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Introducing vectors for engineering applications
Introducing vectors for engineering applications

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1.3 Modelling motion with non-perpendicular vectors

The example of Alice and Bob pushing a block of ice was made simpler by the fact that they were pushing along horizontal and vertical vectors. Let’s look at another example, where the vectors are in arbitrary directions.

Consider the situation in Figure 8, where Alice and Bob have attached ropes to a face of the block of ice and are now pulling it in different directions. If Bob pulls with a force of 130 N at an angle of 47° clockwise from the horizontal, and Alice pulls with a force of 110 N at an angle of 24° anticlockwise from the horizontal, what is the combined force applied to the block, and what is the acceleration of the block?

Described image
Figure 8 Alice and Bob pulling a block of ice

As before, let’s start by creating an abstract drawing, with Alice and Bob replaced by arrows, as illustrated in Figure 9(a). Here, bold a represents the force applied by Alice and bold b represents the force applied by Bob. The vector bold a has a magnitude of 110 N and a direction with an angle of 24° measured anticlockwise from the positive x-axis, while bold b has a magnitude of 130 N and a direction with an angle of 47 super ring operator, measured clockwise from the positive x-axis.

Described image
Figure 9 Representing the vectors bold a and bold b

Now, let’s calculate the magnitude and direction of the resultant vector bold a plus b. Figure 9(b) shows the result of visually adding the vectors bold a and bold b. Unlike the previous example, bold a and bold b are not perpendicular, so the triangle formed by bold a, bold b and the resultant bold a plus bold b is not a right-angled triangle. In this case, we cannot use Pythagoras’ theorem or the trigonometric functions to calculate the magnitude and direction of bold a plus bold b, and instead we need to use other properties of triangles, such as the sine or cosine rules.

If we knew one of the interior angles, then we could use the cosine rule,

a squared equals b squared plus c squared minus two times b times c times cosine of cap a comma

to calculate the magnitude of bold a plus bold b from the magnitudes of bold a and bold b, as follows:

absolute value of a plus b squared equals absolute value of b squared plus absolute value of c squared minus two times absolute value of b times absolute value of c times cosine of theta comma

where theta is the interior angle opposite bold a plus bold b, as illustrated in Figure 10(a).

Described image
Figure 10 Finding the interior angle θ

To find theta we can make use of alternate angles (Z-angles). The angle cap x in Figure 10(b) is an alternate angle with the angle indicating the direction of vector bold a, so cap x equals 24 super ring operator. Also, cap x, theta and the angle indicating the direction of vector bold b all lie on a straight line, so

sum with 3 summands cap x plus theta plus 47 super ring operator equals 180 super ring operator full stop

Using this relation, we can now find the size of angle theta, and use this in the cosine rule to determine the length of bold a plus bold b.

Activity 4

Use the cosine rule to determine the length of edge c in the diagram below, to two decimal places.

If we apply our answer to Activity 4 to the vector diagram in Figure 10, then we have calculated that the magnitude of bold a plus bold b is approximately 195.73 N. The direction of bold a plus bold b is given by the angle phi in Figure 11(a).

Described image
Figure 11 Finding the direction of bold a plus bold b

We already know that the direction of bold a, so if we can find the angle cap b in Figure 11(b), then we can calculate phi from

cap b equals phi plus 24 super ring operator full stop

We can calculate cap b using the sine rule,

equation sequence part 1 a divided by sine of cap a equals part 2 b divided by sine of cap b equals part 3 c divided by sine of cap c comma

as follows:

absolute value of a plus b divided by sine of theta equals absolute value of b divided by sine of cap b full stop

Activity 5

Use the sine rule to calculate angle cap b in the diagram below, to one decimal place.

From Activities 4 and 5, we can therefore say that, approximately, the magnitude of bold a plus bold b is 195.73 N and its direction is given by the angle

equation sequence part 1 phi equals part 2 cap b minus 24 super ring operator equals part 3 38.9 super ring operator minus 24 super ring operator equals part 4 14.9 super ring operator comma

which is measured clockwise from the horizontal, as illustrated in Figure 12. This is the combined force applied by Alice and Bob to the block of ice.

Described image
Figure 12bold a plus bold b

Activity 6

A block of ice has a mass of a metric tonne (one multiplication 10 cubed kg), and the resultant force on the block is described by a vector bold cap f with a magnitude of 195.73 N and a direction of 14.9°, measured clockwise from the positive x-axis. Give the magnitude of the acceleration to two decimal places and the direction to one decimal place.