### 1 Using formulae and substitution

In the study of engineering, physical quantities can be related to each other using a formula. The formula will contain variables and constants which represent the physical quantities. To evaluate a formula we must
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substitute
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numbers in place of the variables.

For example, Ohm’s law provides a formula for relating the voltage, $v$ , across a resistor with resistance value, $R$ , to the current through it, $i$ . The formula states

$\phantom{\rule{2em}{0ex}}v=iR$

We can use this formula to calculate $v$ if we know values for $i$ and $R$ . For example, if $i=13$ A, and $R=5\phantom{\rule{0.3em}{0ex}}\Omega $ , then

$$\begin{array}{rcll}v& =& iR& \text{}\\ & =& \left(13\right)\left(5\right)& \text{}\\ & =& 65\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}& \text{}\end{array}$$The voltage is 65 V.

Note that it is important to pay attention to the units of any physical quantities involved. Unless a consistent set of units is used a formula is not valid.

##### Example 62

The kinetic energy, $K$ , of an object of mass $M$ moving with speed $v$ can be calculated from the formula, $K=\frac{1}{2}M{v}^{2}$ .

Calculate the kinetic energy of an object of mass 5 kg moving with a speed of $2{\text{ms}}^{-1}$ .

##### Solution

In this example $M=5$ and $v=2$ . Substituting these values into the formula we find

$$\begin{array}{rcll}K& =& \frac{1}{2}M{v}^{2}& \text{}\\ & =& \frac{1}{2}\left(5\right)\left({2}^{2}\right)& \text{}\\ & =& 10\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}& \text{}\end{array}$$In the SI system the unit of energy is the joule. Hence the kinetic energy of the object is 10 joules.

##### Task!

The area, $A$ , of the circle of radius $r$ can be calculated from the formula $A=\pi {r}^{2}$ .

If we know the diameter of the circle, $d$ , we can use the equivalent formula $A=\frac{\pi {d}^{2}}{4}$ . Find the area of a circle having diameter 0.1 m. Your calculator will be preprogrammed with the value of $\pi $ .

$\frac{\pi {\left(0.1\right)}^{2}}{4}=0.0079\phantom{\rule{1em}{0ex}}{\text{m}}^{2}$

##### Example 63

The volume, $V$ , of a circular cylinder is equal to its cross-sectional area, $A$ , times its length, $h$ .

Find the volume of a cylinder having diameter 0.1 m and length 0.3 m.

##### Solution

We can use the result of the previous Task to obtain the cross-sectional area $A=\frac{\pi {d}^{2}}{4}$ . Then

$$\begin{array}{rcll}V& =& Ah& \text{}\\ & =& \frac{\pi {\left(0.1\right)}^{2}}{4}\times 0.3& \text{}\\ & =& 0.0024\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}& \text{}\end{array}$$The volume is $0.0024{\text{m}}^{3}$ .