272 Basic Engineering Mathematics
The vertical component of the 15 N force is 15 sin 0
◦ and
the vertical component of the 10 N force is 10 sin 90 ◦ .
The total vertical component of the two velocities,
V = 15 sin 0
◦
+ 10 sin 90
◦
= 0 + 10 = 10
Magnitude of resultant vector
=
√
H 2 + V 2 =
√
15 2 + 10 2 = 18.03 N
The direction of the resultant vector,
θ = tan −1
V
H
= tan −1
10
15
= 33.69 ◦
Thus, the resultant of the two forces is a single vector
of 18.03 N at 33.69 ◦ to the 15 N vector.
There is an alternative method of calculating the resultant vector in this case. If we used the triangle method,
the diagram would be as shown in Figure 29.23.
15 N
10 N
R
Figure 29.23
Since a right-angled triangle results, we could use
Pythagoras’ theorem without needing to go through the
procedure for horizontal and vertical components. In
fact, the horizontal and vertical components are 15 N
and 10 N respectively.
This is, of course, a special case. Pythagoras can only
be used when there is an angle of 90 ◦ between vectors.
This is demonstrated in worked Problem 9.
Problem 9. Calculate the magnitude and
direction of the resultant of the two acceleration
vectors shown in Figure 29.24.
15 m/s 2
28 m/s 2
Figure 29.24
The 15 m/s
2 acceleration is drawn horizontally, shown
as 0a in Figure 29.25.
0
15
a
28
b
R
␣
Figure 29.25
From the nose of the 15 m/s 2 acceleration, the 28 m/s 2
acceleration is drawn at an angle of 90 ◦ to the horizontal,
shown as ab.
The resultant acceleration, R, is given by length 0b.
Since a right-angled triangle results, the theorem of
Pythagoras may be used.
0b =
15 2 + 28 2 = 31.76 m/s
2
and
α = tan
−1
28
15
= 61.82
◦
Measuring from the horizontal,
θ = 180 ◦ − 61.82 ◦ = 118.18 ◦
Thus, the resultant of the two accelerations is a single
vector of 31.76 m/s 2 at 118.18 ◦ to the horizontal.
Problem 10. Velocities of 10 m/s, 20 m/s and
15 m/s act as shown in Figure 29.26. Calculate the
magnitude of the resultant velocity and its direction
relative to the horizontal
20 m/s
10 m/s
15 m/s
158
308
1
2
3
Figure 29.26
The horizontal component of the 10 m/s velocity
= 10 cos30 ◦ = 8.660 m/s,
The vertical component of the 15 N force is 15 sin 0
◦ and
the vertical component of the 10 N force is 10 sin 90 ◦ .
The total vertical component of the two velocities,
V = 15 sin 0
◦
+ 10 sin 90
◦
= 0 + 10 = 10
Magnitude of resultant vector
=
√
H 2 + V 2 =
√
15 2 + 10 2 = 18.03 N
The direction of the resultant vector,
θ = tan −1
V
H
= tan −1
10
15
= 33.69 ◦
Thus, the resultant of the two forces is a single vector
of 18.03 N at 33.69 ◦ to the 15 N vector.
There is an alternative method of calculating the resultant vector in this case. If we used the triangle method,
the diagram would be as shown in Figure 29.23.
15 N
10 N
R
Figure 29.23
Since a right-angled triangle results, we could use
Pythagoras’ theorem without needing to go through the
procedure for horizontal and vertical components. In
fact, the horizontal and vertical components are 15 N
and 10 N respectively.
This is, of course, a special case. Pythagoras can only
be used when there is an angle of 90 ◦ between vectors.
This is demonstrated in worked Problem 9.
Problem 9. Calculate the magnitude and
direction of the resultant of the two acceleration
vectors shown in Figure 29.24.
15 m/s 2
28 m/s 2
Figure 29.24
The 15 m/s
2 acceleration is drawn horizontally, shown
as 0a in Figure 29.25.
0
15
a
28
b
R
␣
Figure 29.25
From the nose of the 15 m/s 2 acceleration, the 28 m/s 2
acceleration is drawn at an angle of 90 ◦ to the horizontal,
shown as ab.
The resultant acceleration, R, is given by length 0b.
Since a right-angled triangle results, the theorem of
Pythagoras may be used.
0b =
15 2 + 28 2 = 31.76 m/s
2
and
α = tan
−1
28
15
= 61.82
◦
Measuring from the horizontal,
θ = 180 ◦ − 61.82 ◦ = 118.18 ◦
Thus, the resultant of the two accelerations is a single
vector of 31.76 m/s 2 at 118.18 ◦ to the horizontal.
Problem 10. Velocities of 10 m/s, 20 m/s and
15 m/s act as shown in Figure 29.26. Calculate the
magnitude of the resultant velocity and its direction
relative to the horizontal
20 m/s
10 m/s
15 m/s
158
308
1
2
3
Figure 29.26
The horizontal component of the 10 m/s velocity
= 10 cos30 ◦ = 8.660 m/s,