260 Higher Engineering Mathematics
158
458
2 m/s
4 m/s
3.5 m/s
308
Figure 24.32
- An object is acted uponby twoforces of mag-
nitude 10N and 8N at an angle of 60◦to each
other. Determine the resultant force on the
object.
[15.62N at 26. 33 ◦to the 10N force] - A ship heads in a direction of E20◦Sata
speed of 20knots while the current is 4knots
in adirection ofN 30◦E.Determinethespeed
and actual direction of the ship.
[21.07 knots, E 9. 22 ◦S]
24.7 Vector subtraction
In Fig. 24.33, a force vectorFis represented byoa.
The vector(−oa)can be obtained by drawing a vector
fromoin the opposite sense tooabut having the same
magnitude, shown asobin Fig. 24.33, i.e.ob=(−oa)
b
2 F o
F a
Figure 24.33
For two vectors acting at a point, as shown in
Fig. 24.34(a), the resultant of vector addition is:
os=oa+ob.
Figure 24.33(b) shows vectors ob+(−oa),thatis,
ob−oaand the vector equation isob−oa=od.Com-
paringodin Fig. 24.34(b) with the broken lineabin
Fig. 24.34(a) shows that the second diagonal of the
‘parallelogram’ method of vector addition gives the
magnitude and direction of vector subtraction ofoa
fromob.
(a) (b)
a a
bsd b
o a o
Figure 24.34
Problem 11. Accelerations ofa 1 = 1 .5m/s^2 at
90 ◦anda 2 = 2 .6m/s^2 at 145◦act at a point. Find
a 1 +a 2 anda 1 −a 2 (i) by drawing a scale vector
diagram, and (ii) by calculation.
(i) The scale vector diagram is shown in Fig. 24.35.
By measurement,
a 1 +a 2 = 3 .7m/s^2 at 126◦
a 1 −a 2 = 2 .1m/s^2 at 0◦
a 1 a 2
a 1 a 2
2.6 m/s^2
Scale in m/s^2
1.5 m/s^2
145
0 1 23
126
a 1
a 2
a 2
a 1
Figure 24.35
(ii) Resolving horizontally and vertically gives:
Horizontal component ofa 1 +a 2 ,
H= 1 .5cos90◦+ 2 .6cos145◦=− 2. 13
Vertical component ofa 1 +a 2 ,
V= 1 .5sin90◦+ 2 .6sin145◦= 2. 99
From Fig. 24.36, magnitude ofa 1 +a 2 ,
R=
√
(− 2. 13 )^2 + 2. 992 =3.67m/s^2
In Fig. 24.36,α=tan−^1
(
2. 99
2. 13
)
= 54. 53 ◦and
θ= 180 ◦− 54. 53 ◦= 125. 47 ◦
Thus, a 1 +a 2 =3.67m/s^2 at 125. 47 ◦