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My professor insists that weight is a scalar. I sent him an email explaining why it's a vector, I even sent him a source from NASA clearly labeling weight as a vector. Every other source also identifies weight as a vector.

I said that weight is a force, with mass times the magnitude of gravitational acceleration as the scalar quantity and a downward direction.

His response, "Weight has no direction, i.e., it is a scalar!!!" My thought process is that since weight is a force, and since force is a vector, weight has to be a vector. This is the basic transitive property of equality.

Am I and all of these other sources wrong about weight being a vector? Is weight sometimes a vector and sometimes a scalar?

After reading thoroughly through his lecture notes, I discovered his reasoning behind his claim:

Similarly to how speed is the scalar quantity (or magnitude) of velocity, weight is the scalar quantity (or magnitude) of the gravitational force a celestial body exerts on mass.

I'm still inclined to think of weight as a vector for convenience and to separate it from everyday language. However, like one of the comments stated, "Definitions serve us."

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    $\begingroup$ Maybe he says weight is a scalar because its direction is always known. $\endgroup$
    – lucas
    Commented Mar 2, 2017 at 3:44
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    $\begingroup$ He might be thinking of weight as the magnitude of the force of weight. After all, nobody ever says "my weight is -170 pounds in the $\hat{z}$ direction". $\endgroup$
    – knzhou
    Commented Mar 2, 2017 at 3:57
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    $\begingroup$ In any case, this is a totally unimportant pedantic point that should have no bearing on your professor's quality as an instructor. $\endgroup$
    – knzhou
    Commented Mar 2, 2017 at 3:57
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    $\begingroup$ @knzhou My doctor always writes a coordinate system when they weigh me... $\endgroup$
    – Ryan Unger
    Commented Mar 2, 2017 at 3:57
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    $\begingroup$ Are you looking for validation or a way to win the argument? Weight is gravity force, that's its modern definition. When I had professors that would say something that didn't make sense, I would ask about the issue. If the professor was the sort that couldn't admit a mistake, then I would just write the issue down in my notes and bring it up my study group. $\endgroup$
    – David Elm
    Commented Mar 2, 2017 at 11:31

20 Answers 20

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On earth, weight of a body is defined as the force by which the body is attracted by the earth towards its center. Weight can thus be considered the same as the gravitational force exerted by the earth on that body. Hence, weight can be deemed a vector since it is a force, irrespective of the planet you consider. $$\vec W=m\vec g=\frac{GMm}{r^2}\hat r$$ As mentioned in the comments, since $g$ has the same direction (directed towards the center of the concerned planet) always, it might be(?) considered a scalar. Thats what your prof is doing. But strictly speaking, weight is a vector.

Hope this helps you.

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    $\begingroup$ Weight is a vector in physics. Weight in 'weights and measures' is a scalar in a variety of other contexts, like selling potatoes. Language is ambiguous. $\endgroup$
    – Whit3rd
    Commented Mar 2, 2017 at 6:45
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    $\begingroup$ @Whit3rd This may be true but the distinction and the difference between the common use of the language and the physics use of concept is usually heavily emphasized in physics courses. $\endgroup$ Commented Mar 2, 2017 at 10:48
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    $\begingroup$ @R.M. By same direction, I meant $\hat r$. Nothing else. And as you have said, it will change from place to place. $\endgroup$ Commented Mar 2, 2017 at 19:56
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    $\begingroup$ The definition you give in your first sentence implies weight is actually a scalar: it is the magnitude of the vector pointing toward the Earth's center. If it were not pointing toward the Earth's center, it would not be "weight". If Earth is exerting the same force towards its center on two objects, those objects have the same weight, regardless of the actual orientation of those force vectors. If weight were a vector, you could not say they had "the same weight" unless they were in the exact same spot. $\endgroup$ Commented Mar 3, 2017 at 0:10
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    $\begingroup$ @SchrodingersCat I don't necessarily disagree, but my point is about your first sentence. Defining weight to be the force in the direction toward Earth's center essentially "factors out" the directional component, making it a scalar. If it were not pointing toward the Earth's center, it would not be weight, so there is no sensible directional degree of freedom by that definition. $\endgroup$ Commented Mar 3, 2017 at 5:21
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We can change the definition of things whenever it is useful. Definitions serve us. If a definition isn't useful, individuals and communities change it, sometimes on the fly, sometimes in context, sometimes explicitly, sometimes implicitly.

In everyday experience, weight is a scalar. You don't write down the direction of the weight of the bananas you buy. Insisting it is a vector is not useful in this context, and definitions exist to both clarify communication and solve problems.

Does adding the direction to the weight of the banana help solve any problems? Or is it noise? Is the scalar weight of the banana a communication problem in this case?

There are going to be other contexts where you will want weight to be a vector; maybe when calculating orbital mechanics of your banana. Even there, weight may not be a useful concept, because there are much better ways to solve orbital mechanics than talking about the directional weight of things (field potentials, say).

In formal mathematics, very specific and exact definitions are used to permit abstractions that don't match any physical situation to be discussed and delt with in a uniform way. Formal mathematics is often pillaged by physics, but physics isn't formal mathematics.

Physicists and Engineers will go off and talk about dirac delta functions whose value is 0 everywhere except at 0, and whose integral from any negative value to a positive one is 1, and then they convolve it with another function.

Now, there are ways to formalize this, but for the most part Physicists and Engineers don't bother. "The Dirac Delta isn't a function" is useful when formalizing it, but isn't nearly as useful when working with it. Knowing the formalization can be useful to avoid possible pitfalls, but it isn't usually useful when trying to use it as a tool to predict behaviour of some system.

Physics is a game of using mathematics (or whatever other tool is handy) to predict (and sometimes explain) the behavior of physical systems. There are often multiple different mathematical games, and you will use different ones for different systems. Newtonian dynamics is a game that works within its domain, and in it velocity is additive. Relativity is a game that is overly complex for some domains, but covers some territory that Newtonian dynamics doesn't cover; in Relativity, velocity isn't additive. Within Newtonian dynamics, velocity is a simple vector in a Euclidean space. Within Relativity, it isn't a simple vector in a Euclidean space.

Weight is a scalar within some games of Physics. In others, it may not be. In almost every reasonable situation you will experience, Weight will be a Scalar, because in almost every game of Physics where the direction of Weight is important, using a vector-based Weight isn't going to be the best tool you have.

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    $\begingroup$ This is the best answer. Physics provides models to understand the world. When one model works, fine. When it does not, use another. Better not to get bogged down by notation. $\endgroup$ Commented Mar 2, 2017 at 21:09
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    $\begingroup$ Usually, when you're dealing with an inclined plane, you need your weight to be vectorial. If you're modeling a pendulum, you may use a Laplacian, but normally using vectors is easier. I agree that maybe most of the time you won't deal with weight as a force (rather, you deal with scalar mass), but we cannot say that "in almost every game of Physics where the direction of Weight is important, using a vector-based Weight isn't going to be the best tool you have". $\endgroup$
    – Wtrmute
    Commented Mar 3, 2017 at 15:07
  • $\begingroup$ You: There are going to be other contexts where you will want weight to be a vector; maybe when calculating orbital mechanics of your banana. Now, when the bananas are orbiting, they are clearly weightless, so now we come to another philosophical question, namely is the zero vector equal to the zero scalar (and does a magnitude of size zero even carry a dimension (unit), such as newtons)? (I am just trying to be humorous). $\endgroup$ Commented Mar 7, 2017 at 9:40
  • $\begingroup$ @jeppe they are weightless in some frames of reference. Solving orbital mechanics in frames of reference where they are weightless seems suboptimal. $\endgroup$
    – Yakk
    Commented Mar 7, 2017 at 13:04
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If he is a PhD and you are a Major in physics there is no reason to take his word above your own in this case. Physicists don't get a rerun of "basic concepts in mechanics" after majoring. Most likely he knew as much about weight when he was a major as he does now. And you will also not get further education on these basic concepts.

However, if instead of arguing about misuse of terms you want to try to understand, ask him "how do we define weight in the context of this course?" For example, it might be that in the particular field it is handy to use the weight's component/projection that is normal to the surface on which the object is. In that case it would be normal to have scalar weight and call it "weight" because of tradition or handiness reasons.

But there is always a chance that your professor is just not very insightful and was never interested to find out and correct his misconceptions of basic subjects. PhD is earned in a subfield, it doesn't automatically make you bright.

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    $\begingroup$ In what sense is this an answer to the question? $\endgroup$
    – Peaceful
    Commented Mar 4, 2017 at 10:12
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    $\begingroup$ @Peaceful The original question had a "This guy has his PhD in physics." I thought it would be useful to address this concern the way I did and indirectly confirm that his knowledge of common definition of weight is correct. Then I pointed out that definitions can be different in different use cases and it would be best to ask how it's defined. I think my answer was the first to say "it can be redefined whatever is useful in the case" idea. If you still aren't convinced this answer carries some value, I will happily accept a downvote :) $\endgroup$
    – Džuris
    Commented Mar 4, 2017 at 15:02
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For any vector physical quantity it may make perfect sense to define a corresponding scalar quantity equal to that vector's magnitude. Unsurprisingly, both those quantities will usually share a common name.

For example, acceleration is a vector, but 9.8 m/s² is certainly a scalar value, there's no notion of direction in it whatsoever.

Of course, you may call the latter acceleration magnitude if you want to be pedantic, but people who are calling it just acceleration (and asserting it's a scalar) are not outright wrong.

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    $\begingroup$ and it points in different directions all over the Earth. $\endgroup$
    – JEB
    Commented Mar 30, 2018 at 20:36
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Weight is a force, so it is a vector.

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  • $\begingroup$ That's what I said but my professor insists that it isn't. $\endgroup$
    – Ryan
    Commented Mar 2, 2017 at 3:36
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    $\begingroup$ @Ryan nothing more to say really: it's a vector. Weight is often confused with mass, which is a scalar. Even if your professor insists that the Sun rises in the West it still rises in the East, unless he/she misspoke or was momentarily confused or there is a misunderstanding. $\endgroup$ Commented Mar 2, 2017 at 3:39
  • $\begingroup$ Forces in different directions affect the motion of an object in different direction. For example, a force in the downward direction causes a different effect than a horizontal or a vecrtical force. For example. for a box on a surface, a downward force might not do anything, but adding a horizontal force chnages its motino differntly. Weight is definitely a force, as $W=mg$. $\endgroup$
    – Sumant
    Commented Mar 2, 2017 at 3:40
  • $\begingroup$ @ZeroTheHero I would hope my physics professor, out of everyone, wouldn't make such a mistake. My only issue is that when I take the test I'll have to answer wrong if he asks such a question just so I can get the points. He's falsely teaching us physics... $\endgroup$
    – Ryan
    Commented Mar 2, 2017 at 3:44
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    $\begingroup$ @Ryan You say that you have argued with your professor often. What is his answer to "Weight is a force, so it is a vector"? $\endgroup$
    – Steeven
    Commented Mar 2, 2017 at 11:31
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As everyone else has said, this debate is mostly a definitional quibble about exactly how the word "weight" is defined. Two thoughts:

(a) Many people have pointed out that in colloquial speech, the word "weight" usually does not include a direction. But in colloquial speech there almost no words that are commonly used to indicate both a magnitude and a direction. E.g. when people colloquially use the word "velocity," they're almost always only referring to a magnitude, not to a direction. That's why the concept of vectors needs to be taught to beginning students - it's not totally intuive. So if you're going by the colloquial rather than the scientific definition of words, then arguably nothing counts a vector.

(b) In the discussion of aerodynamics in particular, there are approximately ten gajillion diagrams that look like this:

enter image description here

It would be hard to make sense of this diagram if you were considering weight to be a scalar.

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  • $\begingroup$ The confusion in colloquial language comes from setting the weight of an object equal to its mass, which is a scalar quantity. "Do the potatoes you put in the bag really weigh 2 kilogrammes?" "Of course Mr Jones, look ", after which the bag is put on a scale which shows indeed a value of two (more or less) kilogrammes. I think that if people use the word force they do connect it with a direction (though not in the case "May the force be with you"). $\endgroup$ Commented Mar 7, 2017 at 14:12
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I wouldn't get too hung up on this. The gravitational force exerted on an object is most definitely a vector, but on (the surface of) the earth, you don't need to make much distinction because for all intents and purposes, it always points in the same direction -- downward.

I suspect the issue may be one of terminology. Yes, if "weight" refers to the force, it is a vector. But your teacher may very well be referring to the magnitude on the force, i.e. the number that you see if you put it on a spring scale. Again, on the surface of the earth, the magnitude conveys essentially the same information as the vector because the direction is known, and you typically don't use the word "weight" in other circumstances.

EDIT: Also, I'm not sure but you might be confusing something.

I said that weight is a force, with mass as the scalar quantity with a downward direction.

Mass, being a scalar, has no direction... did you mean to say something different?

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  • $\begingroup$ Yes, I meant to say that weight is composed of mass, a scalar, and a downward direction. $\endgroup$
    – Ryan
    Commented Mar 2, 2017 at 14:37
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It seems that your professor has confounded weight with mass.

Mass is a scalar, given in (kilo)gram as SI/CGS unit and never changes its value alone by changing the location. 1 kilogram is 1 kilogram on earth, in 10 000 km over earth, on the Jupiter or Sun or in a space station in weightlessness.

Weight is, as you already mentioned, a vector and given in Newton/dyn as SI/CGS unit. It is the force a given mass is accelerated towards another mass. Therefore it changes by the location. If you have a rope, you give its strength in Newton (most prefer daN, dekanewton because it is almost the same force as 1 kg on normal earth). The same rope holding barely a mass on Earth will snap on Jupiter, but can hold six times the same mass on the moon.

So you are right. Yakk's answer is incorrect because the vector component is not neglible. If you make precise measurements, you will see that mountains or regions with high density changes the direction of the weight, you cannot say anymore it is pointing to the barycenter of the earth. Yes, physicists are sometimes also sloppy in everyday life and use mass/weight interchangably on earth or use magnitude of forces as a shortcut, but weight is a vector.

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    $\begingroup$ Especially in the U.S., non-physicists often confuse weight with mass. When something "weighs 1 pound" it also "weighs 0.454 kilograms" even though technically weight isn't the same as mass, weight is mass times gravity. In the U.S. customary system (the world's only non-metric system) mass is measured in "slugs" but no one in everyday life uses that term. They just say "weight" and many people assume that kg is also a measure of weight. $\endgroup$
    – jkdev
    Commented Mar 3, 2017 at 5:25
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    $\begingroup$ @jkdev +1 I thought first "Why, surely 1 pound is 0.454 kilogram ?" and then I realized: pound is a force ?!. I always thought of it as mass unit and never knew of the existence of slug. $\endgroup$ Commented Mar 3, 2017 at 20:30
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    $\begingroup$ It's related to the word "sluggish." When an object has more mass, making the object move faster is more difficult. $\endgroup$
    – jkdev
    Commented Mar 3, 2017 at 20:47
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If weight is not a vector then why is it that there is a position between the Earth and the Moon where your weight is zero.
At this point the gravitational attraction on you due to the Moon (your weight due to the Moon) is equal in magnitude but opposite in direction to the gravitational attraction on you due to the Earth (your weight due to the Earth).

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    $\begingroup$ But that is the case with two like charges separated by a distance too! and Potential is zero at a point between two. Indeed Potential is a scalar! $\endgroup$
    – jonsno
    Commented Mar 2, 2017 at 8:50
  • $\begingroup$ @samjoe What has potential got to do with this question? Weight is equal to mass times gravitational field strength. Note also that the potential at that point can be any value you choose it to be. $\endgroup$
    – Farcher
    Commented Mar 2, 2017 at 8:54
  • $\begingroup$ @samjoe If you define electrical potential to be zero at $\infty$, then the potential between to like charges can't be zero. The vector electric field might be zero, but not the potential. $\endgroup$
    – Bill N
    Commented Mar 3, 2017 at 1:24
  • $\begingroup$ @BillN I meant unlike charges duh! $\endgroup$
    – jonsno
    Commented Mar 3, 2017 at 8:05
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As noted in many answers, weight is a vector.

The subtext of your question - how to deal with an instructor who makes mistakes - however, is harder to answer. Many people, myself included, don't like admitting their mistakes. I certainly wouldn't recommend, however, being confrontational, questioning their qualifications or work, or making remarks about their age. If possible, just understand the matter to your satisfaction, then let it go.

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Everything that we can resolve with some reference axis is a vector

As acceleration can be resolved in $x$,$y$ and $z$ direction is therefore it's a vector quantity.

Vector's have direction.

Force also have direction. Therefore weight is definitely a vector quantity.

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Mass is a scalar; weight is a vector. Mass does not change regardless of gravitation field, but weight (to be precise) is the sum of vector components from all gravitation fields that attract an object.

For example, even on the surface of the Earth the Moon exerts some minute vector component that sums with the Earth's to give you the exact weight of an object, which will depend on the magnitude and direction of the vector component directed toward the Moon's center of mass, as well as on the component directed toward the Earth's center of mass. This becomes even more significant if an object is in space somewhere between the Earth and the Moon.

To specify a precise weight, one must consider all the components of the weight vector. The magnitude and the direction of a weight vector are dependent on its components. If you want to weigh the oceans precisely, you need to specify their tidal positions.

In short, if a quantity is the sum of vectors, the quantity itself must be a vector.

However, practically speaking, it's impossible to solve a 3-body Kepler problem exactly. So, in the absence of certainty about the position of each planet exerting gravitational pull on an object, it would be futile to attempt a precise sum of all the vector components contributing to an object's weight. That (and the insignificance of the influence of other planets in the Solar System on an object at the Earth's surface) may be one reason that extra-terrestrial vector components of an object's weight usually are ignored.

In common usage "weight" is taken to mean only the vector pointing toward the Earth's center of mass. That may be why some people use the word "weight" as though it were solely a magnitude, like a scalar, because the direction of the weight vector is tacitly assumed, and is left unstated, and insignificant other components of the weight vector are ignored.

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I've always defined weight as the magnitude of the force exerted by gravity—the convention @knzhou and @gogators refer to in the comments. Wikipedia also mentions this convention, citing Halliday, Resnick, and Walker's Fundamentals of Physics (8th ed.) as an example of a textbook that defines weight in this way. (It may not be a coincidence my first few physics courses used this book.) I was surprised to learn that this convention isn't the most common one.

If you keep studying physics or math, you'll often run into situations where several conflicting definitions of a term coexist, even though you may think one of them is obviously better than the others. For example:

  • In special relativity, some people define energy and momentum as the time and space components of the 4-momentum, respectively. Others define energy as the magnitude of the 4-momentum, and momentum as the 4-momentum itself.

  • In differential geometry, some people allow manifolds to have boundaries. Others reserve the term manifold for manifolds without boundaries, using the term manifold with boundary when boundaries are allowed.

  • (This isn't really a terminology conflict, but I can't resist mentioning it. When reading physics or math in French, beware of the false cognate positif. It sounds like it means positive, but it actually means nonnegative! The French term for positive is strictement positif.)

These conflicts of convention don't cause any problems, as long as everyone is aware that multiple conventions exist, and everyone is careful to say which convention they're using. They probably aren't going away anytime soon, so I recommend getting used to them.

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I think we just can follow Wikipedia definition:

In science and engineering, the weight of an object is usually taken to be the force on the object due to gravity. Weight is a vector whose magnitude (a scalar quantity), often denoted by an italic letter W, is the product of the mass m of the object and the magnitude of the local gravitational acceleration g.

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In colloquial language, weight is often set equal to mass. For example: "My weight is 70 kilogrammes." No one says: "I weigh 700 Newton" [assuming $g=10(\frac {m} {sec^2})$]. All weight scales show you a number. Gauged for the earth (and standing on the scale on earth) this is your mass. But on the moon, the weight scale shows you a different number. This means that what the number the scale shows you is not your mass (except on earth) but a measure of the force that's acting on you. So what you actually measure with a scale is the force. So weight is a force, but only on earth, you can see directly your mass while standing on a scale. Gauged for the moon, off course you would see the same mass, but multiplied by the moon's gravitational acceleration (which is a vector, and thus weight is one too) you'll weigh less than on earth. Putting an earth gauged scale on the moon, you'll see a smaller number and people are right by saying that you weigh less on the moon, but because weight is confused with mass, many also think their mass is less on the moon.

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  • $\begingroup$ "No one says..." I actually did say it twice. Then I switched to "My mass is..." and now I don't care anymore. $\endgroup$
    – Burkhard
    Commented Mar 7, 2017 at 12:47
  • $\begingroup$ Okay. But I meant to write," Almost nobody says...". And indeed, what does it matter? Almost all people know what you mean by saying "I weigh 73 kilogrammes" (0r x pounds). $\endgroup$ Commented Mar 7, 2017 at 13:54
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If "weight" is understood as a force of gravity, then it is a vector, because force is a vector. Mass is the scalar value that can be used to compute the gravity force.

When gravity force is computed (F = mg), scalar (m) is multiplied to vector (g), making the result a vector.

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The weight $\small\begin{pmatrix} 0 \\ 0 \\-87\text{ N}\end{pmatrix}$ is a vector in 3D space,

whereas the weight $-87$N is a scalar (and a vector in 1D space).

On the other hand, if weight's direction (towards the centre of Earth) is considered to be a given, then it makes sense to specify weight just as a magnitude (i.e., a nonnegative scalar) like $87$N.

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I think your professor is mixing up terms. Mass is scalar, weight is a vector. But many people get into the habit of using the terms interchangeably. Also, don't always believe everything someone in a "superior position" tells you. Sometimes they are wrong, so question everything.

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According to Wikipedia,

"In science and engineering, the weight of an object is related to the amount of force acting on the object, either due to gravity or to a reaction force that holds it in place.

Some standard textbooks define weight as a vector quantity, the gravitational force acting on the object. Others define weight as a scalar quantity, the magnitude of the gravitational force. Others define it as the magnitude of the reaction force exerted on a body by mechanisms that keep it in place: the weight is the quantity that is measured by, for example, a spring scale. Thus, in a state of free fall, the weight would be zero. In this sense of weight, terrestrial objects can be weightless: ignoring air resistance, the famous apple falling from the tree, on its way to meet the ground near Isaac Newton, would be weightless.

The unit of measurement for weight is that of force, which in the International System of Units (SI) is the newton. For example, an object with a mass of one kilogram has a weight of about 9.8 newtons on the surface of the Earth, and about one-sixth as much on the Moon. Although weight and mass are scientifically distinct quantities, the terms are often confused with each other in everyday use (i.e. comparing and converting force weight in pounds to mass in kilograms and vice versa).

Further complications in elucidating the various concepts of weight have to do with the theory of relativity according to which gravity is modelled as a consequence of the curvature of spacetime. In the teaching community, a considerable debate has existed for over half a century on how to define weight for their students. The current situation is that a multiple set of concepts co-exist and find use in their various contexts."

Therefore, weight is a force.

If your physics professor is still refusing to listen, then **Tell your physics professor to check out these websites:

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Here, I would suggest using NASA as more authoritative than you teacher to eliminate opinion. https://www.grc.nasa.gov/www/k-12/airplane/vectors.html

Directly from their opening summary: Scalars were quantities without direction, including length, speed, volume, area, mass, density, pressure, temperature...

Vectors are quantities with direction: displacement, velocity, acceleration, momentum, force, lift, drag, thrust, weight.

I have never been sure why, but for some reason weight and mass are one that is often flipped with people incorrectly claiming that mass is a vector and weight is not and not understanding that weight does have direction: towards the center of force causing it, in our case gravity so down.

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    $\begingroup$ This just repeats the OP’s reasoning and sources. $\endgroup$
    – JDługosz
    Commented Mar 2, 2017 at 17:40

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