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Suppose I have a body under free fall. I wish to write the equation of motion. There is a force of gravity $mg$ acting in the downward direction. Let us consider two unit vectors $\hat{u}$ and $\hat{d}$ corresponding to the up and down directions. Hence $\hat{u}=-\hat{d}$

Suppose I take the downward direction to be positive. In that case I can write:

$$\sum F_d=mg=ma$$

Hence, my equation of motion, in the downward positive case is :

$$m\ddot{x}=mg$$

Hence, $$x=ut+\frac{1}{2}gt^2$$

As you can expect, $x$ is positive, and increases with $t$. This is to be expected, as the body falls downward, the displacement in the downward direction must be positive and increase with time.

Now suppose, I take the upward direction as positive. In the upward direction, the force of gravity is $-mg\hat{u}$ i.e. it takes a negative sign. My new force equation becomes :

$$\sum F_u=-mg=m\ddot{x}$$

Solving this, we get :

$$x=-(ut+\frac{1}{2}gt^2)$$

Again, this is to be expected. If we take the upward direction to be positive, the displacement comes out to be negative, since the body is moving down.

The important thing to notice here is that the equations of motion are different depending on which side I've taken positive. Moreover, if I change the direction, the answer becomes negative of what we obtained before changing the direction.

This all seems well and good, until I run into a different problem, where the force is dependent on the displacement of the body in question. Suppose, the body is now vertically hanging by a spring, and we have pulled it and released.

If we take the downward direction as positive, then the displacement of the body is positive. The resultant restoring force works in the upward direction, while gravity works in the downward direction as before. We can write the gravity as $mg\hat{d} $ or $-mg\hat{u}$. Similarly, we can write the restoring force as $-k(x+x_0)\hat{d}$ or $k(x+x_0)\hat{u}$.

I can now write the force equation in any direction. In the upward direction, I can write : $$\sum F_u = -mg + k(x+x_0)=ma$$

Simplifying, I get :

$$ma=kx$$

In the downward direction, I can write :

$$\sum F_d = mg-k(x+x_0)=ma$$

Simplifying, I get :

$$ma=-kx$$

According to what we did in case of the free fall, this would be correct. However, if we try to solve this equation of motion, we'd get two very different answers for the two cases. In this answer by @Farcher, changing the direction doesn't change the equation of motion, it just changes the boundary condition. According to this answer, if I take the upward direction as positive, the displacement of the body becomes negative, and so, the acceleration must also become negative. Hence, the equation of motion remains the same. The boundary conditions remain the same though. So, if the answer in the downward positive comes out to be some number, in the upward positive condition, the answer would be negative of the same number as expected.

What I can't understand is, in the example of the free fall, I didn't take the direction of acceleration into consideration. As I considered the upward direction to be positive, the acceleration was still $\ddot{x}$. The forces in the RHS got an extra negative sign in front of it, but the acceleration remained the same. The force in the LHS was always $m\ddot{x}$ irrespective of which direction is taken positive. Hence, the equations of motion were different for the upward and downward positive cases.

In the SHM example however, I'm expected to change the sign of acceleration, and make it $-\ddot{x}$ if I make the upward direction positive. This cancels out the negative sign in front of the forces, and so the equation remains the same. In this example, the force on the LHS is no longer $m\ddot{x}$ irrespective of direction. In fact, it changes sign based on direction.

Why do these problems seem to follow different rules, or is there something obvious that I'm missing here ? Any help would be appreciated.

EDIT : In my previous questions on this topic, I had asked how to formulate and solve the equation of motion in a spring-block system, if I change the direction that I'm going to consider positive. Over there, changing the positive direction doesn't affect the equation of motion under question. However, here I'm comparing this to the case of free fall, or other problems where the force doesn't depend on displacement. In these cases, changing the direction does change the equation of motion. My question is, why are the two types of problems behaving differenty.

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    $\begingroup$ This is the third time you've asked the same question today. I don't think you're reading the previous answers, so I'm voting to close. $\endgroup$ Commented Jan 4, 2022 at 5:35
  • $\begingroup$ @Chemomechanics I don't see how this is the same question. I agree it is a question about signs of forces. But it clearly compares two different cases, and asks why is it different. For example, in the answer that I linked, if I simply put $v=0$, I'd get the wrong answer if I change my positive direction and keep the equation of motion the same. $\endgroup$
    – RayPalmer
    Commented Jan 4, 2022 at 5:39
  • $\begingroup$ @Chemomechanics I'm just asking, why in some problems, if I change my positive direction, the equation of motion and boundary conditions changes, and in some other problems, changing the positive direction doesn't change the equation of motion, only the boundary conditions. I don't see how that is a repeat of my original question. $\endgroup$
    – RayPalmer
    Commented Jan 4, 2022 at 5:41
  • $\begingroup$ You can re-edit your previous questions. $\endgroup$ Commented Jan 4, 2022 at 5:56
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    $\begingroup$ In all three problems, you've chosen to use multiple definitions for the same variable $x$ (displacement, positive distance) and then encountered a contradiction. Consensus physics is based on defining variables uniquely. Writing, reading, analyzing and answering these long redundant examples is a waste of everyone's time. You won't encounter contradictions if you apply consistency in what $x$ is meant to measure. $\endgroup$ Commented Jan 4, 2022 at 7:08

2 Answers 2

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. . . or is there something obvious that I'm missing here?

Indeed, you are missing the important ideas as to the difference between a component of a vector and its magnitude.

If you look in textbooks there is only one equation of the type $x=ut+\dfrac 12 gt^2$ and that is it.
The equation is independent of the choice of direction because $x, \, u$ and $g$ are components of three vectors in a chosen direction.

Perhaps working example will help?

A body is thrown down at $5 \,\rm m\,s^{-1}$ in a region where the acceleration due to gravity is $10 \,\rm m\,s^{-2}$ downwards. How far does the body travel in $2\,\rm seconds$?

Using unit vector down, $\hat d$.
$x\hat d=u'\hat dt+\dfrac 12 g\hat dt^2$ with $u'\hat d = +5\hat d$ and $g = +10 \hat d \Rightarrow x\hat d = +30\hat d =30 (+\hat d)$.
So the body travels $30\,\rm m$ downwards in $2\, \rm seconds.$

Using unit vector up, $\hat u$.
$x\hat u=u'\hat ut+\dfrac 12 g\hat ut^2$ (note the same equation as for $\hat d$) with $u'\hat u = -5\hat u$ and $g = -10 \hat u \Rightarrow x\hat u = -30\hat u =30 (-\hat u)$.
So, as calculated before, the body travels $30\,\rm m$ downwards in $2 \,\rm seconds.$

The equations that you have derived require you to put in the magnitudes of $u$ and $g$ into the equations and that is why you have two different equations.

Perhaps you will gain an insight to what I have written by proving to yourself that the equation $ma = -kx$ is good for all definitions of a relevant unit vector having noted that $a$ and $x$ are components of vectors not magnitudes?

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  • $\begingroup$ Yes thank you so much. I realized this, and was in the process of writing a long answer comparing all of your other answers on this topic. This reminds me that, while going through your answers on similar questions, particularly this answer, I think you have missed a negative sign in front of $mg$, when you have written the equation after taking the positive direction to be upwards. $\endgroup$
    – RayPalmer
    Commented Jan 4, 2022 at 8:27
  • $\begingroup$ In this other answer you've correctly shown that if downward direction is positive, then $ma=-bv+mg$, where $a$ and $v$ are components in the downward direction. However if the upward direction is positive, $ma=-bv-mg$, where $a$ and $v$ are components in the upward direction. In the previous link, you have not given this negative sign in front of $mg$ while changing from downward to upward positive direction. $\endgroup$
    – RayPalmer
    Commented Jan 4, 2022 at 8:30
  • $\begingroup$ Ah now I get it, in the first answer, you considered the component of gravity and in the second one, you considered the magnitude. So the sign changed in the second answer. Sorry for this inconvenience. $\endgroup$
    – RayPalmer
    Commented Jan 4, 2022 at 9:11
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    $\begingroup$ @RayPalmer No problem. How else are you going to learn if you do not ask questions? $\endgroup$
    – Farcher
    Commented Jan 4, 2022 at 16:15
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Based on some further digging, and this wonderful answer, I think I can pinpoint the sources of error, and write an answer for myself.

Let us return to our body under free fall. Considering $\hat{d}$ and $\hat{u}$ to be unit vectors in the down and up directions. The force of gravity can be written as $\vec{F_g}=mg\hat{d}/mg(-\hat{u})$.

Let us consider the upward direction to be positive. In the upward direction, let us resolve the forces using Newton's second law : $$\sum\vec{F}_u=mg(-\hat{u})=m\ddot{x}\hat{u}$$

Hence, we have obtained : $m\ddot{x}=-mg$. This shows that the acceleration or more importantly the component of acceleration in the upward direction is negative. This implies, that acceleration must be in the downward direction as expected.

If we now want to consider the downward direction to be positive, then the force relation becomes :

$$\sum\vec{F_d}=mg\hat{d}=m\ddot{x}\hat{d}$$

Hence, we have obtained $m\ddot{x}=mg$. However, as you may notice, here $\ddot{x}$ represents the component of acceleration in the downward direction, and this is positive. In the next examples, if I take $\hat{u}$ to be positive, then I'm going to use $\ddot{x}_u $ to denote component of acceleration in the positive direction, and so on.


Now, let us move on to a second example. This time we shall analyze the restoring force due to a spring and see the equations of motion. Suppose, we now use the unit vectors $\hat{l}$ and $\hat{r}$ in the left and right directions relative to the mean position.

The displacement of the spring can be considered to be $\vec{x}=x_l\hat{l}/x_r\hat{r}$. Remember that $x_l$ and $x_r$ denote the components and not the magnitude. Hence they can take both positive and negative values. For example, if $x_r$ is positive, that means displacement is positive to the right, which implies stretching.

Now, according to Hooke's law, the restoring force $\vec{F}_r=-k\vec{x}$

If I now take the left side to be positive, I can write : $$\sum\vec{F}=-kx_l\hat{l}=m\ddot{x}_l\hat{l}$$

Similarly, if I take the right side to be positive, I can write : $$\sum\vec{F}=-kx_r\hat{r}=m\ddot{x}_r\hat{r}$$

In both cases, I obtain the equation $m\ddot{x}_l=-kx_l$ or $m\ddot{x}_r=-kx_r$. Hence, the solutions are sinusoidal in both cases. However, due to boundary conditions, depending on whether the string was stretched or compressed and so on, we can say that $x_l(t)=-x_r(t)$.


This solution seems slightly different from the one that we obtained in this question. There is a small issue regarding how I have reasoned about, in this linked question. I had considered a compressed spring and taken the right direction to be positive.

Another important thing to mention, is that I had considered the final displacement to be some $-s\hat{r}$ and initial displacement to be, well, $0\hat{r}$. This made sense to me, since the spring was compressed to the left of the origin. The problem here is, the moment I defined the displacement to be $-s\hat{r}$, $s$ no-longer means the component of displacement in the $\hat{r}$ direction. Instead, it now means the magnitude of displacement, or rather the distance from the origin. You can also say that this $s$ now represents the component in the left direction.

Now, this magnitude cannot be negative. So, if we want to solve this, we have to solve the problem in a piecewise manner, as @Chemomechanics suggested in an answer to that question.

In order to solve this problem, there are three ways to procced. Method $1$ is solving the problem piecewise for positive and negative displacement : $s\hat{r}$(extension) and $-s\hat{r}$(compression), and note that $s>0$. Basically $s=|x_r|$, since we have taken right side as positive.

Method $2$ is notice that displacement in the $\hat{r}$ direction is $x_r\hat{r}$, and so, $x_r\hat{r}=-s\hat{r}$(compression) and $x_r\hat{r}=s\hat{r}$ (stretching). By replacing the magnitude $s$ with the component $x_r$, we don't need to solve the equation piecewise anymore. $x_r$ can take both positive and negative values.

Method $3$ is simple re-labelling, where we change the way a vector is defined. If we define displacement in the $\hat{r}$ direction to be negative i.e. $\vec{x}=-s\hat{r}$, then $s$ stops being the magnitude and becomes the component again, as we have redefined displacement. Hence $\vec{\ddot{x}}=-\ddot{s}\hat{r}$ and $\vec{\dot{x}}=-\dot{s}\hat{r}$. One might ask how we can redefine displacement in such a way. The answer is, we can write displacement $\vec{x}$ as $s_l\hat{l}$ or $s_r\hat{r}$, Here $s_r$ and $s_l$ can take both positive and negative values, but since it represents the same vector, the componenents must have opposite sign. So, $\vec{x}=s_r\hat{r}=-s_l\hat{r}=s_l\hat{l}$.

Notice the second one. It is of the form $\vec{x}=-s\hat{r}$. Our restoring force $-k\vec{x}$ now becomes $ks\hat{r}$. However, since there is a negative sign on acceleration, our equation is still that of a harmonic oscillator. However, remember, we are now solving for $s_l$ i.e. component of displacement in the $\hat{l}$ or in this case, the negative direction. Hence, the answer would have a negative sign, because of the flipped boundary conditions.

This third method is used in this answer and this answer


To wrap all of this up, let us look at a final problem. Now we have a mass hanging from a spring, under the force of gravity. We also have a damping force present. Essentially we are looking at a damped harmonic oscillator. A very similar problem has been solved by @Farcher in this answer and this answer, where the problem of free fall in a viscous medium has been dealt with.

In our problem, we are going to use our familiar $\hat{d}$ and $\hat{u}$ vectors. The force of gravity works in the downward direction, and can be written as $mg\hat{d}/mg(-\hat{u})$. However, the form of the restoring force would be slightly different depending on which direction is taken to be positive. This is because, before we set the system in motion, it will first come equilibrium. In equilibrium, the the spring will extend by an amount $x_0$ in the downward direction. If we take the downward direction to be positive, then we would have $\sum \vec{F}=mg\hat{d}+k(-x_0\hat{d})=0$, at equilibrium. If we had taken the upwards direction as positive, then we would have $\sum \vec{F}=-mg\hat{u}+kx_0\hat{u}=0$. Since $x_0$ is component, it would take positive value in $\hat{d}$ and negative value in $\hat{u}$ direction. I' d normally write $\vec{x_0}=x_{0_d}\hat{d}/x_{0_u}\hat{u}$. Here $x_{0_d}=-x_{0_u}$

The reason for showing this was to note that the sign of $x_0$ depends on which direction is taken to be positive.

Now, let us assume the downward direction to be positive. The component of displacement in this direction is now $x_d$ and this can be both positive or negative. Hence $\vec{x}=x_d\hat{d}$. Hence $\vec{\dot{x}}=\dot{x}_d\hat{d}$ and $\vec{\ddot{x}}=\ddot{x}_d\hat{d}$. Moreover, if the downward direction is positive, $x_0$ is also positive. Armed with these, we can write :

$$\sum \vec{F}_d=m\ddot{x}_d\hat{d}=mg\hat{d}-k(\vec{x}+x_0\hat{d})-b\vec{\dot{x}}=mg\hat{d}-k(x_d\hat{d}+x_0\hat{d})-b\dot{x}_d\hat{d}$$

This can be simplified to :

$$m\ddot{x}_d=-kx_d-b\dot{x}_d$$

Now if we had taken the upward direction to be positive, our displacement would be $\vec{x}=x_u\hat{u}$. Hence $\vec{\dot{x}}=\dot{x}_u\hat{u}$ and $\vec{\ddot{x}}=\ddot{x}_u\hat{u}$. Moreover, if the upward direction is positive, then $x_0$ is negative. Hence we can write our equation again :

$$\sum \vec{F}_u=m\ddot{x}_u\hat{u}=-mg\hat{u}-k(\vec{x}-x_0\hat{u})-b\vec{\dot{x}}=-mg\hat{u}-k(x_u\hat{u}-x_0\hat{u})-b\dot{x}_u\hat{u}$$

This can again be simplified to :

$$m\ddot{x}_u=-kx_u-b\dot{x}_u$$

However, since the boundary conditions have changed, we would ultimately obtain $x_d(t)=-x_u(t)$.

So, we can say that if we change the direction, our final result would change. The sign would flip. As for the equations, does it change ? Well, there are several things to consider in that regard. In the above answer by @Farcher, $g$ represents the component of gravitational acceleration, and can take both positive and negative values. That's why even after changing the direction the two equations remained same. The is because $g$ in the above answer is $g_u$ and $g_d$, and so we can write $\vec{F}_g=g_d\hat{d}/g_u\hat{u}$. In that sense, the equations look exactly the same.

However, if we consider the magnitude of $g$, as we have done here, then the sign of $g$ would change as we change directions and our equation would look different. Ofcourse in this example, this is compensated by the sign of $x_0$ and so the final equation looks the same, but in general the equation would appear different. For example, the answer by @Farcher to the free falling body, considers the magnitude of gravity to be $g$, and so, the expression changes.

So, in conclusion, if I consider the components of all vectors in my equation, the expression would remain the same. This would be true even for free fall, because now upward positive equation becomes $m\ddot{x}_u=mg_u$ and downward positive becomes $m\ddot{x}_d=mg_d$, where $g_u=-9.8$ and $g_d=+9.8$. So, if we stick to components, the form remains the same.

However, if we start using magnitudes instead, the sign of the term that uses magnitude, would flip when we flip the positive direction, and thus, the expression would change.


In the example, in the question, I mentioned the case for a hanging spring. If we take $\hat{d}$ as positive, then our equation was something line :

$$m\ddot{x}\hat{d}=mg\hat{d}-kx\hat{d}$$

However, if we make our upward direction positive, then our equation of motion became :

$$m\ddot{x}\hat{u}=-mg\hat{u}+kx\hat{u}$$

This worried us initially, because it does not seem like a harmonic oscillator. However, there is an issue in this equation. On the LHS, $\ddot{x}$ is the component of acceleration in the $\hat{u}$ direction. However, on the right, $x$ is no longer the component of displacement in the $\hat{u}$ direction. Why you may ask ?

Imagine a vector $\vec{x}$. In the right direction, it is $5\hat{r}$. What would this vector be, in the left direction ? Well, obviously $-5\hat{r}$. But now comes the important question. What is the component of this vector in the right direction ? It is $5$. But what is the component in the left direction ? Well, it is $-5$. The component is negative, and has the negative sign built in. In the left direction, we have $\vec{x}=x_l\hat{l}=-5\hat{l}$. As you can see, the negative sign is absorbed into the component. Hence, $x_l=-x_r$.

In our sum, we had a $-kx\hat{d}$ term. In the upward positive direction, we put $\hat{u}=-\hat{d}$.Hence, our term became $-kx(-\hat{u})$. Now instead of combining the negative sign with $x$ to get the component in the upward direction, we cancelled it with the negative sign in front. So, what is this $x$ now ? It is still the component in the downward direction.

Moreover, in the upward positive direction the component would have been negative. So we could have written $\vec{x}=x_u\hat{u}=-|x_u|\hat{u}$. Moreover, when $x_u<0$, we have $|x_u|=-x_u=x_d$. So, in our problem, the $x$ ended up being the magnitude of $\vec{x}$ in this new positive direction.

In the original down direction, the component was positive, and positive component is same as magnitude. Hence, when we used magnitude instead of components in the equation, our terms changed.

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