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

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Discussion Overview

The discussion revolves around the apparent weight of an object in an accelerating elevator, focusing on the signs of gravitational acceleration (g) and the system's acceleration (a). Participants explore the implications of these signs in the context of free body diagrams and the normal force experienced by a person standing on a scale in an elevator. The scope includes conceptual reasoning and technical clarification regarding the application of forces in a non-inertial reference frame.

Discussion Character

  • Exploratory
  • Technical explanation
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • One participant presents the equation for apparent weight and calculates the acceleration of the system based on different apparent weights, questioning the sign of acceleration when the elevator accelerates downwards.
  • Another participant suggests that using free body diagrams and considering pseudo forces can simplify understanding, indicating that the direction of acceleration affects the signs used in calculations.
  • Some participants argue about the direction of forces in free body diagrams, with one asserting that both gravitational force and the acceleration of the elevator should be considered in the same direction, leading to confusion over their signs.
  • There is a discussion about the convention of signs, with some participants stating that gravitational acceleration (g) is always positive, while others suggest that it depends on the chosen coordinate system.
  • One participant emphasizes that the speed of the elevator is irrelevant, focusing instead on the acceleration's effect on apparent weight.
  • Another participant reiterates the confusion over why acceleration (a) and gravitational acceleration (g) have opposite signs despite both acting downwards in certain scenarios.

Areas of Agreement / Disagreement

Participants express differing views on the sign conventions for acceleration and gravitational force, with no consensus reached on the correct interpretation of these signs in the context of the problem. The discussion remains unresolved regarding the relationship between the signs of g and a.

Contextual Notes

Participants highlight the importance of sign conventions and the assumptions made in free body diagrams, indicating that the discussion is dependent on these definitions and may vary based on individual interpretations.

Maxo
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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.
 
Last edited by a moderator:
  • #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!
 

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