2.1 Scalars and vectors  (Page 5/29)

This result means your friend walked $D_{DB}=0.45D_{AB}=0.45(6.0\text{km})=2.7\text{km}$ from the point where he finds his tackle box to the fishing hole.

When vectors $\overrightarrow{A}$ and $\overrightarrow{B}$ lie along a line (that is, in one dimension), such as in the camping example, their resultant $\overrightarrow{R}=\overrightarrow{A}+\overrightarrow{B}$ and their difference $\overrightarrow{D}=\overrightarrow{A}-\overrightarrow{B}$ both lie along the same direction. We can illustrate the addition or subtraction of vectors by drawing the corresponding vectors to scale in one dimension, as shown in [link] .

To illustrate the resultant when $\overrightarrow{A}$ and $\overrightarrow{B}$ are two parallel vectors, we draw them along one line by placing the origin of one vector at the end of the other vector in head-to-tail fashion (see [link] (b)). The magnitude of this resultant is the sum of their magnitudes: R = A + B . The direction of the resultant is parallel to both vectors. When vector $\overrightarrow{A}$ is antiparallel to vector $\overrightarrow{B}$ , we draw them along one line in either head-to-head fashion ( [link] (c)) or tail-to-tail fashion. The magnitude of the vector difference, then, is the absolute value $D=|A-B|$ of the difference of their magnitudes. The direction of the difference vector $\overrightarrow{D}$ is parallel to the direction of the longer vector.

In general, in one dimension—as well as in higher dimensions, such as in a plane or in space—we can add any number of vectors and we can do so in any order because the addition of vectors is commutative    ,

and associative    ,

Moreover, multiplication by a scalar is distributive    :

We used the distributive property in [link] and [link] .

When adding many vectors in one dimension, it is convenient to use the concept of a unit vector    . A unit vector, which is denoted by a letter symbol with a hat, such as $\widehat{u}$ , has a magnitude of one and does not have any physical unit so that $\left|\widehat{u}\right|\equiv u=1$ . The only role of a unit vector is to specify direction. For example, instead of saying vector ${\overrightarrow{D}}_{AB}$ has a magnitude of 6.0 km and a direction of northeast, we can introduce a unit vector $\widehat{u}$ that points to the northeast and say succinctly that ${\overrightarrow{D}}_{AB}=(6.0\text{km})\widehat{u}$ . Then the southwesterly direction is simply given by the unit vector $\text{−}\widehat{u}$ . In this way, the displacement of 6.0 km in the southwesterly direction is expressed by the vector

A ladybug walker

A long measuring stick rests against a wall in a physics laboratory with its 200-cm end at the floor. A ladybug lands on the 100-cm mark and crawls randomly along the stick. It first walks 15 cm toward the floor, then it walks 56 cm toward the wall, then it walks 3 cm toward the floor again. Then, after a brief stop, it continues for 25 cm toward the floor and then, again, it crawls up 19 cm toward the wall before coming to a complete rest ( [link] ). Find the vector of its total displacement and its final resting position on the stick.

Strategy

If we choose the direction along the stick toward the floor as the direction of unit vector $\widehat{u}$ , then the direction toward the floor is $+\widehat{u}$ and the direction toward the wall is $\text{−}\widehat{u}$ . The ladybug makes a total of five displacements:

$\begin{array}{c}{\overrightarrow{D}}_1=(15\text{cm})(+\widehat{u}),\hfill \\ {\overrightarrow{D}}_2=(56\text{cm})(\text{−}\widehat{u}),\hfill \\ {\overrightarrow{D}}_3=(3\text{cm})(+\widehat{u}),\hfill \\ {\overrightarrow{D}}_4=(25\text{cm})(+\widehat{u}),\text{and}\hfill \\ {\overrightarrow{D}}_5=(19\text{cm})(\text{−}\widehat{u}).\hfill \end{array}$

The total displacement $\overrightarrow{D}$ is the resultant of all its displacement vectors.

Solution

The resultant of all the displacement vectors is

$\begin{array}{cc}\hfill \overrightarrow{D}& ={\overrightarrow{D}}_1+{\overrightarrow{D}}_2+{\overrightarrow{D}}_3+{\overrightarrow{D}}_4+{\overrightarrow{D}}_5\hfill \\ & =(15\text{cm})(+\widehat{u})+(56\text{cm})(\text{−}\widehat{u})+(3\text{cm})(+\widehat{u})+(25\text{cm})(+\widehat{u})+(19\text{cm})(\text{−}\widehat{u})\hfill \\ & =(15-56+3+25-19)\text{cm}\widehat{u}\hfill \\ & =\mathrm{-32}\text{cm}\widehat{u}.\hfill \end{array}$

In this calculation, we use the distributive law given by [link] . The result reads that the total displacement vector points away from the 100-cm mark (initial landing site) toward the end of the meter stick that touches the wall. The end that touches the wall is marked 0 cm, so the final position of the ladybug is at the (100 – 32)cm = 68-cm mark.

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