When is the acceleration due to gravity negative and when is it positive?

In summary, the acceleration due to gravity is negative when the rocket is moving upwards and positive when it is moving downwards.
  • #1
potatogirl
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Homework Statement
I do know that it depends on the axis/reference frame chosen. In my class, we always use up as positive and down as negative. However, I am still confused. I know an object in freefall will accelerate at -9.8 m/s^2 with this axis reference...but sometimes in our homework problems, the only way to get a correct answer is for g to be positive.

See the following homework problem:

A rocket of mass 4.50 × 10^5 kg is in flight. Its thrust is directed at an angle of 55.0° above the horizontal and has a magnitude of 7.50 X 10^6 N. Find the magnitude and direction of the rocket's acceleration. Give the direction as an angle above the horizontal.
Relevant Equations
F=ma
I know to break it down into its x and y components and then use Pythagorean:

Acceleration in the x direction is Fx/m ---> (7.50 x 10^6*cos55) / (4.50 x 10^5 kg) = 9.56 m/s^2

Acceleration in the y direction is: (Fy - mg)/m ---> ((7.50 x 10^6*sin55) - (4.5 x 10^5* 9.8 m/s^2)) / (4.5 x 10^5 kg)

Then I can take it from there...but this problem (like many others) hinges on g being positive 9.8m/s^2 and not negative. In the above equation we have it as positive (I saw a solution key online that shows it like that). When I use g as negative I get the wrong answer. So I am just trying to get the concept down so I can keep my neg/positive signs correct for future problems.

I feel like this is a stupid question but I am just not getting it unfortunately. Any help in understanding is appreciated! Is it because the rocket is moving upwards and not downwards? Do we only use -9.8 m/s^2 in freefall??

Thank you!
 
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  • #2
Whenever in doubt, draw a free body diagram.

That being said, in your Fy equation, there is a negative sign in front of the gravity term, so the directionality is being ”applied” there.
 
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  • #3
potatogirl said:
Acceleration in the y direction is: (Fy - mg)/m ---> ((7.50 x 10^6*sin550 - 4.5 x 10^6*sin) - (4.5 x 10^6 * 9.8 m/s^2)) / (4.5 x 10^5 kg
Your equation is treating ##g## as a scalar and subtracting it. That's the same result as treating it as a negative and adding it, which is what you want to do. If you want to be consistent just toss a ##-## in front of the ##g## and change the ##-## between brackets to a ##+##

[edit: subsequent comments and apparently ##g## is a scalar, or at least a vector with a fixed position of "down". Doesn't really change my suggestion, though, except don't change the sign if you're using the letter ##g## in the equation]
 
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  • #4
The standard usage is that "g" is a magnitude, so positive. Thus if up is positive then the acceleration due to gravity is -g.
Personally, I think this convention is unfortunate, but there it is.
 
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  • #5
But consistent in 1, 2, or 3, dimensions. How do you do better?
 
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  • #6
potatogirl said:
When I use g as negative I get the wrong answer. So I am just trying to get the concept down so I can keep my neg/positive signs correct for future problems.
The sign of ##g## is a recurring issue with beginners. One way to sort it out, until you become a practiced expert, is to think of the acceleration of gravity as always being down. You can't go wrong because, after all, that is how "down" is defined. Obviously, "up" is opposite to "down" so if you add two components of forces ##F_{up}## and ##F_{down}##, there must be a relative negative sign between them. By this I mean that their sum can be written in two different ways:
  1. ##F_{sum} = F_{up}-F_{down}.##
  2. ##F_{sum} =- F_{up}+F_{down}=-(F_{up}-F_{down}).##
In the above expressions both ##F_{up}## and ##F_{down}## are positive quantities. Note that expression 1 is the negative of expression 2. This means that if you calculate the magnitude of ##F_{sum}##, you will get the same answer because the negative sign disappears.

Let's look at a specific example. Say ##F_{up}= 10~##N and ##F_{down}= 20~##N. Then the two calculations give
##F_{sum} = 10~\text{N}-20~\text{N}=-10~\text{N}.##
##F_{sum} =-10~\text{N}+20~\text{N}=+10~\text{N}.##

The magnitude of the sum is clearly 10 N. What about the direction? The direction is the same as the direction of the component with the largest magnitude. In the above expression, that is ##F_{down}##. Therefore ##F_{sum}## is directed down. The answer is "The net force has magnitude 10 N and points down".

What if you have to submit an answer to an online homework-grading algorithm where "up" and "down" are not acceptable options but "+" and "-"? Then you have to check the statement of the problem for a statement that specifies whether "down" is positive or negative. If there is no such statement, you have to assume the standard convention that "up" is "positive" and "down" is negative.

So for this example you would proceed as above to find the magnitude, then simply replace "down" with a negative sign and provide the answer "The net force is - 10 N." However, if the problem clearly stated something like "Take direction down as positive", then the answer is "The net force is 10 N."

The gist of all this is that if you replace ##g## with -9.8 m/s2 in an equation like ##~F_{net}=T-mg~## for a mass hanging from a string, you will be changing the relative negative sign between the two termsn. That is not correct because you will be effectively changing the direction of the weight from being opposite to the tension to being in the same direction. Changing the relative sign results in changing the relative orientation of the vectors from antiparallel to parallel. Now that you know about it, preserving the relative orientation of the vectors when you substitute numbers is something to watch for.
 
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  • #7
I absolutely agree that g should always have the value +9.8m/s 2. The rest simply follows from the required free body diagram with vectors in any coordinate dimensionality . I agree it is often a point of confusion, because this is not rigorously enforced.
 
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  • #8
hutchphd said:
I agree it is often a point of confusion, because this is not rigorously enforced.
I think that the confusion is a matter of sloppiness if not laziness. People refer to symbol ##g## as the "acceleration of gravity" instead of "the magnitude of the acceleration of gravity" thus conveying the impression that they mean ##\vec g## instead of ##g##.
 
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  • #9
I would prefer the free body diagram to be marked with a vector arrow labeled Fg and the value (+/_ mg) assigned when the substitution is made into Newton equation.
`
 
  • #10
hutchphd said:
I would prefer the free body diagram to be marked with a vector arrow labeled Fg and the value (+/_ mg) assigned when the substitution is made into Newton equation.
Me too. In the FBD, the label is magnitude of the weight, ##W## not the vector ##\vec W## because the location of the tip of the vector specifies the direction. When the down arrow is transferred into the algebraic equation, ##W## remains as is (scalar independent of choice of axes) while the tip of the arrow is replaced with a "+" or "-" depending on whether "down" is, respectively, positive or negative.
 
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  • #11
Acceleration is a vector. It has components. It has a magnitude.

If an object is in free fall and you've chosen the y-direction to be upward, then ##a_y=-g## and the magnitude of the acceleration is ##g##.
 
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  • #12
I think another reason why students often are confused is because they don't recognize that ##\vec F## and ##F## represent two different quantities, so they think the two symbols are interchangeable. Comparing ##\vec F_{\rm sum} = \vec F_{\rm up} + \vec F_{\rm down}## and ##F_{\rm sum} = F_{\rm up} - F_{\rm down}## then makes it seem like the minus sign showed up out of nowhere.

We also tend to overload some symbols. For example, in one context, we might say ##\vec F = F_x \,\hat i + F_y \,\hat j## where we take ##F_x## and ##F_y## to be the scalar components of ##\vec F## and can therefore be negative. In other places, we say ##F_x## is the magnitude of ##\vec F_x##, the vector component of ##\vec F## in the ##x##-direction, in which case ##F_x## is by definition positive.

For instance, in @kuruman's example above, ##F_{\rm up}## and ##F_{\rm down}## represent the magnitudes of the respective vectors. But ##F_{\rm sum}## comes out negative, so it's apparently the scalar component of ##\vec F_{\rm sum}## and not a magnitude unlike the other ##F##s.
 
  • #13
vela said:
For instance, in @kuruman's example above, ##F_{\rm up}## and ##F_{\rm down}## represent the magnitudes of the respective vectors. But ##F_{\rm sum}## comes out negative, so it's apparently the scalar component of ##\vec F_{\rm sum}## and not a magnitude unlike the other ##F##s.
In my example, ##F_{\text{sum}}## is the difference of magnitudes taken one way and then the other so that one is negative and the other positive. Then I state that the magnitude of the difference taken either way is the magnitude of the sum of the two antiparallel vectors which obviously is less than the magnitude of either vector. Then I point out that the direction of the of the sum, considered as a vector not as a component, is the direction of the larger of the two vectors that make up the sum.

The problem asks to find the magnitude and direction, i.e. give the answer in polar form. I have carefully avoided using the term "component" in my example in order to sidestep the "overload of symbols" that you mentioned. Instead, I attempted to anchor direction on the intuitively understood concept of "down" and provide a recipe for figuring out the direction in a manner that does not require defining which way is positive.
 
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  • #14
vela said:
I think another reason why students often are confused is because they don't recognize that ##\vec F## and ##F## represent two different quantities, so they think the two symbols are interchangeable. Comparing ##\vec F_{\rm sum} = \vec F_{\rm up} + \vec F_{\rm down}## and ##F_{\rm sum} = F_{\rm up} - F_{\rm down}## then makes it seem like the minus sign showed up out of nowhere.

We also tend to overload some symbols. For example, in one context, we might say ##\vec F = F_x \,\hat i + F_y \,\hat j## where we take ##F_x## and ##F_y## to be the scalar components of ##\vec F## and can therefore be negative. In other places, we say ##F_x## is the magnitude of ##\vec F_x##, the vector component of ##\vec F## in the ##x##-direction, in which case ##F_x## is by definition positive.

For instance, in @kuruman's example above, ##F_{\rm up}## and ##F_{\rm down}## represent the magnitudes of the respective vectors. But ##F_{\rm sum}## comes out negative, so it's apparently the scalar component of ##\vec F_{\rm sum}## and not a magnitude unlike the other ##F##s.
You should call vector components not "scalar", although they are, of course numbers. I'd reserve "scalars" strictly to quantities, which are invariant under the choice of the basis you use to describe things in. The vector components are not, because they transform under changes (rotations) of the (Cartesian) basis. Also the vector itself is, by definition, independent of the basis, i.e.,
$$\vec{F}=F_x \hat{i} + F_y \hat{j} + F_z \hat{k} = F_x' \hat{i}' + F_y' \hat{j}' + F_z' \hat{k}',$$
and the rules, how to transform from one Cartesian frame to another is dictated by this "invariance" of the vector on the choice of the basis. It's of course very cumbersome to work this out in this awkward notation. So let's change it in writing
$$\vec{F}=\sum_{j=1}^3 F_j \vec{e}_j,$$
where ##\vec{e}_j## are Cartesian basis vectors with
$$\vec{e}_j \cdot \vec{e}_k=\delta_{jk}=\begin{cases}1 & \text{if} \quad j=k,\\ 0 & \text{if} \quad j \neq k \end{cases}.$$
Then the change between two such bases are described by introducing a transformation matrix ##T_{jk}## via
$$\vec{e}_j' = \sum_{k} \vec{e}_k T_{kj}.$$
To have both bases being Cartesian you must have
$$\vec{e}_{j_1}' \cdot \vec{e}_{j_2}'=\delta_{j_1 j_2} = \sum_{k_1,k_2} \vec{e}_{k_1} \cdot \vec{e}_{k_2} T_{k_1j_1} T_{k_2 j_2} = \sum_{k_1,k_2} \delta_{k_1 k_2} T_{k_1 j_1} T_{k_2 j_2}=\sum_{k_1} T_{k_1 j_1} T_{k_1j_2},$$
which tells you that the matrix ##\hat{T}=(T_{kj})## must be a orthonormal matrix, because the above condition reads in matrix notation
$$\hat{T}^{\text{T}} \hat{T}=\hat{1}.$$
I.e., the transposed matrix must be the inverse matrix.

Now we can easily see, how the vector components have to transform:
$$\vec{F}=\sum_k F_k \vec{e}_k = \sum_j F_j' \vec{e}_j' = \sum_{jk} T_{jk} F_k \vec{e}_j'.$$
Since the decomposition of a vector wrt. the basis ##\vec{e}_j'## is unique you get
$$F_j'=\sum_k T_{jk} F_k.$$
This ensures that the vector stays the same under the basis transformation.
 

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