Basic Engineering Mathematics, Fifth Edition

270 Basic Engineering Mathematics

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

Figure 29.15.

358

0

40.96 N

28.68 N

50 N

a

b

Figure 29.15

(To check: by Pythagoras,

0 b=

√

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

Figure 29.16.

Problem 6. Resolve the displacement vector of

40m at an angle of 120◦into horizontal and vertical

components

308

20m/s^2 10m/s

17.32m/s

b

a

0

Figure 29.16

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

Figure 29.17.

2 20.0N

40N

1208

0

34.64N

a

b

Figure 29.17

29.6 Addition of vectors by calculation

Two forcevectors,F 1 andF 2 ,areshowninFigure29.18,

F 1 being at an angle ofθ 1 andF 2 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 29.18

A method of adding two vectors together is to use

horizontal and vertical components.

The horizontal component of forceF 1 isF 1 cosθ 1 and

the horizontal component of forceF 2 isF 2 cosθ 2 .The