270 Basic Engineering Mathematics
i.e. the horizontal component of F = F cos θ, and
sin θ =
ab
0b
from which, ab = 0b sin θ = F sin θ
i.e. the vertical component of F = F sinθ .
Problem 4. Resolve the force vector of 50 N at an
angle of 35 ◦ to the horizontal into its horizontal and
vertical components
The horizontal component of the 50 N force,
0a = 50 cos35 ◦ = 40.96 N.
The vertical component of the 50 N force,
ab = 50 sin 35 ◦ = 28.68 N.
The horizontal and vertical components are shown in
Figure 29.15.
358
0
40.96 N
28.68 N
50 N
a
b
Figure 29.15
(To check: by Pythagoras,
0b =
40.96 2 + 28.68 2 = 50 N
and
θ = tan
−1
28.68
40.96
= 35
◦
Thus, the vector addition of components 40.96 N and
28.68 N is 50 N at 35 ◦ .)
Problem 5. Resolve the velocity vector of 20 m/s
at an angle of −30 ◦ to the horizontal into horizontal
and vertical components
The horizontal component of the 20 m/s velocity,
0a = 20 cos(−30
◦
) = 17.32 m/s.
The vertical component of the 20 m/s velocity,
ab = 20 sin(−30 ◦ ) = −10 m/s.
The horizontal and vertical components are shown in
Figure 29.16.
Problem 6. Resolve the displacement vector of
40 m at an angle of 120
◦ into horizontal and vertical
components
308
2 0 m / s
210 m/s
17.32 m/s
b
a
0
Figure 29.16
The horizontal component of the 40 m displacement,
0a = 40 cos120 ◦ = −20.0 m.
The vertical component of the 40 m displacement,
ab = 40 sin 120 ◦ = 34.64 m.
The horizontal and vertical components are shown in
Figure 29.17.
220.0 N
40 N
1208
0
34.64 N
a
b
Figure 29.17
29.6 Addition of vectors by
calculation
Two force vectors, F 1 and F 2 , are shown in Figure 29.18,
F 1 being at an angle of θ 1 and F 2 at an angle of θ 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 29.18
A method of adding two vectors together is to use
horizontal and vertical components.
The horizontal component of force F 1 is F 1 cos θ 1 and
the horizontal component of force F 2 is F 2 cos θ 2 . The