Basic Engineering Mathematics, Fifth Edition

276 Basic Engineering Mathematics

R

0

27.67

6.99





Figure 29.38

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

This result is as expected, sincev 2 −v 1 −v 3

=−(v 1 −v 2 +v 3 )and the vector 28.54 units at

194. 18 ◦is minus times (i.e. is 180◦out of phase

with) the vector 28.54 units at 14. 18 ◦

Now try the following Practice Exercise

PracticeExercise 115 Vector subtraction

(answers on page 352)

1. Forces ofF 1 =40N at 45◦andF 2 =30N at
   125 ◦act at a point. Determine by drawing and
   by calculation (a)F 1 +F 2 (b)F 1 −F 2


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

29.8 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 vector of a system.

Two different systems are shown in Figure 29.39, but,

in each of the systems, the resultant vector isad.

ad

b

(a)

a

d

b

c

(b)

Figure 29.39

The vector equation of the system shown in

Figure 29.39(a) is

ad=ab+bd

and that for the system shown in Figure 29.39(b) is

ad=ab+bc+cd

Thus, in vector equations of this form, only the first and

last letters,aandd, respectively, fix the magnitude and

direction of the resultant vector. This principle is used

in relative velocity problems.

Problem 13. Two cars,PandQ, are travelling

towards the junction of two roads which are at right

angles to one another. CarPhas a velocity of

45km/h due east and carQa velocity of 55km/h

due south. Calculate (a) the velocity of carP

relative to carQand (b) the velocity of carQ

relative to carP

(a) The directions of the cars are shown in

Figure 29.40(a), which is called aspace diagram.

The velocity diagram is shown in Figure 29.40(b),

in whichpeis taken as the velocity of carPrela-

tive to pointeon the earth’s surface. The velocity

ofPrelativetoQis vectorpqand the vector equa-

tionispq=pe+eq. Hence, the vector directions

are as shown,eqbeing in the opposite direction

toqe.

(a) (b) (c)

P Q

E

N

W

S

55 km/h

45 km/h

p

e

q

p e

q

Figure 29.40

From the geometry of the vector triangle, the mag-

nitude ofpq=

√

452 + 552 = 71 .06km/h and the

direction ofpq=tan−^1

(

55

45

)

= 50. 71 ◦

That is,the velocity of carPrelative to carQis

71.06km/h at 50. 71 ◦

(b) The velocity of carQrelative to carPis given by

the vector equationqp=qe+epand the vector

diagram is as shown in Figure 29.40(c), havingep

opposite in direction tope.