Chains and Ropes over Edges and Pulleys

Transcript

0:00Introduction

Hello everyone! Okay, one small puzzle before we start. Hang a rope five metres long over a small smooth peg: two metres on one side, and three metres on the other. Now let it go. The longer side has just one extra metre, and gravity pulls that extra metre down. So, does the rope fall with the full g, ten metres per second squared? Hmm, keep your guess in mind. Today: chains and ropes over edges and pegs. In JEE, they come up often in laws of motion. So, let's begin.

0:34A chain hanging over a table edge

So, look at the figure on the right, from our lesson on pulleys. Here, a block of mass capital M lies on a smooth table, and a hanging mass, small m, pulls it along over a pulley. The arrows marked small a show the acceleration. The hanging weight pulls both bodies, right? Now, make the block and the hanging mass out of one single piece: a uniform chain. Its length is capital L, and a length small x hangs over the edge. Uniform means every metre of it has the same mass. Okay? The table is smooth, so there is no friction. And lambda, the Greek letter on screen, is the mass of one metre of chain. With me? Now tell me. What pulls this chain along, and how much mass does that pull have to move? Yes, exactly! Only the weight of the hanging part pulls; the table holds up the weight of the rest. The hanging part is x metres long, so its mass is lambda x, and its weight is lambda x g, where g is the acceleration due to gravity. Clear? But this pull has to move the whole chain, not just the hanging part, because the chain moves together. So the mass moved is lambda times capital L, right? So, a equals lambda x g, over lambda capital L; lambda cancels, and a equals x over capital L, times g. Net pull over total mass, the same idea as the block. Got it? Now one more question. As the chain slides off, does this acceleration stay the same? No, it does not! Every moment, more chain hangs over the edge. So x grows, and a grows with it; the double arrow on screen means implies. This is not a constant acceleration, so be careful before you use the equations for constant acceleration here. Okay?

2:24A rope over a small smooth peg

Alright, now a rope over a peg. The rope is uniform, and the peg is small and smooth, so there is no friction at the peg. Clear? The two sides have lengths l one and l two; the small one and two written below the l are just labels. The little less-than sign says l one is shorter, so the longer side goes down. Right? Now look at the figure again. Which of the two systems does this rope look like? Correct! The Atwood machine on the left. Each side of the rope acts like one hanging mass: m one equals lambda l one, and m two equals lambda l two. With me? And for the Atwood machine, we know the answer: a equals, in brackets, m two minus m one, times g, over m one plus m two. Net pull over total mass, once again! Now put in the masses, and lambda cancels once more. So, a equals l two minus l one, over l one plus l two, times g. On top, the extra length that drives the motion; below, the whole rope that moves. Clear? But remember, this is the acceleration only at this instant. As the rope slides, l two grows, and l one shrinks, so a grows too, just like the chain. Got it?

3:44Topper's trick: tension at any cut

Now, the tension inside a chain or a rope. How do we find the tension, capital T, at one point? Simple: imagine a cut at that point, and take the piece on one side alone. Okay? Then it is just Newton's second law for that piece. The net force on it, with T included, equals its mass times its acceleration. Right? Now think about our chain on the table, cut at the edge. Which side would you choose? The table side, very good! The table holds up its weight, and there is no friction, so only T pulls it along. That is the topper's trick: choose the simpler side, the one with fewer forces. Got it?

4:25Example 1: rope over a peg with 2 m and 3 m sides

Now, remember our puzzle from the start? Let's solve it. Here it is: a uniform rope five metres long hangs over a small smooth peg, with two metres on one side and three metres on the other. Find its acceleration. Pause the video, and try it first. Done? Okay, let's check. l one is two metres, and l two is three metres; the m stands for metres. We take g as ten metres per second squared, written m s to the power minus two. Now, only one extra metre is pulling. So that metre just falls freely, with g, ten metres per second squared, right? No, no, careful! That extra metre is part of the rope, so it has to drag all five metres along. So, use our formula: three minus two, over two plus three, times ten. Okay? That is ten over five: two metres per second squared. See, puzzle solved! Not ten, but two, only one fifth of g, because one metre of pull moves five metres of rope. And this is the acceleration just as it is let go; as it slides, it grows. Clear?

5:36Example 2: tension in a chain sliding off a table

Now a harder one, with letters this time. Read it with me: a uniform chain of mass m and length capital L lies on a smooth table, with a length x hanging over the edge. Find its acceleration, and the tension at the edge. Pause the video, and try both parts. Done? Okay, let's check. The acceleration, we already know. The hanging weight is m x over capital L, times g, and it moves the whole chain, of mass m. So, a equals x g over capital L, as before. Okay? Now, the tension at the edge. The hanging part hangs from the edge, so the tension there is simply its weight, m x g over capital L, right? Oh no, careful! The hanging part is speeding up downwards, so the chain above it pulls it up with less than its weight. Use our trick: take the table side. Its length is capital L minus x, so its mass is m times, in brackets, capital L minus x, over capital L, and on the smooth table, only T pulls it along. With me? So T is its mass times a: m times, capital L minus x, over capital L, times x g over capital L, and that gives T equals m g x, times capital L minus x, over capital L squared; the small two up top means squared. Okay? Now let's check it from the other side. On the hanging part, its weight pulls down, and T pulls up. So, m x over capital L times g, minus T, equals m x over capital L, times a. Clear? So T equals m x over capital L, times, in brackets, g minus x g over capital L; simplify, and you get m g x, times capital L minus x, over capital L squared. The same answer! Cut at the edge, take either side, and you get the same tension. Got it?

7:38Quick revision: chain on an edge, rope on a peg, tension

So, quick revision. Chain over a smooth table edge: a equals x over capital L, times g. Rope over a small smooth peg: a equals l two minus l one, over l one plus l two, times g. Both are just net pull over total mass. And for the tension at a cut: Newton's second law for one side alone, and choose the simpler side. Now go and try the practice problems on this page. You can do them, I am sure. All the best!