I miss. Hello, Jackson. How are you today? Great. Good to the last. Yeah because this lesson is the last one before Christmas. Yes. So I hope you have a nice Christmas with your family. Yeah. No class. No class. That's good. Yes, you've done very well so far. So what we were going to do was finish up materials and look at some exam style questions again today and on the 20 fourth, what do you want to do when we come back to lessons? What do you want to practice? What topic do you think we should look at? Do that. Kinematic equations here. Yeah. Would you like me to give you some questions to do over your Christmas or will we have a break everybody? No, no, I want to have a break. Okay, that's fine. We'll do them when we come back. So sue at quit soe that there's nice bit with that. So then we should with suas with materials we should have paper one pretty well covered. So that's good. So reminding ourselves that we have these equations, if we need them, stress, strain, Young modulus, energy stored in a stretch, spring, etceter. So these are at the back of the paper, one paper. So. Hthat's. Yep, so exams type questions. This question is about measuring the Young modulus of steel using Searls apparatus. So we have two steel wires. We have a leveling bud. Buwe have a fixed MaaS and a variable MaaS. So you you have two long parallel identical steel wires suspended from a rigid bracket in the roof. Attached to a measuring device and microchrometer here at the start of the experiment, the leveling bubble is centered using the micrometer. So you know what a leveling bubble is. It's when you want to make sure something is exactly horizontal. So you have. Liquid here and across here. And you have to make sure that the liquid is exactly in the middle for this to be exactly horizontal. Builders use it a lot. At the start of the experiment, the leveling bubbble is centered here. The variable MaaS can be increased. Which extends the right hand wire. This then requires the micrometer to be adjusted again by exactly the extension of the wire in order to level the bubble. So you keep stretching this wire because, remember, wire is ductile, ductile. So this war gets longer, this one, you need to change the apparatus so that it's horizontal again, and the micrometer will give you the reading of how much it is extended. This enables the extension of the wito be measured accurately as the load is increased. So these are the loads we put here, and the extension can be read off there in millimeexplain. How using two parallel wires can reduce the effect of temperature changes on the accuracy of the experiment. So we have two parallel wires. How might temperature affect the accuracy of the experiment? You see felt temperature changes. Effects of temperature changes to parallel wires because if you're stretching a wire, it's like you're doing work on the wire. Do you think stretching a wire might heat us up? Stretching Yeah maybe because the hand and the wire can make friction and make and and transform some kinetic energy, Amazon Elastic potential energy to some thermal energy. So it will be so temperature will be change. But if there's two wire, maybe the energy will be hot. So the temperature will change. So the temperature change. It's the same across. Both wars. So we're stretching both wires. So they expand together, they stretch together. So this discounts the temperature change plotograph of load against extension using the data provided in the table on the right above. Now, if you're ever asked to plot a graph, you must do it carefully, load against extension. So we can put load here in kilograms and extension in millimeters. And then you plot the graph zero, one, two, three, four, five. And 2.5. So 0.51, 1.52. So this is 2.5. So I'm just going to plot the last point roughly. So I must do a nice small cross. So our graph will look something like that. Determine the grthere's four Marks for plotting a graph. So sensible axes, correct plots and labeling both axes with units and a line of best fit. So that's where you get four Marks. What would the gradient of my graph be? Actually it should be 2.5 not two point. It should be because I've done it. I had done this already, but it seems to have delleaa dish 2.47, which is nearly 2.5. So what would be the gradient of this graph? How would I calculate this? Jackson? The gradient yes that is 2.5 and five so 2.5 mm 2.5 times ten users -3m ingradients with L X. Minthree, so spring constant is a gradient spring constant equal to force divided by extension, so that is five times nine. Maybe your gradients five diviversibility. 2.5 1010 to the minus three, 2000 nution or kilograms per memillimeter to per meter. Yeah. The wire has a diameter of this and a length of this. What is the cross secual area of the wire? Cross section in the 0.6 divided by two, 0.3 times ten plus the minus three. Square. Times five I 2.87, times 2.83 times ten to the minus seven. Times pi. So 2.8I ten to the minus seven. And what would be a suitable unit for the cross sectional area. So the cross sectional area of the wire. What's area measured in? What unit is used for area? It's given in the question meters squared isn't it meters ter squared Yeah I forgot say you needs Yeah. Show that the Young modulus of steel is about 210 gigapascals. Show that the Young modulus, Young modules equis the stress divided by string, and the stress is the force divided by the area, and the spring is extendivided by him. Initial strength, an initial lso the diameter, we know the diameter length I. And we want to know more about so can we use okay, okay. So we can use maybe we can use this graph. Yes, absolutely. Imagine 2.5 and the force is five times 9.8 and straight area across area. Yeah winknow across area. So five times 9.81 divided by 2.8 times ten to the minus seven. And then this number divided by 2.5 times ten to the minus three divided by. Divided by 3m. 2.5 times ten to 14. A note cracks 21o kiaska. Maybe somewhat, maybe something is five times. Let me say again, times. So the force and now just to remind you Jackson, the load was given in kilograms that we want the force. So the maximum force is five times g Yeah 9.81. I don't know why they do this. Times the original length, times three. See, when you divide this by this, this inverts. So it's f times length over area, times extension 2.8 by ten, the minus seven. Ten to the minus seven. Times extension two point. 46. Nearly 2.5 mm. Let's try. So I multiplied five kilograms times G Let's see if this gives us the correct equation, the correct value, 210 by ten to the nine pascals, five times 9.8, one times three. Divided by 2.8, I tend to the minus seven times two points, four, six by ten to the minus three. So I get 2.13 by ten to the minor, by ten to the eleven. It's not minus eleven, it's eleven. Which is. 200 and it's roughly. By ten to the nine. Ascal ascoyeah. MaaS times g times length see when I. Multiply these. This inverts this bottom thing inverts. So force times length over area times extension. Okay, let's have a look at some other types of. Question and pain. We did that. This is another way to measure hooks, Young modulus, so you can stretch one wire and. Stretching the wire under tension, masses were added up to a maximum of 3.3 kilograms. Each time a MaaS was added, the extension of the wire was calculated. So this is your other way of measuring Young modulus. This graph MaaS against extension was obtained. Initially, the extension increased linearly, so you can see linear proportional relationship state, what is meant by linear increase linearly in relation to this graph and what can be concluded about the wifrom this observation. So state what is meant by increased linearly. So which bit of the graph is it talking about increased linearly? Circle the bit of graph we're talking about, Jackson. Talking about this graph for the MaaS H and extension. So maybe that is a part which over the limit of proportionality. This does this bit prove that the wire obeys hooks law? No. Is that this part? I think this part exactly. So the answer they're looking for is is the first part of the graph shows that as MaaS increases, so does extension. Extension. In direct proportion. Proving this. The war is obeying. Okay, so that's where you get two Marks. The first mark shows a direct proportional relationship, and then the second mark says it's obeying hook law. Okay, two Marks use the graph to calculate the maximum energy that the wire could store while behaving linearly. So. Behaving. Linear. How would you work that out? Again? Use the graph. We can find the maximum point in this, in this range, and we can use the energy equation e. Yeah energy ep equal to the half fx good. Elastic strain energy is a half fx. Remember well remember membered though, but it should be in your exam two a half. Effects? But again, remember, this is MaaS and the equation is force. So you have to. So let's see if we can make a triangle as the area under this bit of the graph. Up to maybe let's clean the graph up a bit. So when does it cease to be nonlinear? Probably that point. So one. 233.5. By ten to the minus two is the extension. And then how do we find the force? In this region. We're given MaaS here Jackson. We're not given force Yeah but we can we can times 9.9 point 81Yeah two point. 75, I'd say. Times 9.81. So. So we have to fit that in. What does that work out to be? Half times 2.7, five times 9.8, one times two point, no times 3.5x minus two. So I guess not points. 64 jewels. Despite the behavior of the wire when the added MaaS was greater than 2.9 kilograms. So two Marks. So what is the happening to the wire after 2.8 grams kilograms have been added? So how would you describe the behavior of the wire in this region? And we need two Marks. Up to this point, the force added is directly proportional to the extension. The wigets uniformly longer per unit MaaS or per unit force added to it up to about this point. What happens when you add 2.9 kilograms or just three kilograms? What does the wire do? Now this is a very interesting experiment to see, and it's a pity I couldn't show you a video of it, but. This Halme. We for MaaS MaaS is above 2.9 kilograms. What happens to the extension? The extension will increase, but. MaaS but MaaS will not increase us big us before Yeah I mean they will increase less increase slowly Yeah. For MaaS is above 2.9 kilograms, the extension will increase, but not in direct proportion to the MaaS. So is there a sudden increase in extension? Or is there a sudden decrease in extension for each unit of MaaS added? Is the extension suddenly getting greater or is it suddenly getting less, Jackson? The graph will to save what you see. What is the graph telling us is happening to extension for a slightly bigger MaaS? Yeah so Yeah, I say it over 2.9 kilograms and the extension will increase but I will not direct proportion to the MaaS but to the MaaS, but I don't think there's relations to the. Is there any is is there relevant to the. To is it relevant to the limit of proportionality? Yes. So which point so means which point is limit of proportionality? I would say it's about here. So if you have a wire that stretches slightly and then goes back to its original shape, it's obeying hook law. It's staying in its limit of proportionality here. But if you go above this limit, you get a greater increase in extension for a small increase in MaaS. Above the limit of proportion. There is a greater increase in extension. Unkilogram MaaS increase as we have on B1d. Limits of proportionality. The elastic limit. So let's look at what's happening. The wire stretches a bit, a bit, a bit uniformly, and will still go back to its original shape if we take the masses away up to this point. But beyond that point, if we keep adding masses, the wiwill suddenly stretch much longer, get much weaker. And then if you keep adding weights, itsnap. So this is the breaking point of the wire. So you get this is called the yield point. Yield points, if you yield, it gives way, you give way. So we looked at something like this before. Not understand about this point greater increasing extension. The stress over strain, which is a bit like force over extension. Different materials behave in different ways. Okay. So a ductile wire. Like copper wire. You can stretch wire up to point a, and it behaves in an elastic way. So the wire can be stretched in this region and go back to its original shape if the force is removed. The elastic limit, they say, is here. So it's still stretching in direct proportion. You can still get it to go back to its original shape up to this stretching point. And then there's a yield point where there's a massive stretch of the wire. This is telling you this, how much strength it has under tension, under being stretched tension. And then it gives a bit more. And then this is the breaking force. Ultimate tensile strength, okay? Remember, with copper wire, it's a ductile material. It can be stretched. And when we move the force itgo back to its original length up to the elastic limit. But then if we go beyond the elastic limit. It suddenly gets much longer and thinner and weaker. The yield stress is the force per unit area that will the material extends plastically, so it will stay permanently stretched for a very small increase in MaaS. I got it. Okay, so different materials. So wibehaves like this, that's the ultimate tensile strength. Steel is a bit like our ductile wire. Glass has no behavior like this. Glass is brittle. Then it breaks. It reaches a fracture point, and then it breaks. Rubber nearly obeys hook's law, and then it just breaks. Rubber has the greatest elastic strain energy for a given tensile stress. It has the greatest extension before it breaks too. But rubber is a very elastic material. Glass is a very plastic material. You can't really stretch glass unless it's very hot, whereas steel wire or copper wire is ductile. Okay, doctor. Okay. So our graph is a bit like the graph that we've just looked at. So first we have it obeying holaw. Then it reaches a limit of proportionality. Then for a very slightly more increase in MaaS, it gets much longer and thinner and weaker. The student modifies the investigation, suggests one modification that would produce a greater extension for a given MaaS. What could the student do? To make this extension greater for a given MaaS. So if you increase the MaaS, the extension would be much greater. State one change that would produce a greater extension for each MaaS or load added. Think of your wire, think of the MaaS added to it. Because the more MaaS and the more extension. Below the limit of proportionality. Yeah so produce a greater extension for given MaaS, greater produce produce a greater extension greater extension, we can use more project, more ductile wire. Yes, good. A lower k value. Or a smaller make the wire thinner as well. Good, well worked out. Suggest two measuring techniques that could be used to ensure the accuracy of the measured extensions. So measuring techniques to ensure the accuracy of the measured extension. So to make sure the extension is accurate, let's have a look. So youhave some sort of marker here and some sort of scale here, didn't you? So youneed a pointer. When when I used to do this experiment, I used to stick a piece of sticky tape on the wire. So as the wire moved, the tape, the the pointer, the marker. Moved, need a pointer. Read at I level, so pointer. Yeah, so a marker. I used to stick a label onto the end of the wire, so as the wire got longer, this moved. A marker or a pointer. Read, sorry, read the value on the pointer. Yeah youhave a ruler situated under the wihere in this bit. So the vare, the ruler would be here. And as the vare got longer, the pointer would move. And youread us at I level read the marker at I level. Vertictically down. Okay. So we have two ways of measuring the Young modulus. We have. This way, stretching, stretching this wirelative to this wiand, making sure our wrething apparations is level. So that's vertical arrangement. And then this is horizontal arrangement. Okay, let's try question ten. And walk down in extended spring types process. They send in the spring. Maybe that is walk down. Yeah, walk down. So walk down. The energy, the elastic potential energy equal to a half F X, so entry entry x are half a half entry data ta X I should see very good. Correct question eleven three following properties can be used to scribe copper. Both nylon and copper can be used to make fishing lines going fishing. Copper fishing lines sink faster than those made of nylon. This makes copper fishing lines more suitable for deep water fishing, so going out into the deep water. Fishing line, by considering the forces acting on the submerge line, explain why nylon is less suitable than copper for deep water fishing. Include a suitable calculation in your answer, so both have the same cross sectional area. Density of salt water is that weight of the copper line, is that weight of the nylon line? Is that. Same croarea. Expone non, a White non is less suitable than copper for deep p water fishing include a suitable calculation, you answer so why is less than so? Firstly, the weight of they have the same cross section area. And and the density of salt water is the same. And the difference is that the weight, same distance, copper is not going to two newtons and near long and miss. How to you read this word? Nylon, nylon, nylon, nylon, nylon. Okay, nylon, 0.028 newtons. So it has less weight. But that is formmark. So. Gross and mutons. Suppose if you go deep water fishing. The fish are deep in the water, so you want your line to sink, not to float, because nylon is like a plastic. It's so I think you need to do something with the density. Density equal to density is equal to M, M divided by v. So env the density. Okay, okay, I know how to do it. That's so first stly, it's a copper 0.02 divided by 9.81, Oh sorry, not 0.22 divided by 9.810, 0.022 and this value divided by the cross area 1.3, times ten to the minus seven and then times 20. 8625. That's volume. Which so first three, is it okay anyway? Copper. I'm equal to 0.022. Kilograms. And then another 0.22. Yeah no, no, that's newtons, but m equthe unit of m is not point zero two L I think I think it's not two l. Okay. Not one to do so. And then the v is equal to 1.3 times ten to the minus seven times 20. 2.6 times ten to minus six. Meters q, not one, no two, two divided by 2.6 times ten to the minus six. So the density equal to 8461.5m seven. For a second. And then that is ndle. 0.028 divided by 9.81 equal to 2.8 maybe 2.9. I point. 2.9 times ten to the minus three kilograms. And. Weture. Same 2.6, Yeah. The density is MaaS over volume. Equal to. 1115.4. So all the density is greater than the density of salt water, but the density of copper is more is bigger than nylon. Why is that better? Good, you're correct. Good. Why I don't think I can write write down the answer. Because in the deep water fishing, we should let the hook. Amazing is that hook? Yeah there's a hook. Sink, sink in deep water. To the. Pastor, Yeah, good. That will that will get to the Marks. Well done. A fish is caught on the hook and copper line extends. Calculate the extension produced. So extension produced extension cross section not online, not online, cross section not online and original lines models of. Okay, miss, I have a good idea, a good idea. Stress equal to the. Pressure equal to f over a good so if f if f is equal to 65 newtons and a equto one minus three times ten to the -7m, 1.3 times ten to the minus seven divided by 20. One times ten to the ten. 呃PaaS Pascal。And the strstress in Young models. Was a symbol of Young models. Is that capital e? Yes, it's good. And we know the Young modulus. Gituearlier, and we know the stress, so string equal to stress divided by us. Moduand, so that is one times ten, one times ten to the ten, divided by 1.2, nine times ten to the eleven. And then this number, original times, original number. So the final extension is 1.55m. Good. All worked out. Okay, so we leave it there for today. Jackson, have a well earbreak and enjoy your few days holiday. And we'll talk again 20 fourth, 20 fifth, 26 th, on 20 seventh, I think, and we'll do suothe that. Okay, Jackson, bye bye. All right.