260 Higher Engineering Mathematics
158
458
2 m/s
4 m/s
3.5 m/s
308
Figure 24.32
10. An object is acted upon by two forces of magnitude 10 N and 8 N at an angle of 60 ◦ to each
other. Determine the resultant force on the
object.
[15.62 N at 26.33 ◦ to the 10 N force]
11. A ship heads in a direction of E 20 ◦ S at a
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]
24.7 Vector subtraction
In Fig. 24.33, a force vector F is represented by oa.
The vector (−oa) can be obtained by drawing a vector
from o in the opposite sense to oa but having the same
magnitude, shown as ob in Fig. 24.33, i.e. ob = (−oa)
b
o
2F
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), that is,
ob − oa and the vector equation is ob − oa = od. Comparing od in Fig. 24.34(b) with the broken line ab in
Fig. 24.34(a) shows that the second diagonal of the
‘parallelogram’ method of vector addition gives the
magnitude and direction of vector subtraction of oa
from ob.
(b)
(a)
a
Ϫa
d
b
b
s
a
o
o
Figure 24.34
Problem 11. Accelerations of a 1 = 1.5 m/s
2 at
90 ◦ and a 2 = 2.6 m/s 2 at 145 ◦ act at a point. Find
a 1 + a 2 and a 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 Ϫ a 2
a 1 ϩ a 2
2.6 m/s 2
Scale in m/s
2
1.5 m/s 2
145Њ
1
2
3
0
126Њ
a 1
Ϫa 2
a 2
a 1
Figure 24.35
(ii) Resolving horizontally and vertically gives:
Horizontal component of a 1 + a 2 ,
H = 1.5 cos90 ◦ +2.6 cos 145 ◦ = −2.13
Vertical component of a 1 + a 2 ,
V = 1.5 sin90 ◦ + 2.6 sin145 ◦ = 2.99
From Fig. 24.36, magnitude of a 1 + a 2 ,
R =
(−2.13) 2 + 2.99 2 = 3.67 m/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 = 3.67 m/s
2 at 125.47 ◦
158
458
2 m/s
4 m/s
3.5 m/s
308
Figure 24.32
10. An object is acted upon by two forces of magnitude 10 N and 8 N at an angle of 60 ◦ to each
other. Determine the resultant force on the
object.
[15.62 N at 26.33 ◦ to the 10 N force]
11. A ship heads in a direction of E 20 ◦ S at a
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]
24.7 Vector subtraction
In Fig. 24.33, a force vector F is represented by oa.
The vector (−oa) can be obtained by drawing a vector
from o in the opposite sense to oa but having the same
magnitude, shown as ob in Fig. 24.33, i.e. ob = (−oa)
b
o
2F
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), that is,
ob − oa and the vector equation is ob − oa = od. Comparing od in Fig. 24.34(b) with the broken line ab in
Fig. 24.34(a) shows that the second diagonal of the
‘parallelogram’ method of vector addition gives the
magnitude and direction of vector subtraction of oa
from ob.
(b)
(a)
a
Ϫa
d
b
b
s
a
o
o
Figure 24.34
Problem 11. Accelerations of a 1 = 1.5 m/s
2 at
90 ◦ and a 2 = 2.6 m/s 2 at 145 ◦ act at a point. Find
a 1 + a 2 and a 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 Ϫ a 2
a 1 ϩ a 2
2.6 m/s 2
Scale in m/s
2
1.5 m/s 2
145Њ
1
2
3
0
126Њ
a 1
Ϫa 2
a 2
a 1
Figure 24.35
(ii) Resolving horizontally and vertically gives:
Horizontal component of a 1 + a 2 ,
H = 1.5 cos90 ◦ +2.6 cos 145 ◦ = −2.13
Vertical component of a 1 + a 2 ,
V = 1.5 sin90 ◦ + 2.6 sin145 ◦ = 2.99
From Fig. 24.36, magnitude of a 1 + a 2 ,
R =
(−2.13) 2 + 2.99 2 = 3.67 m/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 = 3.67 m/s
2 at 125.47 ◦