Basic Engineering Mathematics, Fifth Edition

272 Basic Engineering Mathematics

The vertical component of the 15N force is 15sin0◦and

the vertical component of the 10N force is 10sin90◦.

The total vertical component of the two velocities,

V=15sin0◦+10sin90◦= 0 + 10 = 10

Magnitude of resultant vector

=

√

H^2 +V^2 =

√

152 + 102 = 18 .03N

The direction of the resultant vector,

θ=tan−^1

(

V

H

)

=tan−^1

(

10

15

)

= 33. 69 ◦

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

of 18.03N at 33. 69 ◦to the 15N vector.

There is an alternative method of calculating the resul-

tant vector in this case. If we used the triangle method,

the diagram would be as shown in Figure 29.23.

15N

R 10N



Figure 29.23

Since a right-angled triangle results, we could use

Pythagoras’ theorem without needing to go through the

procedure for horizontal and vertical components. In

fact, the horizontal and vertical components are 15N

and 10N respectively.

This is, of course, a special case.Pythagoras can only

beused when there is an angleof 90◦between vectors.

This is demonstrated in worked Problem 9.

Problem 9. Calculate the magnitude and

direction of the resultant of the two acceleration

vectors shown in Figure 29.24.

15m/s^2

28m/s^2

Figure 29.24

The 15m/s^2 acceleration is drawn horizontally, shown

as 0 ain Figure 29.25.

a 15 0

28

b

R

 

Figure 29.25

From the nose of the 15m/s^2 acceleration, the 28m/s^2

acceleration is drawn at an angleof 90◦tothehorizontal,

shown asab.

The resultant acceleration,R, is given by length 0 b.

Since a right-angled triangle results, the theorem of

Pythagoras may be used.

0 b=

√

152 + 282 = 31 .76m/s^2

and α=tan−^1

(

28

15

)

= 61. 82 ◦

Measuring from the horizontal,

θ= 180 ◦− 61. 82 ◦= 118. 18 ◦

Thus,the resultant of the two accelerations is a single

vector of 31.76m/s^2 at 118. 18 ◦to the horizontal.

Problem 10. Velocities of 10m/s, 20m/s and

15m/s act as shown in Figure 29.26. Calculate the

magnitude of the resultant velocity and its direction

relative to the horizontal

20m/s

10m/s

15m/s

158

308

(^1)
(^2)
(^3)
Figure 29.26
The horizontal component of the 10m/s velocity
=10cos30◦= 8 .660m/s,