Higher Engineering Mathematics

VECTORS, PHASORS AND THE COMBINATION OF WAVEFORMS 229

D

5. A load of 5.89 N is lifted by two strings, mak-
   ing angles of 20◦and 35◦with the vertical.
   Calculate the tensions in the strings. [For a
   system such as this, the vectors representing
   the forces form a closed triangle when the
   system is in equilibrium]. [2.46 N, 4.12 N]


6. The acceleration of a body is due to four
   component, coplanar accelerations. These are
   2 m/s^2 due north, 3 m/s^2 due east, 4 m/s^2 to
   the south-west and 5 m/s^2 to the south-east.
   Calculate the resultant acceleration and its
   direction. [5.7 m/s^2 at 310◦]


7. A current phasori 1 is 5 A and horizontal. A
   second phasori 2 is 8 A and is at 50◦to the
   horizontal. Determine the resultant of the two
   phasors,i 1 +i 2 , and the angle the resultant
   makes with currenti 1. [11.85 A at 31.14◦]


8. A ship heads in a direction of E 20◦Sata
   speed of 20 knots while the current is 4 knots
   in a direction of N 30◦E. Determine the speed
   and actual direction of the ship.
   [21.07 knots, E 9.22◦S]

21.4 Vector subtraction

In Fig. 21.11, a force vectorFis represented byoa.
The vector (−oa) can be obtained by drawing a vec-
tor fromoin the opposite sense tooabut having
the same magnitude, shown asobin Fig. 21.11, i.e.
ob=(−oa).

Figure 21.11

For two vectors acting at a point, as shown in
Fig. 21.12(a), the resultant of vector addition
is os=oa+ob. Figure 21.12(b) shows vectors
ob+(−oa), that is,ob−oaand the vector equation
isob−oa=od. Comparingodin Fig. 21.12(b) with
the broken line ab in Fig. 21.12(a) shows that the sec-
ond diagonal of the ‘parallelogram’ method of vector
addition gives the magnitude and direction of vector
subtraction ofoafromob.

Figure 21.12

Problem 6. Accelerations ofa 1 = 1 .5 m/s^2 at

90 ◦anda 2 = 2 .6 m/s^2 at 145◦act at a point. Find

a 1 +a 2 anda 1 −a 2 by (i) drawing a scale vector

diagram and (ii) by calculation.

(i) The scale vector diagram is shown in Fig. 21.13.

By measurement,

a 1 +a 2 = 3 .7m/s^2 at 126◦

a 1 −a 2 = 2 .1m/s^2 at 0◦

Figure 21.13

(ii) Resolving horizontally and vertically gives:

Horizontal component ofa 1 +a 2 ,

H= 1 .5 cos 90◦+ 2 .6 cos 145◦=− 2. 13

Vertical component ofa 1 +a 2 ,

V= 1 .5 sin 90◦+ 2 .6 sin 145◦= 2. 99

Magnitude ofa 1 +a 2 =

√

(− 2. 132 + 2. 992 )

= 3 .67 m/s^2

Direction ofa 1 +a 2 =tan−^1

(

2. 99

− 2. 13

)

and must lie in the second quadrant sinceHis

negative andVis positive.