Understanding the Sign of 'g' and 'a': An Apparent Weight Example

AI Thread Summary
The discussion centers around the apparent weight equation FN = mg + ma, where FN is the normal force, m is mass, g is gravitational acceleration, and a is system acceleration. Participants explore the implications of positive and negative signs for acceleration when an elevator accelerates upwards or downwards, noting that apparent weight increases with upward acceleration and decreases with downward acceleration. The confusion arises from the differing signs of g and a, despite both representing downward acceleration; g is treated as a constant positive value, while a can be negative during downward acceleration. The conversation emphasizes the importance of sign conventions in physics and the distinction between forces and acceleration in free body diagrams. Understanding these concepts is crucial for accurately analyzing motion in systems like elevators.
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In my physics book the equation for apparent weight is given as

FN = mg + ma

where FN is the normal force, m is the mass of the object, g is the gravitational acceleration of the object (= 9.8 m/s2) and a is the acceleration of the system.

For example the system could be someone standing on a scale inside an elevator. Let's for simplicity's sake say this person weighs 100 kg. Let's also for simplicity's sake only consider the cases where the elevator goes from rest and accelerates either upwards or downwards, and not the cases when it is already moving and slows down. When the elevator is standing still then, the acceleration a of the system is 0, so the normal force will be FN = 100*9,8 = 980 N. The normal force is the apparent weight that is shown on the scale and when the elevator is standing still it's equal to the true weight.

Now if the apparent weight is, let's say a) 700 N and b) 1100 N, how large is the acceleration a of the system? According to the equation above it will be

a = FN/m - g

so we will get

a) a = 700/100 - 9.8 = -2.8 m/s2
b) a = 1100/100 - 9.8 = 1.2 m/s2

Now what puzzles me here is the sign of the acceleration. We know that the apparent weight is larger when the elevator goes from rest and accelerates upwards, and less when it goes from rest and accelerates downwards. Hence, the negative sign in a) implicates an elevator going from rest accelerating downwards.

But if a = -2.8 m/s2 implicates something accelerating from rest DOWNWARDS, then why shouldn't the formula have g = -9.8m/s2, when g being gravitation obviously also means acceleration from rest downwards. It doesn't, in the equation we used a positive value for g! If we had used g = -9.8m/s2, both case a) and b) would have given the same sign which is incorrect.

Where is the error? Why doesn't g and a, when both are going in the same direction (from rest to acceleration downwards) have the same sign?
 
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actually the formula seems a bit of no use to me.I always sketch the free body diagram of lift in my brain and then doing some pseudo force work the question is solved ALWAYS!
in the first part for sure lift accelerate downwards. the guys who done with the formula considered "a" (the base of formula) in upward direction. therefore the acceleration is negative but negative with respect to their chosen direction of acceleration.
work on your imagination and consider this as lift please see the FBD




l---------------l
l---------------l a\uparrow
l---\downarrow mg-------l
 
Don't you mean the other way around? If we draw a free body-diagram and have FN upwards and mg and ma both pointing downwards then we get exactly the equation from my book. If we had instead drawn ma upwards, the answers to a) and b) would have reversed signs.

But assuming ma upwards also doesn't make sense to me. I mean, why should we assume mg and ma as pointing in DIFFERENT directions in order to get the same sign for acceleration in the same direction?

I mean the way I think of it, logically it should be: same direction <=> same sign. So why isn't it like that here?
 
Maxo said:
Where is the error? Why doesn't g and a, when both are going in the same direction (from rest to acceleration downwards) have the same sign?
g is just a constant; it's always positive.
 
Maxo said:
Now what puzzles me here is the sign of the acceleration. We know that the apparent weight is larger when the elevator goes from rest and accelerates upwards, and less when it goes from rest and accelerates downwards. Hence, the negative sign in a) implicates an elevator going from rest accelerating downwards.
The speed of the elevator is irrelevant. All that matters it its acceleration. If it accelerating upwards, then the elevator floor is pressing harder into your feet to accelerate you will the elevator--thus the increase in your apparent weight. If it accelerates downward, then it presses less and your apparent weight is less.
 
doc al said:
g is just a constant; it's always positive.

actually it all depends on sign convention
 
namanjain said:
actually it all depends on sign convention
Actually, it doesn't.
 
Doc Al said:
Actually, it doesn't.

AS FAR AS I KNOW IT DEPENDS ON CONVENTION THE VALUE OF g
if you take upward direction as positive it'll be positive
but maybe i am wrong,
i ve a link that supports me (maybe)
https://www.physicsforums.com/showthread.php?t=260060
 
Doc Al said:
The speed of the elevator is irrelevant. All that matters it its acceleration. If it accelerating upwards, then the elevator floor is pressing harder into your feet to accelerate you will the elevator--thus the increase in your apparent weight. If it accelerates downward, then it presses less and your apparent weight is less.

I understand all that. The only thing I don't understand is the sign (+/-) of the acceleration. We have a free body diagram where we have assumed both mg and mg to be pointing downwards. Still, accelerating in the same direction they have opposite signs - even though they point in the same direction. Why? Same direction should be same sign!
 
  • #10
namanjain said:
AS FAR AS I KNOW IT DEPENDS ON CONVENTION THE VALUE OF g
if you take upward direction as positive it'll be positive
but maybe i am wrong,
i ve a link that supports me (maybe)
https://www.physicsforums.com/showthread.php?t=260060
Most standard textbooks use the convention that g is just a constant equal to 9.8 m/s^2.

Thus, if you choose a sign convention such that up is positive the acceleration of a falling object is: a = -g.
 
  • #11
Maxo said:
I understand all that. The only thing I don't understand is the sign (+/-) of the acceleration. We have a free body diagram where we have assumed both mg and mg to be pointing downwards. Still, accelerating in the same direction they have opposite signs - even though they point in the same direction. Why? Same direction should be same sign!
Can you rephrase your question?

On a free body diagram, the weight always points downward. So if up is positive, the weight is -mg.

The acceleration could be positive or negative, depending on the situation.
 
  • #12
Ok, so we have a free body-diagram with FN pointing upwards and both mg and ma pointing downwards. So ma and mg are assumed to point in the same direction.

Let's make it as simple as possible and ONLY consider the following scenario: The elevator is at rest, and then starts accelerating DOWNWARDS.

Now, according to the equation, the normal force will be SMALLER during acceleration downwards than what it is when the elevator is at rest. So we can calculate the acceleration a by rearranging the equation:

a = FN/m - g

Now, when the normal force is smaller than before, the value of a will be NEGATIVE. SO we can conclude that a negative value of a corrsponds to acceleration DOWNWARDS.

Now the question: g is ALSO acceleration DOWNWARDS. But g is positive, not negative like a. But if a and g are pointing in the same direction, WHY DO THEY HAVE OPPOSITE SIGNS?
 
  • #13
namanjain said:
actually the formula seems a bit of no use to me.I always sketch the free body diagram of lift in my brain and then doing some pseudo force work the question is solved ALWAYS!
in the first part for sure lift accelerate downwards. the guys who done with the formula considered "a" (the base of formula) in upward direction. therefore the acceleration is negative but negative with respect to their chosen direction of acceleration.
work on your imagination and consider this as lift please see the FBD




l---------------l
l---------------l a\uparrow
l---\downarrow mg-------l

Maxo said:
Don't you mean the other way around? If we draw a free body-diagram and have FN upwards and mg and ma both pointing downwards then we get exactly the equation from my book. If we had instead drawn ma upwards, the answers to a) and b) would have reversed signs.

But assuming ma upwards also doesn't make sense to me. I mean, why should we assume mg and ma as pointing in DIFFERENT directions in order to get the same sign for acceleration in the same direction?

I mean the way I think of it, logically it should be: same direction <=> same sign. So why isn't it like that here?

the a i made is acceleration of lift. so it's pseudo force acts downwards and equation comes out same. also 'a' is negative ,and the positive direction of 'a' is upwards so 'a' overall in example one NET ACTS DOWNWARDS. don't match it with sign convention because my explanation itself has an convention that positive direction of a is upwards.
the pseudo force of ma acts downward as you want
if one really want a formula he need to decide direction of acceleration, the makers did the same by choosing their direction as upwards. think over it
if you feel me to be wrong then you must consider reverse situation i.e. acceleration downwards and design a formula yourself and you see value of 'a' positive in example 1
new formula

F=mg-ma
 
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  • #14
Maxo said:
Ok, so we have a free body-diagram with FN pointing upwards and both mg and ma pointing downwards. So ma and mg are assumed to point in the same direction.
On a free body diagram, only the forces FN and mg appear. mg always points down.

Let's make it as simple as possible and ONLY consider the following scenario: The elevator is at rest, and then starts accelerating DOWNWARDS.
OK, that means the acceleration is downward.

Now, according to the equation, the normal force will be SMALLER during acceleration downwards than what it is when the elevator is at rest. So we can calculate the acceleration a by rearranging the equation:

a = FN/m - g

Now, when the normal force is smaller than before, the value of a will be NEGATIVE. SO we can conclude that a negative value of a corrsponds to acceleration DOWNWARDS.
OK.

Now the question: g is ALSO acceleration DOWNWARDS. But g is positive, not negative like a. But if a and g are pointing in the same direction, WHY DO THEY HAVE OPPOSITE SIGNS?
Again, g just stands for the magnitude of the acceleration due to gravity.

Let's apply Newton's 2nd law: ∑F = ma

First find the net force. The only forces acting are gravity (down, thus negative) and the normal force (up, thus positive).

So: ∑F = -mg + Fn

Set that equal to ma:
-mg + Fn = ma

So, when the normal force is less than mg, the acceleration is negative. Works out just fine.
 
  • #15
namanjain said:
the a i made is acceleration of lift. so it's pseudo force acts downwards
It's not necessary to introduce the concept of pseudo forces here. That's an advanced topic that is usually not covered at this level.
 
  • #16
namanjain said:
the a i made is acceleration of lift. so it's pseudo force acts downwards and equation comes out same. also 'a' is negative ,and the positive direction of 'a' is upwards so 'a' overall in example one NET ACTS DOWNWARDS. don't match it with sign convention because my explanation itself has an convention that positive direction of a is upwards.
the pseudo force of ma acts downward as you want
if one really want a formula he need to decide direction of acceleration, the makers did the same by choosing their direction as upwards. think over it
if you feel me to be wrong then you must consider reverse situation i.e. acceleration downwards and design a formula yourself and you see value of 'a' positive in example 1
new formula

F=mg-ma

But if we switch direction of ma like you are suggesting then ma and mg are pointing in different directions on the free body diagram. If they are pointing in different directions they should have opposite signs. I still don't understand why the directions of ma and mg doesn't give the same sign for the same direction?

Doc Al said:
On a free body diagram, only the forces FN and mg appear.
Why doesn't ma appear in the free body diagram?
 
  • #17
Maxo said:
Why doesn't ma appear in the free body diagram?
Only forces appear in a free body diagram. "ma" is not a force.
 
  • #18
Doc Al said:
Only forces appear in a free body diagram. "ma" is not a force.
What is it then?

And how can the equation FN = mg + ma be made from the free body diagram if ma doesn't appear on it?
 
  • #19
There is a severe misunderstanding of "F=ma" here.

On the left hand side are all the forces acting on the object. The "resultant" goes on the right hand side of the equation, i.e. it resulted in an acceleration of the object, the dynamical result, the motion, etc. If the object doesn't move, this is where you equate all the forces on the left hand side to zero, such as the case when you deal with statics.

Zz.
 
  • #20
Doc Al said:
It's not necessary to introduce the concept of pseudo forces here. That's an advanced topic that is usually not covered at this level.

Actually I disagree, I think if pseudo forces would explain this better, I would welcome this explanation.

I think this pseudo force made it more understandable. What we have here is a non-inertial refrence frame which is fixed to the accelerating elevator system (including the scale but not including the person standing on the scale). So that's why ma is not considered a force acting on the scale, because it's the reference frame itself that is accelerating.

So if we have this accelerating reference frame, when the elevator accelerates downward, it could be seen as that the gravitational force actually becoming less (in relation to the reference frame). And the FN is only the reaction of the gravitational force, so it also becomes less.

Then for some reason, the fact that ma is the same as reference frame, makes it's signs be inverted. I still don't understand why this happens though. If someone could explain this particular detail, that would be great.
 
  • #21
Maxo said:
What is it then?
"ma" is what you set the net force equal to in Newton's 2nd law. Itself it is not a force and does not belong in a free body diagram.

And how can the equation FN = mg + ma be made from the free body diagram if ma doesn't appear on it?
Read my earlier post on how to apply Newton's 2nd law to this problem. "ma" comes from applying Newton's 2nd law, not from a free body diagram.
 
  • #22
Maxo said:
Actually I disagree, I think if pseudo forces would explain this better, I would welcome this explanation.
Instead of playing with pseudoforces, what you need to do is understand Newton's 2nd law and how to apply it.
 
  • #23
Doc Al said:
Instead of playing with pseudoforces, what you need to do is understand Newton's 2nd law and how to apply it.

You still haven't explained the actual question at hand, which is why accelerations have different signs despite that they are in the same direction.
 
  • #24
Maxo said:
You still haven't explained the actual question at hand, which is why accelerations have different signs despite that they are in the same direction.
What's to explain? The acceleration due to gravity would be negative, but so what? Nothing is in free fall here.

There is only one acceleration: The acceleration of the elevator. The 'g' describes the magnitude of the acceleration due to gravity of something in free fall; it helps us calculate the weight of a mass. But nothing is actually in free fall here.
 
  • #25
Are you saying that g only has a magnitude and not any direction? That can't be right. We all know that g is directed towards the center of the Earth so it does in fact have a direction. So why can't this direction be included in the equations?
 
  • #26
Maxo said:
Are you saying that g only has a magnitude and not any direction? That can't be right. We all know that g is directed towards the center of the Earth so it does in fact have a direction. So why can't this direction be included in the equations?
Once again: The symbol 'g' is just a magnitude!

The direction of gravity is downward. That direction is definitely included in your free body diagram. And that's why the weight is given as -mg when deriving the equation that you used. (Assuming up as positive.)
 
  • #27
Doc Al said:
Once again: The symbol 'g' is just a magnitude!
Why? According to Newtons second law, F=ma. Both F and a are vectors in the same direction. For gravitation we have a=g so F=mg. If F is a vector, then either m or g has to be a vector. m is definitely not a vector, so g has to be vector. How can you then say g is just a magnitude and not a vector?
 
  • #28
Maxo said:
Why? According to Newtons second law, F=ma. Both F and a are vectors in the same direction. For gravitation we have a=g so F=mg. If F is a vector, then either m or g has to be a vector. m is definitely not a vector, so g has to be vector. How can you then say g is just a magnitude and not a vector?
The acceleration of an object in free fall is a vector. Its magnitude is g and its direction is downward.

Weight is a vector. Its magnitude is mg and its direction is downward.

You are insisting on using g as a vector. Your book doesn't seem to do that. (Most do not.) I recommend against doing so.

Rather than blindly use an equation from your textbook (that you quoted in your first post), try to understand how it was derived. If your book used g as a vector then that equation would be different.
 
  • #29
Doc Al said:
The acceleration of an object in free fall is a vector. Its magnitude is g and its direction is downward.

Weight is a vector. Its magnitude is mg and its direction is downward.

You are insisting on using g as a vector. Your book doesn't seem to do that. (Most do not.) I recommend against doing so.

Rather than blindly use an equation from your textbook (that you quoted in your first post), try to understand how it was derived. If your book used g as a vector then that equation would be different.
I have hardly "blindly" used an equation, but I have dedicated this whole thread on just trying to understand how it was derived.

Anyway yeah, I want to use g as a vector. Why wouldn't you do that? It doesn't seem to be too far-fetched, it's also done in this article for example http://en.wikipedia.org/wiki/Gravitational_acceleration#For_point_masses

How would the equation look if g is treated like a vector instead? That would maybe answer my initial question, which still hasn't been answered unfortunately.
 
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  • #30
Maxo said:
I have hardly "blindly" used an equation, but I have dedicated this whole thread on just trying to understand how it was derived. So there's no need for you to try to be an a$$hole by saying such things.
Did your book not derive it? The derivation is just an application of Newton's 2nd law. Reread my previous posts for a derivation.

Anyway yeah, I want to use g as a vector. Why wouldn't you do that? It doesn't seem to be too far-fetched, it's also done in this article for example http://en.wikipedia.org/wiki/Gravitational_acceleration#For_point_masses

How would the equation look if g is treated like a vector instead? That would maybe answer my initial question, which still hasn't been answered unfortunately.
If you insist, then apply Newton's 2nd law as a vector equation. It's the same derivation:
∑F = mg + Fn

Set that equal to ma:
mg + Fn = ma

Done!
 
  • #31
F(external)=ma

mg-N=ma
direction of a downward
or if
a is upwards
N-mg=ma
N=mg+ma
 
  • #32
Doc Al said:
Did your book not derive it? The derivation is just an application of Newton's 2nd law. Reread my previous posts for a derivation.


If you insist, then apply Newton's 2nd law as a vector equation. It's the same derivation:
∑F = mg + Fn

Set that equal to ma:
mg + Fn = ma

Done!
Yeah, that was how I saw it already in the first post. However, it still doesn't explain my question since the first post.
 
  • #33
Maxo said:
Yeah, that was how I saw it already in the first post. However, it still doesn't explain my question since the first post.
Rephrase your question then. (As I think I've answered it.)

Expressed as a vector equation, the equation in your first post is incorrect.
 
  • #34
Maxo said:
What is it then?

And how can the equation FN = mg + ma be made from the free body diagram if ma doesn't appear on it?

The free-body diagram includes two forces: ##F_g##, the downwards force of gravity on the passenger, whose magnitude is ##mg##, and the upwards force from the floor of the elevator on the soles of the passenger's shoes, which is (for now) unknown. The magnitude of this latter force is the "apparent weight" and it's what we're trying to derive.

Adopt the sign convention that positive forces and accelerations act upwards, and let ##F_f## represent the force from the floor and we have:

##F_f + F_g = ma## -- (total force from free-body diagram in Newton's law)
##F_f - mg = ma## -- (get the sign and the magnitude of ##F_g## right)
##F_f = ma + mg##

And, because the magnitude of ##F_f## is the apparent weight, we have your result.
 
  • #35
It seems my question is misunderstood. I understand all this that you write. That's not what I'm asking about. I understand this equation, how it's derived and why. I will try to reformulate it. Please hear my explanation:

What I am trying to find is some way to make so the following criteria is satisfied:

Two vectors pointing in the same direction should have the same sign.

The two vectors I'm talking about here are g and a.

Example. 100 kg on scale in elevator that acclerates downwards from rest. Scale shows 500 N. How large is the acceleration? Answer is 500/100-9.8 = -4.8 m/s2. Both g and a point in the same direction (downwards). But

g = 9.8 m/s2
a = -4.8 m/s2

Here's the thing: Both g and a point in the same direction (downwards). BUT THEY DON'T HAVE THE SAME SIGN.

I want it to be so that same direction = SAME SIGN. So the question is: How can a new equation be derived that satisifies this criteria?
 
  • #36
Maxo said:
It seems my question is misunderstood. I understand all this that you write. That's not what I'm asking about. I understand this equation, how it's derived and why. I will try to reformulate it. Please hear my explanation:

What I am trying to find is some way to make so the following criteria is satisfied:

Two vectors pointing in the same direction should have the same sign.
Obviously, two vectors pointing in the same direction will have the same sign when expressed as components.

The two vectors I'm talking about here are g and a.

Example. 100 kg on scale in elevator that acclerates downwards from rest. Scale shows 500 N. How large is the acceleration? Answer is 500/100-9.8 = -4.8 m/s2. Both g and a point in the same direction (downwards). But

g = 9.8 m/s2
a = -4.8 m/s2
g is a vector; g is not. Expressed in terms of components, the value of g is -g = -9.8 m/s^2. (Taking up as positive, of course.)

Here's the thing: Both g and a point in the same direction (downwards). BUT THEY DON'T HAVE THE SAME SIGN.

I want it to be so that same direction = SAME SIGN. So the question is: How can a new equation be derived that satisifies this criteria?
You need to review how the equation you posted in your first post is derived.

If you want things in vector form, then g appears in the weight: mg. Put in component form, w = -mg.
 
  • #37
Maxo said:
Two vectors pointing in the same direction should have the same sign.

There is your problem. If you allow a vector to have negative magnitude, this statement is not true. A vector of magnitude r and direction phi is the same as one of magnitude -r and direction 180-phi.

To avoid problems like this, it is best to adopt the convention that a vector's magnitude is always positive. It is also critical to distinguish between a vector g and its magnitude g (which under this convention is positive).
 
  • #38
Start from Newton's Second equation: ma=∑Fi.

You stand on scales. It pushes you upward with normal force N, while gravity G pulls you downward. |G|=mg. Note that g is a positive number.

Choose if you consider up or down positive. If "up" is positive, the equation becomes ma= N-mg, and positive acceleration means accelerating upward. Rearranging the equation, N=ma+mg. If the lift accelerates upward, a is positive and the normal force is greater than mg, your apparent weight increased.
If the lift accelerates downward a is negative, N is less than mg. Your apparent weight decreased. If a=-g, you are in free fall, and you are weightless.ehild
 
  • #39
Maxo, in both your original and reformulated question you made the same mistake, and it obviously comes from confusion about signs. If you want to declare g as a vector, no problem. But if you say that downwards direction is positive, g is acceleration, so all downward accelerations will be positive, and upwards negative. This doesn't mean that if lift moves downwards the acceleration is positive.
Example. 100 kg on scale in elevator that acclerates downwards from rest. Scale shows 500 N. How large is the acceleration? Answer is 500/100-9.8 = -4.8 m/s2. Both g and a point in the same direction (downwards). But

g = 9.8 m/s2
a = -4.8 m/s2

Here's the thing: Both g and a point in the same direction (downwards).
No they don't. If they pointed in the same direction, scale would read more than 100 x 9.81 = 981N.
Just knowing the scale reading and mass, you can resolve the sign of acceleration a. You can't however tell if the elevator is moving up or down.
It may be moving, for example, downwards but decelerating or moving up and accelerating. In both cases the acceleration a will be of the same sign and magnitude.
And all this is anambiguously derived from second Newton's Law.
 
  • #40
xAxis said:
No they don't. If they pointed in the same direction, scale would read more than 100 x 9.81 = 981N.
Uh I made a blander here. What I wanted to say is that if your result was correct, the scale would read more than 981N.
 
  • #41
xAxis said:
Maxo, in both your original and reformulated question you made the same mistake, and it obviously comes from confusion about signs. If you want to declare g as a vector, no problem. But if you say that downwards direction is positive, g is acceleration, so all downward accelerations will be positive, and upwards negative. This doesn't mean that if lift moves downwards the acceleration is positive.
No they don't. If they pointed in the same direction, scale would read more than 100 x 9.81 = 981N.
Just knowing the scale reading and mass, you can resolve the sign of acceleration a. You can't however tell if the elevator is moving up or down.
It may be moving, for example, downwards but decelerating or moving up and accelerating. In both cases the acceleration a will be of the same sign and magnitude.
And all this is anambiguously derived from second Newton's Law.

Actually I want to say that downwards direction is negative, and upwards positive. That's why I mean if a is negative when accelerating downwards (please note I'm not saying "simply moving" or "decelerating", but accelerating downwards), then so should g be. Since g is obviously also pointing downwards.

You say there is no problem declaring g as a vector. Then I don't understand why when doing that and considering downwards a negative direction, why g is not negative?
 
  • #42
Since you seem to just ignore all the advice you've been given in this thread, how about you show how this equation is derived using Newton's laws.

Maxo said:
In my physics book the equation for apparent weight is given as

FN = mg + ma
 
  • #43
Maxo said:
You say there is no problem declaring g as a vector. Then I don't understand why when doing that and considering downwards a negative direction, why g is not negative?
You really should start using different symbols for the component of the vector and the magnitude of the vector. Let's say the Z-Axis of the coordinate system is up:

g = 9.81

\vec{g} = (0, 0, -9.81)

Or since it is only 1D:

\vec{g} = -9.81
 
  • #44
A.T. said:
You really should start using different symbols for the component of the vector and the magnitude of the vector. Let's say the Z-Axis of the coordinate system is up:

g = 9.81

\vec{g} = (0, 0, -9.81)

Or since it is only 1D:

\vec{g} = -9.81

Great, that's how I see it aswell! Could you then please explain why using \vec{g} = -9.81 in the equation \vec{F_{N}} = m\vec{g} + m\vec{a} will give a POSITIVE sign for \vec{a} (i.e. which would correspond to positive/upwards direction), when the elevator accelerates DOWNWARD from rest (i.e. \vec{a} should also pointing in the negative/downwards direction, i.e. have a negative sign)?
 
  • #45
Maxo said:
Great, that's how I see it aswell!
Note that he says g = 9.81, not g = -9.81.

Could you then please explain why using \vec{g} = -9.81 in the equation \vec{F_{N}} = m\vec{g} + m\vec{a} will give a POSITIVE sign for \vec{a} (i.e. which would correspond to positive/upwards direction), when the elevator accelerates DOWNWARD from rest (i.e. \vec{a} should also pointing in the negative/downwards direction, i.e. have a negative sign)?

For one thing, this vector equation \vec{F_{N}} = m\vec{g} + m\vec{a} is not valid. Per Newton's 2nd law, it should be:
\vec{F_{N}} + m\vec{g} = m\vec{a}
 
  • #46
Doc Al said:
Note that he says g = 9.81, not g = -9.81.



For one thing, this vector equation \vec{F_{N}} = m\vec{g} + m\vec{a} is not valid. Per Newton's 2nd law, it should be:
\vec{F_{N}} + m\vec{g} = m\vec{a}

Finally! This equation makes sense! Why didn't you just write this in the first place? ;)

To me, this way of writing is more logical. I think vector notation is more clear.

Thanks for your help and patience.
 
  • #47
Maxo said:
Finally! This equation makes sense! Why didn't you just write this in the first place? ;)

To me, this way of writing is more logical. I think vector notation is more clear.

Thanks for your help and patience.

I don't get it.

I said way earlier that all the forces should be on one side of the equation, and the resultant dynamics (ma) should be on the other side of the equation.

And it took 46 posts for this to finally sink in?

Zz.
 
  • #48
Maxo said:
...
You say there is no problem declaring g as a vector. Then I don't understand why when doing that and considering downwards a negative direction, why g is not negative?
Again, you can treat g a vector but it has to be constant. Once you decide on sign convention then g's sign is determined. Makes no sense to ask question why g is not negative. It simply is.
You use 2nd Newton's to calculate ma.
And you always add all the forces in your diagram and only then you can talk about what a is. You just have to be careful with signs. In your example from page 2, for instance, the normal force is positive and g is negative which means that upwards is positive. Lift is accelerating downwards which means a is negative, which it is.
If you chosen g to be positive, than the normal force would be negative and a positive. Again everything is consistent.
 
  • #49
ZapperZ said:
I don't get it.

I said way earlier that all the forces should be on one side of the equation, and the resultant dynamics (ma) should be on the other side of the equation.

And it took 46 posts for this to finally sink in?

Zz.
You didn't use vector notation, which was what I was looking for. Maybe I didn't manage to express clearly what I was looking for, so you don't have to feel bad for misunderstanding. You were apparently not the only one.

xAxis said:
Again, you can treat g a vector but it has to be constant. Once you decide on sign convention then g's sign is determined. Makes no sense to ask question why g is not negative. It simply is.
You use 2nd Newton's to calculate ma.
And you always add all the forces in your diagram and only then you can talk about what a is. You just have to be careful with signs. In your example from page 2, for instance, the normal force is positive and g is negative which means that upwards is positive. Lift is accelerating downwards which means a is negative, which it is.
If you chosen g to be positive, than the normal force would be negative and a positive. Again everything is consistent.

Thanks, that makes sense.

Thanks to everyone else who posted and I tried to help me also. I appreciate it my friends
 
  • #50
A pretty confusing thread. The people who attempted to help Maxo didn't seem to succeed. For some reason I thought that this might actually be interesting, from the point of view of studying the pedagogical issues.

My answer to the original question would have been this: The apparent weight was defined so that g must be positive, and a must be positive for upwards acceleration, and negative for downwards acceleration.

Then, if the question still remains as why it is this way, the answer remains the same: The apparent weight was defined so that that's the way signs go! It can be seen from the definition!

The apparent weight was defined here:

Maxo said:
In my physics book the equation for apparent weight is given as

FN = mg + ma

You can define it with different sign conventions too, if you want, of course.

When a discussion like this drifts to new confusing questions such as possible "signs of vectors", the side track should be discouraged. Maxo should have been told that vectors usually don't have signs, and if he insists on them having them, that would be a new concept which would require a new definition. Writing down a definition for this new concept would not help in the original problem though.
 
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