# Sign conventions with gravity

• B
Gold Member

## Main Question or Discussion Point

We say that, for gravity, F=mg, where g≈-9.8 m/s2. Then, we have U=mgh. But, it seems like that equation would only work if g were positive 9.8 m/s2. That is, because ##F=-\frac{dU}{dr}##, and ##\frac{dU}{dr}## is essentially ##\frac{dU}{dh}##, we should get F=-mg. So, for the two equations to work, it seems like the g in the first equation should have the opposite sign as the g from the second equation. But that seems like a really weird convention to have. So, if the convention is that g is positive then we should have F=-mg, and if g is negative we should have U=-mgh, right?

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Orodruin
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Be careful with your sign conventions. If you define your coordinate to increase when going against the field, then indeed F=-mg, where g is positive, this is just telling you the force goes in the opposite direction relative to the coordinate increase, just as expected.

• AlphaLearner and Isaac0427
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Be careful with your sign conventions. If you define your coordinate to increase when going against the field, then indeed F=-mg, where g is positive, this is just telling you the force goes in the opposite direction relative to the coordinate increase, just as expected.
Ok. And in that case, U=mgh, right? And if you define the coordinates to increase when going with the field, you would have F=mg and U=-mgh where g is positive, right? But, why do most sources I have seen use both F=mg and U=mgh? Isn't that inconsistent?

Orodruin
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Ok. And in that case, U=mgh, right? And if you define the coordinates to increase when going with the field, you would have F=mg and U=-mgh where g is positive, right? But, why do most sources I have seen use both F=mg and U=mgh? Isn't that inconsistent?
That depends on the definition of h relative to the coordinate.

Dale
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But, why do most sources I have seen use both F=mg and U=mgh?
The g in these two cannot be the same. The g in the first equation is a vector and the g in the second equation is a scalar.

It is absolutely up to the reader/student to learn what each term in an equation means. You should never assume that the same symbol means the same thing in a different equation, nor should you assume the opposite!

Orodruin
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The g in these two cannot be the same.
Of course they can. You just study one direction (ie, one component of the force) and name the coordinate ##x## in the down direction, leading to the relation ##h = x_0-x## and ##dh/dx=-1##. You just have to be careful with your conventions.

• Dale
Dale
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Of course they can.

If I wrote those two equations then g in each equation would not be the same thing. The OP needs to read the text to find out how the textbook author defined them and not assume that the symbols are the same.

Last edited:
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The g in these two cannot be the same. The g in the first equation is a vector and the g in the second equation is a scalar.

It is absolutely up to the reader/student to learn what each term in an equation means. You should never assume that the same symbol means the same thing in a different equation, nor should you assume the opposite!
I understand that last part, but in these equations, both represent the vector gravitational acceleration (for the second equation I could have said U=mg⋅h). It seems like the gravitational acceleration takes on two different values.

Doc Al
Mentor
My advice is to always use g to be just the magnitude of the acceleration due to gravity. Thus g = 9.8 m/s^2.

So then: The acceleration due to gravity is g, downward. The weight of an object is mg, downward.

Depending on the sign convention you choose for any particular application, those vector components can be positive or negative.

• AlphaLearner, PeroK and Dale
Dale
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both represent the vector gravitational acceleration (for the second equation I could have said U=mg⋅h)
That is a very unusual convention. Given that, you need to make doubly sure that you do not assume anything about the meaning of the symbols from one equation to another. The author is using different notation from the community, so it is likely that they use different notation from section to section also

then indeed F=-mg, where g is positive
But in the equation F= -mg, 'g' is the only value which can be negative. But you say 'g' is positive, then what makes 'mg' negative when 'g' is positive?

Doc Al
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But in the equation F= -mg, 'g' is the only value which can be negative. But you say 'g' is positive, then what makes 'mg' negative when 'g' is positive?
mg is just the magnitude of the force; the direction is downward. With a sign convention that up is positive, the vertical component of that force would be -mg.

jtbell
Mentor
With a sign convention that up is positive, the vertical component of that force would be -mg.
This is normally written with some notation indicating that we are dealing with a component, e.g. Fy = -mg.

mg is just the magnitude of the force; the direction is downward. With a sign convention that up is positive, the vertical component of that force would be -mg.
Yes, even I agree that 'mg' is always act downwards but then depends on direction of object being displaced in. If object displaces upwards, gravity act downward, work done by gravity is negative. If body displace along gravity, gravity does positive work. I understand. But in F = -mg, If 'g' is taken positive, it means that body was moving along gravity. When 'mg' is negative and 'g' is positive in it based on above equation. Means gravitational force acting on freely falling body is negative, does gravity do negative work on freely falling body? How can we simply say since 'mg' act downward means force is -mg? I did not understand.

Orodruin
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But in F = -mg, If 'g' is taken positive, it means that body was moving along gravity.
It has nothing to do with how the body is actually moving, it only depends on which direction is defined as being the positive direction.
Means gravitational force acting on freely falling body is negative, does gravity do negative work on freely falling body?
The sign of the work done by gravity depends on whether or not the object moves against or in the same direction as the gravitational field, not on the way you define your coordinate.
How can we simply say since 'mg' act downward means force is -mg?
This is only true if you define the coordinate as increasing when the object goes up. It all depends on the choice of coordinates.

PeroK
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But in the equation F= -mg, 'g' is the only value which can be negative. But you say 'g' is positive, then what makes 'mg' negative when 'g' is positive?
Reality makes it so! The maths must follow the physics. Gravity acts downwards/towards the centre of the Earth. All equations and sign conventions are geared to respect that physical fact.

There's nothing mathematical that is going to jump up and tell you that you have the signs mixed up. It's up to you to align the mathematics with the physical reality.

• AlphaLearner
Doc Al
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Means gravitational force acting on freely falling body is negative, does gravity do negative work on freely falling body? How can we simply say since 'mg' act downward means force is -mg?
The work done by a force depends on the direction of the force and the direction of the displacement. If both force and displacement are in the same direction -- as in a falling body -- the work done is positive. (Note that force and displacement are both negative in that case, using our sign convention.)

It has nothing to do with how the body is actually moving, it only depends on which direction is defined as being the positive direction.

The sign of the work done by gravity depends on whether or not the object moves against or in the same direction as the gravitational field, not on the way you define your coordinate.

This is only true if you define the coordinate as increasing when the object goes up. It all depends on the choice of coordinates.
Thank you sir, I have got some clarity. But in first reply, positive direction of object is defined as direction along the gravity right? And in kinematics, in projected bodies 'g' is considered negative since goes against 'g' and freely falling bodies 'g' is considered positive as goes along 'g', we applied same here. Am I right?

Doc Al
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Thank you sir, I have got some clarity. But in first reply, positive direction of object is defined as direction along the gravity right? And in kinematics, in projected bodies 'g' is considered negative since goes against 'g' and freely falling bodies 'g' is considered positive as goes along 'g', we applied same here. Am I right?
No, you're just confusing yourself.

Just remember this: The direction of the force of gravity and the direction of the acceleration due to gravity (for a freely falling object) are both down. Whether that's positive or negative is entirely up to you and your sign convention.

• AlphaLearner
Just remember this: The direction of the force of gravity and the direction of the acceleration due to gravity (for a freely falling object) are both down. Whether that's positive or negative is entirely up to you and your sign convention.
I understand, 'g' always acts downwards followed by 'mg'. But can't we blindly take as a body moves up, 'g' act downward so 'g' is taken negative and as body moves down 'g' as positive since it act downwards in all cases?

This is not a confusion for me since today, for an year since I learnt it. I once quarreled with my room mates regarding this! Even I said 'g' always act downward. But they told me this. Later my physics teacher drew the positive x,y axes beside each case: Projected and freely falling. In projected, he normally defined positive y-axis upwards and since 'g' is downward, told to take 'g' as negative. In freely falling, he just turned the positive y-axis downwards and since 'g' is also downward, told to take 'g' as positive.
So now I understand, 'g' is always downwards, no matter the case is. But its sign depends on how we define the coordinate system.

Thanks for help.

Doc Al
Mentor
I understand, 'g' always acts downwards followed by 'mg'. But can't we blindly take as a body moves up, 'g' act downward so 'g' is taken negative and as body moves down 'g' as positive since it act downwards in all cases?
You can choose a sign convention such that the acceleration of a falling body is negative, thus a = -g. (But I recommend that the symbol g just stand for the magnitude of that acceleration.)

So now I understand, 'g' is always downwards, no matter the case is.
The acceleration due to gravity is always downwards, and its magnitude is g.

But its sign depends on how we define the coordinate system.
Right. You can choose down to be positive, in which case the acceleration of a falling body is a = +g.