Higher Engineering Mathematics, Sixth Edition

Vectors 257

10N

15N

Figure 24.22

The horizontal component of the 15N force is 15cos0◦
and the horizontal component of the 10N force is
10cos90◦
The total horizontal component of the two velocities,

H=15cos0◦+10cos90◦= 15 + 0 = 15

The vertical component of the 15N force is 15sin0◦
andthevertical component ofthe10Nforceis10sin90◦
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, then the diagram would
be as shown in Fig. 24.23.

15N

R 10N



Figure 24.23

Since a right-angled triangle results then we could use
Pythagoras’s theoremwithout 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. Pythagorascan only be

used when there is an angle of 90◦between vectors.

This is demonstrated in the next worked problem.

Problem 9. Calculate the magnitude and

direction of the resultant of the two acceleration

vectors shown in Fig. 24.24.

15m/s^2

28m/s^2

Figure 24.24

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

as 0 ain Fig. 24.25.

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

acceleration is drawn at an angleof 90◦tothehorizontal,

shown asab.

a 015

28

b

R

 

Figure 24.25

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 ◦