energy to air resistance ratio, and so be more likely to hit its target. To make the
game of darts fair rules exist on the construction of darts! Note the data has not ...
Exploring the range of materials Question 10S: Short Answer 1. Properties important for food packaging materials include: cost, weight, hygiene, whether recyclable, durability nature of materials and content. Steel coated with tin (pasta, syrup, beans) glass (pickled cabbage, vinegar, coffee) low density polyethylene (LDPE) (thyme, bread, tube of dessert sauce, oil) polypropylene (lids except pickled cabbage & vinegar which are steel coated with tin), cardboard/paper/plywood (Oxo, sugar, cheese, honeyflakes, lo-salt), polyester (PET) (peppercorns). 2. Remind students of the formula density = mass/volume. They could do a rough estimate of volume either by eye, or by measuring diameter and length, assuming a square cross section. If volume was estimated to be 2 cm3, then the mass of a dart would be 38.2 g. The corresponding mass for the steel dart 15.6 g. The tungsten dart might be expected to have a higher kinetic energy to air resistance ratio, and so be more likely to hit its target. To make the game of darts fair rules exist on the construction of darts! Note the data has not been given in SI units here, to make this quick calculation easier. 3. Aluminium is light (density 2.71 g cm–3) and a good conductor of heat which makes it useful for saucepans. Its low density also makes it useful in aircraft parts (alloyed with magnesium). 4. High pressure steam could corrode the blades. 5. Stress wide range of textiles used in clothing – polyester, Lycra, cotton, nylon, wool, leather, acrylic. Mention new materials like Tencel. Some students may not like to wear fur or leather because of animal welfare concerns. 6. Pineapple fibres are clearly seen by scraping the outer green layer from the leaves with a sharp knife. Better known natural textiles are silk, cotton, wool, linen, viscose, straw, leather and the wood pulp padding of disposable nappies. 7. Cork makes a useful hat because it is light, waterproof and keeps you cool. Today it is used mainly to stopper wine bottles, for flooring and soundproofing. 8. Many office buildings have big glass frontages – you could also look how glass is used in your school/college to create a feeling of light and space. Perhaps discuss safety features (toughening) and where glass is not really glass e.g. double glazing, which is Perspex (a polymer). Another modern use of glass is for optical fibres used in telecommunications. 9. Steel is magnetic and so can be lifted from a pile of non-ferrous material by an electromagnet. This gives a chance to introduce the recycling concept, perhaps discuss local recycling schemes and to what extent the students themselves/school/college recycle materials. 10. The lorry and bus tyres are 75% natural rubber, and the car tyre 100% synthetic on the outside, 50% natural on the inside. The material selected reflects the demands made. Lorries and buses must bear a heavier load than a car. This provides an opportunity to stress that natural is sometimes best, when it comes to materials, and there is nothing, as yet, to match the performance of natural rubber. But we cannot use natural rubber all the time because demand exceeds supply.
Analysis of tensile testing experiments 1
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Question 40S: Short Answer 1. Perspex 1260 N, polypropylene 970 N, aluminium 300 N, brass 1080 N, mild steel 2300 N. 2. Perspex 3070 N, polypropylene 1230 N, aluminium 1470 N, brass 4100 N, mild steel 3440 N. 3. Steel – the gradient of the straight part of the line is steeper and the Young modulus is the slope.
Calculations on stress, strain and the Young modulus Question 45S: Short Answer 1. strain
extension original length
25 mm 75 mm
strain = 0.33 This is sometimes expressed as a strain of 33%. 2. Strain has mm/mm. These cancel out to give a quantity with no units. 3. stress
force area
so F = stress A = 109 N m–2 10–8 m2 F = 10 N 4. stress
force area
so area
force stress
15 kN 75 MN m – 2
area = 2 10–4 m2 A
2
d 2 4
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so d
d 2 4
d
4 2 10 4 m 2 3.14
d = 1.6 10–2 m or 1.6 cm 5. A
d 2 4
3.14 (50 10 3 m ) 2 4
=1.96 10–3 m2 stress
force area
4 kN 1.96 10 -3 m 2
stress = 2.0 106 N m–2 or 2.0 MPa 6. A
d 2 4
3 .14 ( 3 .6 10 3 m ) 2 4
= 1.0 10–5 m2 stress
force area
20 N 1.0 10 – 5 m 2
stress = 2 MPa 7. The stress in the suspension cable and fishing line is the same. The Young modulus for nylon is 2–4 GPa, while the Young modulus for steel is 200 GPa. So the strain in a nylon fishing line would be very much higher than that in the steel cable. The roadway must be held horizontal. If the suspension cables were nylon they would sag. 8. A
3
d 2 4
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3.14 (3.6 10 3 m ) 2 4
=1.02 10–3 m2 E
F A l l
so l
F l AE
20 10 –3 N 25 m 1.02 10 – 3 m 2 2 1011 N m – 2
l = 2.5 10–3 m or 2.5 mm
Measuring the Young modulus Question 50S: Short Answer
1. A = r 2 = (0.12 mm)2 = 4.5 10-2 mm2 = 4.5 10-8 m2 2. Stress = F / A = 10 N / 4.5 10-8 m2 = 2.2 108 N m-2 or Pa 3. Strain = 3.5 mm / 1600 mm = 2.2 10-3 4. Young modulus = stress / strain = 2.2 108 N m-2 / 2.2 10-3 = 1.0 1011 N m-2 5. Area is 4 times less, so stress is 4 times greater. Stress = 4 2.2 108 N m-2 = 8.8 108 N m-2 6. Stress is 4 times greater, length is same, so extension is 4 times greater. Extension = 4 3.5 mm = 14 mm 7. Extension = 3.5 mm / 2 = 1.8 mm to two significant figures 8. Length = 1.6 m 2 = 3.2 m
Electrical properties Question 70S: Short Answer 1. R
4
l 2 A and A r
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3 2 so A 3 .14 (0 .5 10 )
1 4
m2
=1.96 10-7 m2 l
then
RA
6 1.96 10 3 m 2 5 10 7 m
l = 2.4 m 2. A
d 2 4
3 .14 (0.46 10 3 m ) 2 4
A = 1.66 10-7 m2 A l
G
9.1 10 5 S m -1 1.66 10 7 m 2 7.3 m
G = 0.02 S 3. R
l A
1.69 10 8 m 0.01 m 35 10 6 m 0.16 10 - 2 m
R = 3.0 10-3 4. A
d 2 4
A 3.14 (0.05 10 3 ) 2
1 4
m2
=1.96 10-9 m2 l
A G
1.0 10 7 S m -1 1.96 10 9 m 2 0.01 S
l = 2.0 m 5
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5.
V r 2 l Al
1.96 10–9 m2 2.0 m V = 3.9 10–9 m3 6. m = V = 3.92 10–9 m3 21.45 103 kg m–3 m = 8.4 10-5 kg cost = 8.41 10–5 kg 3.5 104 GBP kg–1 cost = £2.94 7. A
l R
9.81 10 8 m 0.81m 100
A = 7.95 10-10 m2 t
A
7.95 10 10 m 2 4.0 10 6 m
t = 0.20 mm 8. V = A l = 7.95 10-10 m2 0.81 m V = 6.44 10-10 m3 m V
= 21.45 103 kg m-3 6.44 10–10 m3 m = 1.38 105 kg cost = 1.38 10–5 kg 3.5 104 GBP kg–1 cost = £0.48 or 48 p
Resistivity and conductivity calculations Question 80S: Short Answer
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1. 6.25 107 S m–1 2. 20 3. 30 m 4. Wire A 5. The same 6. 20 7. 0.028 8. 138 m–1, 1.1 10–6 m
Conductance and conductivity Question 100S: Short Answer 1. A = r 2 = (0.25)2 = 0.20 mm2 = 2.0 10-7 m2 2. G = A / L = (5.8 107 S m-1 1.96 10-7 m2 ) / 25 m = 0.45 S 3. R = 1 / G = 1 / 0.45 S = 2.2 4. Conductivity = 1 / = 1 / 1.1 106 m = 9.1 105 S m-1 5. A = 2.0 0.25 mm = 0.50 mm2 = 5.0 10-7 m2 6. G = A / L = 9.1 105 S m-1 5.0 10-7 m2 / 2.5 m = 0.18 S 7. G = 2 0.18 S = 0.36 S 8. R = 1 / G = 1 / 0.36 S = 2.8
Making estimates about the mechanical behaviour of materials Question 10E: Estimate 1. Mass of a car is more than 100 kg, maybe up to 1000 kg. Guess 1000 kg, or weight 10 000 N. Cross-section of cable to be more than weight / tensile strength, 10000 N 10 9 Pa
10 5 m 2 10 mm 2
so diameter d given roughly by d 2 / 4 = 10 mm2, and is of the order 3 mm. 2. If the tensile strength of nylon is of the order 100 MPa, then it can carry a load of up to 100 MPa cross-sectional area = 108 Pa ( / 4) (10-3 m)2 100 N or 10 kg weight. 3. Tension T in a guitar string is of the order 100 N (10 kg weight). Diameter of the order 1 mm, with 7
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cross-section A of order 10-6 m2. Young modulus E = stress / strain, so strain = stress / E = 100 N / (10-6 m2 1011 Pa) = 10-3. If the string is of order 1 m long, extension = 1 mm. 4. Suppose the Young modulus of rope is of the order 1 GPa. If the rope diameter is 10 mm, so having a cross-section about 10-4 m2, and the climber's weight is 100 kg weight, or 1000 N, then the stress is 107 Pa. The strain, is given by stress / Young modulus = 107 Pa / 109 Pa = 10-2. If the rope is 10 m long, the extension is 10-2 10 m = 0.1 m.
Making estimates about the electrical behaviour of materials Question 20E: Estimate 1. Resistance is 1 km-1. Use R
l A
with length l = 1000 m and R = 1 . Cross-section of cable to be at least A
l 10 8 m 1000 m R 1
= 10-5 m2 = 10 mm2. Diameter d given roughly by d 2 / 4 = 10 mm2, so is of the order 3 mm. 2. Estimate diameter of fine wire as 1 / 10 mm. Cross-section is then of order 10-2 mm2, or 10-8 m2. Length needed is then l
AR 10 -8 m 2 100 100 m 10 -8 m
3. Guess insulator is 50 cm long, diameter 10 cm. Resistance is then R
l 0.5 m 1010 m A 0.1 m2
which is between 1011 and 1012 . 4. Guess that the film is 10 times narrower than its length and 100 times thinner. Cross-section is then 0.1 m 0.01 m = 0.001 m2. Resistance is then R
l 10 -6 m 10 8 m A 0.001 10 -12 m 2
= 10 Resistivity of silicon is 1010 times larger, so resistance is 1011 .
The Bronze Age Question 20C: Comprehension 8
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1. Smelting – a chemical method for extracting a metal from its ore, alloy – a mixture of two or more pure metals, casting – shaping a metal by pouring it into a mould in the molten shape, hammering and working – manipulating a hot metal mechanically with tools and shaping it into an object. 2. It can be shaped into objects, of both utility and beauty. 3. Stone. 4. It is soft and cannot be made into a hard cutting edge. 5. An alloy of copper and tin. 6. It is harder than copper or tin alone and stronger than both. 7. In combination with the discovery of ceramics and fixed writing, it marked the start of Chinese civilisation and we have the evidence in the form of artefacts they left behind, like the Shang bronzes. These bronzes formed an art which grew spontaneously from its own technical skill. 8. It became universal, its use spreading across the globe, and it replaced stone for a multitude of applications, as plastic replaced metal, wood and paper in the 20th century. 9. The Iron Age.
Portraits in plastic Question 30C: Comprehension 1. Ordinary polystyrene, high impact polystyrene, expanded polystyrene. 2. Yoghurt pots, fridge interiors, eye-shadow compacts, toys, disposable drinking glasses, tape cassette and CD cases, envelope windows. 3. By the addition of up to 10% polybutadiene or styrene-butadiene copolymer. 4. It is resistant to fats and oils; the ordinary form is not. 5. High impact is opaque and stronger. 6. Good insulator, good shock protection, low density 7. Unsightly litter, CFCs used in its manufacture. 8. Assessing the cost in terms of energy and resources used in a material from its manufacture, through its use and disposal, and setting these against the benefits the material offers. 9. It has improved; CFCs no longer used in its manufacture, it can now be recycled, energy can be extracted from material not recycled. 10. Packaging, cassette cases, plant pots, manufacture of building materials.
Stress, strain and the Young modulus Question 50D: Data Handling
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1.
stretching copper 5000 4500 4000 3500 3000 2500 2000 1500 1000 500 0 0
1.0
0.5
1.5
extension / m
2. The values of stress and strain for copper are Extension x / mm
Force / N
Stress / Pa
Strain
0.00
0.0
0
0
0.15
1.0
1.32 10
0.35
2.0
2.65 10
0.51
3.0
3.97 10
0.66
4.0
5.30 10
0.86
5.0
6.62 10
1.01
6.0
7.95 10
7
0.85 10
-4
7
1.98 10
7
2.83 10
7
3.67 10
7
4.80 10
7
5.65 10
-4 -4 -4 -4 -4
using cross-sectional area A = d 2 / 4 = 7.85 10-9 m2, with d = 0.10 mm and original length 1.79 m. 3. The ratios of stress to strain are:
10
Stress / Pa
Strain
0
0
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(Stress / strain) / GPa
Stress / Pa
Strain
(Stress / strain) / GPa
7
0.85 10
7
1.98 10
7
2.83 10
7
3.67 10
7
4.80 10
7
5.65 10
1.32 10 2.65 10 3.97 10 5.30 10 6.62 10 7.95 10
-4
156
-4
134
-4
141
-4
144
-4
138
-4
141
The values vary within the range 134 GPa to 156 GPa, reflecting the uncertainties in the extension. The slope of the best fit straight line through the origin is 6.0 N mm-1, giving a value of 142 GPa for the Young modulus. 4. Average ratio stress / strain = Young modulus E = 142 GPa. 5. The value is systematically higher than the data book value 130 GPa. A possible source of error is an underestimate of the diameter of the wire, which is very difficult to measure accurately. 6.
stretching nylon 0.70 0.60 0.50 0.40 0.30 0.20 0.10 0 0
10
20
30 40 extension / mm
7. The values of stress and strain for nylon are:
11
Stress / MPa
Strain
0.00
0.00
12.7
0.0042
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50
60
70
Stress / MPa
Strain
using cross-sectional area A = d 2 / 4 = 0.0078 mm2 and original length 2.62 m. 8. The ratios of stress to strain are: Stress / MPa
Strain
(Stress / strain) / GPa
0.00
0.00
12.7
0.0042
3.03
25.5
0.0080
3.18
38.2
0.0111
3.45
50.9
0.0156
3.25
63.7
0.0191
3.34
76.4
0.0233
3.28
The ratios are approximately constant, consistent with the graph being a straight line through the origin.
9. The average value of the ratios stress / strain = 3.25 GPa. The Young modulus is, to two significant figures, E = 3.3 GPa.
Resistivity and conductivity Question 60D: Data Handling 1.
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The resistance is directly proportional to the length, since the graph is a straight line through the origin. 2. The ratios of resistance to length are: Resistance / length 1.36 10 m
-1
1.36 10 m
-1
1.36 10 m
-1
6 6 6
1.35 10 m 6
-1
The ratio of resistance to length may be expected to be constant if the resistivity is constant in the relationship ρ=
RA L
is constant, since R / L is constant. The average value is 1.36 106 m-1. 3. Cross-sectional area A = 30 m 2.4 m = 72 m2 = 72 10-12 m2. 4. The average value of R / L = 1.36 106 m-1, with area A = 72 10-12 m2, then = 1.36 106 m-1 72 10-12 m2 = 9.8 10-5 m. 5. The conductance will be directly proportional to the area if the conductivity in the relationship σ=
13
GL A
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is constant, so that the ratio G / A is constant. 6. The values of cross-sectional area and conductance are: Cross-sectional area / m2
Conductance / S
72
0.44 10
144
0.87 10
216
1.31 10
288
1.74 10
-3 -3 -3 -3
The graph of conductance against area is:
variation of conductance with area 0.0020 0.0018 0.0016 0.0014 0.0012 0.0010 0.0008 0.0006 0.0004 0.0002 0 0
50
100
150 200 area / m2
250
300
350
The conductance is directly proportional to the cross-sectional area. 7. The ratios of conductance to cross-sectional area are: Conductance / area 6.06 10 S m 6
-2
6.04 10 S m 6
-2
6.06 10 S m 6
-2
6.04 10 S m 6
-2
The graph is a straight line through the origin, so that the ratio G / A will be constant. The average value of the ratio G / A = 6.05 106 S m-2. 14
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8. Since sin n
n b
and the average value of G / A = 6.05 106 S m-2 with length L = 1.68 103 m, then conductivity = 6.05 106 S m-2 1.68 103 m = 1.02 104 S m-1. 9. The two values are = 9.8 10-5 m and = 1.02 104 S m-1. Since = 1 / , we can check that 1 / = 1 / (9.8 10-5 m) = 1.02 104 S m-1. The values are consistent.
Review questions Question 90X: Explanation–Exposition 1. Mechanical (tensile strength, compressive strength, hardness, toughness, density), optical (transparency, refractive index), electrical (conductivity). There are many possibilities for uses. 2. Most modern buildings include steel, steel-reinforced concrete, glass. They may also include stone, as an exterior facing or for floors and staircases. The interior partition walls are likely to use aluminium studs faced with plasterboard. Ceramics are commonly used for toilets, though sometimes stainless steel is used. You may want to describe the materials used for electrical wiring, or for plumbing. 3. Clothing made be made from natural fibres such as cotton, linen and silk, or artificial fibres of which there are a vast selection. 4. Food packaging is easier to describe than foods themselves. The following are commonly used as packaging materials: polythene, polystyrene, glass, tin-coated steel cans, aluminium cans, cardboard, paper. You should be able to describe any of these materials using the properties listed in question 1. 5. Refer to the A–Z for the meaning of these terms. With some structures, the materials making it are considered to have failed if there is any permanent deformation; for example, a bridge, or railway tracks – here yield stress is important. With others, the deformation is less important but breaking is certainly failure; for example, a floor – here breaking stress matters. 6. If the frame were not stiff, it would certainly worry parents and children using the apparatus; for example, the height of the swings from the ground may depend on how many swings were in use! Any material which carries a variable load does this by deforming in a Hooke's Law manner: when a force is applied, it deforms to the point that its own tension (or compressional) force matches the applied force. The forces are then in equilibrium. 7.
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Ductile
plastic region
elastic
strain
Brittle
elastic
strain
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Rubber
elastic
strain
8. There are many possibilities here. Clearly it is essential to use electrical insulators and conductors appropriately, but mechanical properties and perhaps aesthetic criteria will be important too. A book such as 'How Things Work' may help you to identify the specific materials commonly selected for these uses. 9. Refer to the Advancing Physics AS student's book. 10. Since resistivity and conductance are inverses, high conductivity is the same as low resistivity. 11. First, why silicon? Silicon is one of the most commonly occurring elements in the Earth's crust, so as a raw material it is cheap. The techniques for growing ultra-pure silicon crystals, and for doping silicon with impurities to make it either n-type or p-type are well understood. Once this is done, components such as resistors, diodes and transistors can all be manufactured reliably and at a microscale. The electrical properties of the resulting miniature circuits can be controlled by voltage changes so that circuits can be made to behave as required.
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