How Things Move: Exploring Forces and 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