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 horizontal15
3 2 130 10m/s20m/s15m/s
Figure 29.11When 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.b195 105 030 1 3 2
raFigure 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.FFigure 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