Higher Engineering Mathematics

VECTORS, PHASORS AND THE COMBINATION OF WAVEFORMS 231

D

Ta n−^1

(

− 6. 99

− 27. 67

)

= 14. 18 ◦, hence the required

angle is 180◦+ 14. 18 ◦= 194. 18 ◦.

Thusv 2 −v 1 −v 3 = 28 .54 units at 194. 18 ◦.

This result is as expected, since

v 2 −v 1 −v 3 =−(v 1 −v 2 +v 3 )

and the vector 28.54 units at 194.18◦is minus

times the vector 28.54 units at 14.18◦.

Now try the following exercise.

Exercise 94 Further problems on vector

subtraction

1. Forces ofF 1 =40Nat45◦andF 2 =30 N at
   125 ◦act at a point. Determine by drawing and
   by calculation (a)F 1 +F 2 (b)F 1 −F 2
   [
   (a) 54.0 N at 78.16◦
   (b) 45.64 N at 4.66◦

]

2. Calculate the resultant of (a)v 1 +v 2 −v 3
   (b) v 3 −v 2 +v 1 whenv 1 =15 m/s at 85◦,
   v 2 =25 m/s at 175◦andv 3 =12 m/s at 235◦.
   [
   (a) 31.71 m/s at 121.81◦
   (b) 19.55 m/s at 8.63◦

]

21.5 Relative velocity

For relative velocity problems, some fixed datum
point needs to be selected. This is often a fixed point
on the earth’s surface. In any vector equation, only
the start and finish points affect the resultant vec-
tor of a system. Two different systems are shown in
Fig. 21.15, but in each of the systems, the resultant
vector isad.

Figure 21.15

The vector equation of the system shown in
Fig. 21.15(a) is:

ad=ab+bd

and that for the system shown in Fig. 21.15(b) is:

ad=ab+bc+cd

Thus in vector equations of this form, only the first

and last letters,aandd, respectively, fix the mag-

nitude and direction of the resultant vector. This

principle is used in relative velocity problems.

Problem 8. Two cars,PandQ, are travelling

towards the junction of two roads which are at

right angles to one another. CarPhas a veloc-

ity of 45 km/h due east and carQa velocity of

55 km/h due south.

Calculate (i) the velocity of carPrelative to car

Q, and (ii) the velocity of carQrelative to carP.

(i) The directions of the cars are shown in

Fig. 21.16(a), called aspace diagram. The

velocity diagram is shown in Fig. 21.16(b), in

whichpeis taken as the velocity of carPrelative

to point e on the earth’s surface. The velocity

ofPrelative toQis vectorpqand the vec-

tor equation ispq=pe+eq. Hence the vector

directions are as shown,eqbeing in the oppo-

site direction toqe. From the geometry of the

vector triangle,

|pq|=

√

(45^2 + 552 )= 71 .06 km/h

and argpq=tan−^1

(

55

45

)

= 50. 71 ◦

Figure 21.16

i.e., the velocity of carPrelative to carQis

71.06 km/h at 50.71◦.

(ii) The velocity of carQrelative to carPis given by

the vector equationqp=qe+epand the vector

diagram is as shown in Fig. 21.16(c), havingep