Vectors 255
From trigonometry (see Chapter 11),
cosθ=
0 a
0 b
from which, 0a= 0 bcosθ
=Fcosθ
i.e. the horizontal component ofF=Fcosθ
and sinθ=
ab
0 b
from which, ab= 0 bsinθ
=Fsinθ
i.e. the vertical component ofF=Fsinθ
Problem 4. Resolve the force vector of 50N at an
angle of 35◦to the horizontal into its horizontal and
vertical components.
The horizontal component of the 50N force,
0 a=50cos35◦=40.96N
The vertical component of the 50N force,
ab=50sin35◦=28.68N
The horizontal and vertical components are shown in
Fig. 24.15.
358
0
40.96N
50N 28.68N
a
b
Figure 24.15
(Checking: by Pythagoras, 0b=
√
40. 962 + 28. 682
=50N
and θ=tan−^1
(
28. 68
40. 96
)
= 35 ◦
Thus, the vector addition of components 40.96N and
28.68N is 50N at 35◦)
Problem 5. Resolve the velocity vector of 20m/s
at an angle of− 30 ◦to the horizontal into horizontal
and vertical components.
The horizontal component of the 20m/s velocity,
0 a=20cos(− 30 ◦)=17.32m/s
The vertical component of the 20m/s velocity,
ab=20sin(− 30 ◦)=−10m/s
The horizontal and vertical components are shown in
Fig. 24.16.
308
20m/s^2 10m/s
17.32m/s
b
a
0
Figure 24.16
Problem 6. Resolve the displacement vector of
40m at an angle of 120◦into horizontal and vertical
components.
Thehorizontal componentof the 40m displacement,
0 a=40cos120◦=−20.0m
The vertical componentof the 40m displacement,
ab=40sin120◦=34.64m
The horizontal and vertical components are shown in
Fig. 24.17.
2 20.0N
40N
1208
0
34.64N
a
b
Figure 24.17
24.6 Addition of vectors by calculation
Two force vectors,F 1 andF 2 , are shown in Fig. 24.18,
F 1 being at an angle ofθ 1 andF 2 being at an angle
ofθ 2.
F 1 F 2
F^1
sin
1
F^2
sin
2
H
V
1
2
F 2 cos 2
F 1 cos 1
Figure 24.18