9781118230725.pdf

49

the vector forms the hypotenuse. We find the magnitude and
angle of netwith Eq. 3-6. The magnitude is

dnet (3-18)

13.9 m. (Answer)

To find the angle (measured from the positive direction of x),
we take an inverse tangent:

tan^1 (3-19)

tan^1 12.7°. (Answer)

The angle is negative because it is measured clockwise from
positivex. We must always be alert when we take an inverse



–3.07 m

13.60 m



dnet,y

dnet,x

2 (13.60 m)^2 (3.07 m)^2

2 d^2 net,xd^2 net,y

d

: tangent on a calculator. The answer it displays is mathe-

matically correct but it may not be the correct answer for

the physical situation. In those cases, we have to add 180°

to the displayed answer, to reverse the vector. To check,

we always need to draw the vector and its components as

we did in Fig. 3-16d. In our physical situation, the figure

shows us that 12.7° is a reasonable answer, whereas

12.7°180°167° is clearly not.

We can see all this on the graph of tangent versus angle

in Fig. 3-12c. In our maze problem, the argument of the in-

verse tangent is 3.07/13.60, or 0.226. On the graph draw

a horizontal line through that value on the vertical axis. The

line cuts through the darker plotted branch at 12.7° and

also through the lighter branch at 167°. The first cut is what

a calculator displays.

3-2 UNIT VECTORS, ADDING VECTORS BY COMPONENTS

KEY IDEA

We can add the three vectors by components, axis by axis,

and then combine the components to write the vector

sum.

Calculations:For the xaxis, we add the xcomponents of

and to get the xcomponent of the vector sum :

rxaxbxcx

4.2 m1.6 m 0 2.6 m.

Similarly, for the yaxis,

ryaybycy

1.5 m2.9 m3.7 m2.3 m.

We then combine these components of to write the vector

in unit-vector notation:

(Answer)

where (2.6 m)iˆis the vector component of along the xaxis

and (2.3 m)jˆis that along the yaxis. Figure 3-17bshows

one way to arrange these vector components to form.

(Can you sketch the other way?)

We can also answer the question by giving the magnitude

and an angle for. From Eq. 3-6, the magnitude is

(Answer)

and the angle (measured from the xdirection) is

(Answer)

where the minus sign means clockwise.

tan^1 

2.3 m

2.6 m 

 41 ,

r 2 (2.6 m)^2 (2.3 m)^2 3.5 m

:r

:r



:r

:r(2.6 m)iˆ(2.3 m)jˆ,

:r

b :c, :r

:

,

:a,

:r

Sample Problem 3.04 Adding vectors, unit-vector components

Figure 3-17ashows the following three vectors:

and

What is their vector sum which is also shown?:r

:c(3.7 m)jˆ.

b

:

(1.6 m)iˆ(2.9 m)jˆ,

:a(4.2 m)iˆ(1.5 m)jˆ,

Additional examples, video, and practice available at WileyPLUS

x

y

–3 2 –2 –1 3 4

–3

–2

–1

1

x

y

–3 2 –2 –1 3 4

–3

–2

–1

2

3

1

1

(a)

2.6i

(b)

r

r

a

c

b

ˆ

–2.3jˆ

To add these vectors,

find their net x component

and their net y component.

Then arrange the net

components head to tail.

This is the result of the addition.

Figure 3-17Vector is the vector sum of the other three vectors.:r