to spend time temporarily expunging 'force' from your mental concepts and instead think in terms of momentum and energy.

Difficult, but you may be right.

Ah. Now I understand your confusion.

The formula F = ma (or F = dp/dt) is, for various reasons, very subtle. First, it's important to realize that we never measure forces. There's no such thing as a 'force-o-meter'. This is not the case for acceleration (or velocity, or position)- we have rulers and stopwatches to measure these things.

Good concept?

Andy wrote

There may be objections to this- surely a spring measures force? Or a strain gauge? Or a pressure gauge? None of those things really do, they rely on constitutive (material properties) relationships which relate a force (dynamic quantity) to a displacement (kinematic quantity).

F=ma conceals this important point- it sets equal a *kinematic* quantity (acceleration), with a *dynamic* quantity (force). We do not ultimately know what causes a force. We have lots of models, to be sure, in as much detail and complexity as you can stomach. But in the end, we still have no force-o-meter.

Not so difficult now, if I intuitively consider your premise above.


Mapes wrote

"The concept of force as a fundamental quantity in the study of mechanics has been criticized by various scientists and philosophers of science from shortly after Newton's enunciation of the laws of motion until the present time. Briefly, the idea of a force, and a field force in particular [Greenwood defines field forces as those involved with action at a distance], was considered to be an intellectual construction which has no real existence. It is merely another name for the product of mass and acceleration which occurs in the mathematics of solving a problem.

Can yougive some references please Mapes

By this premise and Andy's above my question becomes a mute point. "considered to be an intellectual construction which has no real existence". If force is a construct for, or a byproduct of, the calculation of other parameters of kinematics, then force is imaginary and cannot cause motion. So what does?


Mapes

the idea of force as a cause of motion should be discarded since the assumed cause and effect relationships cannot be proved.

Tell me more, Tell me more! Please Please!

This is all good for me

Russ

Hmm, maybe this will help. In order to apply a force to an object and not have it move, you need two force pairs on it, not one. If you have an object bolted to the ground and you push on it, you and the object exchange a force and the object and the bolt exchange a force. The forces sum to zero, with two positive and two negative.

If you have only one force pair, one of those forces manifests itself as ma in order to maintain the balance. If you start with F1=F2 and know that there is no corresponding force pair to stop the acceleration, then F2 must be a force due to inertia and acceleration. F2=ma.

Re reading this I see what you are staying (i think) So if I push against someone who is pushing back we each have Inertial forces (1 pair) and apllied forces (2 pair). Remove one of the applied force and we have motion. This means we have unbalanced forces and so one inertial force must accelerate to give us the balanced equation. Hmmm! Have I got that right. Not heard that concept before, I'll have to think about that one

Cheers all Dave

 

[regor60]

Dear all

When I first learned about forces and motion I was taught F=ma in terms of apply a force to a mass and the acceleration of that mass = x. I was quite clear about how something moved ie apply a force to a mass - it moves or tends to move.

But then I came onto the D' Alembert variation of F-ma=0 ie everything in equilibrium.
Very useful it is too.

In terms of F= ma so ma = F and F-F = 0. So the I thought well how can things move then? I was told by many that the insertion of inertia into the equation so that F-ma = 0 is just a mathematical trick and useful for resolving forces and moments in a mechanical system.

However to me at least inertia is real and easily proved. You can't apply a force unless an equivalent force pushes back.

So I decided that motion is caused by the transfer of kinetic energy of momentum in the direction of the force applied. In terms of conservation of momentum, so that even though there is equal and opposite inertial force there is not equal and opposite momentum and the motion carries on in the direction of the force.

But I can't quite resolve that concept! Since the force applied to the mass does work on the mass as it accelerates therefore (in my mind) the inertial force must do equal and opposite (negative) work to the applied force. As Ek = 1/2mv^2 and work = fd theta and this becomes W = 1/2mv^2theta then Ek of applied force should equal negative Ek of inertial force.

So how do things move??

Can anyone explain? have I got my concepts mixed up?


Thanks Dave Smith

F is an external force on the object, ma is not an external force...

 

[Cyrus]

This is your problem, you are simply looking at F-ma=0 and saying, it equals zero, so it must be in equilibrium. But now you are assuming there is some OTHER m'a'=0.

in other words,

F-ma=0=m'a'

and concluding this a'=0, and therefore it does not move. You have imposed a m'a' term to satisfy a FALSE NOTION you made up in your head that it must stay at rest. There is no such, m'a' =0.

The fact is F-ma=0. This does NOT mean a=0! Very important you realize this. THIS a, is the acceleration, and it is NOT zero. This a' term I made up is just that, something YOU made up without realizing, its not equal to zero, and infact it does not even exist.

 

[russ_watters]

In my degree we were always taught that 'forces and moments in a mechanical system , whether stationary or in motion,are always in equilibrium' IE D'Alemberts version of Newtons law F=ma => F-ma=0.

Are you a biomedical engineer?? I'm a mechanical engineer and our concepts of this should be identical.

1A: Just accept it (My reply: Can't) Russ wrote- Maybe I should Russ but isn't it a complete statement of how we explain acceleration in terms of force?

Physicists may see this differently, but I see this as a needless complication of something that can mathematically be reduced to f=ma anyway.

confused! - so are you agreeing with me then, ie that forces are always in equilibrium and do not cause motion??

No, the forces are in equilibrium which means there must be motion. The equilibrium (the reaction force) is provided by ma. If you didn't have F2=ma in the equation and you applied a force of, say, 1N, you'd have 1-0=0 for F1-F2=0 -- an untrue statement.

[cyrus was basically saying the same thing]

 

[rcgldr]

> >When calculating accelerations, reaction forces are not included.

> Oh! so there aren't any reaction forces, which is it? Is it valid (even if useful)
> to ignore a parameter just for convenience?

Reaction forces are the result of acceleration, they are real, but they are not used to calculate acceleration. Acceleration = force (or torque) divided by inertia, and reaction forces aren't included in this acceleration equation.

 

[david smith]

Ok so thanks for your tenacity in staying with me, I expect some have lost the will to live with this question

So Jeff wrote

Reaction forces are the result of acceleration, they are real,

Most agree with that, Good!

Cyrus wrote

This is your problem, you are simply looking at F-ma=0 and saying, it equals zero, so it must be in equilibrium. But now you are assuming there is some OTHER m'a'=0.
in other words,
F-ma=0=m'a'
and concluding this a'=0, and therefore it does not move. You have imposed a m'a' term to satisfy a FALSE NOTION you made up in your head that it must stay at rest. There is no such, m'a' =0.

The fact is F-ma=0. This does NOT mean a=0! Very important you realize this. THIS a, is the acceleration, and it is NOT zero. This a' term I made up is just that, something YOU made up without realizing, its not equal to zero, and infact it does not even exist.

and then Russ added

No, the forces are in equilibrium which means there must be motion. The equilibrium (the reaction force) is provided by ma. If you didn't have F2=ma in the equation and you applied a force of, say, 1N, you'd have 1-0=0 for F1-F2=0 -- an untrue statement.

AaaaaH! Light going on, small epiphamy

So if I read F=ma is the same as ma = ma then mass * acceleration of one object is equal to the mass * acceleration of another object acting on it, then, without considering ma as a force (as suggested by Andy) this makes sense. OK! Hooray!

Can I say then that:
Force is a useful intuitive construct of ma and is not necessarily real. F=ma indicates force is correlative with acceleration of a body but not necessarily causative.

So to me this looks like conservation of momentum ie change in momentum1 (inversely)= change in momentum2, (assuming no losses) is it?

So can I say that a change in momentum = a transfer of energy?

Can I also say change in momentum1 = change in momentum2 in the direction of the force applied?

Therefore could I make the statement that:

Movement is due to transfer of energy from one body to another and in the direction of the force applied.


Much appreciated Dave Smith

 

[Andy Resnick]

All, Russ, Jeff, Mapes, Andy, Cyrus, thakyou for you time and effort and your input.

Aha! this is more like it, these are the types of answers I've been looking for.

Let me explain, I work as a Podiatrist and deal with pathological problems of gait and posture in terms of biomechanics. Therefore I am dealing with kinetics and kinematics -forces and movement.

<snip>

Cheers all Dave

Consider this simple action: standing still while raising one arm up. It's clear that there are local unbalanced forces from the unapposed muscle contracting, causing the bones to pivot around joints, etc. What may be less clear is that in order to remain standing still during this, there are also unbalanced forces distributed through the entire skeleton, causing small shifts in balance and ultimately, weight distribution on the feet.

So far so good? Now think about what would happen if an astronaut did the same thing. Because there's no opposing force at his feet (from being up in space an in a microgravity environment), the astronaut's body must conserve momentum. So, as (s)he raises the arm, the entire body is likely to be set into rotation to ensure that the center of mass remains stationary.

Is that helpful?

 

[Dale]

Hi David, I think you are getting some basic concepts mixed up.

But then I came onto the D' Alembert variation of F-ma=0 ie everything in equilibrium.

In what sense does this imply any kind of equilibrium? Usually equilibrium is defined as a=0 which does not follow from F-ma=0.

In terms of F= ma so ma = F and F-F = 0.

And further 0=0. In general you can reduce any equation to the trivial 0=0. That doesn't imply anything whatsoever about the original relationship other than that it was true (algebraically).

 

[russ_watters]

Can I say then that:
Force is a useful intuitive construct of ma and is not necessarily real.

You most certainly cannot.

F=ma indicates force is correlative with acceleration of a body but not necessarily causative.

I'm not quite sure what you mean there. You can have a force without an acceleration, but you cannot have an acceleration without a force. If there is an acceleration, force is the cause.

So to me this looks like conservation of momentum ie change in momentum1 (inversely)= change in momentum2, (assuming no losses) is it?

[snip] Can I also say change in momentum1 = change in momentum2 in the direction of the force applied?

I'm not sure why the word "inverseley" is in there, but the change in momentum 1 is equal to the change in momentum 2 (the second statement is true). Remember, velocity is a vector.

So can I say that a change in momentum = a transfer of energy?

No, momentum and energy are not the same thing. A transfer of or the consumption of energy does not necessarily imply a change in momentum.

Therefore could I make the statement that:

Movement is due to transfer of energy from one body to another and in the direction of the force applied.

No, you cannot. A change in momentum does not imply a transfer of energy. Consider an orbit. Two bodies are constantly applying forces to each other and being accelerated by each other, yet there is no change in kinetic or potential energy.

You're going back to the idea in your original post, so let's revisit something else (same issue, really):

kinetic energy of momentum

As was already stated, that's just word salad. It doesn't actually mean anything. Kinetic energy and momentum are two different things. There is no such thing as "kinetic energy of momentum".

 

[david smith]

Russ

Thanks for your help Russ and all, I guess I'll just have to learn to accept what I was taught and accept defeat on this eh!

All the best Dave

 

[rcgldr]

I think the issue here is that although reaction forces peform work, it doesn't mean that acceleration doesn't occur.

Take the simple case where a spring accelerates a block on a frictionless surface. The integral of force times distance supplied by the spring equals the work done and the increase in kinetic energy of the block. At the same time, the potential energy of the spring has gone to zero. So the springs force increased the kinetic energy of the block, and the blocks reaction force reduced the potential energy of the spring to zero. So potential energy in the spring was converted into kinetic energy of the block. The spring peformed work on the block and the block peformed (negative) work on the spring, resulting in the block being accelerated to some specific speed, and the spring ending up with zero potential energy.

Or in the case of a 100% elastic head on collision between a moving body and a stationary body, both with equal masses. During the dwell time of the collsion, the force from the moving body accelerates the staionary body, while the reactive force of the stationary body decelerates the moving body to zero velocity.

For another example, take the case of a car accelerating (ignoring aerodynamic drag). The car's tires push back on the pavement, which creates a very small amount of angular acceleration of the earth. The reaction force from the pavement results in the forwards acceleration of the car, which has a much smaller mass and gains most of the kinetic energy. (Note that angular momentum of the Earth and car (consider the car to be in low orbit) remain constant.)

As another example, take the case of a rocket in outer space. The rocket engine accelerates spent fuel up to a high rate of speed in one direction, and the reaction force from the fuel accelerates the rocket in the other direction.

I'm not sure how to describe reaction forces in cases like gravity, where there are equal and opposite attactive forces, and it's not clear to me if either of these forces could be considered to be a reactive force.

 

[david smith]

Dale

=DaleSpam;1639357]Hi David, I think you are getting some basic concepts mixed up.In what sense does this imply any kind of equilibrium? Usually equilibrium is defined as a=0 which does not follow from F-ma=0.

Applied force + inertial force = 0 inertial force being equal and opposite to applied force

And further 0=0. In general you can reduce any equation to the trivial 0=0. That doesn't imply anything whatsoever about the original relationship other than that it was true (algebraically).

I don't agree with the 2nd part, not that that means much but here are some references that have a different view.

Cheers Dave

 

[Dale]

Applied force + inertial force = 0 inertial force being equal and opposite to applied force

Then that is the key. This definition of equilibrium does not mean a=0, which is the usual definition. In fact, this definition of equilibrium does not imply anything about the motion of the body under consideration. IMO, this is a completely useless concept of equilibrium since it is always true and does not simplify the problem.

here are some references

Yes, and from the first paragraph of the first reference:

Many people believe that D’Alembert’s approach to mechanics, an alternative to the momentum balance approach, should not be taught at this level. Students attempting to use D’Alembert methods make frequent mistakes. We do not advise the use of D’Alembert mechanics for first-time dynamics students.

You are demonstrating exactly why they included this caution as the initial paragraph. My recommendation is to use standard analysis techniques and the standard definition of equilibrium. You obviously are getting confused by this approach which adds nothing of value that I can see until you get into Lagrangian mechanics.

 

[david smith]

Jeff wrote

Jeff Reid;1639820]I think the issue here is that although reaction forces perform work, it doesn't mean that acceleration doesn't occur.

Agreed

Jeff

Take the simple case where a spring accelerates a block on a frictionless surface. The integral of force times distance supplied by the spring equals the work done and the increase in kinetic energy of the block. At the same time, the potential energy of the spring has gone to zero. So the springs force increased the kinetic energy of the block, and the blocks reaction force reduced the potential energy of the spring to zero. So potential energy in the spring was converted into kinetic energy of the block. The spring peformed work on the block and the block peformed (negative) work on the spring, resulting in the block being accelerated to some specific speed, and the spring ending up with zero potential energy.

Russ wrote

A change in momentum does not imply a transfer of energy.

Jeff are you saying that it can?



Or in the case of a 100% elastic head on collision between a moving body and a stationary body, both with equal masses. During the dwell time of the collsion, the force from the moving body accelerates the staionary body, while the reactive force of the stationary body decelerates the moving body to zero velocity.

Agreed, so that's change of momentum 1 = change of mometum2 isn't it?

As we are carrying on this discussion (much to my delight) can you explain what

Russ wrote IE

Kinetic energy and momentum are two different things.

and so

"There is no such thing as "kinetic energy of momentum".

Kinetic energy is 1/2mv^2 (scalar) and momentum (vector) is mv (p=mv)so by increasing the momentum the kinetic energy is increased (isn't it?)

The standard unit of energy is the Joule, such that 1 Joule is equal to the work done by a force of 1 Newton in moving a distance of 1 metre - in the direction of the force.

So that energy is scalar but in terms of a vector ie force.

The unit of momentum is kgm/s + direction so clearly eergy and mometum are not the same but


Is Russ saying energy is the ability to do work F*d but momentum p=mv does no work since there is no force involved so energy can't be equated to momentum.

But doesn't a body with momentum have the ability to do work?

When a body with momentum acts on a stationary mass doesn't the energy transfer from one body to the other to conserve momentum and so there is motion of the second body, which now has momentum. In the interaction of the two boies there were forces and accelerations. So aren't all these directly conected and dependent on each other?

Is this what jeff is saying?

Jeff

For another example, take the case of a car accelerating (ignoring aerodynamic drag). The car's tires push back on the pavement, which creates a very small amount of angular acceleration of the earth. The reaction force from the pavement results in the forwards acceleration of the car, which has a much smaller mass and gains most of the kinetic energy. (Note that angular momentum of the Earth and car (consider the car to be in low orbit) remain constant.)

As another example, take the case of a rocket in outer space. The rocket engine accelerates spent fuel up to a high rate of speed in one direction, and the reaction force from the fuel accelerates the rocket in the other direction.

I'm not sure how to describe reaction forces in cases like gravity, where there are equal and opposite attactive forces, and it's not clear to me if either of these forces could be considered to be a reactive force.

Not quite sure what you are trying to say here?

Cheers Dave

 

[david smith]

You are demonstrating exactly why they included this caution as the initial paragraph. My recommendation is to use standard analysis techniques and the standard definition of equilibrium. You obviously are getting confused by this approach which adds nothing of value that I can see until you get into Lagrangian mechanics

.

Dale, you are right in many ways. However I have no problems with the maths, at the level which I work. (which is low compared to you physicists) One can manipulate an expression or group of expresion by precicely applying the rules of algebra and get a correct outcome without understanding any concepts. I'm just trying to get a better understanding of the concepts. I regularly use the form of this - that = 0 in my work, papers and thesis, and find it more useful than this = that * something.

I question my understanding of the concepts when someone says something like
" it is the plantarflexion moments about the ankle joint that lift the heel and not the velocity of the Centre of mass moving about joint.

So are they are saying it is not the momentum of the body that carries you forward, when moving forward necessitates the ankle plantarflexion and heel lift to raise the CoM and increase its potential energy for use in the next step.

I have to have an explore of my understanding of principles. and so I have with all your help. For which I am most grateful.

Cheers Dave
Cheers Dave

 

[russ_watters]

Jeff

Russ wrote

Jeff are you saying that it can?

What I meant was that it does not automatically imply it. There are lots of situations where it is helpful to use conservation of energy to find a change in velocity. But there are also lots of situations where it does not.

That doesn't change anything anyone has told you. You are trying to draw a broad general conclusion about a connection between energy and momentum and I provided a specific example to show how that general conclusion doesn't fit.

Kinetic energy is 1/2mv^2 (scalar) and momentum (vector) is mv (p=mv)so by increasing the momentum the kinetic energy is increased (isn't it?)

Again, again, again, again, again: kinetic energy is a scalar and momentum is a vector. In a circular orbit, a force is applied, momentum changes, and kinetic energy does not.

Not quite sure what you are trying to say here?

He's saying that when a car accelerates, the change in momentum of the Earth and the car is the same, but the car gains virtually all of the kinetic energy. Since kinetic energy is mv^2 and the Earth gains less velocity than the car, it gains less kinetic energy even though they both gain/lose the same momentum.

 

[Dale]

I'm just trying to get a better understanding of the concepts.

OK, I can appreciate that.

So, aside from the trivial algebraic manipulation of the equation, the only actual conceptual change of the D'Alembert approach is to consider inertia (-ma) to be a force. This is equivalent to working in the non-inertial rest frame of the object under consideration. In this frame there exists a ficticious force equal to -ma that must be introduced in order to have Newton's first and second laws hold.

Are you familiar with the concepts of non-inertial reference frames (e.g. rotating reference frame) and ficticious forces (e.g. centrifugal force)? If not, then I would recommend you begin studying those topics in order to understand the concepts in the D'Alembert approach.

The inertial force is like any other ficticious force. In particular it violates Newton's 3rd law and it is always proportional to the mass of the object. Ficticious forces can do work in the non-inertial coordinate system and accelerometers cannot detect acceleration due to ficticious forces.

 

[rcgldr]

Regarding momentum versus kinetic energy, take the case of non-elastic collisons. In this case some or all of the energy is consumed by deformation of the masses involved (the deformation process also produces some heat). Say 2 cars of equal mass and opposing velocity collide head on. All of the energy is consumed as both cars collapse and end up with 0 velocity. The sum of momentum for the two cars going in opposite directions at the same speed is zero before and after the collision. The kinetic energy for the 2 cars was equal to 2 times 1/2 mass (of each car) times (each car's) speed^2 before and 0 after the collision.

 

[Andy Resnick]

.
<snip>
I question my understanding of the concepts when someone says something like
" it is the plantarflexion moments about the ankle joint that lift the heel and not the velocity of the Centre of mass moving about joint.

So are they are saying it is not the momentum of the body that carries you forward, when moving forward necessitates the ankle plantarflexion and heel lift to raise the CoM and increase its potential energy for use in the next step.

<snip>
Cheers Dave

I've heard of the concept contained in your second paragraph- walking is essentially controlled falling. I admit it's an attractive hypothesis, but I'm not familiar enough with the literature to think it's been validated.

In order to walk, many things have to happen in synchrony to maintain balance. Think about walking on slippery ice (or a layer of ball bearings, etc.). This should give you an idea of how the foot and ankle act to control the reaction force of the ground, to provide a frictional force allowing walking. For example, I have to shorten my stride when walking on ice.

 

[david smith]

I've heard of the concept contained in your second paragraph- walking is essentially controlled falling. I admit it's an attractive hypothesis, but I'm not familiar enough with the literature to think it's been validated.

In order to walk, many things have to happen in synchrony to maintain balance. Think about walking on slippery ice (or a layer of ball bearings, etc.). This should give you an idea of how the foot and ankle act to control the reaction force of the ground, to provide a frictional force allowing walking. For example, I have to shorten my stride when walking on ice.

Yes longer strides induce higher GRF which then overcome the frictional forces of the shoe - ice interface and you slip and slide into a heap.


Andy, When doing gait analysis in a lab we usually measure the ground reaction forces and combine this with kinematic data about the limbs of interest, usually using 3D video camera systems that log relative positions of markers on the body thru time and space.
We use inverse dynamics to characterise the joint, limb and muscle actions. The model is a linked segment rigid body from which we can derive forces moments and powers.
This is a link to a good summary of the technique.http://www.sportsci.com/adi2001/adi/services/support/tutorials/gait/chapter2/2.3.asp

Therefore, to summarise the gait cycle, in normal gait (walking) the Centre of Mass has a sinusodial progression in 3 dimensions. Considering the 2D saggital plane ie forward and vertical action, As the foot strikes there is a three rocker sequence of events IE first over the heel then about the ankle and finally about the ball joints of the toes (MPJ's). This is called the stance phase. At the progression from heel to MPJ's the heel lifts and raises the CoM. After proceeding over the MPJ's the CoM progresses forwards and downwards toward the next heel strike.

Side to side and twisting actions come from the need to balance the CoM on one leg during contralateral swing thru and the torque produced by leg swing. The leg swing is balance by arm swing counter torque and these two action build up torque in the spinal complex, which also helps to drive the hips and legs and augments leg muscle action.

So forward progression is a complexity of the above, the amount of forward drive from each component depends on the individual gait style, of which there are many and various within the range of 'normal'. However many would argue that the energy for forward progression is mainly supplied by the momentum of the CoM (I've probably worded that wrong considering the comments on this thread regarding the relationship of momentum to energy!)continuing forward and falling in an arc to the ground. The heel lift supplied by the muscle action of the calf muscles, + stored elastic energy in the large Achilles tendon, gained as it opposes the momentum of the CoM, lifts the CoM. This raising of the CoM increases the potential energy to later become kinetic energy and augment the loss of kinetic energy thru the stance phase. As the body become inclined forward the swing leg is forward and contacting the ground and the action of the achilles becomes propulsive just before the rear leg starts its swing thru. The whole action repeats. There are also varying amounts of action from hip and knee extensors and flexors, which also augment forward progression depending on gait stlye. The use of ElectroMyography, to measure muscle unit action potential (MUAP), can give clues as to the the timing and perhaps magnitude of muscle actions thru the whole stride.

The full stance phase gives a sinus wave output characterisation to the vertical GRF applied to the foot. IE A high peak at braking after heel strike, a low trough as the CoM rolls over the planted foot thru the ankle joint and another high peak as the heel rises and propulsion occurs. If heel raise occurs early before the CoM has passed over the ankle joint then the CoM tends to be relatively accelerated backwards and so retards the mometum.

One of the main objectives of rehabilitation in terms of gait is to reinstate an effiecient transition from one step to the next with as little retarding of the momentum of the CoM as possible, which will reduce the need for auxillary actions of muscle power and so reduce the possibility of increased tissue stress and reduce physiological and metabolic energy requirements for the individual.

It has been thought that reduction of the sine wave amplitude increases walking efficiency. Recently however research appears to show that there is an optimum frequency to the sine wave, which indicates the use of stored elastic energy in the achilles tendon and perhaps the spinal complex (spinal engine) is an important input.
In other words walking like Groucho Marks defines a low amplitude, almost flat, sine wave action, mechanically more efficien in theory but physiologically requires more effort and is not an efficient way of walking. Which makes sense, since because nature loves to conserve energy, most of us would walk that way all the time.

Early slowing of the CoM, as in a pathological gait (abnormal), reduces the kinetic energy available to the body.(Perhaps, I'm not sure now) This results in reduced gait velocity or alternatively greater muscle action or some trick joint action compensation is then required to continue at the same velocity, which potentially increases internal forces and stress and increases metabolic energy requirements. However low metabolic and physiological energy use is optimal.

So in the efficient gait the body would use less metabolic and physiological energy to do the same work in the same time as the inefficient gait would achieve.

Therefore impeded CoM momentum = higher physiological energy required
Unimpeded CoM momentum = Low physiological energy required

Can you now see how I (wrongly it seems) relate momentum of the CoM to kinetic energy requirements of the human mechanical system.
Since higher forces do not necessarilly indicate faster walking, can you also see how I queried, is it the forces applied or the energy transfer that cause motion?

Now I've written that I can see that higher forces or higher energy use does not indicate faster gait. Both are being wasted trying to accelerate the Earth in various directions.


Cheers Dave

 

[david smith]

OK, I can appreciate that.

So, aside from the trivial algebraic manipulation of the equation, the only actual conceptual change of the D'Alembert approach is to consider inertia (-ma) to be a force. This is equivalent to working in the non-inertial rest frame of the object under consideration. In this frame there exists a ficticious force equal to -ma that must be introduced in order to have Newton's first and second laws hold.

Are you familiar with the concepts of non-inertial reference frames (e.g. rotating reference frame) and ficticious forces (e.g. centrifugal force)? If not, then I would recommend you begin studying those topics in order to understand the concepts in the D'Alembert approach.

The inertial force is like any other ficticious force. In particular it violates Newton's 3rd law and it is always proportional to the mass of the object. Ficticious forces can do work in the non-inertial coordinate system and accelerometers cannot detect acceleration due to ficticious forces.

Dale

I had learned about centrifugal / centripedal and coriolis force/effect but I did some reading to refresh and very useful it was too.

However at the end of the day centrifugal and coriolis forces are effects of perception due to the observers viewpoint being within the inertial rest frame or the rotating frame.
They are not forces at all as can be acertained by the observer outside the rotating frame who see that it is the rotating frame that chages direction and experiences acceleration, whereas the particle within it continues unhindered in the same direction.IE Newtons first law.
The centrifugal force feel as if it is at 90dgs to the centre of rotation beacuse the inside viewers perception of the acceleration changes at the same angular velocity as the objects.
Centripedal is a real force reaction to centrifugal force and it appears to me that Inertial force is a real reative force to acceleration.

So I read this paper, On the Origin of Inertial Force, C. Johan Masreliez Redmond, WA
jmasreliez@estfound.org. Which gives a very thorough explanation of the origin of inertial force, much of which was beyond me but essentially he is saying inertial force is a reation to acceleration within the inertial reference frame. IE unlike gravity and almost opposite to gravity it cannot be felt at rest whereas gravity cannot be felt in freefall.

So as such if a body requires a force to accelerate it then there are two questions.

You can't apply a force without an opposing force. So if a body with momentum collides with another body at rest, the body at rest has to accelerate and invoke a reactive inertial force. However the first body cannot apply any force to accelerate the second body until the second body has accelerated so then how can the applied force accelerate it. This appears to be a chicken and egg situation.

And

If two bodies act on each other in the way explained above then there are two inertial forces. One with a negative acceleration and one with a positive acceleration repectively related to their mass. With the other ficticious forces there was no acceleration and no change of direction IE no force.However with this configuration there are two ficticious forces that do cause acceleration and change of direction. Therefore they cannot be ficticious. Can they?

Obviously I still can't see inertial force from your perspective even tho I understand what you are saying. F=ma but ma is not an opossing force like a = f/m gives an equivalent sum but is not equal to a. Is this because you always view from the perspective of acceleration and I from the perspective of force. Therefore I always see balanced forces and you see unbalanced forces.


Ayway I've done many hours of reading and you have helped me to understand more, so its all good.


Cheers Dave

 

[rcgldr]

Cetrifugal force is just as real as any reaction force. It's the centripital force that accelerates an object "inwards", and centrifugal force is the reaction force, equal to mass times acceleration as it always is.

 

[david smith]

Cetrifugal force is just as real as any reaction force. It's the centripital force that accelerates an object "inwards", and centrifugal force is the reaction force, equal to mass times acceleration as it always is.

Jeff

True, but isn't the centrifugal force an inertial force also. EG A car travel in a straight lime at 10m/s and object on the dash board is also traveling in the same direction at 10m/s. The car suddenly turns thru 90dgs but the object that has negligible friction on the dashboard interface continues on its original course at 10m/s at that point it appears to the passenger that something / someforce has accelerated it sideways toward the sidewindow. This is the apparent centrifugal force but it is actually the car, not the object, that experiences accelleration as it changes direction to go thru a turn.
The same principle applies even when the car goes around a long bend.

At the point that the object hits the window there is a 'real' centripedal force that accelerates it in the opposite direction and the object also accelerates the car thru its action on the window but as the car is relatively massive it does not change direction. Therefore the centrifugal force also becomes real in terms of the inertial reference frame.

I think this is how Dale concludes that inertial force is fictitious by the fact that it can only exist as a reaction to acceleration. However to my way of thinking both forces are the result of reactive accelerations and so both are either ficticious or real depending how you think about them.

In the case of a continuous rotation at a constant angular velocity such as a roundabout.
If the roundabout was enclosed so there was no reference to the resting frame then the observer, who is also the object within the rotational frame, would have no way of knowing about their angular velocity. At the same time they would feel the force that pushes them against the walls of the roundabout. This is the centipetal force holding them into the roundabout and their reactive inertial force, which again is the tendency to carry on in a straight line but because the tangent vector of the linear acceleration infinitely changes directly proportionaly to the angular deflection of the rotation it always appear to be pushing straight in and have a reaction straight in IE at 90dgs to the centre of rotation.

If the side wall suddenly disappeared then the person would fly out at a tangent to the arc but to another observer inside the roundabout it would appear that the first person had flown out at a right angle to the centre of rotation. The tangental vector can be thought of as having two vectors of origin in the x, y, directions. Again the second observers perspective does not allow them to see the true tangental motion since they both have a similar velocity in onedirection (eg x vector) and so the second observer only sees the velocity in the direction that has a vector at 90dgs to the centre (eg y vector).

That's my understanding at least

Cheers Dave

 

[Andy Resnick]

Yes longer strides induce higher GRF which then overcome the frictional forces of the shoe - ice interface and you slip and slide into a heap. Andy, When doing gait analysis in a lab we usually measure the ground reaction forces and combine this with kinematic data about the limbs of interest, usually using 3D video camera systems that log relative positions of markers on the body thru time and space.
We use inverse dynamics to characterise the joint, limb and muscle actions. The model is a linked segment rigid body from which we can derive forces moments and powers.
This is a link to a good summary of the technique.http://www.sportsci.com/adi2001/adi/services/support/tutorials/gait/chapter2/2.3.asp

<snip>

Can you now see how I (wrongly it seems) relate momentum of the CoM to kinetic energy requirements of the human mechanical system.
Since higher forces do not necessarilly indicate faster walking, can you also see how I queried, is it the forces applied or the energy transfer that cause motion?

Now I've written that I can see that higher forces or higher energy use does not indicate faster gait. Both are being wasted trying to accelerate the Earth in various directions.Cheers Dave

This was a highly informative post- thank you!

Edit: How is this model modified for walking up or down steep inclines?

 

[david smith]

Andy

In answer to your question:

Edit: How is this model modified for walking up or down steep inclines?

It is unusual to study the gait on an incline, I have not done so formaly. Most studies / gait analysis are done on a flat surface of a lab or clinic. There obviously is a limitation here but the flat ground gives us a universal reference frame for 'normal gait' or as I prefer a 'gold standard' gait. Most of us are normal in that we can ambulate perfectly well for our daily lives and do so without pathology. Most of us are do not attain the gold standard which is just an arbritary bench mark for reference. These bench marks contain parameters such as, Joint angular range of Motion, Joint compliance to a force, Joint neutral position, 3D GRF, muscle firing timing, moments, powers step length, cadence, etc.

The main differences for stair climbing V's normal ammbulation would be the greater range of motion required by the joints to progress upwards and forwards, this requires greater physiologial energy and forces and ultimately more metabolic energy requirements.
In this case gravity is a greater consideration in the equation where there is likely to be large sinusodial amplitude and an overall raising of the mass to a higher potential energy level.

Gait analysis for the purposes of rehabilitation might be done on an incline or on stairs where the individual requires this. However it is often quite plainly obvious why someone might have difficulty or pain when engaging in these activitie. EG they have restriction, for whatever reason of hip knee or ankle flexion / dorsiflexion. If these Ranges of motion are restricted in terms of angular deflection or stiffeness then clearly it inhibits the progression of the centre of mass to the optimum position for continued efficient forward progress.

Sometimes there is a need to study certain activities when considering optimum designs for prosthetics or orthotics.

The fact is that ambulation looses forward momentum thru a gait cycle. In any activity where forward motion is required, and in terms of this thread and the CoM, the CoM must be in the optimum position at the optimum time to be accelerated forward by the muscular action. If it is not it will tend to be accelerated sub optimally and forward progression will be inhibited.


I have attached some PDF's of papers that may interest you.


As a point of interest, I live I Folkestone Kent and have seen many Nepalese soldiers from the Gurka regiment. They live / come from the mountains of Nepal and I was suprised to discover that they have a huge range of dorsiflexion (flexion of the top of the foot toward the front of the leg) of the ankle. Norm = 10dgs (on a good day) they had 40dgs+
which means their ability to get the CoM ahead of their feet are not restricted by the stiffness of the ankle joint when walking up steep hills. In theory this means they should be able to walk uphill easier. If you watch films of Nepalese sherpas climing mountains they appear to walk with ease uphil even with large loads, they walk as if on flat ground (almost).


Cheers Dave Smith

 

[david smith]

Russ wrote

As was already stated, that's just word salad. It doesn't actually mean anything. Kinetic energy and momentum are two different things. There is no such thing as "kinetic energy of momentum".

Russ, I have been doing some reading on energy, work and momentum.
How does your statement above correlate with the following

I realize that Ke cannot be mv since Ke is a scalar and mv is a vector and is like saying 'what direction is the weight of this book'.
However Ke is scalar because velocity is squared. And velocity is squared because of the premise that energy = work and work = F * d => 1/2mv^2.

Can I ask why is energy determined a F*d and not F*time. Usually F*t is the impulse that momentum can be passed from one body to another. Therefore Ft=mv, and by determining the change in momentum of one body in relation to another we can find the total energy used or lost.
If EnergyK = F*t it might make the problem of the link between physiological work and mechanical work of the human body easier to resolve. I guess that's a mute point since we are stuck with Work = F*d = 1/2mv^2 = Ek

Wouldn't that (Ek=F*t) leave us with conservation of momentum and energy in all cases?
IE momentum is energy.

E.G. A 10-gram bullet is fired from a rifle at a speed of 700 m/sec into a 1.50-kg wooden block suspended by a string that is two meters long.


After the collision, through what vertical distance (h) does the block rise?
0.91 meters
During the collision we can only use conservation of momentum.

mbullet(vo bullet) + Mblock(vo block) = (m + M)vc
(0.01)(700) + (1.5)(0) = (0.01 + 1.5) vc
7 = (1.51) vc
vc = 4.64 m/sec

After the collision, the block and bullet are now behaving as one object. We can use conservation of energy to determine how high the pendulum swings.

KEbottom + PEbottom = KEtop + PEtop
KEbottom = PEtop
½mvc2 = mgh
½(1.51)(vc)2 = (1.51)(9.8)h
½(4.64)2 = 9.8h
h = 0.91 m


How much KE is lost during the collision?
2434 J
To determine how much KE is lost, compare the total KE before the collision to the total KE afterwards.

KEbefore = ½(mbullet)(vo bullet)2 + ½(Mblock)(vo block)2
= ½(0.01)(700)2 + ½(1.5)(0)2
= 2450 J

KEafter = ½(mbullet + Mblock)(vc)2
= ½(1.51)(4.64)2
= 16.3 J

This collision lost 2434 J of energy, or 99.3% of the bullet's original energy! http://dev.physicslab.org/Document.aspx?doctype=3&filename=Momentum_MomentumEnergy.xml

Where did it go?

Both momentum and kinetic energy are in some sense measures of the amount of motion of a body. http://galileoandeinstein.physics.virginia.edu/lectures/momentum.html

The difference comes from the fact that most collisions are not perfectly elastic and so while momentum is always conserved energy is not. http://en.wikipedia.org/wiki/Momentum

What if you take this premise

Part 1: Kinetic energy is not ½mv².


A 4kg object dropped 1m (meter) has the same amount of ½mv² as a 1kg object dropped 4m, because force times distance equals ½mv² for an accelerating mass. But a rocket accelerating the masses to those velocities requires twice as much energy as fuel for the large mass as for the small one.

Therefore, both masses do not have the same energy; the rocket does not transform energy in proportion to ½mv²; ½mv² is not kinetic energy; and a gallon of fuel does not produce a consistent amount of ½mv².

Part 2: Kinetic energy is mv.


A 4kg object dropped for 1s (second) has the same amount of mv (momentum) as a 1kg object dropped for 4s, because force times time equals mv for an accelerating mass. A rocket accelerating the masses to those velocities uses the same amount of energy as fuel for both masses.

Therefore, both masses have the same amount of energy; the rocket transforms energy in proportion to mv; mv is kinetic energy; and a gallon of fuel produces a consistent amount of mv.



This proof shows that momentum is all there is to energy. ½mv² is just an equation. How could they both be conserved at the same time when they have different dynamics, which was the original question? The decision to conserve them both was rationalism for convenience, not a law of nature. ref Gary Novak http://nov55.com/ener.html

Also

E=mc^2 Therefore m = E/c^2 energy increases as speed increases and as mass is energy mass increases as speed increases http://galileo.phys.virginia.edu/classes/252/energy_p_reln.html

Velocity is speed with a vector so as mass or velocity increases so does energy

Kinetic energy of a body can be related to the momentum by the equation.

Ek = p^2/2m http://en.wikipedia.org/wiki/Kinetic_energy



I know that energy is like fuel (storage, amount, scalar) and momentum is motion (velocity, mass, time, distance, vector) but the catch is that as velocity increases so does energy and mass. So they are bound together unlike petrol and a car where the faster a car goes the less fuel it will have.

I know I have worn you to a frazzle on this one but one more answer would be appreciated.
Please

Cheers Dave

 

[Dale]

However at the end of the day centrifugal and coriolis forces are effects of perception due to the observers viewpoint being within the inertial rest frame or the rotating frame.
They are not forces at all as can be acertained by the observer outside the rotating frame who see that it is the rotating frame that chages direction and experiences acceleration, whereas the particle within it continues unhindered in the same direction.IE Newtons first law.

Yes, exactly the same as with the inertial force.

Centripedal is a real force reaction to centrifugal force and it appears to me that Inertial force is a real reative force to acceleration.

A reaction force is part of an action-reaction pair satisfying Newton's 3rd law. The key point about an action-reaction pair is that they act on two different bodies. So the centripetal force cannot be a reaction force to the centrifugal force since they both act on the same body.

So as such if a body requires a force to accelerate it then there are two questions.

You can't apply a force without an opposing force. So if a body with momentum collides with another body at rest, the body at rest has to accelerate and invoke a reactive inertial force. However the first body cannot apply any force to accelerate the second body until the second body has accelerated so then how can the applied force accelerate it. This appears to be a chicken and egg situation.

Even when considering more than one body the inertial force is not a reaction force. For example, consider three electrons (initially at rest) located at the points of an equilateral triangle. In the D'Alembert approach each electron experiences 3 forces, two electrostatic, and one inertial. Each electrostatic force on one electron is matched by an equal and opposite electrostatic reaction force on another electron, thus satisfying Newton's 3rd law as all real forces do. In contrast, the inertial forces are not matched by any equal and opposite force on any other electron, thus violating Newton's 3rd law as all ficticious forces do.

However with this configuration there are two ficticious forces that do cause acceleration and change of direction. Therefore they cannot be ficticious. Can they?

Ficticious forces certainly can cause (or prevent) acceleration in their non-inertial frame. Ficticious forces can also do work. Sometimes they are even conservative so they can be associated with a potential field.

Please do not think that the word "ficticious force" is a derogatory statement that I am using to belittle the D'Alembert approach. On the contrary, the term "ficticious force" is simply the technical term, and I find ficticious forces can be quite useful when properly understood and applied in an appropriate problem. My dislike of the D'Alembert approach is not that the inertial force is a ficticious force, but rather that the approach is very confusing and adds little value.

 

[Dale]

Can I ask why is energy determined a F*d and not F*time. Usually F*t is the impulse that momentum can be passed from one body to another.

Why would we need another name for the same thing? As you say F*t is already (change in) momentum, so why would use the word energy to also refer to the same thing? If we didn't call F*d energy then what would you prefer we call it?

 

[david smith]

Why would we need another name for the same thing? As you say F*t is already (change in) momentum, so why would use the word energy to also refer to the same thing? If we didn't call F*d energy then what would you prefer we call it?

Work --- Energy and work are defined by the same units, why not energy and momentum instead?

Although I'm not really arguing it should be changed, like some do. However they appear to be paranoid conspiracy theorists, we are right and everybody else is wrong type, when you read their work. Then again this was said about Galileo and Socrates etc and they were right, mostly. Intuitively momentum is very much like energy and you know what they say, if it smells like cr--, looks like cr-- and tastes like cr--, chances are it is cr--, Which some might say is what I'm talking anyway

LoL Dave

 

[david smith]

Dale

Originally Posted by david smith
So as such if a body requires a force to accelerate it then there are two questions.

You can't apply a force without an opposing force. So if a body with momentum collides with another body at rest, the body at rest has to accelerate and invoke a reactive inertial force. However the first body cannot apply any force to accelerate the second body until the second body has accelerated so then how can the applied force accelerate it. This appears to be a chicken and egg situation.

Even when considering more than one body the inertial force is not a reaction force. For example, consider three electrons (initially at rest) located at the points of an equilateral triangle. In the D'Alembert approach each electron experiences 3 forces, two electrostatic, and one inertial. Each electrostatic force on one electron is matched by an equal and opposite electrostatic reaction force on another electron, thus satisfying Newton's 3rd law as all real forces do. In contrast, the inertial forces are not matched by any equal and opposite force on any other electron, thus violating Newton's 3rd law as all ficticious forces do.

Originally Posted by david smith
However with this configuration there are two ficticious forces that do cause acceleration and change of direction. Therefore they cannot be ficticious. Can they?

Ficticious forces certainly can cause (or prevent) acceleration in their non-inertial frame. Ficticious forces can also do work. Sometimes they are even conservative so they can be associated with a potential field.

Please do not think that the word "ficticious force" is a derogatory statement that I am using to belittle the D'Alembert approach. On the contrary, the term "ficticious force" is simply the technical term, and I find ficticious forces can be quite useful when properly understood and applied in an appropriate problem. My dislike of the D'Alembert approach is not that the inertial force is a ficticious force, but rather that the approach is very confusing and adds little value.

I see what you are saying here, even tho I don't know much about electrostatic force other than its a type of gravity between electrons and reading up about it it is difficult to get a definition of the Electromagnetic force (EMF) only definitions of its action, EG Coulombs law.

It seems to me that the two EMF vectors equal one resultant force vector, which is resisted by an inertail force in the opposite direction to the resultant force vector. Obviuosly the electron can only accelerate in one linear direction so there can only be one linear reaction vector.

My interpretation of your explanation:

One body A acts on another the second body B accelerates and a particle within the inertial frame apparently accelerates in the opposite direction until it hits the wall of the second body B. As it apparently moves within the inertial frame there is an apparent force actin on it which is ficticious. OK.

The same as centrifugal force, as you say, which would mean we should have a name for the force that acts on the particle when it hits tha second B bodies wall and accelerates in the same direction (same direction as body A and B). Like centripedal force, but there isn't. Can you explain why?


If I accept what you say then I still have to ask, since there is some time lag between acceleration of body B and the particle hitting the second bodies wall, how is the acceleration of the second body B started and I would have to conclude that it is the transfer of momentum which would equal a transfer of a certain amount of energy.

The transfer of energy however, requires a force, F*t, so I would conclude that there is no time lag and therefore the ficticious force acts like a real force and resists the applied force instantaneously, simultaneously and oppositely IE F(A)=ma(particle B) and in fact the same happens in reverse to body A. Just as Newton said equal and opposite reactions.

Therefore while an inertial force can be considered ficticious in the term you speak of it can be also considered real since it always acts as a real force.

I think that probably settles my mind on the subject Dale, thanks very much but I would be interested in your response tho.

Also it occurs to me that some of these arguments in this thread appear tautalogical.

IE you can argue that Ek is not momentum as long as you stick to the rule that Ek = F*d and not F*t. To disprove this by making the counter argument that Ek is F*t and is equivalent to momentum is invalid and therefore nonesense.

Also
Inertial force is ficticious as long as you stick to the rule that there are equal and opposite accelerations and not forces. I see equal and opposite forces.

Which kind of goes back to this query from the same post that you did not answer before:

If two bodies act on each other in the way explained above then there are two inertial forces. One with a negative acceleration and one with a positive acceleration repectively related to their mass. With the other ficticious forces there was no acceleration and no change of direction IE no force.However with this configuration there are two ficticious forces that do cause acceleration and change of direction. Therefore they cannot be ficticious. Can they?


Cheers Dale, I am enjoying this voyage of discovery.

 

[Dale]

Regarding your energy and momentum comments, if you ever get interested in relativity you should investigate the four-momentum. It is a concept that unites energy and momentum in the same way that time and space are united into spacetime. Energy is not momentum any more than time is space, but they are intimately connected and that connection is very interesting. It is much more illuminating than simply trying to get rid of energy by redefining it to be the same as momentum.

they appear to be paranoid conspiracy theorists, we are right and everybody else is wrong type

It has nothing to do with paranoia. If someone were to suggest that we use the word "red" to refer to the taste of sugar you would say "we already have a perfectly good word for that, 'sweet', and what word would you use to describe the color of an apple?". Your suggestion is silly for all the same reasons. It is not about conspiracies or right or wrong, it is about clear communication.

Also it occurs to me that some of these arguments in this thread appear tautalogical. IE you can argue that Ek is not momentum as long as you stick to the rule that Ek = F*d and not F*t. To disprove this by making the counter argument that Ek is F*t and is equivalent to momentum is invalid and therefore nonesense.

The idea that everyone else in the world should drop all standard definitions at your every whim is patently absurd. Use the standard definitions of terms. If you have a truly new idea then use a new term for it, but don't change established terms. Anything else only leads to communication breakdown and serves no useful purpose.

To help you understand how silly your argument sounds perhaps this translation will help. To me this comment:

IE you can argue that Ek is not momentum as long as you stick to the rule that Ek = F*d and not F*t.

Sounds like this: "IE you can argue that red is not sweet as long as you stick to the rule that red = the color of an apple and not the taste of sugar."

 

[Dale]

As it apparently moves within the inertial frame there is an apparent force actin on it which is ficticious. OK.

Almost. "As it apparently stays stationary within the non-inertial frame there is an apparent force acting on it which is ficticious."

Therefore while an inertial force can be considered ficticious in the term you speak of it can be also considered real since it always acts as a real force.

There are several features that ficticious forces have that distinguish them from real forces:
1) they appear only in non-inertial reference frames
2) they are undetectable by accelerometers
3) they are proportional to the mass
4) they violate Newton's 3rd law

The inertial force satisfies all four of these. It is ficticious. Why don't you stop making pointless objections and start thinking about what it implies and how you can use it.

Inertial force is ficticious as long as you stick to the rule that there are equal and opposite accelerations and not forces. I see equal and opposite forces.

In the attached image I have drawn the 3-electron situation that I was describing earlier, and I have drawn it according to the D'Alembert approach with the inertial forces considered as actual forces. Consider electron A. The electrostatic force x acting on A is equal and opposite to the electrostatic force -x acting on B. They consititute an action-reaction pair satisfying Newton's 3rd law. Similarly with force y acting on A and -y acting on C. So, which force acting on B or C is equal and opposite to z? There is none, it does not satisfy Newton's 3rd law, it is a ficticious force.

 

[david smith]

Dale

From your electron diagram I would say that I am able to see more clearly what you mean.
Where there is more than one force acting on a body then each force has an equal and opposite reaction but the resultant acceleration also results in a fictitious force in the opposite direction IE INERTIAL FORCE. So therefore the same must be true even where there is one force acting on a body even though it is not so apparent.

So is this right?: Imagine if there were three planets A,B,C, and A and C each have 400 times the mass of B.

They come to be arranged in a triangular formation; whereupon they are each acted upon by each other’s gravity. A and C have equal and opposite forces, BC and AB also have equal and opposite forces. They all tend to accelerate toward each other. However B has a much greater acceleration in a direction toward the centre of AC and therefore has an inertial force in the opposite direction to the acceleration, which has no equal and opposite reaction force. Have I got it at last?

So what about the case where two bodies (A & B) act on each other with a non-elastic collision.

Ex1) They both have the same mass, A has momentum X and collides with B which has zero momentum. They stick together and the resultant velocity of A+B is the original momentum divided by the resultant mass. In the time it took for mass AB to achieve the final velocity, i.e. ½X, there was force applied = F*t (force impulse). Is it correct to say that during the force impulse there was equal and opposite reactions between A and B and that this action reaction was the inertial force of each. Furthermore that it is not possible to define whether one or the other is action or reaction.

Or is it? If one body has momentum and the other does not or there is a relative momentum differential then there must always be a resultant momentum in the direction of the greatest momentum. IE (mv1)100kg * 20m/s V’s (mv2)50kg * 30m/s = resultant velocity in direction of mv1, so perhaps the larger momentum = the applied force. Which intuitively makes sense. So could I say that F*t = the relative change in momentum in the direction of the largest momentum (with respect to theta)

If this is true then it is also true to say that A did work on B in terms of F*d.
The Kinetic energy of AB is 1/2mv^2 => (m(A+B)/2)* (½ X)^2 = ½ EkA
After A did the work on B, AB had half the original energy. Momentum conserved but energy isn’t in a non-elastic collision. How do we account for that lost energy? Especially when there is no energy lost in an elastic collision. The force time integral stays the same (I think, but I may have the maths wrong). IE with the same mass the force time integral (area under the curve) is no larger as the time reduces. So if a 100kg mass decelerates from 20m/s to 0m/s in 1 sec then this is a force of 2000N. If the same mass decelerates to 0m/s in 0.5s then this equals a force of 4000N. (2000*1) / (4000*2) = 1 = same integral. Lost energy doesn’t seem to be related to the F*t?

Does the Earth have momentum with respect to an object free falling by gravity towards it?
If two bodies have equal and opposite force in terms of gravity then the object, say 200kg mass, attracts the Earth with the same force as the Earth attracts the object. However the Earth has a infinitesimally smaller acceleration than the object but they both would have the same momentum at time of impact.?

One more question – If as mass accelerates the force increases (F=ma) then if a given mass decelerates to zero velocity in zero time, is this equal to infinite force or no force?

Dale wrote

It has nothing to do with paranoia. If someone were to suggest that we use the word "red" to refer to the taste of sugar you would say "we already have a perfectly good word for that, 'sweet', and what word would you use to describe the color of an apple?". Your suggestion is silly for all the same reasons. It is not about conspiracies or right or wrong, it is about clear communication

I think you might be getting the wrong end of the stick Dale. I was pointing out that there are some people, who appear to be highly qualified in physics, that argue that the energy equation Ek = 1/2mv^2 (F*d) is wrong. But then when you read other work by the same people they also think many things are wrong and that there is some kind of conspiracy going on and perpetuated by the establishment to suppress the truth. However when you read their arguments they sound quite logical. But then I'm not educated enough in physics to see the 'wood for the trees' as it where. (Example below) Also that the arguments are tautological in both ways until you can show why you use the original defining equation ie F*d or F*t.

Cheers Dave
Example

Simple and Unquestionable

Mathematical Proof

Kinetic Energy is not ½mv² but mv


Nitty Gritty Perspective

Physicists designed a definition of energy which erased some difficult questions which they could not answer. They cannot understand force amplification through a lever. But they know that force decreases as distance increases on a lever. So defining energy as the combination of force and distance assures that the total is always the same, and energy is conserved through a lever.

What if force is one of the forms of energy? Then physicists cannot explain how force amplification conserves energy. The truth is, force is one of the forms of energy regardless of definitions, because force and motion are interchangeable.

But physicists get around that problem by referring to force as potential energy, when it absorbs energy from motion. Potential energy is a self-contradictory word game. Potential energy supposedly is not energy, because the concepts are too mysterious to explain, while it supposedly is energy in conserving the energy which it absorbs. Like all word games, it is played in two contradictory ways.

When the definition of energy is corrected, the mysteries cannot be swept under a rug. When force times distance is not energy, then energy is not conserved through a lever. This means conservation laws must be modified to include amplification of energy where levers amplify force. And potential energy disappears, as force is one of the forms of energy.

A model for concepts is an elastic collision. As two objects move toward each other, they have energy of motion, and no force exists between them. As they collide, force between them develops and reduces the motion. At some point, all motion stops, and then the objects are pushed apart. When the motion stops, there is no kinetic energy, and all of the energy has been converted to force. Therefore, force is one of the forms of energy. Physicists call it potential energy, but that word game is dishonest.

So the driving force of the erroneous definition of energy is that physicists do not want a complex form of conservation laws which allows energy to be amplified through a lever. Not wanting correct concepts is not science. Trying to simplify away complexities destroys the ability of science to determine what the laws of nature really are.

Small minded persons will say no one sat down and conspired this. It took a few hundred years, but the decisions were arbitrated along the lines described above. For example, the Leibniz analysis which defined the existing concept of kinetic energy began with the force-distance combination for no explainable reason. Leibniz obviously liked the force-distance combination, because it does magical things for the analysis of levers.



The Test

A 4kg object dropped 1m (meter) has the same amount of ½mv² as a 1kg object dropped 4m, because force times distance equals ½mv² for an accelerating mass. But a rocket accelerating the masses to those velocities requires twice as much energy as fuel for the large mass as for the small one.

Therefore, both masses do not have the same energy; the rocket does not transform energy in proportion to ½mv²; ½mv² is not kinetic energy; and a gallon of fuel does not produce a consistent amount of ½mv².

--------------------------------------------------------------------------------
A 4kg object dropped for 1s (second) has the same amount of mv (momentum) as a 1kg object dropped for 4s, because force times time equals mv for an accelerating mass. A rocket accelerating the masses to those velocities uses the same amount of energy as fuel for both masses.

Therefore, both masses have the same amount of energy; the rocket transforms energy in proportion to mv; mv is kinetic energy; and a gallon of fuel produces a consistent amount of mv.

All logic and evidence of energy points to the same conclusion. The logic created the need to derive the mathematical proof.

About ninety percent of physics is corrupted by the error

Defining Energy

After showing mathematically that the energy equation is in error, there needs to be an explanation of what the error means. This page explains some of that.

What is the Definition of Energy.

Energy takes many forms. The first to be identified was kinetic energy. It is somewhat of a reference for defining energy, because it is tangible and easy to conceptualize.

Energy has been misdefined in that the formula for kinetic energy is incorrect as shown mathematically on other pages. On this page, the logic is described apart from the mathematics.

The formula is KE = ½mv². It indicates that the energy of motion is in proportion to mass times velocity squared. Squaring the velocity is the problem, because no mass can move at velocity squared. As a result, the formula is an abstraction apart from the motion of the mass. explanation

A similar contradiction in logic shows up in the force-distance form of the analysis. Supposedly, kinetic energy is proportional to force times distance for an accelerating mass. However, the force does not move through any distance relative to the mass it acts upon. Distance relates to the starting point, which the force does not act upon.

As indicated elsewhere (including collision analysis) transformations of kinetic energy need to be analyzed relative to the point where the energy acts (impact points), while force times distance creates a reference frame relative to the starting point. Errors result from the incorrect reference frame.

Kinetic energy should be represented as mv, which is called momentum. It is proportional to force times time for an accelerating mass.

This material is reduced to simple logic not to be pedantic but to show how the logic changes as the concepts are corrected. It also allows some original analyses to be derived from the logic.

Original Concepts.
The original basis for defining kinetic energy in terms of mv² was that momentum is supposedly not a conserved quantity. Later, momentum was considered to be conserved. So the historical basis for the definition of energy contradicts modern concepts.

It also contradicts the logic of Newton's laws which indicate that the force-time combination must be conserved through all interactions, because otherwise forces would not be equal and opposite. Force times time produces a defined amount of mv, not ½mv².


Shortly before Newton's time (1642-1727), Rene Descartes (1596-1650) drew the conclusion that there is a fixed amount of momentum in the universe, because it is conserved during interactions (momentum being mass times velocity, mv). From his studies of collisions, he concluded that changes in motion are produced by force, and force is quantitated in proportion to momentum divided by time (F = mv/t).

Newton then extended the concept stating three important laws. They are:



1. Force equals mass times acceleration.

2. For every force, there must be an equal and opposite force.

3. An object at rest stays at rest, or an object in motion maintains its motion unless a force acts upon it. (Which is inertia.)


At around that time, Leibniz (Gottfried Wilhelm von -) published a paper (in 1686) claiming to prove that momentum is not the conserved quantity of motion—mv² is. Eventually, Leibniz's view became the concept of kinetic energy in spite of contradictions with Newton's laws.

The issue was how to relate force to motion in order to get a quantity which is conserved. While force creates motion, one additional factor must be known to quantitate the results. One must know how long the force acts upon the mass. There are two alternatives for completing the quantitation. One alternative is to measure the time that the force acts; and the other is to measure the distance that the mass moves while the force is acting upon it. To multiply the force times time yields the resulting momentum (Ft = mv). To multiply the force times the distance yields the resulting ½mv² (Fs = ½mv²).

Eventually, the conclusion was drawn that both quantities are conserved, even though Leibniz directly stated that his analysis does not conserve momentum. During elastic collisions, both mv and ½mv² are conserved. But during inelastic collisions, both cannot be conserved simultaneously. The applications of energy function by the dynamics of inelastic collisions, because the force moves with the mass it is acting upon.

Some persons claimed that the disagreements were nothing but semantics. But in fact, the issue is not semantics, because energy is used in discrete quantities as fuel. Fuel will only produce a definable amount of mv, not ½mv².

For example, the rate at which energy is added to a system (energy divided by time) is called power. With mv divided by time, power becomes force only (mv/t = F). With ½mv² divided by time, power becomes force times velocity (½mv²/t = Fv).

 

[Dale]

From your electron diagram I would say that I am able to see more clearly what you mean.
Where there is more than one force acting on a body then each force has an equal and opposite reaction but the resultant acceleration also results in a fictitious force in the opposite direction IE INERTIAL FORCE. So therefore the same must be true even where there is one force acting on a body even though it is not so apparent.

So is this right?: Imagine if there were three planets A,B,C, and A and C each have 400 times the mass of B.

They come to be arranged in a triangular formation; whereupon they are each acted upon by each other’s gravity. A and C have equal and opposite forces, BC and AB also have equal and opposite forces. They all tend to accelerate toward each other. However B has a much greater acceleration in a direction toward the centre of AC and therefore has an inertial force in the opposite direction to the acceleration, which has no equal and opposite reaction force. Have I got it at last?

Yes, this is correct.

So what about the case where two bodies (A & B) act on each other with a non-elastic collision.

In this force the normal contact forces are real forces that are equal and opposite. Using the D'Alembert approach there are also (ficticious) inertial forces. Since the contact force is the only real force then the inertial forces are each equal and opposite to their respective contact forces, and therefore equal and opposite to each other, but this is just a coincidence by considering such a simple problem. That is why I started with the 3 electron example, it makes the distinction clear.

Ex1) They both have the same mass, A has momentum X and collides with B which has zero momentum. They stick together and the resultant velocity of A+B is the original momentum divided by the resultant mass. In the time it took for mass AB to achieve the final velocity, i.e. ½X, there was force applied = F*t (force impulse). Is it correct to say that during the force impulse there was equal and opposite reactions between A and B and that this action reaction was the inertial force of each.

No, the action reaction pair was the contact forces, as I mentioned above. In general the inertial forces are not equal and opposite.

Furthermore that it is not possible to define whether one or the other is action or reaction.

Yes, that is true in general for any action-reaction pair.

Or is it? If one body has momentum and the other does not or there is a relative momentum differential then there must always be a resultant momentum in the direction of the greatest momentum. IE (mv1)100kg * 20m/s V’s (mv2)50kg * 30m/s = resultant velocity in direction of mv1, so perhaps the larger momentum = the applied force. Which intuitively makes sense. So could I say that F*t = the relative change in momentum in the direction of the largest momentum (with respect to theta)

I don't know what you are saying here. Especially, what is theta?

If this is true then it is also true to say that A did work on B in terms of F*d.
The Kinetic energy of AB is 1/2mv^2 => (m(A+B)/2)* (½ X)^2 = ½ EkA
After A did the work on B, AB had half the original energy. Momentum conserved but energy isn’t in a non-elastic collision. How do we account for that lost energy? Especially when there is no energy lost in an elastic collision.

That is correct. Momentum is always conserved in any isolated collision, but kinetic energy is not conserved in a non-elastic collision. Instead, in a non-elastic collision some of the kinetic energy is used to do work deforming the colliding bodies, generating heat, etc.

The force time integral stays the same (I think, but I may have the maths wrong).

I'm not 100% sure, I have never worked that out explicitly, but my guess would be that the force time integral would be different. The total (vector) momentum of the system is conserved, but in the elastic collision each individual particle experiences a greater change in momentum than in the non-elastic collision. Since it experienced a greater change in momentum then it must have experienced a greater force time integral.

Does the Earth have momentum with respect to an object free falling by gravity towards it?

Sure, the Earth has momentum in any reference frame where it is moving.

If two bodies have equal and opposite force in terms of gravity then the object, say 200kg mass, attracts the Earth with the same force as the Earth attracts the object. However the Earth has a infinitesimally smaller acceleration than the object but they both would have the same momentum at time of impact.?

Only in their center of momentum frame. In any other frame they will have different momenta. However, their change in momentum will be the same magnitude in all reference frames.

One more question – If as mass accelerates the force increases (F=ma) then if a given mass decelerates to zero velocity in zero time, is this equal to infinite force or no force?

I will leave this one up to you to answer. What is the acceleration in this case? Remember acceleration is a change in velocity divided by a change in time.

I was pointing out that there are some people, who appear to be highly qualified in physics, that argue that the energy equation Ek = 1/2mv^2 (F*d) is wrong. But then when you read other work by the same people they also think many things are wrong and that there is some kind of conspiracy going on and perpetuated by the establishment to suppress the truth.

They are crackpots, as you should have been able to tell immediately from their conspiracy theories.

In any case, the energy equation cannot be wrong, it is a definition so it is correct by definition! I must say that your repeated failure to grasp this simple concept and your gullibility in accepting a conspiracy theorist as authoritative are much more worrisome than your understandable confusion by the D'Alembert approach. Use your own mind, how rational are the conspiracy theorists that you know personally, specifically when talking about the subject of their conspiracy theory?

 

[david smith]

Hey Dale

I suppose that concludes this thread. Just to say Isn't that the beauty of forums, one can ask or explore any question, whether profound, simple, stupid or silly but within the forum's theme and anyone can choose to answer, in any way they see fit. As you did very well, and I thank you for enlightening me.

All the best Dave


All the best Dave