Mathematical logic equations, Lecture notes of Mathematics

Download Mathematical logic equations and more Lecture notes Mathematics in PDF only on Docsity!

Engineering Principles

<v140988_ph_001>

DRAFT

UNIT 1

2 Engineering Principles

Getting to know your unit

To make an effective contribution to the design and development of

engineered products and systems, you must be able to draw on the

principles laid down by the pioneers of engineering science. The

theories developed by the likes of Newton and Ohm are at the heart of

the work carried out by today’s multi-skilled engineering workforce.

This unit covers a range of both mechanical and electrical principles and

some of the necessary mathematics that underpins their application to

solve a range of engineering problems.

How you will be assessed This unit is externally assessed by an unseen paper-based examination. The examination is set and marked by Pearson. Throughout this unit you will find practice activities that will help you to prepare for the examination. At the end of the unit you will also find help and advice on how to prepare for and approach the examination. The examination must be taken under examination conditions, so it is important that you are fully prepared and familiar with the application of the principles covered in the unit. You will also need to learn key formulae and be confident in carrying out calculations accurately. A scientific calculator and knowledge of how to use it effectively will be essential. The examination will be two hours long and will contain a number of short- and long-answer questions. Assessment will focus on applying appropriate principles and techniques to solving problems. Questions may be focused on a particular area of study or require the combined use of principles from across the unit. An Information Booklet of Formulae and Constants will be available during the examination. This table contains the skills that the examination will be designed to assess.

Assessment objectives AO1 Recall basic engineering principles and mathematical methods and formulae AO2 Perform mathematical procedures to solve engineering problems AO3 Demonstrate an understanding of electrical, electronic and mechanical principles to solve engineering problems AO4 Analyse information and systems to solve engineering problems AO5 Integrate and apply electrical, electronic and mechanical principles to develop an engineering solution

This table contains the areas of essential content that learners must be familiar with prior to assessment. Essential content A1 Algebraic methods A2 Trigonometric methods B1 Static engineering systems B2 Loaded components C1 Dynamic engineering systems D1 Fluid systems D2 Thermodynamic systems E1 Static and direct current electricity E2 Direct current circuit theory E3 Direct current networks F1 Magnetism G1 Single-phase alternating current theory

Assessment This unit is externally assessed using an unseen paper-based examination that is marked by Pearson.

DRAFT

4 Engineering Principles

▸ ▸ When using common logarithms, there is no need to include the base 10 in the notation, so where N = 10 x , you can write simply log N = x. This corresponds to the log function on your calculator.

Natural logarithms In a similar way, when dealing with natural logarithms with base e: ▸ ▸ Where N = e x , then loge N = x. ▸ ▸ Natural logarithms with base e are so important in mathematics that they have their own special notation, where loge N is written as ln N. So where N = e x , then ln N = x. This corresponds to the ln function on your calculator.

The laws of logarithms There are a number of standard rules that can be used to simplify and solve equations involving logarithms. These are the laws of logarithms. They are summarised in Table 1..

▸ ▸ Table 1.2 The laws of logarithms

Operation Common logarithms Natural logarithms Multiplication log AB = log A + log B ln AB = ln A + ln B

Division (^) log __ A B = log^ A^ − log^ B^ ln^

A __ B = ln^ A^ − ln^ B Powers log An^ = n log A ln An^ = n ln A Logarithm of 0 log 0 = not defined ln 0 = not defined Logarithm of 1 log 1 = 0 ln 1 = 0

P A U s E P O I N T

Use your calculator to practise finding the common and natural logarithms of a range of values. Be sure to include whole numbers and decimal fractions less than 1. What happens when you try to find the logarithm of 0 or a negative number using your calculator? Hint (^) From the relationship between logarithms and indices you know that if log 0 = x , then 0 = 10 x. What value must x take if 10 x^ = 0? Take a few moments to think about why log 0 and log −1 are not defined and will produce an error on your calculator.

Extend

Worked Example

You can use the laws of indices to simplify expressions containing indices.

• a^2 a^4 a^0 a −3.2^ = a 2 + 4 + 0 − 3.2^ = a 2.
• (√

__ a ) 3 a −1^ = ( a 0.5)^3 a −1^ = a 1.5^ a −1^ = a 1.5−1^ = a 0. or √

__ a

• a^

3 __ (^2) b (^2) a −

---

b

= a 1.5 b^2 a −1 b −1^ = a 1.5 − 1 b 2 – 1^ = a 0.5 b^1

or b √

__ a

• a __________−1^ b −1^ a^^1 __^2 b −^

= a −1 b −1 a 0.5 b^2 = a −1 + 0.5 b −1 + 2^ = a −0.5 b^1

or ___ √ b __ a

• a −3^ ( b −2)^2 a 3.5^ = a −3 + 3.5 b −4^ = a 0.5 b −4^ or √

__ ___^ a b^4

Logarithms Logarithms are very closely related to indices. The logarithm of a number ( N ) is the power ( x ) to which a given base ( a ) must be raised to give that number. ▸ ▸ In general terms, where N = a x , then loga N = x. In engineering, we encounter mainly common logarithms , which use base 10, and natural logarithms , which use base e. The Euler number (e) is a mathematical constant that approximates to 2.718. You will come across this again later in the section dealing with natural exponential functions.

Common logarithms Common logarithms are logarithms with base 10. ▸ ▸ Where N = 10 x , then log 10 N = x.

DRAFT

Learning aim A UNIT 1

Engineering Principles

Exponential growth and decay Exponential functions where x and y are variables and N is a constant take the general form y = N x. Some common systems found in engineering and in nature – such as charging capacitors, radioactive decay and light penetration in oceans – can be defined by exponential functions. Several important growth and decay processes in engineering are defined by a special type of exponential function that uses the Euler number (e) as its base ( N ). This is known as the natural exponential function and takes the basic form y = e x. If you plot the function y = e x^ as a graph (see Figure 1.1 ), then the curve it describes has two special characteristics that other exponential functions do not have. ▸ ▸ At any point on the curve the slope or gradient of the graph is equal to e x. ▸ ▸ At x = 0 (where the curve intersects the y -axis) the graph has a slope or gradient of exactly 1.

y

O x

▸ ▸ Figure 1.1 Graph of the natural exponential function y = e x Examples of engineering formulae containing natural exponential functions are shown in Table 1.3 for charging/ discharging a capacitor. ▸ ▸ Table 1.3 Example engineering formulae containing natural exponential functions Description Formula Capacitor charge voltage (^) V = V 0 (1 − e − t / RC ) Capacitor discharge voltage (^) V = V 0 e− t / RC

Key term Tangent – a straight line with a slope equal to that of a curve at the point where they touch.

Worked Examples

1 Use the laws of logarithms to solve the equation log 14 (1 −^ x )^ = log 9 ( x^ + 3). solution log 14 (1 −^ x )^ = log 9 ( x^ + 3) (1 − x ) log 14 = ( x + 3) log 9 log 14 − x log 14 = x log 9 + 3 log 9 log 14 − 3 log 9 = x log 9 + x log 14 log 14 − 3 log 9 = x (log 9 + log 14)

x = _________________log 14 − 3 log 9 log 9 + log 14

= −0.817 (rounded to 3 significant figures (s.f.)) Always check your solution by substituting the unrounded value of the solution back into the original equation: log 14 (1 −^ x )^ = log 14 [1 − (−0.817…)]^ = 2.08 (to 3 s.f.) log 9 ( x^ + 3)^ = log 9 (−0.817… + 3)^ = 2.08 (to 3 s.f.) 2 Use the laws of logarithms to solve the equation ln 3 t = 2 ln^12 ___ t + 2. solution ln 3 t = 2 ln^12 ___ t + 2 ln 3 + ln t = 2(ln 12 – ln t ) + 2 ln 3 + ln t = 2 ln 12 – 2 ln t + 2 3 ln t = 2 ln 12 – ln 3 + 2

ln t = ___________________2 ln 12 − ln 3 + 2 3 = 1.957… You can now use the general relationship that where ln N = x , then N = e x. In this case, ln t = 1.957…, so t = e1.957…^ , which you can use your calculator to evaluate. t = 7.08 (to 3 s.f.) Always check your solution by substituting the unrounded value of the solution back into the original equation: ln 3 t = ln 21.235… = 3.056 (to 4 s.f.) 2 ln___^12 t + 2 = 2 ln 1.695… + 2 = 2 × 0.5278… + 2 = 3.056 (to 4 s.f.)

P A U s E P O I N T Examine the characteristics of exponential functions by plotting a graph of^ y^ = 2

x (^) for values of x between 1 and 10. Hint (^) Tabulate values of x and y. Ensure that your axes are of appropriate size and scale. The gradient of the tangent at any point on the curve reflects the rate of change in y with respect to x at that point. What happens to this rate of change as x increases?

Extend

DRAFT

Learning aim A UNIT 1

Engineering Principles

y

O x

▸ ▸ Figure 1.2 The graphical representation of a linear equation is a straight line The gradient of a linear equation can be: ▸ ▸ positive – the y values increase linearly as the x values increase (the line slopes up from left to right) ▸ ▸ negative – the y values decrease linearly as the x values increase (the line slopes down from left to right) ▸ ▸ zero – the y values stay the same as the x values increase (the line is parallel to the x -axis).

solving pairs of simultaneous linear equations Sometimes in engineering it is necessary to solve systems that involve pairs of independent equations that share two unknown quantities. There are several examples of this in the assessment activity practice questions at the end of this section. When you solve simultaneous equations, you are determining values for the unknowns that satisfy both equations. This is easiest to see by considering two independent linear equations in x and y that are plotted graphically on the same axes (see Figure 1.3 ). The only position where the same values of x and y satisfy both equations is where the lines intersect. You can read these values from the x -axis and the y -axis. In this example, the solution is x = 2 and y = 4.

Key terms subject of an equation – a single term becomes the subject of an equation when it is isolated on one side of the equation with all the other terms on the other side. For example, in the equation y = 4 x + 3, the y term is the subject. Gradient – also called ‘slope’, measures how steep a line is. It is calculated by picking two points on the line and dividing the change in height by the change in horizontal distance, or ___________________change in^ y^ value change in x value

Linear equations

Linear equations contain unknowns that are raised to their first power only, such as x raised to its first power or x^1 (which is usually written simply as x ). Linear equations take the general form y = mx + c.

Linear equations containing a single unknown quantity can be solved by rearranging to make the unknown quantity the subject of the equation.

Worked Example – Rearranging a linear equation

Rearrange 4 x + 3 = 12 to make x the subject of the equation. solution 4 x = 12 − 3 = 9 Subtract 3 from both sides of the equation. x =^9 __ 4

Divide both sides of the equation by 4.

If the same unknown quantity occurs more than once in a linear equation, then all the terms containing the unknown quantity need to be isolated on one side of the equation with all the other terms on the other side. This is called ‘gathering like terms’.

Worked Example – Gathering like terms in a linear equation

Rearrange 9 b + 6 = 14 − 2 b to make b the subject of the equation. solution

9 b + 2 b = 14 − 6

Add 2 b and subtract 6 on both sides of the equation. 11 b = 8 Combine like terms. b = ___^8 11

Divide both sides of the equation by 11.

straight-line graphs

Graphs in mathematics usually use a horizontal x -axis and a vertical y -axis (known as Cartesian axes).

A linear equation with two unknowns can be represented graphically by a straight line as shown in Figure 1..

In the general formula used to describe a linear equation, y = mx + c , m is the gradient of the line and c is the value of y where the line intercepts the y -axis (when x = 0).

DRAFT

8 Engineering Principles

y

O x

▸ ▸ Figure 1.3 The solution of the simultaneous linear equations y = x + 2 and 2 y = − x + 10 is given by the intersection point of the two lines

Of course you could always plot graphs of the linear equations when solving simultaneous equations, but this is a time-consuming approach and less accurate than the alternative algebraic methods. The two main algebraic methods – substitution and elimination – are best explained by working through examples.

Worked Example – Substitution method

Solve this pair of simultaneous equations: y = x + 2 (1) 2 y = − x + 10 (2)

solution Substitute equation (1) into equation (2) to get 2( x + 2) = − x + 10. So 2 x + 4 = − x + 10 Multiply out the brackets. 2 x + x = 10 – 4 Collect like terms. 3 x = 6 x = 2 Divide both sides by 3. Now substitute x = 2 into equation (1) in order to find y : y = 2 + 2 = 4. To check your solution, substitute x = 2 into equation (2) and see if you get the same value for y : 2 y = −2 + 10 = 8 y = 4

Worked Example – Elimination method

Solve the same pair of simultaneous equations as before: y = x + 2 (1) 2 y = − x + 10 (2)

solution Multiply equation (1) by 2 so that the y terms in both equations become the same: 2 y = 2 x + 4 (3)

Subtract equation (2) from equation (3) to eliminate the identical terms in y : (2 x + 4) − (− x + 10) = 0 3 x – 6 = 0 3 x = 6 x = 2 As before, substitute x = 2 into equation (1) to find y = 4, and then check your solution in equation (2).

Coordinates on Cartesian grid Any point on a plotted linear equation can be expressed by its coordinates ( x , y ). In the example shown in Figure 1.3 , the two lines cross at the point where x = 2 and y = 4. Its coordinates are (2, 4). Any point on the x -axis has a y -coordinate of zero. For example, the point (4, 0) lies on the x -axis. Similarly, any point on the y -axis has an x -coordinate of zero, so the point (0, 4) lies on the y -axis. The point (0, 0), where both x - and y -coordinates are zero, is called the origin and is represented by ‘ O ’.

Quadratic equations Quadratic equations contain unknowns that are raised to their second power, such as x^2. They take the general form y = ax^2 + bx + c. In this equation x and y are unknown variables, and coefficients a , b and c are constants that will be discussed later. A quadratic equation with two unknowns can be represented graphically by a curve. An example is shown in Figure 1..

DRAFT

10 Engineering Principles

Finding the terms to put in the brackets is then often a case of trying different values until you identify those that meet the required criteria: x^2 + 2 x − 8 = ( x − 2)( x + 4) ▸ ▸ Equate each of the factors to zero to obtain the roots of the equation. ( x − 2) = 0, so x = 2 ( x + 4) = 0, so x = − ▸ ▸ Check that each root satisfies the original equation by substitution. 3 x + 8 = 5 x + x^2 When x = −4: −12 + 8 = −20 + 16 or −4 = − When x = 2: 6 + 8 = 10 + 4 or 14 = 14 ▸ ▸ Check to ensure that each solution is reasonable in the context of the question and clearly state the solution. There are two values of x for which the quadratic equation 3 x + 8 = 5 x + x^2 is true. These are x = 2 and x = −4. See the worked example on page 11. Some examples of engineering formulae describing quadratic relationships are given in Table 1..

▸ ▸ Table 1.5 Examples of engineering formulae describing quadratic relationships

Description Formula Standard general form ax^2 + bx + c = 0 Displacement (^) s = ut + 1 __ 2 at^2

Total surface area of a cylinder (^) A = 2π r^2 + 2π rh

solving quadratic equations using the quadratic formula Sometimes it is difficult to use factorisation to find the roots of a quadratic equation. In such cases, you can use an alternative method. The formula for the roots of a quadratic equation arranged in the standard form is:

x = −^ b^ ±^

√

---

_________________^ b^2 − 4 ac 2 a

The second example can be factorised further by using the methods for factorising quadratics described in the next section.

solving quadratic equations using factorisation Let us take an example of a quadratic equation that you might need to solve as part of an engineering problem: 3 x + 8 = 5 x + x^2 ▸ ▸ First, rearrange the equation (if necessary) to make one side of the equation zero. 0 = 5 x + x^2 – 3 x – 8 Move all the terms to one side of the equation. 0 = x^2 + 2 x − 8 Arrange into the standard form for quadratic equations. ▸ ▸ The next step is to factorise the right-hand side of this equation.

To do this, you need to find two expressions that when multiplied together give x^2 + 2 x − 8. This looks difficult, but there are a few general guidelines that will help: ▸ ▸ If the coefficient of the x^2 term ( a ) is 1, then the coefficient of each of the x terms in the factors will also be 1. ▸ ▸ When the number terms in the two factors are multiplied together, the product must equal the number term ( c ) in the quadratic expression. ▸ ▸ If the coefficient of the x^2 term ( a ) is 1, then the coefficient of the x term ( b ) in the quadratic expression is equal to the sum of the number terms in the two factors. Applying these guidelines to x^2 + 2 x − 8, you know that: ▸ ▸ the coefficients of the x terms in the factors will be 1 ▸ ▸ the product of the number terms in the factors will be − ▸ ▸ the sum of the number terms in the factors will be 2.

Key term Coefficient – a number or symbol that multiplies a variable. For example, in the expression 3 x , 3 is the coefficient of the variable x.

DRAFT

Learning aim A UNIT 1

Engineering Principles

Worked Example

A train runs along a level track with a velocity of 5 m s−1. The driver presses the accelerator, causing the train to increase speed by 2 m s−2^. The motion of the train is defined by the equation s = ut + 1 __ 2 at^2 where: s is displacement (the distance travelled) (m) u is initial velocity (m s−1) t is time (s) a is acceleration (m s−2). Calculate the time the train takes to travel a distance of 6 m.

solution Rearrange the equation into the general form for a quadratic, making one side equal to zero: __^1 2 at

(^2) + ut − s = 0 Substitute the values u = 5, s = 6 and a = 2 into the equation: __^1 2 (2) t

(^2) + 5 t − 6 = 0 t^2 + 5 t − 6 = 0 The left-hand side can be factorised by inspection to give ( t + 6)( t − 1) = 0. This equation is true when either of the factors is equal to zero, so t = −6 or t = 1. It is important to check these solutions by substituting back into the original quadratic equation given in the question: s = ut + 1 __ 2 at^2 = 5 t + t^2. You will find that when t = 1, s = 6 and when t = −6, s = 6. This confirms that the solutions determined by finding the roots of the quadratic equation are mathematically correct, because they both give the distance required in the question. However, you must now consider whether both solutions actually fit the practical situation in the real world. In this case t = −6 doesn’t make sense as a solution, because it would mean that the train reached the required distance 6 seconds before it started accelerating. The roots can be seen more clearly when illustrated graphically, as in Figure 1.. Finally, you should clearly state the solution to the problem: The time taken for the train to travel a distance of 6 m is 1 s.

▸ ▸ Train running along a track

y

O x

▸ ▸ Figure 1.5 Graph showing roots of the quadratic expression t^2 + 5 t − 6 at t = −6 and t = 1

DRAFT

Learning aim A UNIT 1

Engineering Principles

▸ ▸ Table 1.6 General formulae for circular measurements (see Figure 1.7 ) Arc length = rθ Circumference of a circle (^) = r (2π) = 2π r Area of a sector (^) = __^12 r^2 θ Area of a full circle = __^12 r^2 (2π) = π r^2

r

r

arc length

θ

sector of a circle ▸ ▸ Figure 1.7 Arc length and sector of a circle

Triangular measurement In right-angled triangles we name the three sides in relation to the right angle and one of the other two angles, θ (see Figure 1.8 ).

Opp^ Hyp

Adj

θ

▸ ▸ Figure 1.8 Trigonometric naming conventions for a right- angled triangle

▸ ▸ The side opposite the right angle is the hypotenuse (hyp). ▸ ▸ The side next to the angle θ is the adjacent (adj) side. ▸ ▸ The side opposite the angle θ is the opposite (opp) side. The ratios of the lengths of these sides are given specific names and are widely used in engineering (see Figures 1.9 – 1.11 ):

▸ ▸ sine (sin), where

sin θ =

opp_____ hyp

P A U s E P O I N T Discuss with a colleague or as a group why methods of finding the roots of quadratic equations are important and useful mathematical tools for engineers. Hint (^) Consider how else you might solve problems that involve quadratic equations. Suppose that you want to solve the quadratic equation y = x^2 + 4 x + 6 to find values of x when y = 1. This would mean finding the roots of the quadratic expression x^2 + 4 x + 5. Can you solve this? Draw a graph of the function y = x^2 + 4 x + 5 to help explain why not.

Extend

A2 Trigonometric methods

Angular measurement

You are already familiar with angular measurements made in degrees. In practical terms, this is the most common way to define an angle on an engineering drawing that a technician might use when manufacturing a component in the workshop.

However, there is another unit of angular measurement, called the radian , which is used extensively in engineering calculations.

Circular measurement

One revolution of a full circle contains 360° or 2π radians.

It is reasonably straightforward to convert angles stated in degrees to radians and vice versa.

Given that 2π radians = 360°

1 radian = 360 °______ 2 π ≈ 57.3° (to 3 s.f.)

and 1° = _____ 3602 π ≈ 0.0175 rad (to 3 s.f.)

The use of radians makes it straightforward to calculate some basic elements of circles with the general formulae shown in Table 1.6 , where the angle θ is measured in radians.

Key terms Degree (symbol: °) – one degree is ___ 3601 th of a complete circle. A complete circle contains 360°. Radian (symbol: rad or c) – one radian is the angle subtended at the centre of a circle by two radii of length r that describe an arc of the same length r on the circumference. A complete circle contains 2π rad. subtend – to form an angle between two lines at the point where they meet.

DRAFT

14 Engineering Principles

When plotted graphically, both the sine and the cosine functions generate periodic waveforms. Both functions vary between a maximum of 1 and a minimum of −1 and so are said to have an amplitude of 1. Both functions have a period of 360° or 2π radians, after which the cycle repeats. The tangent function does not generate a smooth waveform, although the graph is still periodic with a period of 180° or π radians. Some values for the trigonometric ratios are given in Table 1..

▸ ▸ Table 1.7 Values of the trigonometric ratios for angles between 0° and 360°

θ ° θ rad sin θ cos θ tan θ 0 0 0 1 0 30 0.52 0.50 0.87 0. 60 1.05 0.87 0.50 1. 90 1.57 1.00 0 ± ∞ 120 2.09 0.87 −0.50 −1. 150 2.62 0.50 −0.87 −0. 180 π 0 −1.00 0 210 3.67 −0.50 −0.87 0. 240 4.19 −0.87 −0.50 1. 270 4.71 −1.00 0 ± ∞ 300 5.24 −0.87 0.50 −1. 330 5.76 −0.50 0.87 −0. 360 2 π 0 1.00 0

sine and cosine rules The basic definitions of the trigonometric functions sine, cosine and tangent only apply to right-angled triangles. However, the sine and cosine rules can be applied to any triangle of the form shown in Figure 1.. ▸ ▸ The sine rule: ______^ a sin A

= _____ b sin B

= ______ c sin C

▸ ▸ cosine (cos), where

cos θ = _____adj hyp

▸ ▸ tangent (tan), where

tan θ = _____opp adj From these definitions it can also be deduced that

tan θ = ______cossin^ θθ

Graphs of the trigonometric functions

0.5π

π

1.5π

2 0.5 2 π rad 2 1.

y

x

▸ ▸ Figure 1.9 Graph of y = sin θ

0.5π

π

1.5π

2 0.5 2 π rad 2 1.

y

x

▸ ▸ Figure 1.10 Graph of y = cos θ

0.5π

π

1.5π

(^2202) π rad 240 260 280

y

x

▸ ▸ Figure 1.11 Graph of y = tan θ

P A U s E P O I N T

Check that you can set up and use a scientific calculator in both degree and radian modes. Hint (^) Look for the set-up screen on your calculator and select Rad. You will find that the D (for degrees) usually displayed at the top of the screen changes to R (for radians). Before carrying out any work involving trigonometry, you should check that you are using the appropriate setting on your calculator. Use some simple examples to show what might happen if you get it wrong.

Extend

DRAFT

16 Engineering Principles

At any fixed instant in time ( t ), the phasor will have a phase angle ( θ ) and its vertical component will be equal to the instantaneous voltage ( v ) on the corresponding sine wave. So the instantaneous voltage ( v ) at any point on the waveform is v = V sin θ In terms of angular velocity, this gives v = V sin( ωt ) This relationship is true when the waveform begins its cycle when t = 0. However, it is common to have a waveform that is said to lead or lag the standard waveform. This phase difference (see Figure 1.16 ) is expressed as an angle ( Φ ), so that v = V sin ( ωt + Φ ) The angular velocity of the phasor ( ω ) is related to the frequency of the waveform ( f ) by ω = 2π f

P A U s E P O I N T Explain the difference between a vector and a phasor.

Hint (^) Consider the different types of systems they are used to describe. Use a diagram to explain the relationship between a phasor and the sinusoidal waveform it describes.

Extend

Mensuration It is very common for an engineer to need to calculate the surface area or volume of three-dimensional shapes (see Figure 1.17 ); for example, to determine the number of tiles required to line a swimming pool or to find the capacity of a cylindrical storage tank. There are several important formulae that can help you do this (see Table 1.8 ).

Cylinder Sphere Cone

r

r

r

l h h

▸ ▸ Figure 1.17 Some commonly encountered regular solids and their dimensions

▸ ▸ Table 1.8 Standard formulae for the surface area and volume of some regular solids

Regular solid Curved surface area ( CSA ) Total surface area ( TSA ) Volume Cylinder 2 π rh 2 π rh + 2π r^2 = 2π r ( h + r ) π^2 rh Cone π rl π r^2 + π rl = π r ( r + l ) __^13 π r^2 h

Sphere 4 π r^2 __^43 π r^3

Often, an apparently complex object can be broken down into a series of regular solids.

V

▸ ▸ Figure 1.16 Phasor representing the sinusoidal waveform v = V sin ( ωt + Φ ), where Φ is the phase shift from the standard waveform

DRAFT

Learning aim A UNIT 1

Engineering Principles

Assessment practice 1.

1 A small boat is capable of a maximum velocity in still water of v km h−1. On a journey upriver against a current with velocity v c km h−1, the boat travels at its full speed for 2 hours and covers 16 km. On a journey back downriver with the same current flow, the boat travels at full speed for 1 hour and 20 minutes and covers 18 km. Establish and then solve a pair of simultaneous linear equations to determine the maximum velocity of the boat v and the current v c. (4 marks)

2 Scrap transformers have been collected for recycling. In total, 203 transformers weigh 403.4 kg. It is found that the transformers come in two different types, which weigh 1.8 kg and 2.3 kg, respectively. Establish and then solve a pair of simultaneous linear equations to determine the number of each type of transformer collected. (4 marks)

3 The motion of two vehicles is described by the linear equations 27 = v – 3 t 13 = v – 4 t By solving this pair of simultaneous equations determine: a) the time t at which both vehicles will be travelling with the same velocity b) the corresponding velocity v at that point. (2 marks)

4 The time t (years) taken for a radioactive isotope contained in stored nuclear waste to decay to 5% of its original quantity is given by the equation 5 = 100 × 2 ____^1622 − t Solve the equation to find the time t. Show evidence of the use of the laws of logarithms in your answer. (2 marks)

5 An engineer has been given a drawing (see Figure 1.18 ) of a triangular plate with minimal dimensional detail.

A B

C

6 m

▸ ▸ Figure 1.18 Engineering drawing of a triangular plate

Calculate the lengths of the sides A , B and C. (4 marks)

6 A pair of dividers used for marking out has legs that are 150 mm long, as shown in Figure 1.19. The angle α between the legs can be set to a maximum of 60° by rotating the adjustment screw. a) Use the sine rule to calculate the maximum distance that can be set between the points of the dividers when α = 60°. b) Use the cosine rule to calculate the value of angle α when the points of the dividers are set to a distance of 38 mm. (4 marks)

7 The displacement s of an accelerating body with respect to time t is described by the equation

s = t^2 − 7 t + 12 Using factorisation, solve this equation to determine the values of t for which s = 2. (3 marks)

8 Use the formula for solving quadratic equations to determine the radius r of an enclosed cylinder which has a total surface area ( TSA ) of 12.25 m^2 and a height h of 1.2 m, where TSA = 2π rh + 2π r^2 (4 marks)

150 mm^ a

▸ ▸ Figure 1.19 A pair of dividers

DRAFT

UNIT 1

Engineering Principles

Learning aim B

One way to add or subtract vector quantities is to split each vector into components acting in specific, perpendicular directions. A force acting in any direction can be resolved into a vertical and a horizontal component. The horizontal components of multiple forces can be simply added to calculate a single horizontal force. Similarly, the vertical components of multiple forces can be simply added to calculate a single vertical force. These can then be recombined into a single resultant force that represents the combined effect of all the original individual forces.

Splitting any vector into its horizontal and vertical components will involve the use of trigonometry and/or Pythagoras’ theorem.

Figure 1.24 shows how a single force F can be resolved into a horizontal component Fh and a vertical component Fv.

θ Fh

Fv F

▸ ▸ Figure 1.24 Force vector resolved into vertical and horizontal components

In the arrangement shown in Figure 1.24 : Fv = F sin θ Fh = F cos θ ___^ Fv Fh

= tan θ

F^2 = Fv^2 + Fh^2

When several forces act on a body, apply the same principles to each of the forces.

Key term Resultant – the force that represents the combined effect of all the forces in a system.

A free body diagram is a sketch containing just the forces acting on the system or the part of the system that you are interested in, as shown in Figure 1..

F 2

Fg

F 1

θ 1 θ 2

▸ ▸ Figure 1.22 Free body diagram showing the forces acting on the suspended mass (which is represented by a point)

A vector diagram is a sketch in which the lengths of the lines representing the force vectors correspond to the magnitudes of the respective forces. The vectors may also have been rearranged to form a triangle of forces as shown in Figure 1.23. (Similarly, larger numbers of force vectors can be arranged into a polygon of forces.)

F 2

Fg F 1 θ 1

θ 2

▸ ▸ Figure 1.23 Vector diagram with the three forces rearranged into a triangle

Resolution of forces in perpendicular directions

When analysing a system in which multiple forces act on a body, it may not be obvious what the overall effect of those forces will be. What we need is a method of adding the forces together to find their combined effect, known as the resultant force acting on the body. We can combine forces graphically by drawing a vector diagram, but here we will consider an alternative approach using the resolution of forces.

P A U s E P O I N T Explain the difference between a space diagram, a free body diagram and a vector diagram. Hint (^) Draw an example of each for the same system. How will the diagrams differ between concurrent and non-concurrent systems of forces?

Extend

DRAFT

20 Engineering Principles

Conditions for static equilibrium A system of forces in static equilibrium will have no tendency to move because all the forces acting in the system are perfectly balanced. The resultant force for a system in static equilibrium is zero. This also means that the horizontal and vertical components of all the forces are such that: ΣF v = 0 ΣF h = 0 For a system of concurrent forces, these equations are enough to define static equilibrium. However, in a non- concurrent system of forces, where the lines of action do not intersect at the same point, there could be a tendency for the system to rotate. The moment of a force describes the tendency of a force to produce rotation about a pivot point or centre of rotation.

The moment of a force ( M ) about a pivot is calculated by multiplying the magnitude of the force ( F ) by the perpendicular distance ( s ) from the pivot point to the line of action of the force. M = Fs So, for a system of non-concurrent forces, there are three conditions that must be met for static equilibrium: ΣF v = 0 ΣF h = 0 ΣM = 0

Worked Example

Find the equilibrant for the system of non- concurrent forces shown in Figure 1..

100 cm^ A

100 cm

53 cm

5 kN 6 kN

2 kN

▸ ▸ Figure 1.27 System of three non-concurrent coplanar forces

Key terms Equilibrant – the force that when applied to a system of forces will produce equilibrium. This force will be equal in direction and magnitude to the resultant but have the opposite sense. Moment – the tendency of a force to rotate the object on which it acts.

Worked Example

Find the resultant and equilibrant for the system of concurrent forces shown in Figure 1..

30 N

20 N

10 N

A

B

C

▸ ▸ Figure 1.25 System of three concurrent coplanar forces solution First, establish the convention that the positive vertical direction is up and the positive horizontal direction is to the right. Resolve each vector into its vertical and horizontal components using the formulae on the previous page. Find the sum of the vertical components (the Greek letter Σ is shorthand notation for ‘sum’): ΣF v = A v + B v + C v = 10 sin 40 + 20 sin 25 − 30 = −15.12 N Find the sum of the horizontal components: ΣF h = A h + B h + C h = −10 cos 40 + 20 cos 25 + 0 = +10.47 N Figure 1.26 shows that the resultant force F r is acting in the positive sense in a direction θ ° below the horizontal, where: tan θ = 15. 12_______10. 47 = 1.

θ = 55.3° (using unrounded values) The magnitude of the resultant can be calculated using Pythagoras’ theorem: F r = √

---

15. 12 2 + 10. 47 2 = 18.39 N So the resultant of this system of forces has magnitude 18.39 N and acts in a direction 55.3° below the horizontal with a positive sense. The equilibrant is the force required to bring this system into equilibrium, so it also has magnitude 18.39 N and direction 55.3° below the horizontal, but it will have a negative sense. ▸ ▸ Figure 1.26 Vector diagram showing how the sums of the vertical and horizontal components are combined to give the resultant

Σ Fh 5 10.

Σ F (^) v 5 15.

Fr 5 (^) 18. (^) N

DRAFT