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

[namanjain]

F(external)=ma

mg-N=ma
direction of a downward
or if
a is upwards
N-mg=ma
N=mg+ma

 

[Maxo]

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.

 

[Doc Al]

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.

 

[Nugatory]

What is it then?

And how can the equation F_N = 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.

 

[Maxo]

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?

 

[Doc Al]

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.

 

[Vanadium 50]

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).

 

[ehild]

Start from Newton's Second equation: ma=∑F_i.

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

 

[xAxis]

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.

 

[xAxis]

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.

 

[Maxo]

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?

 

[Doc Al]

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.

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

F_N = mg + ma

 

[A.T.]

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

 

[Maxo]

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)?

 

[Doc Al]

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}

 

[Maxo]

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.

 

[ZapperZ]

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.

 

[xAxis]

...
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.

 

[Maxo]

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.

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

 

[jostpuur]

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:

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

F_N = 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.

 

[Doc Al]

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

Actually, I did:

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

(I used bold to represent vectors.)

 

[Maxo]

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.

I agree that would have been good if I had been told that. I wish I would have been able to formulate myself that was in fact was I was asking for already in the beginning but I guess I couldn't see it from such a "meta" perspective when I was struggling in the middle of the question so to speak.

Anyway it's an interesting pedagogical analysis and if you could find some way to implement this way of thinking on the forum, in schools/universites and/or generally I would welcome that. I would like to help with this if possible, if you have some further ideas on the subject please share them on the forum (perhaps in a new thread since it's off topic here) or on in a private message.

Anyway. The original question has been solved. Carry on, gentlemen!