Maths for science and technology

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5 Units, significant figures and scientific notation

Have you come across an assignment question or an exam question where you think everyone is setting out to make life as difficult as possible for you? You know which formula to use but none of the figures you have been given can be plugged in directly.

Here is a question from an exam paper on a technology course:

A simply supported square-section beam has a load of 5.0 kN hung from it centrally. The beam is constructed from solid aluminium with 50 mm × 50 mm cross-section and is 0.4 m in length between the end supports. What is the central deflection when the load is present? Take Young's Modulus to be 72 GN m−2.

Don't worry if this question doesn't make sense to you ? it is purely an example to illustrate the use of scientific notation and significant figures.

Throughout this section we shall use the following symbols for the values referred to in the question.

Force F
Length l
Young's Modulus E
Breadth b
Depth d
Deflection Δ

The formula for central deflection, which we will be using later in this section is

Units

The first issues is that of units. The values are not given using a coherent system of units. The system used in science and engineering problems is the Système Internationale, usually called the SI system (Table 1).

Table 1 Some common SI units

SI Unit What it measures Symbol
Kilogram mass kg
Metre length m
Second time s
Joule energy J
Newton force N
Kelvin temperature K

Because of the magnitude of values, it is sometimes appropriate to use another letter alongside that used to define the unit (Table 2). This makes calculations more manageable.

Table 2 Some common prefixes used with SI units

Prefix What it stands for Definition
m milli  × 10−3
k kilo  × 10+3
M mega  × 10+6
G giga  × 10+9

Now look at the units in the question. Only the distance between the supports is given in an SI unit. The remainder need to be converted. The kN needs to be converted to N, the mm need to be converted to m and the GN m−2 to N m−2. This necessitates a lot of multiplying and dividing by multiples of 10 and gives scope for getting the order of magnitude wrong.

Another point we would like you to note at this stage is that the data in the question is given to no more than two-figure accuracy. When we work through the calculation, we will find that the central deflection is 1.77777 m. Is it reasonable to quote to five decimal places? Surely we can't justify an accuracy of 1/10 000 m when some of the data is accurate to only 1/10 m. What accuracy can we quote in the answer?

The ideas of significant figures and scientific notation enable us to deal with these issues systematically.

Significant figures

The number of significant figures in a number is found by counting all the digits from the first non-zero digit on the left.

Leading zeros are not significant figures. Trailing zeros are significant figures only when they occur after the decimal point.

Let's look at a two examples.

  1. 765.430 has six significant figures. You start counting from the 7 which is the first non-zero digit on the left. The trailing zero is a significant figure; if it were not, it would not be necessary to include it.

  2. 0.04321 has four significant figures. The leading zeros are essential to give the magnitude of the number. The first non-zero digit on the left is 4 and counting the significant digits then gives us a total of four.

Example 5

Write the following values correct to the number of significant figures given beside each.

(a) 723.6792 (4 sig figs)
(b) 0.0763 (1 sig fig)
(c) 6382 (2 sig figs)

Discussion

  1. 723.6792

    First count four digits from the left. This gives 723.6

    Now look at the next digit.

    This is 7 which indicates that the number is actually closer to 723.7 than 723.6

    Rounding to four significant figures gives 723.7

  2. 0.0763

    Counting from the first non-zero digit on the left, we have 0.07 Now examine the next digit.

    This is 6 which indicates that the number is closer to 0.08 than 0.07 Rounding to one significant figure gives 0.08

  3. 6382

    Counting from the left, the first two significant figures are 6 and 3 The next digit is 8 so we would round up the 3 to 4 This gives a value of 6400 to two significant figures.

Activity 12

Now here are some exercises for you to try.

  1. How many significant figures do these numbers have?

    1. 0.6540

    2. 760.45

    3. 10002

    4. 700.302

    5. 1.9

    6. 0.00029

  2. Write each of these values to the number of significant figures stated in brackets.

    1. 0.00328 (2)

    2. 974.721 (3)

    3. 0.79312 (4)

    4. 1,654.769 (4)

Answer

Discussion

  1. The number of significant figures in each case is given below.

    1. 4

    2. 5

    3. 5

    4. 6

    5. 2

    6. 2

  2. The solutions are given below.

    1. 0.0033

    2. 975

    3. 0.7931

    4. 1655

Back to the original problem

A simply supported square-section beam has a load of 5.0 kN hung from it centrally. The beam is constructed from solid aluminium with 50 mm × 50 mm cross-section and is 0.4 m in length between the end supports. What is the central deflection when the load is present? Take Young's Modulus to be 72 GN m−2.

Now let us identify the numerical data given in the original problem and convert it into SI units

Symbol Value from question Value in SI units
F 5.0 kN 5.0 × 103N
l 0.4 m 0.4 m
E 72 GN m−2 72 × 109N m−2
b 50 mm 50 × 10−3 m
d 50 mm 50 × 10−3 m

The SI units could now be substituted into the formula for deflection, but they are not all in the same format. This is where the idea of scientific notation comes in.

Scientific notation

In order to make such a calculation simpler, we usually work in scientific notation. Scientific notation is really quite a simple idea, which allows us to write all of our numbers in the same format.

Scientific notation uses the format a.b × 10c

In other words, there is always only one digit in the range 1 ≤ digit ≤ 9 (i.e. one digit in the range 1 to 9) before the decimal point regardless of the magnitude of the number.

So, for example, 674 is written as  6.74 × 102

  58.4 is written as  5.84 × 101

  0.0756 is written as  7.56 × 10−2

Activity 13

Write the following numbers in scientific notation.

  1. 56432

  2. 0.00091

  3. 222

  4. 11.645

  5. 7543.46

  6. 0.054321

Answer

Discussion

  1. 56432 = 5.6432 × 104

    The decimal point in 56432 lies after the final digit. It is not usual to include the decimal point when there are no digits following it. We have moved the decimal point four places to the left, which is the same as dividing by 104. In order to keep the magnitude of the number correct, we must now multiply by 104.

  2. 0.00091 = 9.1 × 10−4

    In this example, the decimal point has to be moved four places to the right, which is equivalent to multiplying by 104. We must divide the resultant number by 104, which is the same as multiplying by 10−4.

    (See Section 3 on indices for clarification if necessary.)

  3. 222 = 2.22 × 102

    We have moved decimal point two places to the left, thus dividing by 102. We must now multiply by 102 to maintain correct value.

  4. 11.645 = 1.1645 × 101

    In moving decimal point one place to the left, we have divided by 101 and must now multiply by 101 to keep original value.

  5. 7543.46 = 7.54346 × 103

    The decimal point is moved three places to the left, hence we have divided by 103. To keep original value we must then multiply by 103.

  6. 0.054321 = 5.4321 × 10−2

    To write this value in scientific notation, the decimal point is moved two places to the right, multiplying the value by 102. It is necessary to divide by 102 to keep order of magnitude correct; this is the same as multiplying by 10−2.

Back to the original problem

We already have a value for F in scientific notation: F = 5.0 × 103. However, l, E, b and d are not in the correct format as they do not have one digit before the decimal point.

Let us look at l first. The decimal point needs to be moved one place to the right. If we move the decimal point one place to the right, we are multiplying the number by 101. We must then multiply the resulting value by 10−1 to ensure that the magnitude of the number is not changed.

If we look at the values for E, b and d we can see that there is more than one digit in front of the decimal point. So we have to move the decimal points to the left. If we move the decimal point to the left, we are dividing by 101 so we must then multiply by 101 to keep the magnitude of each value correct.

Remember: 10a × 10b = 10a + b (see Section 3 on indices for clarification if necessary).

The value for E is now written in scientific notation and SI units.

The values of b and d are identical and may be written as

Now we can produce a new set of values that are in both SI units and scientific notation.

F 5.0 × 103 N
L 4.0 × 10−1 m
E 7.2 × 1010 N m−2
b 5.0 × 10−2 m
d 5.0 × 10−2 m

Now we should be able to calculate an answer to the problem posed at the outset of this section.

You may not be familiar with the equation but the procedures are the same regardless of the formula so don't worry if this is not your area of study.

The formula for deflection of a beam in this situation was given as

We do not have a value for I. This entity is the second moment of area of the beam and is given by the formula

Let us calculate I.

Remember: 10a ÷ 10b = 10a−b and (10a)ab (see Section 3 on indices for clarification if necessary).

As we are going to substitute this into the formula for deflection, it is best to leave the value for I in this format. This will lead to a more accurate answer.

Now substituting in the formula for deflection gives

If we look at the number of digits following the decimal point and compare this to the data given in the question, we would appear to have a value that is more accurate than the data used to produce it. Rounding this value to two significant figures, gives a value of 1.8 × 10−4 m. To give us an idea of what the deflection might look like, let us convert this to mm: 1.8 × 10−4 m = 0.18 mm, which is a very small deflection in the beam.

To help you grasp the procedure, we are going to work through another example.

Example 6

Vertical strings support two metal spheres so that they are touching. Sphere 1, which has a mass m1 of 50 g, is pulled aside to the right until it reaches a height h1 of 20 cm. It is then released and swings to undergo an elastic collision with sphere 2, which has a mass m2 of 120g.

  1. What is the velocity of the two spheres immediately after the collision?

  2. To what height do the two spheres swing just after the collision?

Assume g, the acceleration due to gravity, is 9.81 m s−2.

Discussion

First, identify the numerical values and convert them into SI units and scientific notation.

Symbol Value Value in SI units Value in scientific notation
m1 50 g 50 × 10−3 kg 5.0 × 10−2 kg
g 9.81 m s−2 9.81 m s−2 9.81 m s−2
h1 20 cm 20 × 10−2 m 2 × 10−1 m
m2 120 g 120 × 10−3 kg 1.20 × 10−1 k0g

Initially both spheres have no velocity.

When sphere 1 is raised to the right it has potential energy, which is converted to kinetic energy just before the collision. This enables us to calculate the velocity of sphere 1 prior to the collision.

Let u1 be the velocity of sphere 1 before the collision and u2 the velocity of sphere 2 before the collision.

Conservation of energy gives us

Rearranging this equation gives

Note: 10−3 ÷ 10−2 = 10(−3−(−2)) = 10−1

So, u1 = 1.980908882 m s−1

Conservation of momentum gives us

where v is the velocity of the spheres immediately after collision.

But since m2 is at rest before the collision, u2 = 0

Hence

It is good practice to work through calculations without rounding until the final value is arrived at. This avoids cumulative errors occurring.

The kinetic energy immediately after the collision is given by

When the spheres swing to their maximum height, the velocity is zero and all the kinetic energy is converted into potential energy. Hence

(m1 + m2)g h2 = 28.852941 17 × 10−3

1.7 × 10−1 × 9.81 × h2 = 28.85294117 × 10−3

Rearranging to find h2

Giving the value to 2 significant figures, h2 = 1.7 cm.

We hope you have grasped the procedure now.

Procedure for solving problems involving SI units, scientific notation and significant figures

Now follow the above procedures to solve this next problem.

Activity 14

Two cars are stopped side by side at traffic lights. When the lights change to green both cars accelerate. The Jaguar XJS reaches a speed of 45 km per hr in 30 seconds while the Nissan Micra takes minutes to reach the same speed.

  1. Assuming uniform acceleration in both cars, calculate how many kilometres each will have travelled in five minutes?

  2. What is the distance between them at this point?

Acceleration, a, is given by v = u + at and distance travelled, s, is given by

where

v = final velocity

u = initial velocity

t = time.

Answer

Discussion

Identify numerical values from the question

Jaguar initial velocity u = 0 m s−1
final velocity v = 45 km per hr
time t = 30s
Micra initial velocity u = 0 m s−1
final velocity v = 45 km per hr
Time minutes
Final time T = 5 minutes

Convert to SI units and write in scientific notation

Symbol value Numerical SI units Scientific notation
Jaguar Initial speed u 0 m s−1 0 m s−1 0 m s−1
Final speed v 45 km per hr 12.5 ms−1 1.25 × 101 m s−1
Time t 30 s 30 s 3.0 × 101 s
Micra Initial speed u 0 m s−1 0 m s−1 0 m s−1
Final speed v 45 km per hr 12.5 ms−1 1.25 × 101 ms−1
Time t minutes 150 s 1.5 × 102 s
Final time T 5 minutes 300 s 3.0 × 102 s

Substitute in formula, separate numbers and calculate

To find the acceleration, a, use v = u + at which gives

(If you are not sure about rearranging equations, refer to Section 4)

In both cases u =  0 m s−1.

So, for the Jaguar

Acceleration for the Jaguar is 0.416666666 m s−2. For the Micra

Acceleration for the Micra is 0.083333333 m s−2.

To find the distance travelled, use

In both cases

For the Jaguar

For the Micra

So the Jaguar has travelled 18.750 km and the Micra has travelled 3.750 km. The distance between them is 15 km.

All values are in km as the question required and the distance between them is exactly 15 km when the values are not rounded during the calculation.

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