2An old riddle!Which is heavier, a pound of lead or a pound of feathers?People who are confused by this riddle do so because they do not understand the difference between mass and density,
3Density.Density depends upon the mass of the atoms inside a substance and how closely packed together they are.We can calculate density using:Density = massvolumeρ = mVS.I. units are kgm-3 but density can also be given in gcm-3
4Examples The density of water is 1000kgm-3 This means that a volume of 1m3 has a mass of 1000kgThe density of ice is 917kgm-3. Ice floats on water because its density is less.The densities of solids and liquids overlap but gases have much lower values of density. This is due to the larger spacing of the moleculese.g. Air has a density of just 1.2kgm-3
5CalculationsQ Calculate the mass of air in a typical lab measuring 8 x 6 x 3mA Volume = 144m3mass = density x volume= 1.2 x 144= 172.8kg
6Measuring densityA regular shaped object: Calculate its volume after measuring its dimensions using a ruler, micrometer or vernier callipers.Measure its mass using an electric balance and then calculate its density using ρ = mV
7An irregular shaped object: Find the volume of the object using the displacement of water.(This method was first used by Archimedes when asked to checkwhether the King’s crown wasmade of pure gold)Measure the mass using an electric balance and then calculate the density
8A liquid: Measure its volume using a measuring cylinder. Find the mass of the liquid and the cylinder and then subtract the mass of the empty cylinder to find the mass of the liquid.Calculate the density of the liquid usingρ = mV
9Further exampleQ The crown that Archimedes had to investigate contained 12.5 x 10-5m3 of gold and 2.2 x 10-5m3 of silver. Calculate its mass and density. (Density of gold is 19300kgm-3 and density of silver is 10500kgm-3).A Mass of gold= ρ x V = x12.5 x 10-5= 2.41kgMass of silver = ρ x V = x 2.2.x 10-5= 0.23kgTotal mass = 2.64kgTotal volume = 14.7 x 10-5m3Density of alloy = m = = 18000kgm-3V x 10-5
10Stretching Springs Measure the original length of the spring Set up the apparatus and apply a force of 1N. This gives a tension of 1N in the spring.Record the new length and find the extension by subtracting the original length.Repeat until you have a total force of 8NPlot a graph of Force (F) against extension (ΔL).
11Hooke’s lawThe graph should be a straight line through the origin showing direct proportion.Hooke’s law states thatthe force applied to a springis directly proportional to itsextension provided it doesnot pass the elastic limit.Force = constant x extensionF = k(ΔL)(Springs can be used to measure forces using a Newton meter)
12Spring Constantk = spring constant or stiffness constant and is found from the gradient of the graph of F against ΔLA typical spring used in school has k = 50Nm-1The stiffer the spring,the greater the value of kSpring X is stiffer thanspring YF/ NExtension / mXY
13ExampleQ A spring has a length of 20mm and stretches to a length of 60mm with a load of 2N. Calculate its spring constant.F = kΔLk = F = = 50Nm-1ΔL 40 x 10-3Q Calculate the length of the spring when a load of 7N is added to itΔL = F = 7 = 0.14mkNew length = = 0.16m
14Elastic limitIf the spring is overstretched it does not return to its original length. It is permanently distorted.The elastic limit is the point beyond which the material is no longer elastic and will not return to its original length when the load is removed.Fx x = elastic limitΔL
16Springs in parallelRepeat the experiment for stretching a spring but use 2 springs (P and Q) in parallel.Calculate the gradient of the graph to find the combined spring constant (k).Theory (see P165) shows that:k = kp + kQ where kp and kQ are the spring constants for springs P and Q ( the springs are stiffer and more difficult to stretch)Find the percentage difference between your value of k and the theoretical value
17Springs in seriesRepeat the experiment for stretching a spring but use 2 springs (P and Q) in series.Calculate the gradient of the graph to find the combined spring constant (k).Theory (see P165) shows that:1 = 1 + 1k kP kQwhere kp and kQ are the spring constants for springs P and Q (the springs are now easier to stretch)Find the percentage difference between your value of k and the theoretical value
18Energy stored Work is done when stretching a spring and this equals the elastic potential energy stored in it.Since work = force x distance= area under graph= 1 F ΔL = 1 kΔL2Elastic potential energystored in a stretched spring = 1 kΔL22F/NΔLExtension/m
19ExampleQ How much energy is stored in a spring which is 32cm long and stretches to 54cm when a force of 6.0N is applied to it.A ΔL = 22cm = 0.22mEp = 1 FΔL2= 1 x 6.0 x 0.22= 0.66J
20Tensile Stress Tensile stress = tension in wire x-sectional area of wireσ = TA Units:Nm-2 or PaStress is a property of the material rather than the individual object being stretchedQ Calculate the stress in a wire of x-sectional area 4 x 10-8m2 when it is stretched with a force of 200NA 5 x 109 Pa
21Tensile strain Tensile strain = extension of wire original length Since it is a ratio it has no units and is once again a property of the material.Q Calculate the strain in a wire of length 2.3m which stretches by 1.5mmA 6.5 x 10-4
22Stretching different materials A rubber band and a polythene strip can be stretched using the same apparatus as used with the spring.A copper wire can be stretched along a bench.In all cases plot a graph of stress against strain.The shape of a stress-strain graph is the same as the shape of a force-extension graph but is common to the material used and not the individual wires.
23Examples Copper is an example of a ductile material. BrittleStressStrainCopper is an example ofa ductile material.Initially it obeys Hooke’s lawand returns to its original lengthif the load is removed.There is then a large increase in its extension and if the load is removed it does not return to its original length.Glass is an example of a brittle material
24More definitionsLimit of proportionality: Point beyond which Hooke’s law is no longer obeyed.Elastic limit: Point beyond which the material is no longer elastic and will not return to its original length when the load is removed.Yield point: Point at which there is marked increase in extension
25Elastic deformation: The material will return to its original shape when the load is removed. Plastic deformation: The material is permanently stretched when the load is removed.Ductile materials like copper can be drawn into wires and undergo large plastic deformation.Brittle materials fracture suddenly without any noticeable yield e.g. glassStrong materials require a high stress to break themUltimate tensile stress: maximum stress that can be applied. This is sometimes called the Breaking stress
26Typical graph E = elastic limit Y = yield point P = limit of proportionalityUTS = ultimate tensile stressS = wire snaps
27A is a brittle materialwhich is also stronge.g glassB is a strong materialwhich is not ductilee.g steelC is a ductile materiale.g copperD is plastic material
28Young modulus of elasticity Young modulus = stressstrainE = σε= T/ΔLA LTLAΔLE is measured in Pa, the same as stress.It is a measure of how difficult it is to change the shape of the material and can be found from the gradient of a stress-strain graph
29ExamplesQ1 The Young modulus for cast iron is 2.1 x 1011Pa and is snaps when the strain reaches What is the stress needed to break it?A 1.1 x 108PaQ2 A wire made of a particular material is loaded with a load of 500 N. The diameter of the wire is 1.0 mm. The length of the wire is 2.5 m, and it stretches 8 mm when under load. What is the Young Modulus of this material?A 2.0 x 1011Pa
30Experiment to measure Young modulus Measure the original length of the wireMeasure the diameter of the wire at different points along its length using a micrometer and find is average diameter. Calculate the x=sectional area of the wire.Load the wire as before and measure its extension.Plot a stress-stain graph and measure its gradient = Young modulus
31Area under stress-strain graph Area under graph =average stress x strain= F x ΔLA L= work donevolumeThe area gives us the energy stored per unit volume and is a measure of the toughness of the material. A tough material needs a large amount of energy to break it.
32Loading and unloading materials A material can be stretched as before by increasing the load on it and the extension measured.It can the unloaded gradually and the extension measured again.A graph can then be plotted showing both the loading and unloading.The shape of this graph will depend upon the type of material used and the size of the load.
33Metal wireThe loading and unloading give the same straight line provided the elastic limit has not been passed.The atoms in the metal are pulled slightly further apart and then return to their original positions when the load is removed.If taken beyond this point the unloading will give a line parallel to the first but leaving a permanent extension.
35Rubber band The rubber band returns to its original length when the load is removed butthe unloading curve isbelow the loading curve.The molecules are like longstrands of tangled spaghetti that become untangled when the rubber is stretched and the become tangled again when the load is removed.Shape.BCAStrainStress/ N
36internal (heat) energy by the rubber band when it is unloaded. The shaded arearepresents the energystored in the form ofinternal (heat) energyby the rubber band whenit is unloaded.The rubber used in car types should give a small area to reduce the energy lost by heat.Shape.BCAStrainStress/ N
38PolythenePolythene does not return to its original length when the load is removed.Like rubber, polythene is a polymer, but when it is stretched new cross links form which prevent the molecules from returning to their original state.