B Why Does Hooke's Law Use a Negative Sign in Scalar Form?

[Heisenberg7]

Assume that we have a block connected to a spring. Also, assume that there is no friction, the spring is massless and ideal. If we were to pull on the block with some force $\vec{F_{pull}}$, we are going to get the spring force $\vec{F_{s}}$ in the opposite direction. Assume that we are moving the block at a constant velocity. Let's now write the Newton's second law for the block: $$\vec{F_{pull}} + \vec{F_{s}} = m\vec{a_x} \implies \vec{F_{pull}} + \vec{F_{s}} = 0 \implies F_{pull} - F_{s} = 0 \implies F_{pull} - (-kx) = 0 \implies x = - \frac{F_{pull}}{k}$$ Now, here comes the problem. $x$ should be positive, not negative (right side of the 0; look at the picture). By, Hooke's law we have that $\vec{F_{s}} = -k\vec{x}$. This makes sense in the vector notation, but after converting it to the scalar notation, why do we need to preserve the minus(it just messes up the numbers)?

 

[berkeman]

Please rewrite this question using vectors and a coordinate system. Are you familiar yet with how to do that?

 

[Heisenberg7]

Please rewrite this question using vectors and a coordinate system. Are you familiar yet with how to do that?

Uhm, I'm sorry, I'm not sure I understand what you are asking me to do. I mean, I've written the equations using vectors. Do you mean like, dividing them into components?

 

[berkeman]

Uhm, I'm sorry, I'm not sure I understand what you are asking me to do. I mean, I've written the equations using vectors. Do you mean like, dividing them into components?

$\overrightarrow{F_{pull}}$, we are going to get the spring force $\overrightarrow{F_{s}}$ in the opposite direction. Assume that we are moving the block at a constant velocity. Let's now write the Newton's second law for the block: $$\overrightarrow{F_{pull}} + \overrightarrow{F_{s}} = m\overrightarrow{a_x} \implies \overrightarrow{F_{pull}} + \overrightarrow{F_{s}} = 0 \implies F_{pull} - F_{s} = 0 \implies F_{pull} - (-kx) = 0 \implies x = - \frac{F_{pull}}{k}$$ Now, here comes the problem. $x$ should be positive, not negative (right side of the 0; look at the picture). By, Hooke's law we have that $\overrightarrow{F_{s}} = -k\overrightarrow{x}$.

Use the bold vector notation in LaTeX instead of the over-arrow, and be consistent with the vector notation through your work. Why did your attempted vector notation disappear as your equation moved along?

 

[Heisenberg7]

Use the bold vector notation in LaTeX instead of the over-arrow, and be consistent with the vector notation through your work. Why did your attempted vector notation disappear as your equation moved along?

Oh, okay. The reason why it disappeared is because I simply summed up the magnitudes of the vectors, or in this case, it's gonna be a minus.

 

[berkeman]

$$\vec F = m \vec a$$
$$\vec F = \vec F_s + \vec F_p$$
$$\vec F = -F_s \hat x + F_p \hat x$$
and so on...

 

[kuruman]

Assume that we have a block connected to a spring. Also, assume that there is no friction, the spring is massless and ideal. If we were to pull on the block with some force $\vec{F_{pull}}$, we are going to get the spring force $\vec{F_{s}}$ in the opposite direction. Assume that we are moving the block at a constant velocity. Let's now write the Newton's second law for the block: $$\vec{F_{pull}} + \vec{F_{s}} = m\vec{a_x} \implies \vec{F_{pull}} + \vec{F_{s}} = 0 \implies F_{pull} - F_{s} = 0 \implies F_{pull} - (-kx) = 0 \implies x = - \frac{F_{pull}}{k}$$ Now, here comes the problem. $x$ should be positive, not negative (right side of the 0; look at the picture). By, Hooke's law we have that $\vec{F_{s}} = -k\vec{x}$. This makes sense in the vector notation, but after converting it to the scalar notation, why do we need to preserve the minus(it just messes up the numbers)?

When you write
$\vec{F_{pull}} + \vec{F_{s}} = 0 \implies F_{pull} - F_{s} = 0$, you are replacing the vectors with their magnitudes and you put a negative sign in them when they are both on the same side of the equation. It is the same as saying $F_{pull} = F_{s}$ or $10 =10$. That's OK so far.

What you cannot do is say that the positive number $F_s$ is the same as the negative number $-kx.$ What you can say is that $F_s=k|x|.$

 

[Chestermiller]

The tension in the spring should be $T=k\Delta L$ where $\Delta L$ is the change in length relative to the unextended configuration. The tension always pulls in the direction away from the block.

 

[hutchphd]

Oh, okay. The reason why it disappeared is because I simply summed up the magnitudes of the vectors, or in this case, it's gonna be a minus.

Hookes Law is a vector equation. In one dimension it is a 1D vector equation, not a scalar equation. Also in 1D Newton's Law is a 1D vector equation. That is actually all that needs to be said.
If you understand what a vector is, the signs follow without handwaving.

 

[kuruman]

Hookes Law is a vector equation. In one dimension it is a 1D vector equation, not a scalar equation.

The vector equation for Hooke's law is,
$\mathbf F=(-kx)~\mathbf {\hat x}$
and has vectors on the two sides.

Since the two vectors are equal, the x-component on the left-hand side must be equal to the x-component of the vector on the right-hand side,
$F_x=-kx.$
Components of vectors are scalars which makes the latter a scalar equation.

 

[hutchphd]

Components of vectors are scalars

We seem to have foundered on the Semantic Sea. The good news is that we are not alone out here

https://www.physicsforums.com/threads/what-the-hell-are-vector-components.372639/

I think the temptation to incorrectly extract the "scalar equation" makes for unnecessary confusion.

 

[A.T.]

We seem to have foundered on the Semantic Sea. The good news is that we are not alone out here

https://www.physicsforums.com/threads/what-the-hell-are-vector-components.372639/

I think the temptation to incorrectly extract the "scalar equation" makes for unnecessary confusion.

The OP's error was correctly pointed out in post #7: A vector magnitude, which is always positive, was replaced by a signed vector component.

 

[Chestermiller]

For a solid rod experiencing a homogeneous axial extension or compression, the stress tensor (in dyadic tensor notation) is $$\boldsymbol{\sigma}=E\frac{\Delta L}{L}\bf{i_xi_x}$$The force exerted by the portion of the rod at $x^+$ on the portion of the rod at $x^-$ is $$\mathbf{F}^+=(A\boldsymbol\sigma)\centerdot \mathbf{i_x}=AE\frac{\delta L}{L}\mathbf {i_x}$$. The force exerted by the portion of the rod at $x^-$ on the portion of the rod at $x^+$ is $$\mathbf{F}^-=(A\boldsymbol\sigma)\centerdot (\mathbf{-i_x})=-AE\frac{\delta L}{L}\mathbf {i_x}$$

The same type of formalism can be used for a spring to automatically guarantee that the sign of the force is correctly determined.

 

[Herman Trivilino]

This makes sense in the vector notation, but after converting it to the scalar notation, why do we need to preserve the minus(it just messes up the numbers)?

You are correct. This problem arises in physics textbooks and lectures quite often, and also in other contexts. The source of the problem is failure to distinguish between vectors, vector components, and vector magnitudes.

If $\vec{F}=-k\vec{x}$ then the correct relation between components would be $F_x=-kx$, assuming one dimension along the  x-axis, and the correct relation between magnitudes would be $F=k|x|$.

 

[kuruman]

If $\vec{F}=-k\vec{x}$ then the correct relation between components would be $F_x=-kx$, assuming one dimension along the  x-axis, and the correct relation between magnitudes would be $F=k|x|$.

Your clarification is confusing because your notation does not treat the two vectors consistently. You have adopted the convention of writing the force vector as $\vec F$, its component as $F_x$ and its magnitude as $F$. That's fine.

However, consistency of notation would require you to write the displacement vector as $\vec x$, its component as $x_x$ and its magnitude as $x$. Yes, I know, the x-component looks silly but how can one otherwise distinguish $x$ the component from $x$ the magnitude? Perhaps using $x$ to denote both the component and the magnitude is the reason behind some people's confusion with the interpretation of $F=−kx$. So how does one get around this?

Clearly, component subscripts are redundant when one deals with one-dimensional vectors. Nevertheless it should be clear what one means by a symbol without an super-arrow since the magnitude of a one-dimensional vector is the same as the magnitude of its component.

The vector equation $\vec F=−k\vec x$ is unambiguous.

If we call $F$ the 1-D component of $\vec F$ and $x$ the 1-D component of $\vec x$ (both of which could be positive or negative), then $F=−kx$ is unambiguous and says the same thing as the vector equation. It is a relation between the scalar components of the 1-D vectors.

However, the non-subscripted $F$ or $x$ cannot also be viewed as magnitudes because magnitudes are always positive. The unambiguous expression for the relation between the magnitudes of the 1-D vectors is either $|\vec F|=k|\vec x|$ or $|F|=k|x|$ but not $F=kx.$

 

[Herman Trivilino]

However, consistency of notation would require you to write the displacement vector as x→, its component as xx and its magnitude as x.

But I've not seen it written that way, anywhere.

Yes, I know, the x-component looks silly but how can one otherwise distinguish x the component from x the magnitude?

By not writing the displacement vector as $\vec{x}$. No college-level introductory physics textbook does that, that I've seen. The first and only time I saw it is in Jackson in grad school, and I thought it stupid. He used it in such a way that its  x-component is $x$, its  y-component is $y$, and its z-component is $z$.

 

[kuruman]

By not writing the displacement vector as $\vec{x}$. No college-level introductory physics textbook does that, that I've seen.

The confusion arises when textbooks abandon the convention when they deal with one-dimensional vector equations. They abandon the notation they have established for 2D and 3D and write $F=-kx$ without clarifying how one thinks of the symbols. By the time they get to springs, almost everybody knows that forces are vectors. However, one may naturally wonder whether symbol $F$ stands for a vector or for a magnitude.

At the introductory level I think it is sufficient to not mention vectors and explain that $F=-kx$ says that "if $F$ is positive, $x$ must be negative and if $F is negative$, $x$ must be positive. In other words, the two point in opposite directions regardless of whether one has chosen to the left or to the right as positive." It might not hurt to add, "this is similar to something you've seen before. The force of kinetic friction is always opposite to the velocity regardless of whether one has chosen to the left or to the right as positive." If, at that point, an inquisitive student asks, "Yes, but is $F=-kx$ a vector equation since $F$ is a vector?", my answer will be. "Yes, and one would write it formally as $\vec F = -k\vec {x}$ which says that the vector $\vec F$ points in the opposite direction as the vector $k\vec{x}.$"

The first and only time I saw it is in Jackson in grad school, and I thought it stupid. He used it in such a way that its  x-component is $x$, its  y-component is $y$, and its z-component is $z$.

Yes, Jackson uses $\mathbf x$ instead of $\mathbf r$ which felt strange when I saw it in grad school like you. I got over it because a symbol is a symbol. Maybe Jackson's intention was to liberate generations of grad students from thinking inside the box where lower case letter x can only stand for a Cartesian component of a vector.

 

[renormalize]

By not writing the displacement vector as $\vec{x}$. No college-level introductory physics textbook does that, that I've seen. The first and only time I saw it is in Jackson in grad school, and I thought it stupid. He used it in such a way that its  x-component is $x$, its  y-component is $y$, and its z-component is $z$.

Yet using that notation is quite common when dealing with Green's functions. From https://en.wikipedia.org/wiki/Green's_function:

Is there any good reason to not introduce this notation in introductory textbooks? What is sacrosanct about using (say) r vs. x?

 

[kuruman]

Is there any good reason to not introduce this notation in introductory textbooks? What is sacrosanct about using (say) r vs. x?

One has to recognize that translating the world around us into mathematical equations is an unnatural act to most people. When novices are asked to do this translation in introductory courses, it is comforting to them to stick to unvarying conventions like "$x$, $y$ and $z$ always stand for position coordinates." The line has to be drawn somewhere, but it's moveable within reason as one becomes more sophisticated mathematically.

In principle, nothing is sacrosanct and one can choose any symbol for any quantity as long as one defines it properly. However, there would be utter confusion if one defines $\mathbf P$ as "Force", $m$ as "mass" and $\mathbf v$ as "acceleration" and writes Newton's second law as $$\mathbf P_{\text{net}}=m~\mathbf v.$$

 

[PeroK]

Your clarification is confusing because your notation does not treat the two vectors consistently. You have adopted the convention of writing the force vector as $\vec F$, its component as $F_x$ and its magnitude as $F$. That's fine.

However, consistency of notation would require you to write the displacement vector as $\vec x$, its component as $x_x$ and its magnitude as $x$. Yes, I know, the x-component looks silly but how can one otherwise distinguish $x$ the component from $x$ the magnitude?

There are many examples where we use the flexibility of our intelligence in order to used simplified or slightly inconsistent notation. In this case, we always have to write $|\vec x|$, as $x$ is already used. But, for most of us it's more natural to keep using $x, y,z$ as the components of $\vec x$.

For example, when we have a function on the real numbers, we write $f(x)$ and in general this applies where the domain of $x$ is any set. Except the natural numbers. In which case, we call it a sequence (not a function) and write $a_n$ instead of the more consistent $a(n)$.

Most examples have the common theme that they use an inconsistent notation for a special case. This seems to be in tune with the way many of us think naturally. Although, you could make a pedagogical case for avoiding these inconsistencies altogether.

 

[renormalize]

When novices are asked to do this translation in introductory courses, it is comforting to them to stick to unvarying conventions like "$x$, $y$ and $z$ always stand for position coordinates."

Agreed. But when spherical coordinates are first encountered in an introductory text, the position notation $\mathbf{r}\equiv\left(r,\theta,\phi\right)$ is often used without causing confusion. Why not also introduce $\mathbf{x}\equiv\left(x,y,z\right)$ in anticipation of possible future advanced courses?

 

[kuruman]

There are many examples where we use the flexibility of our intelligence in order to used simplified or slightly inconsistent notation. In this case, we always have to write $|\vec x|$, as $x$ is already used. But, for most of us it's more natural to keep using $x, y,z$ as the components of $\vec x$.

Agreed. But who is "we"? I make a distinction between experienced practitioners who've seen and taught it and novices who see it for the first time. The latter may be as intelligent as the former but they lack the flexibility you mentioned because they haven't seen it all yet and don't know what lies ahead. I think that the incremental approach to learning is to first put them in a box and let them become familiar with it, then open windows to show that there is a world outside and finally open a poor and let them out to explore on their own.

Agreed. But when spherical coordinates are first encountered in an introductory text, the position notation $\mathbf{r}\equiv\left(r,\theta,\phi\right)$ is often used without causing confusion. Why not also introduce $\mathbf{x}\equiv\left(x,y,z\right)$ in anticipation of possible future advanced courses?

Because students tend to forget what they have seen and used in one course, e.g. the work-energy theorem in mechanics, when they take the next course, e.g. E&M. They cannot be trusted to remember something that they saw but never used. The usual procedure is to strike while the iron is hot. You explain the material, you do an example or two, you assign homework, you put it on the next test and hope for the best.

 

[PeroK]

Agreed. But who is "we"? I make a distinction between experienced practitioners who've seen and taught it and novices who see it for the first time. The latter may be as intelligent as the former but they lack the flexibility you mentioned because they haven't seen it all yet and don't know what lies ahead. I think that the incremental approach to learning is to first put them in a box and let them become familiar with it, then open windows to show that there is a world outside and finally open a poor and let them out to explore on their own.

There may be no best way to do it. Some students might throw a wobbler at something like $x_x$, however rational that might be. They might prefer $(x, y,z)$.

The limited evidence I have from this forum is that the students who have to be spoon fed everything never really progress. Whereas, the bright ones take this sort of thing in their stride and move on. If a student is stumped by $\vec x = (x, y, z)$, then I suspect there is little hope for them.

 

[Herman Trivilino]

Your clarification is confusing because your notation does not treat the two vectors consistently. You have adopted the convention of writing the force vector as $\vec{F}$, its component as $F_x$ and its magnitude as $F$. That's fine.

Yes. I always stressed to my students that we have to keep these three things distinct and in the fore:

1. Vectors.
2. Vector components.
3. Vector magnitudes

Strictly speaking we should always follow the mathematician's habit of writing the magnitude of $\vec{F}$ as $|\vec{F}|$, but we don't, and our textbooks don't either.

I stress the following:

1. Vectors are neither negative nor positive.
2. Vector components can be either negative or positive.
3. Vector magnitudes are never negative.

So,
1. $\vec{F}$ is a vector. As such, it is neither negative nor positive, and it is a mistake to substitute a numerical value for it in any equation, whether it be a positive number or a negative number.

2 . $F_x$ is a vector component. As such, a student can substitute a negative or positive number for it in any equation.

3. $F$ is a magnitude. As such, as student can never a substitute a negative number for it.

However ,consistency of notation would require you to write the displacement vector as $\vec{x}$, its component as $x_x$ and its magnitude as $x$.

Well, one could write $\vec{F}=−k \vec{r}$, $F_x=−kr_x$, and $F=k|r_x|$; but it is not conventional to write $r_x$ as the x-component of $\vec{r}$, but is instead conventional to write $x$ as the x-component of $\vec{r}$.

My experience has taught me that the most meaningful expression for students is $F_x=−kx$. They understand that $F_x$ is a vector component; and it can be substituted with zero, a positive number, or a negative number. On the other hand, $x$ is a coordinate and therefore it can be substituted with zero, a positive number, or a negative number. Thus the minus sign in the relation $F_x=−kx$ means that $F_x$ and $x$ will always have opposite signs.

IIRC I have seen you use the expression "one-dimensional vectors". I assume you are taking that from the study of rectilinear motion with constant acceleration, for example, in relations such as $v=v_o+at$. Well, first of all, all vectors can be made one-dimensional with the proper choice of coordinate axes. But more to the point equations like this are not, to my way of thinking, vector equations. They are relations between vector components. Thus equations like the above are actually component equations. They can't be vector equations because students substitute numerical values for the variables, and they can't be magnitude relations because the students often have to substitute negative values for the variables.

This relation, for example, should be written as $v_x=v_{ox}+a_x t$, as Randy Knight does in his textbook. But I teach rectilinear motion before vectors, and I've found that the use of all these subscriptsis too confusing. An alternative would be to teach vectors first, and then move on to rectilinear motion, making use of these subscripts in these relations between vector components, but that means students learn vectors, then don't see their utility until after rectilinear motion is taught, and that delay causes problems.

Anyway, the vector form, for example, of the above equation is $\vec{v}=\vec{v_o}+\vec{a} t$. When studying two-dimensional motion, this vector equation is written as two separate component equations: $v_x=v_{ox}+a_x t$ and $v_y=v_{oy}+a_y t$, and students are finally exposed to a formalism where the utility of the distinction between vectors and vector components is meaningful.

My goal was always the improvement of student comprehension for the largest number of students possible.

Of course, this is all according to the conventions I've developed based on my way of thinking. I'm not advocating that others have to adopt them. I'm just explaining what I do and why. Before retiring in 2020 I had 34 consecutive years of teaching introductory college-level physics in a full-time capacity. I have a Master's Degree in Physics, attended an intensive and excellent faculty-enhancement program for two consecutive summers with AAPT meetings attended and supplemented each fall and spring semester, and have taken over 400 hours of NSF-sponsored faculty enhancement workshops.

I'm not trying to toot my own horn here, I'm just trying to convey the level of thought and experience I've put into my way of thinking and the conventions I've adopted.

 

[Herman Trivilino]

What is sacrosanct about using (say) r vs. x?

Nothing sacrosanct. It's just an issue of trying to get as many students as possible to comprehend. It's all a matter of preference.

 

[kuruman]

IIRC I have heard you use the expression "one-dimensional vectors". I assume you are taking that from the study of rectilinear motion with constant acceleration, for example, in relations such as v=vo+at.

Here is what I mean and how I think of a one-dimensional vector. Consider vector $\mathbf A$ fixed in space. It has three components relative to an arbitrary set of axes. Rotate the axes so that one of them is along $\mathbf A$. Vector $\mathbf A$ is now a one-dimensional vector.

Whether we realize it or not, we perform this preparatory rotation every time we write down one-dimensional kinematic equations. Clearly, it's not something that introductory physics students need to know at that time. The smart ones will be able to put it together when they learn that vector components depend on the choice of axes. The idea is reinforced and mastered when they see the value of choosing one axis parallel to the acceleration (and the net force) in FBDs.

 

[PeroK]

Although there are three spatial dimensions, the use of 2D or 1D vectors for systems confined to a plane or line must be valid. Mathematically, a vector need not have a minimum of three components.

 

[Herman Trivilino]

Whether we realize it or not, we perform this preparatory rotation every time we write down one-dimensional kinematic equations.

I don't. When I write the one-dimensional kinematic equations I'm writing down the relations between components of that vector. And there are cases where we use those same component equations for motions that are not one-dimensional. See the example I gave in my last post.

You're having your students substitute numerical values for some of the quantities in those equations, and having them solve for numerical values for others. One can never validly substitute a numerical value for a vector quantity, because vector quantities have both a magnitude (always non-negative) and a direction. IMO they are forming a bad habit that will confuse many of them as their education proceeds, even if they are unable to articulate that confusion.

 

[kuruman]

I don't. When I write the one-dimensional kinematic equations I'm writing down the relations between components of that vector.

But you do. When you write "down the relations between components of that vector" why don't you do it along an axis that is at some angle $\theta$ with respect to the direction of motion? Because things need to be kept simple. Nevertheless, you have made the choice of an axis because you cannot write vector components unless you first specify axes, one here but there could be two or three as needed.

When you write $~y=y_0+v_0t-\frac{1}{2}gt^2$, you have already (a) chosen the axis parallel to the acceleration of gravity, as the one in which to write the vector components of the initial position, velocity and acceleration and (b) chosen the negative direction to be the same as the direction of the acceleration of gravity. Note that this choice is local to where the motion takes place because the direction of the acceleration of gravity near the surface of the Earth depends on longitude and latitude.

And then there is the recurrent question, "if I substitute numbers for $g$ in the equation, do I use +9.8 m/s² or -9.8 m/s² ?"

 

[Herman Trivilino]

When you write "down the relations between components of that vector" why don't you do it along an axis that is at some angle θ with respect to the direction of motion?

I do, in the case of two-dimensional motion. Of course I choose $\theta$ to be zero in the case of one-dimensional motion, so I take your point there. But I'm still writing a relation between components.

 

[Herman Trivilino]

And then there is the recurrent question, "if I substitute numbers for g in the equation, do I use +9.8 m/s² or -9.8 m/s² ?"

Years ago I was using a textbook that forced the students to make that choice. In other words, $g=\pm 9.8 \ \mathrm{m/s^2}$. That was a huge mistake.