1.1 Modelling motion with perpendicular vectors

1.1.2 Calculating the magnitude of a combined force

The net effect of these two forces by combining them visually using arrows, as illustrated in Figure 4. Figure 4(a) shows an abstraction of the drawing in Figure 3, with Alice and Bob replaced by arrows representing the forces applied by Alice and Bob to the block of ice. Here, represents the force applied by Alice, who is below the block, and represents the force applied by Bob, who is to the left of the block. Notice that the vectors are shown to be acting on the centres of the faces of the block. This is because if forces are applied away from the centres, this can create a rotation, and that is a more complicated situation to model. Such rotational effects are outside the scope of this module.

Figure 4 Combining the vectors and

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In the image on the left, Vector b pushes the block of ice to the right whilst vector a pushes the block forwards. On the right is a vector diagram showing how these two vectors combine so the block moves diagonally in the direction of a + b.

Figure 4 Combining the vectors MathJax failure: MathML - Unexpected text node: '

Figure 4(b) shows the result of visually adding the vectors and . . In this example, and are perpendicular, so the triangle formed by , and the resultant is a right-angled triangle, with as the hypotenuse. So we can use Pythagoras’ theorem to find the magnitude of , written , from the magnitudes of and , written and .

Activity 1

If is a positive vertical vector that has magnitude 110 N and is a positive horizontal vector that has magnitude 130 N, what is the magnitude of the resultant to two decimal places?

1.1.3 Calculating direction of motion of a combined force

Using the fact that , and the resultant form a right-angled triangle, we can also use trigonometric functions to find the direction of . This is illustrated in Figure 4.5, where the direction of is represented by the angle . But is outside the triangle formed by , and , so we can’t directly calculate its size using trigonometry; we also need to use our knowledge of angles and triangles.

Figure 5 Finding the direction of the resultant

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Finding the direction of the resultant vector a + b. Vector a points forward whilst vector b points to the right. The resultant vector points diagonally in the direction a + b at an angle θ. Opposite, corresponding and alternate angles green box figure 1: Two straight lines intersect. The angles opposite to one another are known as opposite angles and are labelled as θ and φ. Opposite angles are equal. Opposite, corresponding and alternate angles green box figure 2: On the left, a line intersects a set of parallel lines. Angles α and ϒ are alternate angles and lie opposite each other on the inner side of each of the two lines. Alternate angles are equal. On the right, the angles β and μ are corresponding angles. They lie on the same side of each line and are also equal.

Figure 5 Finding the direction of the resultant MathJax failure: MathML - Unexpected text node: '

In particular, we can use the properties of angles on lines, as summarised here.

Opposite, corresponding and alternate angles

Where two lines intersect, opposite angles are equal. This is commonly referred to as the X-angles rule. For example, in the following diagram, .

Where a line intersects parallel lines:

• Alternate angles are equal. This is commonly referred to as the Z-angles rule. For example, in the diagram below, .

• Corresponding angles are equal. This is commonly referred to as the F-angles rule. For example, in the diagram below, .

For example, to find the angle in Figure 5, we can use alternate angles (Z-angles). So in Figure 6, angles and are equal, and to find we use trigonometry, e.g. the tangent function.

Figure 6 Identifying corresponding angles

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Vector a points forward whilst vector b points to the right. The resultant vector points diagonally in the direction a + b. The vectors b and a + b form an angle A which is the alternate angle with θ.

Figure 6 Identifying corresponding angles

1.1.4 Calculating acceleration from a combined force

With the magnitude and direction of the resultant vector now calculated, we can use Newton’s second law to determine the acceleration of the block. Recall that Newton’s second law states that

The block has a mass of a metric tonne () and because it is made of ice we will ignore any forces due to friction. The force applied by Alice and Bob is , which has the magnitude calculated in Activity 1 and the direction calculated in Activity 2.