Higher Engineering Mathematics, Sixth Edition

262 Higher Engineering Mathematics

R

0

2 27.67

2 6.99





Figure 24.38

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 ◦

Usingcomplex numbers,

v 2 −v 2 −v 3 = 40 ∠ 190 ◦− 22 ∠ 140 ◦− 15 ∠ 290 ◦

=(− 39. 392 −j 6. 946 )

−(− 16. 853 +j 14. 141 )

−( 5. 130 −j 14. 095 )

=− 27. 669 −j 6. 992

= 28. 54 ∠− 165. 82 ◦ or

28. 54 ∠ 194. 18 ◦

Now try the following exercise

Exercise 104 Further problems on vector

subtraction

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.
   [(a) 54.0N at 78. 16 ◦(b) 45.64N at 4. 66 ◦]


2. Calculate the resultant of (a) v 1 +v 2 −v 3
   (b)v 3 −v 2 +v 1 whenv 1 =15m/sat 85◦,v 2 =
   25m/s at 175◦andv 3 =12m/s at 235◦.
   [(a)31.71m/sat 121. 81 ◦
   (b) 19.55m/s at 8. 63 ◦]

24.8 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 Fig. 24.39, but in each

of the systems, the resultant vector isad.

ad

b

(a)

a

d

b

c

(b)

Figure 24.39

ThevectorequationofthesystemshowninFig.24.39(a)

is:

ad=ab+bd

and that for the system shown in Fig. 24.39(b) is:

ad=ab+bc+cd

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

last letters, ‘a’and‘d’, 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 (i) the velocity of carP

relative to carQ, and (ii) the velocity of carQ

relative to carP.

(i) The directions of the cars are shown in

Fig. 24.40(a), called aspace diagram.Theveloc-

ity diagram is shown in Fig. 24.40(b), in which

peis taken as the velocity of carPrelative to

pointeon the earth’s surface. The velocity ofP

relative toQis vectorpqand the vector equa-

tion ispq=pe+eq. Hence the vector directions

are as shown,eqbeing in the opposite direction

toqe.