Higher Engineering Mathematics, Sixth Edition

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