Higher Engineering Mathematics, Sixth Edition

256 Higher Engineering Mathematics

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 total horizontal component of the two forces,

H=F 1 cosθ 1 +F 2 cosθ 2

The vertical component of forceF 1 isF 1 sinθ 1 and the

vertical component of forceF 2 isF 2 sinθ 2

The total vertical component of the two forces,

V=F 1 sinθ 1 +F 2 sinθ 2

Since we haveHandV, the resultant ofF 1 andF 2

is obtained by using the theorem of Pythagoras. From

Fig. 24.19, 0 b^2 =H^2 +V^2

i.e. resultant=

√

H^2 +V^2 at an angle

given byθ=tan−^1

(

V

H

)

V

b

Resultant

a

H

0 

Figure 24.19

Problem 7. A force of 5N is inclined at an angle

of 45◦to a second force of 8N, both forces acting at

a point. Calculate the magnitude of the resultant of

these two forces and the direction of the resultant

with respect to the 8N force.

The two forces are shown in Fig. 24.20.

458

8N

5N

Figure 24.20

The horizontal component of the 8N force is 8cos0◦

and the horizontal component of the 5N force is

5cos45◦

The total horizontal component of the two forces,

H=8cos0◦+5cos45◦= 8 + 3. 5355

=11.5355

The vertical component of the 8N force is 8sin0◦

and the vertical component of the 5N force is 5sin45◦

The total vertical component of the two forces,

V=8sin0◦+5sin45◦= 0 + 3. 5355

=3.5355

From Fig. 24.21, magnitude of resultant vector

=

√

H^2 +V^2

=

√

11. 53552 + 3. 53552 =12.07N



Resultant

H11.5355N

V3.5355N

Figure 24.21

The direction of the resultant vector,

θ=tan−^1

(

V

H

)

=tan−^1

(

3. 5355

11. 5355

)

=tan−^10. 30648866 ...= 17. 04 ◦

Thus,the resultant of the two forces is a single vector

of 12.07N at 17. 04 ◦to the 8N vector.

Perhaps an easier and quicker method of calculating

the magnitude and direction of the resultant is to use

complex numbers(see Chapter 20).

In this example, theresultant

= 8 ∠ 0 ◦+ 5 ∠ 45 ◦

=(8cos0◦+j8sin0◦)+(5cos45◦+j5sin45◦)

=( 8 +j 0 )+( 3. 536 +j 3. 536 )

=( 11. 536 +j 3. 536 )Nor12.07∠ 17. 04 ◦N

as obtained above using horizontal and vertical

components.

Problem 8. Forces of 15N and 10N are at an

angle of 90◦to each other as shown in Fig. 24.22.

Calculate the magnitude of the resultant of these

two forces and its direction with respect to the

15N force.