Vectors 269
(iii) The resultant force is shown asRand is measured
as18Nand angleθis measured as 34 ◦.
Thus,the resultant of the two force vectors is 18N at
34 ◦to the 15N force.
Problem 3. Velocities of 10m/s, 20m/s and
15m/s act as shown in Figure 29.11. Determine, by
drawing, the magnitude of the resultant velocity and
its direction relative to the horizontal
15
3
2
1
30
10m/s
20m/s
15m/s
Figure 29.11
When more than 2 vectors are being added the nose-to-
tail method is used. The order in which the vectors are
added does not matter. In this case the order taken isν 1 ,
thenν 2 ,thenν 3. However, if a different order is taken
the same result will occur.
(i) ν 1 is drawn 10 units long at an angle of 30◦to the
horizontal, shown as 0 ain Figure 29.12.
b
195
105
0
30
1
3 2
r
a
Figure 29.12
(ii) From the nose ofν 1 ,ν 2 is drawn 20 units long at
an angle of 90◦to the horizontal, shown asab.
(iii) From the nose ofν 2 ,ν 3 is drawn 15 units long at
an angle of 195◦to the horizontal, shown asbr.
(iv) The resultant velocity is given by length 0 rand
is measured as22m/sand the angle measured to
the horizontal is 105 ◦.
Thus,the resultant of the three velocities is 22m/s at
105 ◦to the horizontal.
Worked Examples 1 to 3 have demonstrated how vec-
tors are added to determine their resultant and their
direction. However, drawing to scale is time-consuming
and not highly accurate. The following sections demon-
strate how to determine resultant vectors by calculation
using horizontal and vertical components and, where
possible, by Pythagoras’ theorem.
29.5 Resolving vectors into horizontal
and vertical components
A force vectorFis shown in Figure 29.13 at angleθ
to the horizontal. Such a vector can be resolved into
two components such that the vector addition of the
components is equal to the original vector.
F
Figure 29.13
The two components usually taken are ahorizontal
componentand avertical component.Ifaright-angled
triangle is constructed as shown in Figure 29.14, 0ais
called the horizontal component ofFandabis called
the vertical component ofF.
(^0) a
F
b
Figure 29.14
From trigonometry(see Chapter 21 and remember SOH
CAH TOA),
cosθ=
0 a
0 b
,from which 0a= 0 bcosθ=Fcosθ