Ropes, Chains and Spring Balances

Transcript

0:00Introduction

Hello everyone! One small puzzle before we start. Take a long, heavy rope, and hang it from a hook in the ceiling. So far, our strings were light, and the tension was the same all along the string, right? But this rope is heavy. So tell me: is the pull the same at every point of this rope? And if not, which point is pulled the hardest? Hmm, keep your answer in mind. Today we see what changes when a rope or a chain has mass, and what a spring balance really reads. In JEE, questions on this come up often. So, let's begin.

0:38A hanging rope: the weight of everything below

So, first, a quick recall. For a light string, that means a massless string, the tension, capital T, is the same all along it. Okay? Now our rope has mass, and it hangs at rest. Pick any point on it, and look only at the part below that point. That lower part is not falling. So who holds it up? Yes, exactly! The rope just above the point pulls it up, and that pull is the tension at the point. So, for a hanging rope at rest, the tension at a point equals the weight of everything below it: the rope below, and anything hung from it. Write this down. Clear? Now, back to our puzzle. Which point of the hanging rope has the most weight below it? The top, very good! The top point holds up the whole rope, and whatever hangs from it. As we go down, less and less is left below, so the tension falls. So the top of the rope is pulled the hardest. Puzzle solved! Right? And this explains spring balances too. Now, a light spring balance reads the tension in it. So, when it hangs at rest, it reads the weight of everything below it. In a line of light balances, each one reads the weight of everything below it, not just the mass hooked right under it. Got it?

1:58A rope pulling a block on a smooth floor

Now the rope lies flat. Here, a rope of mass capital M pulls a block of mass small m on a smooth floor, so no friction. And we pull the free end of the rope with a force F, okay? The rope and the block move together. So, what is their acceleration? Correct! Treat them as one body of mass M plus m. The only horizontal force on it is the pull F, so a equals F over M plus m, the total mass. With me? Now a quick question. We pull with a force F, so the tension is F everywhere in the rope, right? No, no, careful! That is true only for a light string. Cut this rope at any point, and look at everything beyond the cut, on the block's side. It accelerates with a, and only the tension at the cut pulls it forward. So, the tension at a point equals the mass beyond that point, times a. Clear? So, at the block, the mass beyond is just m. The tension there is m F over M plus m, less than the pull F itself. Why? Because part of your pull goes into accelerating the rope itself. Right? And with no block, just a uniform rope of length capital L, pulled by F on the same smooth floor? At a distance small l from the pulled end, the tension is F times, in brackets, one minus l over L, okay? Let's check the two ends. At the pulled end, l is zero, so T equals F, the full pull. At the far end, l equals L, so T equals zero. In between, it falls steadily. Got it?

3:35A chain hanging between two supports

Last idea. Picture a chain hanging between two supports, sagging in the middle. At each end, the support pulls along the chain, at a slant. So, which part of these slanted pulls holds the chain up? The vertical parts, yes, exactly! The vertical components of the two end tensions together carry the weight of the chain, m g, its mass times g. With me? And sideways? The weight is vertical, so it does nothing sideways. So the horizontal component of the tension is the same all along the chain. Keep these two facts ready for example two. Okay?

4:12Example 1: tension at the top and middle of a hanging rope

Alright, example one, and it is our puzzle with numbers. Here, a uniform rope of mass two kilograms and length four metres hangs from a ceiling; k g is kilogram, and the small m is metre. There is a three kilogram block at its lower end. Find the tension at the top, and at the middle of the rope. Pause the video, and try it first. Done? Okay, let's check. We take g, the acceleration due to gravity, as ten metres per second squared, the m s to the minus two on screen. Now the top first. What hangs below the top point? Everything, yes! The whole rope, two kilograms, and the block, three kilograms. Total, five kilograms. Right? So T top, that is T with the small word top written below it, equals five times ten: fifty newtons. The capital N after the number is newton, the unit. Clear? Now the middle. It is the middle, so the tension is half of fifty, twenty five newtons, right? Oh no, careful! The block does not become half. Below the middle hangs the lower half of the rope, and the rope is uniform, so that is one kilogram. Plus the full block, three kilograms. Total, four kilograms. Okay? So T mid equals four times ten: forty newtons. See, fifty at the top, forty at the middle. The tension is largest at the top, just as we found in our puzzle. That's it!

5:43Example 2: chain tension at the supports and lowest point

Now a harder one, JEE level. Here, a uniform chain of mass m hangs between two supports at the same level. At each support, it makes an angle theta with the horizontal; theta is the Greek letter on screen. Find the tension at the supports, and at the lowest point. Hint: use the two chain facts we just learnt. Pause and try. Done? Okay, let's check. The chain makes the same angle theta at both ends. And the horizontal part of the tension is the same all along the chain. So the tension is the same at both supports; call it T, okay? Now, which direction should we balance to find T: horizontal or vertical? Vertical, very good! Each end pulls up with T sine theta. Two ends, so two T sine theta equals m g. The double arrow means implies: T equals m g over two sine theta. That is the tension at each support. Right? Now the lowest point. What is special about the chain there? It is horizontal, exactly! So the tension there has no vertical part at all; it is all horizontal. And the horizontal part is the same all along the chain. So T nought, that is T with a small zero below it, equals T cos theta. Clear? Now substitute for T: m g over two sine theta, times cos theta. Cos theta over sine theta is cot theta, the cotangent. So T nought equals m g over two, times cot theta. Two facts, two lines, and the chain is done. That's how simple it is!

7:25Quick revision: hanging rope, pulled rope and chain

So, quick revision. One: in a hanging rope, the tension at a point is the weight of everything below it. Two: in a pulled rope, it is the mass beyond the point, times a; and a light spring balance reads the tension in it. Three: for a chain, the vertical parts of the end tensions carry its weight, and the horizontal part is the same all along. Now go and try the practice problems on this page. All the best!