Rough Inclines

Transcript

0:00Introduction

Hello everyone! One small puzzle before we start. You launch a block up a rough ramp at ten metres per second. It slows, stops, and slides back down to you. Does it come back at ten metres per second? Hmm, are you sure? Today we find the exact speed. Friction on an incline comes up often in JEE, so, let's begin.

0:25Forces on a block on a rough incline

Look at the figure. Here is a block on a rough incline at angle theta, the Greek letter at the bottom. The green arrow is the weight, m g: mass times g, the acceleration due to gravity. So, how much of m g pulls the block down the slope? Yes, exactly! Only the part along the slope: m g sine theta. The other part, m g cos theta, presses the block into the slope. Okay? Capital N, the blue arrow, is the normal reaction. If nothing else presses on the block, N just balances that push: N equals m g cos theta. Right? Now friction, small f, the orange arrow. On the left, the block tends to slide down, so friction acts up the slope. On the right, it moves up; v, the dark arrow, is its velocity. Now which way does friction act? Still up the slope? No, no, careful! Down the slope. Friction opposes the relative motion, or the motion it is about to make. So when the block moves up, or is just about to move up, friction acts down. Drawing it up the slope every time is a common trap. Got it?

1:36When does the block stay at rest by itself?

Now the block sits at rest, with no other force on it. To hold it, friction must balance m g sine theta. But static friction has a limit. Mu, the Greek letter, with a small s below: mu sub s, the coefficient of static friction. The limit is mu s times N, which is mu s m g cos theta. With me? So tell me, when does the block stay at rest by itself? Correct! When the friction needed is within the limit. m g sine theta, less than or equal to, mu s m g cos theta; that sign means less than or equal to. Divide by m g cos theta: tan theta less than or equal to mu s. No m! Heavy or light, it starts sliding at the same angle. Clear?

2:25Sliding down, going up

Now the block slides down, as on the left. Down the slope, m g sine theta; up the slope, friction, mu times N, with mu now the kinetic coefficient. Newton's second law: m times a equals m g sine theta, minus mu m g cos theta. Okay? Cancel the mass. The acceleration a equals g times, sine theta minus mu cos theta. No mass again! Right? Now the right picture: the block is projected up the slope. What slows it down now? Both forces, yes, exactly! m g sine theta and friction both act down the slope. So the retardation, the slowing down, is g times, sine theta plus mu cos theta. Plus, not minus! Remember this for our puzzle. Clear?

3:14Holding the block with a force P along the slope

Now, a steep incline, where the block would slide down by itself. We hold it at rest with a force, capital P, up along the slope. For the least such P, which way does friction act? Up the slope, very good! The block is about to slide down, so friction helps the push with its full limiting value. So the least P is m g times, sine theta minus mu cos theta. With me? Now increase the push until the block is about to move up. Friction turns around and acts down the slope. So the greatest P is m g times, sine theta plus mu cos theta. Got it? So any P from the least to the greatest keeps it at rest. In between, static friction simply adjusts itself. Okay?

4:01Topper's trick: rough versus smooth

Now a topper's trick. Say a block slides from rest down a smooth incline, then down a rough one, same length, same angle. Distance s equals half a t squared, so the time t is proportional to one over root a; the curly sign means proportional to, and the tick is the square root. Right? Smooth: acceleration g sine theta. Rough: g times, sine theta minus mu cos theta. So t sub rough over t sub smooth is the root of the smooth acceleration over the rough one. Divide by g sine theta: one over root of one minus mu cot theta; cot is one over tan. It needs tan theta more than mu, so the block really slides. Clear?

4:45Example 1: acceleration down a rough 37 degree incline

Now, example one. Here, a two kilogram block slides down a thirty seven degree incline, with mu k equal to zero point five. Find its acceleration. On screen, k g is kilograms, the small circle is degrees, and mu sub k is the kinetic coefficient. Pause and try it first. Done? Okay, let's check. For thirty seven degrees we take sine as zero point six, cos as zero point eight. And g is ten metres per second squared, the m s to the power minus two. Got it? Now, the question gives the mass, two kilograms. Do we need it? No! The mass cancels, as we saw. The block slides down, so friction acts up the slope. So a equals g times, sine thirty seven degrees minus mu k cos thirty seven degrees. Okay? Put in the numbers; the cross means times. Zero point five times zero point eight is zero point four. So ten times, zero point six minus zero point four: two metres per second squared, down the slope. The two kilograms? Never used! That's it.

6:00Example 2: block projected up a rough incline and back

Now the harder one: our puzzle! Here, a block is projected up a thirty seven degree incline at ten metres per second; m s to the power minus one is metres per second. Mu is zero point five. Find how far it goes, and its speed when it returns to the start. Hint: going up and coming down are different. Pause and try. Done? Okay, let's check. Going up, gravity and friction both slow it. Zero point five times zero point eight is zero point four, so the retardation is ten times, zero point six plus zero point four: ten metres per second squared. Right? How far? Use v squared equals u squared minus two a s, with u the starting speed and v the final. At the top v is zero, so zero equals ten squared minus two times ten times s. The double arrow means this gives: s equals five metres. With me? Now the block stops for a moment at the top. Does it stay there, or slide back? Check the condition for rest. Tan thirty seven degrees is zero point seven five, greater than mu, zero point five; that sign means greater than. So friction cannot hold it; it slides back. Okay? Now, the speed at the bottom. It went up at ten, so it comes back at ten metres per second, right? Oh no, careful! Coming down, friction fights gravity, so the acceleration is only two metres per second squared, as in example one. Over the same five metres from rest, v squared is two times two times five: twenty. So v is two root five, about four point five metres per second; the wavy sign means approximately. Less than half! Puzzle solved. Got it?

7:53Quick revision: rest, sliding down and going up an incline

So, quick revision. One: at rest by itself if tan theta is at most mu s, and N is m g cos theta when nothing else presses. Two: sliding down, friction acts up, and a equals g times, sine theta minus mu cos theta. Three: going up, friction acts down, and the retardation is g times, sine theta plus mu cos theta. Now try the practice problems on this page. All the best!