Connected Bodies with Friction

Transcript

0:00Introduction

Hello everyone! Okay, one small puzzle before we start. Picture a two kilogram block on a rough slope of thirty degrees, with mu equal to zero point five. It is tied by a string, over a pulley at the top, to a hanging one kilogram mass, and it sits still. So how much friction does the rough slope give it? Hmm. Most students write mu times N straight away, right? What if I told you the answer is zero? Rough surface, zero friction! How? Today: bodies tied together, with friction. Questions on this come up often in JEE, so let's begin.

0:39First question: does anything move?

So, look at the figure on the right, from our lesson on pulleys; the left one is the Atwood machine, and we don't need it today. Capital M is a block on a table, and small m hangs from a string over a pulley at the edge. The green arrow marked small a shows the acceleration, if there is one. Take the string as light and the pulley as frictionless, so the tension is the same on both sides. But today, the table is rough. Okay? Now, before any calculation, one question. Forget friction for a moment. What force is trying to move this system? The weight of the hanging mass, m g, yes, exactly! Here g is the acceleration due to gravity, so m g is small m times g. Step one: this net pull, with friction ignored, is called the driving force. Right? Step two: the largest friction the surfaces can give. For one surface, it is mu times N, with mu, this Greek letter, the coefficient of friction, and capital N the normal reaction. If several surfaces rub, add up mu N for each one. With me? Now compare the two. If the driving force is smaller than the largest friction, what happens? Nothing moves, correct! This sign means less than or equal to. And here is the key point. Then the friction need not be mu N at all. Each friction takes just the value that keeps everything in equilibrium. Okay? And if the driving force is larger, this sign means greater than, the bodies move. Now kinetic friction acts, and it opposes the motion of each body relative to its surface. Got it?

2:19At rest: friction is only what is needed

Now, back to our puzzle; it is one of the course examples. Here, a two kilogram block, k g for kilograms, sits on a rough slope at thirty degrees; the small circle means degrees. Mu is zero point five, and over a pulley at the top, a one kilogram mass hangs from the string. Okay? Step one, the driving force. Take g as ten metres per second squared; that is the m s to the power minus two on screen. The hanging mass pulls with its weight: one times ten, ten newtons, up the slope. The cross means times, and the upright N after a number is the newton, the unit of force. Right? And part of the block's own weight acts along the slope: two times ten times sine thirty degrees. That is also ten newtons, but down the slope. Clear? So the friction is mu times N: zero point five times twenty times cos thirty degrees, about eight point six six newtons; the wavy equals sign means about. Right? No, no, careful! The two pulls already balance, ten against ten. The block is not even trying to slide, so the friction is zero. Zero! Mu N is only the largest friction available, not the friction that acts. Puzzle solved! Got it?

3:36Example 1: acceleration and tension, block on a rough table

Now, example one, straight from our picture. Here, a four kilogram block on a table is tied over a pulley to a hanging two kilogram mass. Mu k, with the small k below, is the coefficient of kinetic friction, and here it is zero point two five. Find the acceleration and the tension. Pause the video, and try it first. Done? Okay, let's check. Step one, does it move? The driving force is the hanging weight: two times ten, twenty newtons. The table is flat, so N is the block's weight, forty newtons. Only mu k is given, so we use it here too: the friction is zero point two five times forty, ten newtons. Right? Twenty is greater than ten, so the block moves, and the ten newtons of friction act against the motion. Treat both bodies as one system: net force twenty minus ten, total mass four plus two. So a equals ten over six. Okay? That is five over three, about one point six seven metres per second squared. Clear? Now, capital T, the tension in the string. The string holds up the two kilogram mass, so T is just its weight, twenty newtons, right? Oh no, careful! The hanging mass is accelerating downwards, so the string pulls less than its weight. Newton's second law for the hanging mass: two times ten, minus T, equals two times a. With me? So T equals two times, ten minus a. Ten minus five over three is twenty five over three, and two times that is fifty over three: about sixteen point seven newtons. Less than twenty, as it must be. Perfect!

5:19Example 2: range of hanging mass on a rough incline

Now a harder one. Here, a five kilogram block sits on a thirty seven degree incline, with mu equal to zero point five. It is tied, over a pulley at the top, to a hanging mass, small m. For what values of m does the block stay at rest? Pause the video, and try it. Done? Okay, let's check. For thirty seven degrees, we take sine as zero point six and cos as zero point eight. Okay? The block's weight along the slope is five times ten times zero point six: thirty newtons, pulling it down the slope. Right? Nothing else presses on the block, so N is five times ten times zero point eight, forty newtons. So f max, the largest friction, is zero point five times forty: twenty newtons. With me? Now think. If the hanging mass is small, which way does the block tend to move, and which way does friction act on it? Yes, exactly! The thirty newtons win, so the block tends to slide down, and friction acts up the slope, helping the string. At rest, the tension is just m g. So m g plus twenty must reach thirty: m g is at least ten newtons; this sign means greater than or equal to. Okay? Now many students stop here, as if friction always acts up the slope. No, no, wait! Make m large, and the block tends to move up. Then friction turns around and acts down the slope, against the string. So m g can be at most thirty plus twenty: fifty newtons. Right? So m g lies from ten to fifty newtons. Divide by ten: m from one kilogram to five kilograms. In between, static friction adjusts itself, and at exactly three kilograms it is zero, just like our puzzle. Topper's trick: the pull must lie within the weight along the slope, plus or minus f max. That's it!

7:13Quick revision: does it move, static vs kinetic friction

So, quick revision. First, compare the driving force with the largest friction available. If nothing moves, each friction takes only the value needed, maybe even zero. And if the bodies move, kinetic friction opposes the motion of each body relative to its surface. Now go and try the practice problems on this page. You can do this. All the best!