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Solution. If a scale drawing of the two forces is made as in figure b, the diagonal of the rectangle will be the resultant force R acting on the toy, which has the same effect on the toy as the two forces of 120 and 90 N. This means that a single force R can be used to find out what is happening to the movement of the toy. The two forces 120 and 90 N are said to be the components of the resultant force R. Using Pythagoras’ theorem in the shaded right-angled triangle, we can find the magnitude of R. The magnitude is the size. Write
R2 = 902 + 1202 R2 = 8100+ 14,400 R2 = 22,500
R = 150 Square root of 22,500
The resultant pull on the toy is 150 newtons.
96 COMPONENTS OF A VECTOR
Finding 9, using trigonometry in the same right-angled triangle, will enable us to state the direction of the resultant of 150 N (see figure c):
The resultant makes an angle of 53.1° with Jacob’s force of 90 N.
In general terms, a set of two components of a vector in two dimensions can now be defined, using the concepts developed in the example. Suppose the vector R has vector components H horizontally and V vertically, and the angle between these components is 9 degrees (see figure d).
The horizontal component of R is H = R cos 9. The vertical component of R is V = Rsin6>.
Example 1. A small aircraft is taking off and at a certain instant its velocity is 150 km h-1, and it is flying at an angle of 10° with respect to the horizontal runway, (see figure e). Find its horizontal and vertical components of velocity.
_ opposite side n — adjacent side
If tan a = b, then a = tan 1b
9 — 53.1° (1 dp) Using inverse tan on the calculator
COMPOSITE BAR GRAPHS 97
H = 150 x cos 10° Horizontal component
V = 150 x sin 10° Vertical component
H = 147.72 (to 2 dp) and V = 26.05 (to 2 dp)
The horizontal and vertical components of velocity are 147.27 and 26.05 km hr1, respectively. This means that the aircraft is gaining altitude at a velocity of 26.05 km h_1.
The components of a vector do not have to be horizontal and vertical, as shown in the following example.
Example 2. Tom is pulling his toboggan with a force of 90 newtons up a slope. The rope fastened to the toboggan makes an angle of 30° with the slope (figure f). Find the components of his pulling force that are parallel to the slope and at a right angle to the slope.
Solution. Let the components of his pulling force be P and Q as shown in figure f. Write
P = 90 x cos 30° and Q = 90 x sin 30°
The components of Tom’s pulling force are 77.94 newtons (to 2 dp) parallel to the slope and 45 newtons at a right angle to the slope.
References: Pythagoras’ Theorem, Square Root, Trigonometry, Vector.
COMPOSITE BAR GRAPHS
When two or more bar graphs showing similar information are drawn together on the same axes the resulting graph is called a composite bar graph. In this example Ann,
98 COMPOSITE SHAPES
Helen, and Liz are the three strikers for the Ramblers soccer team, and the composite bar graph in the figure shows their goal-scoring feats over three seasons. Composite bar graphs are useful for showing trends or changes that have taken place over a period of time. For example, it is clear from the graph that Liz is the only striker who is scoring more goals in successive years.
Goals scored by Ann, Helen, Liz
Reference: Bar Graph.
These are shapes that are made up of two or more other shapes. The processes of finding the area and the perimeter of a composite are explained in the example.
Example. The composite shape of a running track is made up of a rectangle, which is shown shaded in the figure, and two semicircles. All the dimensions are in meters. Find the area and perimeter of the composite shape.
Solution. The perimeter of this composite shape is made up of two straight lines, each of length 100 meters, and the circumference C of a whole circle of diameter 70 meters. Write
C = ttD Formula for circumference of a circle
C = 7T x 70 Substituting D = 70
C = 219.9 (to 1 dp) Using it in a calculator
COMPOSITE TRANSFORMATIONS 99
Total perimeter = 100 + 100 + 219.9 = 419.9 m
The perimeter of the composite shape is 419.9 meters.
The area of this composite shape is made up of the area of a whole circle and the area of a rectangle. Write, for the circle,
Area = ttR2 Formula for area of a circle
Area = 7t x 352 R = 70 ^ 2
Area = 3848.5 m2 (to 1 dp)
and for the rectangle,
Area = length x width Formula for area of a rectangle
Area = 100 x 70 Length = 100 and width =70
Area = 7000 m2
Area of composite shape = 3848.5 + 7000 = 10,848.5 m2
References: Area, Circumference, Circle, Perimeter, Rectangle, Semicircle, Square. COMPOSITE TRANSFORMATIONS
These are sometimes called combined transformations. When a final image is formed after two or more transformations such that the first image becomes the object for the second transformation, then a composite transformation has taken place. Often capital letters are used to represent transformations. For example, R is often used to represent a rotation and M a reflection, since m is the first letter of the word mirror. The transformations that may be combined with each other are reflection, rotation, translation, and enlargement. The composite transformations that are explained here are reflection in two parallel mirrors and reflection in two perpendicular mirrors.