One Block on Another

Transcript

0:00Introduction

Hello everyone! Okay, one small puzzle before we start. Put a heavy book on a sheet of paper, on a table. Pull the paper gently, and the book comes along with it. Now yank the paper, hard and sudden. The paper flies out, and the book is left behind! Same paper, same book. So what changed? Hmm, think about it. Nobody ever touched the book, right? Today: one block resting on another. Questions on this come up often in JEE, so let's begin.

0:34Pull the lower block: who moves the upper one?

So, here is our setup. Small m is the upper block, resting on capital M, the lower block. The floor is smooth, so there is no friction from the floor, okay? Capital F is a horizontal force that pulls only the lower block. Nobody touches the upper block at all. So tell me. If nobody touches the upper block, why does it move forwards with the lower one? Friction, yes, exactly! The lower block tries to slide out from under the upper one. So friction, small f, drags the upper block forwards. And by Newton's third law, the same friction pulls the lower block backwards. Right? Now, the limit. The upper block presses down on the lower one with a normal force equal to its weight, m g, where g is the acceleration due to gravity. So the largest friction is mu m g. Mu, this Greek letter, is the coefficient of friction between the two blocks, and this sign means less than or equal to. Friction is the only horizontal force on the upper block, so its acceleration, small a, is at most mu m g over m, that is, mu g. Clear? Now the big question. For what values of F do the two blocks move together? Many students say: F must be at most mu m g, the friction limit, right? No, no, careful! F is not the friction. F has to accelerate both blocks together, a total mass of M plus m, in brackets. So F equals M plus m, times a, and a can be at most mu g. So they move together only while F is at most mu g times M plus m. With me?

2:13The whole picture

Now let's see all of this in one picture. Small m on top, capital M below, on a smooth floor. Capital F, the dark arrow, pulls only the lower block. The green arrows are the friction: f on small m, forwards, and f on capital M, backwards. And the text on the right sums it up: they stay together while f is at most mu m g, so a is at most mu g, and F is at most mu g times M plus m. Okay? And that solves our puzzle! The book is the upper block, and friction can give it at most mu g. This sign means greater than. Yank the paper with an acceleration greater than mu g, and the book simply cannot keep up. It is left behind. Got it?

3:00The two-block method

Alright, what we just did is a method that works for every pair like this. Step one: assume the blocks move together, and find the common acceleration, a, from the external forces only. The friction between the blocks is internal, so it drops out. Okay? Step two: look at the block that is not pulled directly. Only friction moves it. So find how much friction it needs, and compare that with mu m g, the most friction can give. Right? Now think. What if that block needs more friction than mu m g? Correct, they slip! Friction simply cannot give more than its limit, so the blocks cannot move together. With me? Once they slip, the friction is kinetic, and it equals mu m g. It acts on each block, the same size, but in opposite directions: forwards on one, backwards on the other. Got it? And now there is no common acceleration. Write Newton's second law for each block separately, and you get two different accelerations. Using one acceleration for both after slipping is a very common mistake. Clear?

4:08Pull the upper block instead

Now, the same two blocks, but this time F pulls the upper block. Which block is not pulled directly now? Yes, exactly, the lower block! The floor is smooth, so nothing else pushes it sideways. Friction from the upper block is the only horizontal force on capital M, okay? And that friction is still at most mu m g. So the lower block's acceleration is at most mu m g over capital M, this fraction. Right? Both blocks move together with this acceleration, so F is at most M plus m, times it. So F is at most mu m g, times M plus m over M, and only on a smooth floor. Look, two different limits, depending on which block you pull. Don't mix them up; that is a common trap. Got it?

4:54Example 1: largest force for two blocks to move together

Example one, straight from our picture. Read it with me: a two kilogram block rests on a four kilogram block, on a smooth floor, and mu between them is zero point four. Find the largest horizontal force on the lower block for which they move together. Pause the video, and try it first. Done? Okay, let's check. The upper block is not pulled directly, so friction alone moves it, and its acceleration is at most mu g. Take g as ten metres per second squared; that is the m s to the power minus two on screen. So zero point four times ten: four metres per second squared. Okay? Now both blocks share this acceleration, and the total mass is four plus two, six kilograms. So F equals M plus m, times a: six times four, twenty four newtons. The upright N after a number is the newton, not the normal reaction. Right? Quick check. The upper block needs friction equal to small m times a: two times four, eight newtons. And mu m g is zero point four times two times ten, also eight newtons. Exactly at the limit! And see, the answer is twenty four, not the friction limit of eight. That was our trap. Perfect!

6:11Example 2: acceleration of each block once they slip

Now a harder one. Same pair: two kilograms on four kilograms, mu zero point four, smooth floor. But now a force of thirty newtons pulls the lower block. Find the acceleration of each block. Pause the video, and try it. Done? Okay. Total mass six kilograms, so thirty over six, five metres per second squared for both blocks, right? No, no, wait! First, check the friction, step two of our method. Remember Example one: they move together only up to twenty four newtons. Thirty newtons is greater than twenty four newtons, so they slip. Clear? Once they slip, the friction is kinetic: mu m g, zero point four times two times ten, eight newtons. Same size on both blocks, but in opposite directions. Okay? The upper block first. Only friction acts on it, forwards. So a one, where the small one below a just labels the upper block, equals eight over two: four metres per second squared, forwards. Right? Now the lower block. Thirty newtons pull it forwards. Does the friction on it help, or oppose? Oppose, correct! Friction holds the lower block back. So a two, the lower block's acceleration, equals thirty minus eight, over four: twenty two over four, which is five point five metres per second squared. With me? So a two is greater than a one. Both blocks move forwards, but the lower block runs ahead, and the upper block slides back across it. That is our book and paper puzzle exactly: yank too hard, and the paper runs out from under the book. That's it!

7:50Quick revision: assume together, check friction, then slip

So, quick revision. Only friction moves the block that is not pulled, and it can give at most mu m g. Assume the blocks move together, find a, then check the friction needed. And if they slip, kinetic friction mu m g acts on each block, in opposite directions, and you find two separate accelerations. Now go and try the practice problems on this page. You can do this. All the best!