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Introducing vectors for engineering applications
Introducing vectors for engineering applications

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1.1 Modelling motion with perpendicular vectors

Let’s consider, where two people, Alice and Bob, are pushing a block of ice, which has a mass of a metric tonne ( one multiplication 10 cubed times kg ). Alice and Bob each push on a different face of the block, as illustrated in Figure 3, and the direction in which the block moves is a consequence of the combination of the forces they apply. If Bob applies a force of 130 N to the left face of the block, and Alice applies a force of 110 N to the bottom face, what is the combined force applied to the block, and what is the acceleration of the block?

Described image
Figure 3 Alice and Bob pushing different sides of a block of ice

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, bold a represents the force applied by Alice, who is below the block, and bold b 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.

Described image
Figure 4 Combining the vectors bold a and bold b

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

Activity 1

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

1.1.3 Calculating direction of motion of a combined force

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

Described image
Figure 5 Finding the direction of the resultant bold a plus bold b

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, theta equals phi .

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

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

Described image
Figure 6 Identifying corresponding angles

Activity 2

In Figure 6, if bold a is a positive vertical vector that has magnitude 110 N and  bold b is a positive horizontal vector that has magnitude 130 N, what is the direction of the resultant bold a plus bold b to one decimal place? Use the fact that  cap a and  theta are alternate angles to help you.

1.1.4 Calculating acceleration from a combined force

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

force equals mass multiplication acceleration or cap f equals m times a full stop

The block has a mass of a metric tonne ( one multiplication 10 cubed times kg ) and because it is made of ice we will ignore any forces due to friction. The force applied by Alice and Bob is bold a plus bold b , which has the magnitude calculated in Activity 1 and the direction calculated in Activity 2.

Example 1 Finding acceleration from a combined force

A block of ice has a mass of a metric tonne ( one multiplication 10 cubed times kg ). If Bob applies a force of 130 N to the left face of the block while Alice applies a force of 110 N to the bottom face, what is the acceleration of the block? Give the magnitude of the acceleration to two decimal places and the angle to one decimal place.

Solution

Newton’s second law gives us

cap f equals m times a comma

so acceleration is given by

a equals cap f divided by m full stop

In the following calculations we will use the vector c to represent acceleration, to avoid confusion with the vector a which is the force applied by Alice. The mass of the block of ice is one multiplication 10 cubed times kg and, when we ignore friction, the only force acting on the block is the resultant force due to Alice and Bob pushing the block, so

equation sequence part 1 m equals part 2 one multiplication 10 cubed equals part 3 10 cubed comma
cap f equals a plus b comma

and the acceleration of the block is given by

equation sequence part 1 c equals part 2 a plus b divided by m equals part 3 a plus b divided by 10 cubed equals part 4 left parenthesis a plus b right parenthesis multiplication 10 super negative three full stop

Multiplying a vector by a positive scalar does not change the direction of the vector, so the direction of the acceleration is the same as the direction of bold a plus bold b , and is given by theta equals 40.2 super ring operator (to 1 d.p.).

We also know that when multiplying a vector by a positive scalar, its magnitude is changed through multiplication. So, the magnitude of the acceleration is

equation sequence part 1 absolute value of a plus b multiplication 10 super negative three equals part 2 170.29 horizontal ellipsis multiplication 10 super negative three equals part 3 0.17 left parenthesis to two d full stop p full stop right parenthesis

So, the block accelerates at 0.17 times m s super negative two (to 2 d.p.) in a direction that is 40.2 super ring operator (to 1 d.p.) from the positive horizontal direction.