# If a force is a vector quantity, thus having direction and magnitude, why do we express pounds/newtons, a unit of weight, as a scalar?

It is in my understanding that forces are vector quantities, and thus have both magnitude and direction. Since weight is a force of gravity, it also must have magnitude and direction. Why do we define the weight of an object (through SI newtons or customary pounds) as a scalar, such as 5 newtons? Why, despite weight being a force/vector quantity, is the direction not specified?

• Because the direction is downwards, a fundamental part of the experience of all human beings. Perhaps when babies are born in space your question will start to be meaningful. – Suzu Hirose Sep 29 '16 at 4:25

## 2 Answers

It's just lazy language, the direction is implicit. If I say "the weight is 4 Newtons", then it's implied, because we're talking about weight, that the direction is "toward the center of the Earth". Similarly, if we say "the thrust on the airplane from the engines is 11,000 lbs", it's implied that the direction is "in the direction the airplane is going".

• I see, thank you the answer. One side-note question that spins off your answer, what if a case was encountered where we would need to add (or any other operation) forces of different directions. How would the implicit directions of multiple forces work? – Mario Ishac Sep 29 '16 at 4:34
• @MarDev: there is a special procedure for adding vectors called (unsurprisingly) vector addition. This takes into account both the direction and magnitude. – John Rennie Sep 29 '16 at 4:45
• @MarDev Did John's comment answer your question? – DanielSank Sep 29 '16 at 6:36
• @DanielSank in addition to some other research I did, yes. Thank you and John for the answer! – Mario Ishac Sep 29 '16 at 20:37

The units we measure a vector in are the units of its magnitude. For instance, magnetic field is measured in Teslas, which is also the unit for the magnitude of the magnetic field.

If we think of the unit as a multiplicative constant to a vector, this makes sense on a component-by-component basis:

\begin{align} \vec{v} & = \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} \frac{\text{meters}}{\text{seconds}} = \begin{pmatrix} 1 \frac{\text{meter}}{\text{second}} \\ 2 \frac{\text{meters}}{\text{seconds}} \\ 3 \frac{\text{meters}}{\text{seconds}} \end{pmatrix} \end{align}