Work and line integrals

[Rick16]

TL;DR

Taylor, example 4.1

I wonder what the work actually represents in this example. Normally, I would say that $W=\int{\vec{F} \cdot d \vec{r}}$ is the work done by the force F as it moves a particle along a path, where the path is determined by this very force. In this case, however, the force points in a direction that has nothing to do with the path of the particle. The particle obviously moves under the influence of some other force that constrains it to stay on its path. The force F does not push the particle along its path, it rather tries to push the particle away from its path. In doing so, it reduces the particle's kinetic energy, and by the work-KE-theorem this means that the force is doing work on the particle. But there is something strange about this work.

 

[wrobel]

There is nothing strange about this. For example, gravity also tries to pull a simple pendulum off its circular path, but that doesn't mean we can't calculate its work.

 

[Rick16]

Yes. Once again I posted a question without thinking it through. Everything is clear, sorry for the bother.

 

[kuruman]

Although I used Taylor many times to teach intermediate mechanics, I did not use his method. I showed my students how to do line integrals from first principles. The idea is simple and takes three steps:

1. Write force $\mathbf F$ and displacement $d\mathbf r$ in unit vector notation.
2. Form the dot product.
3. Integrate over variable $x.$

Example (modified from Taylor's 4.1)
Find the work done by force $\mathbf F = Ay~\mathbf{\hat x} +Bx~\mathbf{\hat y}$ along parabola $y=x^2$ from $x=0$ to to $x=x_0.$

Step 1
$\mathbf F = Ay~\mathbf{\hat x} +Bx~\mathbf{\hat y}$
$d\mathbf r = dx~\mathbf{\hat x}+dy~\mathbf{\hat y}=dx\left(\mathbf{\hat x}+\dfrac{dy}{dx}\mathbf{\hat y}\right)$
Along the given parabola these become,
$\mathbf F = Ax^2~\mathbf{\hat x} +Bx~\mathbf{\hat y}$
$d\mathbf r =dx\left(\mathbf{\hat x}+2x\mathbf{\hat y}\right)$

Step 2
$dW=\mathbf F\cdot d\mathbf r=Ax^2dx+2Bx^2dx=(A+2B)x^2dx$

Step 3
$W=(A+2B)\int_0^{x_0}x^2dx=\frac{1}{3}(A+2B)x_0^3.$

It works with any given path written as $y=f(x).$

 

[Rick16]

Although I used Taylor many times to teach intermediate mechanics, I did not use his method. I showed my students how to do line integrals from first principles. The idea is simple and takes three steps:

Thank you for your detailed explanation. This is not even new to me, I have done it like this myself before.

My question was actually due to a very stupid misconception. I first learned about work in 8th grade, where work was defined as force times distance and it was usually done by a guy pushing a cart or lifting a crate. The displacement was always in the direction of the force and it was always caused by this one and only force. And although I have since done calculations with work as a dot product, I apparently never thought it through what this actually meant. I still had this old 8th grade picture of work sitting in the back of my head, and when I looked at Taylor’s example, this old picture came up and acted as a conceptual barrier. That spell is now finally broken, I hope.

 

[Mike_bb]

Thank you for your detailed explanation. This is not even new to me, I have done it like this myself before.

My question was actually due to a very stupid misconception. I first learned about work in 8th grade, where work was defined as force times distance and it was usually done by a guy pushing a cart or lifting a crate. The displacement was always in the direction of the force and it was always caused by this one and only force. And although I have since done calculations with work as a dot product, I apparently never thought it through what this actually meant. I still had this old 8th grade picture of work sitting in the back of my head, and when I looked at Taylor’s example, this old picture came up and acted as a conceptual barrier. That spell is now finally broken, I hope.

You can't understand why work is dot product of force and displacement, right?

 

[Rick16]

You can't understand why work is dot product of force and displacement, right?

No, I do understand it. I just had this misconception about it, which I have finally clarified for myself.

 

[wrobel]

There’s a great concept called 'definition.' A definition is a definition.
The definition of work done is
$$\int_{t_1}^{t_2}(\boldsymbol F,\boldsymbol{\dot r})dt$$

 

[Rick16]

There’s a great concept called 'definition.' A definition is a definition. Sometimes a banana is just a banana. And a definition is always just a definition.
The definition of work done is
$$\int_{t_1}^{t_2}(\boldsymbol F,\boldsymbol{\dot r})dt$$

I am glad for you that you never have misconceptions.

 

[Mike_bb]

There’s a great concept called 'definition.' A definition is a definition.
The definition of work done is
$$\int_{t_1}^{t_2}(\boldsymbol F,\boldsymbol{\dot r})dt$$

I don't agree with this approach. I can say that by definition 2*2=5.

There is always a reason why this or that thing has meaning behind it.

 

[Rick16]

There is always a reason why this or that thing has meaning behind it.

Yes. I actually ran into my conceptual problem with Taylor's example, because I always blindly applied the definition of work as a dot product, without really thinking about what it concretely means. I know too many definitions that I have never thought through. I think the main reason why physics has to be learned through problem solving is exactly this: knowing definitions is by far not enough. It is only by applying the definitions that you begin to understand what they really mean.

 

[A.T.]

The force F does not push the particle along its path, it rather tries to push the particle away from its path. In doing so, it reduces the particle's kinetic energy, and by the work-KE-theorem this means that the force is doing work on the particle.

Applying a force "away from the path" as in "perpendicular to the path" is not doing any work. That is for example what the constraint force does, that constrains the object to stay on a path, no matter what other forces act on it. It's neither acting to reduce nor to increase the particle's kinetic energy.

Applying a force along the path but opposite to velocity is doing negative work (and reducing kinetic energy if no other forces do sufficient positive work to compensate the negative work done). But I don't see for which of the three paths (a, b, c) the force F would do negative work and act to reduce kinetic energy here.

 

[Rick16]

But I don't see for which of the three paths (a, b, c) the force F would do negative work and act to reduce kinetic energy here.

Neither do I. I completely misinterpreted the whole situation.

 

[kuruman]

And although I have since done calculations with work as a dot product, I apparently never thought it through what this actually meant.

This is a good example of why you don't really master a concept until you have to explain it to others, which means that first you have to examine it from all possible angles in order to explain it to yourself before you attempt to explain it to others.

In this particular case, a seed could have been planted by your 8th-grade teacher when you covered projectile motion. "When the acceleration and the velocity are in the same direction, the speed increases; when they are in opposite directions the speed decreases; when they are perpendicular to each other the speed doesn't change." This obvious idea blossoms into work, energy and the work-energy theorem when Newton's 2nd law comes along and establishes net force as mass times acceleration.

 

[wrobel]

There is always a reason why this or that thing has meaning behind it.

Most pointless debates arise precisely from searching for a 'meaning behind it'.
In fact, there are definitions and theorems that connect them, for example:

$$T\mid_{t=t_2}-T\mid_{t=t_1}=\int_{t_1}^{t_2}(\boldsymbol F,\boldsymbol v)dt,$$
$$ T:=\frac{m\vert{}\boldsymbol v\vert{}^2}{2}$$

 

[Matterwave]

Most pointless debates arise precisely from searching for a 'meaning behind it'. In fact, there are definitions and theorems that connect them, for example:

I could perfectly well define $H:=\frac{5}{3}m^2v^3$ and indeed I'm pretty sure I can prove some "theorems" about this, but who cares?

 

[kuruman]

I could perfectly well define $H:=\frac{5}{3}m^2v^3$ and indeed I'm pretty sure I can prove some "theorems" about this, but who cares?

Theorists?

 

[Matterwave]

Theorists?

I'd like to see some chime in to tell me they care about that particular expression.

But even if they did, it just means I didn't make my example absurd enough.

 

[wrobel]

I could perfectly well define H:=53m2v3 and indeed I'm pretty sure I can prove some "theorems" about this, but who cares?

That seemed like an objection, but I didn't quite catch it. Could you please clarify your point?

 

[Matterwave]

That seemed like an objection, but I didn't quite catch it. Could you please clarify your point?

There are reasons why we define things the way we do. Trying to understand those reasons is not a fools errand, but in fact a major point of learning and understanding in the first place.

Your post #15 seemed to suggest otherwise: that looking for meaning (I read this loosely as reasoning) in definitions is just the cause of confusion. If you meant meaning strictly as in "not a tautology" then fine, but if that's the case, I would say you made your point confusingly.

 

[wrobel]

There are reasons why we define things the way we do. Trying to understand those reasons is not a fools errand, but in fact a major point of learning and understanding in the first place.

Explain this to me with an example. In #15, I wrote down a formula derived from Newton's second law. This formula is useful for solving problems.
Beyond that (other than obtaining useful formulas), what reasons do you see to define kinetic energy and work in this specific way and no other?

 

[Matterwave]

Explain this to me with an example. In #15, I wrote down a formula derived from Newton's second law. This formula is useful for solving problems.
Beyond that (other than obtaining useful formulas), what reasons do you see to define kinetic energy and work in this specific way and no other?

For example, that expression of kinetic energy is the one that shows up in (non-relativistic) energy conservation laws. That expression of the work gives rise to conservative vs non conservative forces.

They don't have to be the only possible expressions. All I am claiming is that it is not a fools errand to look for reasons behind why we define things the way we do.

 

[Rick16]

This is a good example of why you don't really master a concept until you have to explain it to others

Should one infer from this that people who don't teach physics don't really understand it?

 

[wrobel]

For example, that expression of kinetic energy is the one that shows up in (non-relativistic) energy conservation laws. That expression of the work gives rise to conservative vs non conservative forces.

So it's all the same thing. Until you write a formula that connects kinetic energy with some other quantity, the formula $T=mv^2/2$ makes no more sense than that gibberish you came up with above.

 

[Matterwave]

So it's all the same thing. Until you write a formula that connects kinetic energy with some other quantity, the formula $T=mv^2/2$ makes no more sense than that gibberish you came up with above.

What do we mean by "meaning" if not the relation of ideas?

 

[wrobel]

What do we mean by "meaning" if not the relation of ideas?

I do not know what you understand by 'meaning'. You started by objecting to my phrase 'A definition is a definition.' Perhaps you see in a definition something more than just a definition. That is your point, not mine. I formulate theorems; I do not search for a sacred meaning in definitions as such.

 

[Matterwave]

I do not know what you understand by 'meaning'. You started by objecting to my phrase 'A definition is a definition.' Perhaps you see in a definition something more than just a definition. That is your point, not mine. I formulate theorems; I do not search for a sacred meaning in definitions as such.

You certainly didn't use that phrase. And I certainly didn't argue for "sacred meaning" in definitions.

But I will drop the argumentation here. It doesn't seem like the point is getting across.

 

[A.T.]

Neither do I. I completely misinterpreted the whole situation.

As a general note: You should not confuse "net work done" (which equals the change in energy) with "work done by an individual force" (which is its path integral). When only one force is doing work (like in this scenario), there is no difference, but scenarios are not always that simple.