Young's Modulus: Elongation Problems

Transcript

0:00Introduction

Hello everyone! Okay, here is a small puzzle. Hang a long rubber cord from the ceiling. It stretches a little under its own weight, right? Now compare: imagine the cord weighed nothing, and hang a load equal to its whole weight at the bottom end. Same weight, same cord. Same stretch? Keep your answer ready. Today: how much a wire stretches, under a load, under its own weight, and when two wires work together. These problems come up often in JEE, so let's begin.

0:32Elongation under a load

So, recall Young's modulus Y: stress over strain, that is F over A, divided by delta L over L, right? F is the stretching force, capital A the area of cross-section, L the original length, and delta L the stretch. The little triangle is the Greek capital delta; it means change in. And all this holds only within the elastic limit. Okay? Now, in elongation problems, we want the stretch itself. So rearrange this. What is delta L? Yes, exactly! Delta L equals F L over A Y, and this one line solves most of today's problems. Right? Quick check. Same material, same thickness, same load, but the wire is twice as long. What happens to the stretch? It doubles, correct! Read the formula: delta L is proportional to F, proportional to L, and proportional to one over capital A, the area. This curly sign means proportional to. So a thicker wire stretches less. Clear? Now the trap. Questions often give the diameter, small d. The area is pi r squared, which is pi d squared over four; pi is the Greek letter for the circle constant. Use the diameter as r, and your area is four times too big. Oh no, careful! Got it?

2:03A stretched wire is a spring

Now turn it around. F equals Y A over L, in brackets, times delta L, right? Force proportional to stretch. Where have you seen this before? In a spring, yes! For a spring, F equals k x, with k the force constant and x the stretch. So the wire behaves exactly like a spring, with force constant k equals Y A over L, nothing more. With me?

2:30Stretching under its own weight

Now, our puzzle. Take a uniform wire of weight capital W, hanging from one end, with no load. Is the tension the same all along it? No, it is not! At the free end, the bottom, there is nothing below to hold, so the tension is zero. At distance x from the free end, the wire holds only the part below it, which weighs W x over L, right? So the tension, capital T, is W x over L, and it grows steadily up to the full weight at the top. Okay? So, compare with the full weight hung at the end of the wire. More stretch, or less? Less, exactly! Only half. The tension rises evenly from zero to the full weight, so on average the wire carries half of it. Delta L equals W L over two A Y: half of W L over A Y, the stretch with the weight hung at the end. So our puzzle is settled! Forgetting this two is a very common mistake. Right? Now, weight is mass times g, and mass is density times volume. Density is rho, the Greek letter that looks like a p, and volume is area times length. So W equals rho A L g. Put it in, and the area cancels! Delta L equals rho g L squared, over two Y: the thickness does not matter at all. Got it?

3:57Two wires: series and parallel

Now two wires together. First, joined end to end, one below the other, with a load at the bottom: this is series. Which wire carries more tension, the upper or the lower? Neither, as long as the wires are light! Their own weight is negligible here, so the same tension F passes through both wires, one after the other. Okay? So each wire stretches by its own F L over A Y, and the stretches simply add. Delta L equals F times, in brackets, L one over A one Y one, plus L two over A two Y two. The small ones and twos below the letters are subscripts; they tell you which wire. Clear? Now parallel: two wires side by side, holding one load together. When the load hangs level, both wires stretch by the same amount. Right? They share the load in proportion to A Y over L; the small i below means any one wire. Look: this A Y over L is just the spring constant k! So the stiffer wire takes the bigger share. With me?

5:05Example 1: extension of a steel wire holding 10 kg

Example one, from your course. Take a steel wire of length four metres and diameter two millimetres, supporting ten kilograms. Find the extension. Y is two times ten to the power eleven pascal, and g is nine point eight metres per second squared; the minus two up there means per second squared. Pause the video, and try it first. Done? Okay, let's check. Ten kilograms is a mass, not a force. So the load is m g: ten times nine point eight, which is ninety eight newtons; the capital N stands for newton. Right? Now the area. The diameter is two millimetres. So the area is pi times two millimetres, squared, correct? No, no, careful! That is the trap we just saw. The radius is half the diameter: one millimetre, which is ten to the power minus three metre. So capital A equals pi r squared: pi times ten to the power minus six square metres. Got it? Now put everything in. Delta L equals ninety eight times four, over pi times ten to the power minus six, times two times ten to the power eleven. Right? On top, three hundred and ninety two; below, about six point two eight times ten to the power five. So delta L is about six point two times ten to the power minus four metre; the wavy equals sign means approximately. Okay? That is about zero point six two millimetre, for a four metre wire holding ten kilograms. Steel is really stiff! Clear?

6:44Example 2: rubber cord stretching under its own weight

Now example two, a little harder: our puzzle with real numbers. Here, a rubber cord eight metres long hangs from a ceiling. Its density is fifteen hundred kilograms per cubic metre; the minus three up there means per cubic metre, and Y is five times ten to the power eight pascal. Find its extension under its own weight, with g equal to ten metres per second squared. Pause the video, and try it. Done? Okay, let's check. First, look: the question gives no area at all. Is that a problem? Not at all! Under its own weight, delta L equals rho g L squared, over two Y, and the area has already cancelled. So no area is needed. Okay? Now the pieces. L squared is eight squared, sixty four square metres. And two Y is two times five times ten to the power eight, which is ten to the power nine pascal. Right? So delta L equals fifteen hundred times ten times sixty four, over ten to the power nine. On top, that is nine hundred and sixty thousand, which is nine point six times ten to the power five. Okay? Divide by ten to the power nine, and delta L is nine point six times ten to the power minus four metre. With me? That is zero point nine six millimetre. Forget the two, and you get one point nine two millimetres, double: the stretch if the cord's weight hung at its end. So, under its own weight, only half the stretch. See how neat that is?

8:26Quick revision: load, own weight, series and parallel

So, quick revision. Under a load, delta L equals F L over A Y: a spring with k equals Y A over L, its force constant. Under its own weight, rho g L squared over two Y: half the end-load stretch. Series: same tension, stretches add. Parallel: same stretch. Now go and try the practice problems on this page. You can do them. All the best!