Pulleys

Transcript

0:00Introduction

Hello everyone! One small puzzle before we start. Hang a three kilogram mass and a two kilogram mass on a light string over a light pulley, and hang the pulley from a hook in the ceiling. Taking g as ten, the two masses weigh fifty newtons together. Now let them go. While they move, does the hook still carry fifty newtons? Hmm, keep your answer in mind. Today, the standard pulley systems; in JEE, they come up often. So, let's begin.

0:30The Atwood machine

So, look at the left half of the figure: the Atwood machine, two masses on one string over a pulley. The masses are m one and m two; the small one and two written below the m are just labels. Now our conditions: the string is light, that means massless, and the pulley is light and frictionless. Clear? Now tell me. Is the pull of the string the same on both sides of the pulley? Yes, exactly! For a light string over a light, frictionless pulley, the tension, capital T, is the same on both sides. And both masses have accelerations of the same size, small a; those are the green arrows in the figure. With me? In the figure, m two goes down and m one goes up. So what is the net pull that drives the motion, and how much mass does it move? Correct! The weight m two g pulls one way, and m one g pulls back; g is the acceleration due to gravity. So, a equals, in brackets, m two minus m one, times g, over m one plus m two, the total mass. The less-than sign in the last brackets says m one is less than m two, okay? Now the tension. Take m one alone. It speeds up upwards, so T minus m one g equals m one a. The double arrow means implies: put in a, and T equals two m one m two g, over m one plus m two. Right? Now, back to our puzzle. The string pulls the pulley down on both sides. So, what must the support of the pulley hold up? Two T, very good! Each side pulls the light pulley down with T, so the support carries two T, not simply the total weight. Got it?

2:11Block on a smooth table, pulled by a hanging mass

Now the right half of the figure. Here, a block of mass capital M lies on a smooth table, so no friction. And small m hangs from the string, over a pulley at the edge. Okay? So, what pulls this system along? Does the weight of capital M help? No, it does not! The table holds that weight up. Only the hanging weight, m g, drives the motion, and it moves both bodies, capital M plus m. So, a equals m g over capital M plus m. Net pull, over total mass. Clear? Now, a common mistake. The string holds the hanging mass, so the tension is just m g, right? No, no, careful! That mass is speeding up downwards, so the string holds up less than its weight. Easier way: only the string pulls capital M along the table. So T equals capital M times a, which is capital M m g, over capital M plus m. Less than m g, as it must be. Got it?

3:13Smooth inclines: the weight along the slope

Alright, now we tilt the table. Here, a block of mass capital M sits on a smooth incline; its angle is theta, the Greek letter on screen. It is tied, over a pulley at the top, to a hanging mass, small m. Okay? The full weight of the block is capital M g. How much of it pulls along the slope? Very good, only the part along the slope! That is capital M g sine theta, and it pulls the block back down the slope. With me? Net pull over total mass again: a equals, in brackets, m minus capital M sine theta, times g, over capital M plus m. If this comes out positive, small m goes down. Clear? And the double incline: blocks m one and m two on two smooth faces, at angles theta one and theta two, tied over a pulley at the top. Same idea. So, a equals, in brackets, m two sine theta two, minus m one sine theta one, times g, over m one plus m two. Okay? So here is the topper's trick. In every system today, the acceleration equals the net pull along the string, over the total mass. Pulls that help the motion are plus, pulls that oppose it are minus. One line, and you can rebuild all these formulas in the exam. Got it?

4:34Example 1: 2 kg and 3 kg Atwood, tension and hook force

Now, example one, and it is our puzzle. In an Atwood machine, the masses are three kilograms and two kilograms; k g on screen is kilogram. Find the acceleration, the tension, and the force on the support of the pulley. Pause the video, and try it first. Done? Okay, let's check. We take g as ten metres per second squared; that is the m s to the minus two on screen. The smaller mass, m one, is two kilograms, and m two is three kilograms. Right? So, a equals, in brackets, three minus two, times ten, over two plus three. Ten over five: two metres per second squared. Okay? Now the tension. T equals two times two times three times ten, over five. One hundred and twenty over five: twenty four newtons; capital N is newton. Clear? And the support? The masses weigh fifty newtons together, so the support carries fifty newtons, right? Oh no, careful! The support carries two T, as we learnt. Two times twenty four: forty eight newtons, not fifty. So, puzzle solved: while the masses accelerate, the hook carries a little less than their weight. That's it!

5:53Example 2: block on a 30 degree incline, hanging mass

Now a slightly harder one. Here, a five kilogram block on a smooth thirty degree incline is tied, over a pulley at the top, to a hanging five kilogram mass; the small circle means degrees. Find the acceleration and the tension. Hint: first decide which way it moves. Pause and try. Done? Okay, let's check. Both masses are five kilograms. Equal masses, so the system stays at rest, right? No, no, wait! Along the slope, only five times ten times sine thirty degrees pulls the block back: twenty five newtons. The hanging mass pulls with five times ten, fifty newtons. So the hanging mass wins. With me? So, a equals, in brackets, five minus five sine thirty degrees, times ten, over five plus five. Twenty five over ten: two point five metres per second squared, and the hanging mass goes down. Okay? Now the tension. The string holds up a five kilogram mass, so the tension is fifty newtons, right? Careful! That mass is speeding up downwards, so the tension is less than its weight. For the hanging mass, five g minus T equals five a. So T equals five times, in brackets, ten minus two point five: thirty seven point five newtons. Got it? Quick check with the block. Up the slope, T minus twenty five is thirty seven point five minus twenty five, which is twelve point five. And the block's mass times a, five times two point five, is also twelve point five. See how clean that was?

7:34Quick revision: same T, support 2T, net pull over mass

So, quick revision. One: with a light string over a light, frictionless pulley, the tension is the same on both sides, and the support carries two T, okay? Two: the acceleration is the net pull along the string, over the total mass; on an incline, only capital M g sine theta counts. Three: for the tension, apply Newton's second law to one body; it is not simply m g. Now go and try the practice problems on this page. All the best!