Further algebra
12.4 Resolution of vectors
Two vectors can be added together to give a single vector called the result
ant. There is a reverse process of taking a single vector a and expressing it in terms of two component vectors v and h at right angles to each other.
The sum of the components is the original vector, i.e. a = v + h. These components are referred to as the resolved parts of the vector.
Consider the vector a shown in Figure 12.17. This vector acts along AC, with triangle ABC being a right-angled triangle. If we take the v vector as being along BC and the h vector along AB, then for the magnitudes of the vectors we have, for the v vector
|v|
Vertical
In Figure 12.17 the triangle rule was used to resolve a vector into its two components. An alternative, but absolutely equivalent method, would have been to use the parallelogram rule. Thus, suppose we have a force of 5 N at an angle of 40° to the horizontal. Then using the parallelogram rule we can draw the horizontal and vertical components as the sides of the parallel
ogram, as shown in Figure 12.18. Then we have, using the equation derived above,
horizontal component = 5 cos 40° = 3.8 N vertical component = 5 sin 40° = 3.2 N
We could therefore replace the 5 N force by the two components of 3.8 N and 3.2 N and the effect would be precisely the same.
Example
Determine the horizontal and vertical components of the 120 kN force acting on the beam shown in Figure 12.19(a).
Figure 12.19(b) shows the parallelogram of vectors which can be used to determine the components:
horizontal component = |F| cos θ = 120 cos 45° = 84.9 kN vertical component = |F| sin θ = 120 sin 45° = 84.9 kN Example
Determine the resultant force acting on the bracket shown in Figure 12.20 due to the three forces indicated.
This problem could be solved by drawing the polygon of vectors.
However, an alternative method of determining the resultant is to resolve all the forces into their vertical and horizontal components. We can then easily determine the sum of the vertical components and the sum of the vertical components. We then have replaced all the forces by just two components and from these can determine the resultant.
For the 3 kN force we have:
horizontal component = 3.0 cos 60° = 1.5 kN vertical component = 3.0 sin 60° = 2.6 kN For the 2.0 kN force we have
horizontal component = 2.0 cos 30° = 1.7 kN
vertical component = 2.0 sin 30° = 1.0 kN For the 5.0 kN force we have
horizontal component = 5.0 cos 70° = 1.7 kN vertical component = -5.0 sin 70° = -4.7 kN
The minus sign is because this force is acting downwards and in the opposite direction to the other vertical components which we have taken as being positive. All the horizontal components are in the same direction. Thus
sum of horizontal components = 1.5 + 1.7 + 1.7 = 4.9 kN sum of vertical components = 2.6 + 1.0 - 4.7 = -1.1 kN
Figure 12.21 shows how we can use the parallelogram rule to find the resultant with these two components. Since the two components are at right angles to each other, the resultant can be calculated using the Pythagoras theorem. Thus:
(resultant)2 = 4.92 + l.l2
Hence the resultant has a magnitude of 5.0 kN. The resultant is at an angle 9 downwards from the horizontal given by:
tan θ = 1.1 4.9 Thus θ = 12.7°.
Example
An object of weight 30 N rests on an incline which is at 35° to the horizontal. What are the components of the weight acting at right angles to the incline and along the incline?
Figure 12.22(a) describes the situation and Figure 12.22(b) how the triangle rule can be used to determine the components. Thus
component down incline = 30 sin 35° = 17.2 N
component at right angles to incline = 30 cos 35° = 24.6 N Example
A projectile has an initial velocity of 10 m/s at an angle of 60° to the horizontal. What are the horizontal and vertical components of this velocity?
(a) 35°
30 N
35° 35°
30 N
(b)
Figure 12.22 Example
4.9 kN θ
1.1 kN
Figure 12.21 Example
Figure 12.23 Example
60°
10 m/s
60°
Figure 12.23 shows how the triangle rule can be used to determine the velocity components. Thus
vertical component = 10 sin 60° = 8.7 m/s horizontal component = 10 cos 60° = 5.0 m/s Revision
17 Determine the horizontal and vertical components of the following forces:
(a) 10 N at 40° to the horizontal, (b) 15 kN at 70° to the horizontal, (c) 12 N at 20° to the horizontal, (d) 30 kN at 80° to the horizontal.
18 An object of weight 30 N rests on an incline which is at 20° to the horizontal. What are the components of the weight at right angles to the incline and parallel to the incline?
19 For each of the following systems of forces as a result of considering the components of each force, determine the resultant force:
(a) 2 N in an easterly direction, 3 N at 60° west of north and 2 N due south,
(b) 4 N in a north-easterly direction, 3 N due west and 5 N at 30° south of east,
(c) 2.8 N in a north-easterly direction, 4 N at 60° south of west and 6 N at 30° south of east.
20 An object has a velocity of 40 m/s in a north-easterly direction. What are the components of this velocity in the north and south directions?
21 A rocket starts its motion from the surface of the earth with a velocity of 400 m/s at an angle of 70° to the horizontal. What are the components of the velocity in the vertical and horizontal directions?
22 An object has a displacement of 10 m in a north-easterly direction.
What are the components of this displacement in the easterly and northerly directions?
1 For the vectors a of magnitude 2 in a direction due east, b of magnitude 3 in a direction due north and c of magnitude 4 in a north-westerly direction, determine
(a) a + b, (b) b + c, (c) a + c, (d) a - b, (e) b - c, (f) a - c.
Problems
2 Determine the resultants for the following forces acting at the same point on an object:
(a) 2 kN in a northerly direction and 4 kN in a westerly direction, (b) 10 N in a vertical direction and 20 N in a horizontal direction, (c) 40 N and 50 N if the angle between the forces is 45°,
(d) 20 N vertically downwards and 10 N in a downward direction at 50°
to the vertical,
(e) 10 N and 12 N if the angle between them is 105°, (f) 20 kN and 12 kN if the angle between them is 120°.
3 An object of weight 30 N is suspended by a string from the ceiling.
What horizontal force F must be applied to the object if the string is to become deflected and make an angle of 25° with the vertical (Figure
12.24)?
4 An object of weight 20 N is supported from the ceiling by two cables inclined at 40° and 70° to the ceiling (Figure 12.25). Determine the tensions in the cables.
5 An object of weight 5 N hangs on a vertical string. At what angle to the vertical will the string be when the object is held aside by a force of 3 N in a direction which is 20° above the horizontal?
6 Determine the resultant force acting on an object when two forces of 20 kN and 40 kN are applied to the same point on the object and the angle between the lines of action of the forces is 90°.
7 Three girders in the same plane meet at a point. If there are tensile forces of 70 kN, 80 kN and 90 kN in the girders, what angles will the girders have to be at if there is equilibrium?
8 What is the resultant force acting on the gusset plate shown in Figure 12.26 as a result of the forces shown?
9 Determine the resultant velocity of an object if it is subject to the following velocities:
(a) 4 km/h in a northerly direction and 3 km/h in an easterly direction, (b) 10 m/s vertically downwards and 4 m/s horizontally,
(c) 20 m/s and 10 m/s if the angle between the velocities is 60°,
(d) 10 km/h in a westerly direction and 20 km/h in a north-easterly direction,
(e) 20 m/s in a southerly direction and 10 m/s in a northerly direction.
10 An object is thrown vertically upwards with a velocity of 10 m/s. If there is a horizontal wind blowing of 5 m/s, what will be the resultant velocity of the object?
11 A pilot sets a course due south with an air speed of 300 km/h. If a 100 km/h wind blows from the south-west, what is the actual velocity of the plane?
12 A cyclist travels 5 km in an easterly direction followed by 7 km in a northerly direction. How far, and in what direction, is the cyclist from the start position?
13 A person walks 6 km in a south-westerly direction and then 4 km in a westerly direction. How far, and in what direction, is the walker from the start position?
14 Determine the resultant force acting on an object if it is acted on by four forces acting in the same plane of 1 N in a westerly direction, 3 N in a south-westerly direction, 6 N in a north-easterly direction and 5 N in a northerly direction.
15 Ship A is moving in a north-easterly direction at 15 km/h. Ship B is moving due west at 8 km/h. To ship A, with what velocity does ship B appear to be moving?
16 Object A moves with a velocity of 4 m/s due north and object B moves with a velocity of 3 m/s due east. What is the velocity of B relative to A?
17 An object of weight 10 N rests on an incline which is at 30° to the horizontal. What are the components of this weight in a direction at right angles to the incline and parallel to the incline?
18 Three forces act in the same plane on the same point on an object. If the forces are 4 N in a direction due north, 7 N in a south-easterly direction and 4 N in a direction 60° south of west, what is the resultant force?
19 A cable exerts a force of 15 kN on a bracket. If the cable is at an angle of 35° to the horizontal, what are the horizontal and vertical compo-nents of the force?
20 Forces of 10 N, 12 N and 20 N act in the same plane on an object in the directions west, 30° west of north, and north respectively. Determine the resultant force.
21 Forces of 1 N, 2 N, 3 N, 4 N and 5 N act in the same plane on an object in the directions north, north-east, east, 60° west of south, and due west respectively. Determine the resultant force.