A question from Resnik about g-force

[rudransh verma]

Suppose there is one force gravitational force $\vec{f_g}$. We can relate this downward force and downward acceleration with Newton sec law. This law can be written as $F_{net,y}=ma_y$ which becomes $$-F_g=m(-g)$$ or $$F_g=mg$$
$$\vec{F_g}=-F_g \hat j=-mg \hat j=m\vec g$$.
Is it right? Because there shouldn’t be a -sign with F in first eqn.
What is the first eqn telling?

 

[hutchphd]

$$\vec{F_g}=-F_g \hat j=-mg \hat j=m\vec g$$. is correct.

 

[rudransh verma]

$$\vec{F_g}=-F_g \hat j=-mg \hat j=m\vec g$$. is correct.

Please tell me about First eqn

 

[hutchphd]

That is the equation for the y component where y is positive upwards and the scalars are magnitudes.

 

[rudransh verma]

That is the equation for the y component where y is positive upwards and the scalars are magnitudes.

It should be F=-mg not -F=-mg

 

[hutchphd]

No, it should not.

 

[rudransh verma]

No, it should not.

Why? The g-force should be in downward direction represented by F=-mg. F=mg says F is positive which means in upward direction.

 

[Bandersnatch]

Why? The g-force should be in downward direction represented by F=-mg. F=mg says F is positive which means in upward direction.

The first equation in the OP is the same as this bit from the third:
$$-F_g \hat j=-mg \hat j$$
What you have in mind is this:
$$\vec{F_g}=-mg \hat j$$
That is, the first one above includes the information about the direction of y vectors, while F and g are just magnitudes. In the second one you treat F as an unknown vector, and let the right hand side of the equation inform you about its direction.

 

[rudransh verma]

The first equation in the OP is the same as this bit from the third:
−Fgj^=−mgj^

Are you saying $-F_g= m(-g)$ is same as $-F_g \hat j=-mg\hat j$. Ok so - sign is telling us same thing as $-\hat j$ in -y direction . So if we cancel -sign we get the eqn in magnitude form.

What you have in mind is this:

Here left side is unknown and we are informed about it via right side.

 

[Bandersnatch]

Yes, that's what I'm saying.

 

[rudransh verma]

Yes, that's what I'm saying.

I have a couple of questions.
1. If $F_g=mg$ is magnitude form how do you represent in +y direction assuming g-force works the other way around?

2. Is this eqn $F=-ma$ possible?

3. Is $-F_g\hat j= -mg\hat j$ same as $F_g\hat j= mg\hat j$ because we can cancel -sign?

4. What does $-F_g=-mg$ mean in words?

I am confused between eqns with -sign on both sides and -sign on one side.

 

[hutchphd]

Those equations are algebraic relations.
In order for them to be physics one must define exactly what each variable means.
Please re-state each equation with corresponding exact definitions of each variable.

/

 

[rudransh verma]

Those equations are algebraic relations.
In order for them to be physics one must define exactly what each variable means.
Please re-state each equation with corresponding exact definitions of each variable.

/

F is force, a is acceleration, $F_g$ is gravitational force and g is gravitational acceleration.

 

[hutchphd]

I said exact.

 

[rudransh verma]

I said exact.

What is exact

 

[Mister T]

F is force, a is acceleration, $F_g$ is gravitational force and g is gravitational acceleration.

No, they aren't. They are the magnitudes of those quantities. For example, F is the magnitude of the force, a is the magnitude of the acceleration.

Magnitudes are never negative.

 

[rudransh verma]

Those equations are algebraic relations.
In order for them to be physics one must define exactly what each variable means.
Please re-state each equation with corresponding exact definitions of each variable.

/

Wait a minute that's what I am asking you.

I don’t understand corresponding exact definition of each variable. I have started force chapter recently and this is just a paragraph from that chapter not some question to be solved.

Is $-F_g=-mg$ like -6 = -2*3. Just magnitudes not variables. Because I am getting confused by seeing -sign on both sides and thinking of them as variables which can be solved further.

Are you saying $F_g$ can be gravitational force where F and g are in same direction by the same signs here and F and mg are equal or it can be a variable whose value we need to find, $-F_g=-mg$ vs $F_g=-mg$ respectively.
The first eqn is stating the second law of Newton whereas the second eqn is where we are trying to find the value of F by the stated law.
@Steve4Physics Can you verify this post?

 

[hutchphd]

First you are overthinking this: In any situation where it matters to you, the context will tell you the sign. And you will check the result for reasonableness.
In general assume all quantities are vectors. In 1D a vector has magnitude and can be + or -. If the equation defiines the object as a magnitude (either verbally or by directly $A=|\vec A|$) it is always positive.
For instance the equation $$F=-ma$$ can be true if you have defined $a$ to be the magnitude of the acceleration but F is understood as a 1D vector. More formally it would be written as $$F=-m|a|$$ but that gets cumbersome. I truly believe you are overthinking this. Most calculations I do without too much concern for the signs but I always look carefully at the result in various limiting cases to make sure I haven't screwed up.

 

[rudransh verma]

For instance the equation F=−ma can be true if you have defined a to be the magnitude of the acceleration but F is understood as a 1D vector. More formally it would be written as

So in $-F_g=-mg$, $-F_g$ is 1D vector, m is scalar mass and $-g$ is again a vector. And according to second law this is how we write the relation between force of gravity and downward acceleration. Force vector is equal to -mg vector.
It does not mean we can cancel the -signs. We can omit it so that we can say F=ma(magnitude form). This is just stating the relation between gravitational force and it’s acceleration g via second law.

This also doesn’t mean the value of $-F_g$ is $-mg$.

 

[hutchphd]

You need to define whether g is a vector or the magnitude of the gracvitational constant. Please read what is written.

I wish you good luck.

 

[rudransh verma]

You need to define whether g is a vector or the magnitude of the gracvitational constant. Please read what is written.

I wish you good luck.

I already said it -g is a vector. I don’t know if g is a vector. But as we all take acceleration due to gravity as -g so it’s a vector. Please tell me if I am wrong at anything in post #19.

 

[hutchphd]

It is wrong.

If g is a vector and F is a vector then absent other forces $$F=mg$$ and any other algebraic manipulation of the equation is also true.

I am finished with this topic

 

[rudransh verma]

It is wrong.

If g is a vector and F is a vector then absent other forces $$F=mg$$ and any other algebraic manipulation of the equation is also true.

I am finished with this topic

You cannot leave me in doubt. Please cooperate.

If it’s written in book $-F_g=-mg$ then $F_g$ and $g$ are magnitudes or they can be vectors as well. - with a scalar is a vector as a vector with - sign.

 

[jbriggs444]

You cannot leave me in doubt. Please cooperate.

If it’s written in book $-F_g=-mg$ then $F_g and g$ are magnitudes or they can be vectors as well. - with a scalar is a vector as a vector with - sign.

I think that you trying to obtain a precise formality that is not there. Overthinking things.

When working in one dimension, there is precious little difference between $-\vec{F}$, $-F\ \hat i$, $F (-\hat i)$ and $-F$. Or between $-mg\ \hat i$, $mg (-\hat i)$, $m (-g) \hat i$ and $-mg$.

You have one bit for sign. How you distribute it among the factors is of no particular interest to anyone.

 

[rudransh verma]

I think that you trying to obtain a precise formality that is not there. Overthinking things.

I don’t understand what I am doing wrong? What is annoying? Is it a useless talk?

 

[rudransh verma]

e. I truly believe you are overthinking this

I apologise if I did something wrong

 

[jbriggs444]

I don’t understand what I am doing wrong? What is annoying? Is it a useless talk?

What are you doing wrong? Overthinking. Obsessing over a simple concept. Expending more thought than it is worth.

What is annoying? I did not think that I had expressed annoyance. However, the phrase "You cannot leave me in doubt" in boldface was not emotionally neutral.

It is useless talk? In my opinion, yes.

 

[Mister T]

4. What does $-F_g=-mg$ mean in words?

The opposite of $F_g$ equals the opposite of $mg$.

Therefore, $F_g$ equals $mg$.

I am confused between eqns with -sign on both sides and -sign on one side.

If you have an equation like $a=-b$ it means that $a$ equals the opposite of $b$.

Therefore the opposite of $a$ equals $b$.

The source of your confusion arises for two reasons. First, you are confused by the minus signs. Second, you are confused about the difference between vectors, vector components, and vector magnitudes.

A vector has both a magnitude and a direction. A vector is neither positive nor negative.

A vector component can be either positive or negative.

A vector magnitude can never be negative.

So, let's take your equation $\vec{F}_g=m\vec{g}$ as an example. This is a vector equation. It tells us that the magnitude of the vector $\vec{F}_g$ is equal to the magnitude of the vector $m\vec{g}$. It also tells us that the two vectors have the same direction. We can also express the same relationship between these two vectors by writing a vector equation in unit vector form. $\vec{F}_g=-mg\hat{j}$. In addition to telling us that the vectors are equal in magnitude and have the same direction, it specifies that that direction is the $-y$ direction.

 

[sophiecentaur]

I have searched this thread for the word "frame". The frame of reference is what defines the signs and vector directions and I can just see you all shouting at each other (or at the OP) because that hasn't been explicitly stated or appreciated.
A number of PF contributors have a (good) habit of regularly suggesting that an OP with a problem should present a Free Body Diagram. That would sort it all out and the signs would all agree then.

 

[vanhees71]

First you are overthinking this: In any situation where it matters to you, the context will tell you the sign. And you will check the result for reasonableness.
In general assume all quantities are vectors. In 1D a vector has magnitude and can be + or -. If the equation defiines the object as a magnitude (either verbally or by directly $A=|\vec A|$) it is always positive.
For instance the equation $$F=-ma$$ can be true if you have defined $a$ to be the magnitude of the acceleration but F is understood as a 1D vector. More formally it would be written as $$F=-m|a|$$ but that gets cumbersome. I truly believe you are overthinking this. Most calculations I do without too much concern for the signs but I always look carefully at the result in various limiting cases to make sure I haven't screwed up.

All this trouble can be avoided when using vectors for vector quantities from day 1 of learning physics. Even the physics-didactics community agrees with this nowadays. A good side effect is that you get used to vectors on a very intuitive level early on, and this helps to understand the more abstract level necessary for more advanced topics like electrodynamics or continuum mechanics, where it gets full vector calculus.

 

[vela]

All this trouble can be avoided when using vectors for vector quantities from day 1 of learning physics. Even the physics-didactics community agrees with this nowadays. A good side effect is that you get used to vectors on a very intuitive level early on, and this helps to understand the more abstract level necessary for more advanced topics like electrodynamics or continuum mechanics, where it gets full vector calculus.

I think the problem isn't really about the concept of vector vs. scalar; it's about precision in mathematical notation. In my experience, most students don't really get that $\vec F$ and $F$ are two different mathematical objects and will randomly interchange them, so they'll write, for example,
$$\vec F_y = \vec T \sin\theta - mg = ma.$$

 

[vanhees71]

Sigh, yes, looks all too familiar :-(.

 

[rudransh verma]

I think the problem isn't really about the concept of vector vs. scalar; it's about precision in mathematical notation. In my experience, most students don't really get that $\vec F$ and $F$ are two different mathematical objects and will randomly interchange them, so they'll write, for example,
$$\vec F_y = \vec T \sin\theta - mg = ma.$$

Exactly. I am one of those. The thing is when you are presented with lot of right information at once then brain doesn't grasp anything at all. Its the mistakes we do makes us learn anything. Spending time working out all the things which are wrong or not possible helps very much. Removing the wrong fog in the brain helps us to clearly see the truth.
Solving problems is a great way to do it not just merely reading the texts.

 

[vanhees71]

It's the only way! Just passively reading a physics book or, even worse, listening to a lecture or videos won't help to learn the material at all. You have to already read a physics book with a pencil and paper and then do the problems thinking yourself without looking at the solutions too early. In short words: get active!

 

[hutchphd]

An adjunct to this process for, shall we say, the less precise thinker (moi) is to cultivate that little inner voice that continuously asks "does that make sense?" in as many ways as is possible. This includes reflexively looking at units (dimension) and thinking about simplified limiting cases with every major calculational step. It is a very useful skill and has managed to keep me from becoming a complete lunatic worrying about signs. Purely defensive on my part but very useful for everyone.