Rough Horizontal Surfaces

Transcript

0:00Introduction

Hello everyone! Okay, one small puzzle before we start. There is a ten kilogram box on a rough floor, with a rope tied to it. Pull the rope straight along the floor, and you need seventy five newtons just to get it moving. Now tilt the rope upwards a little. Will you need more force, or less? Most students say more: part of your pull is wasted going upwards, right? Hold on to your answer; Example two will settle it. Today: blocks on rough horizontal floors. Questions on these come up often in JEE, so let's begin.

0:36Sliding freely on a rough floor

First, the simplest case. Picture a block of mass m sliding on a level, rough floor, and nothing else pushes it. Vertically, the floor's push, capital N, the normal reaction, balances the weight, m g. And the friction, small f, is mu times N, that is, mu m g. Mu, this Greek letter, is the coefficient of friction; while the block slides, it is the kinetic one. With me? Now tell me. Take a heavier block, same floor, same speed. More weight means more friction. So it stops faster, right? No, no, careful! More friction, yes, but also more mass to slow down. Retardation, which means deceleration, is friction over mass: mu m g over m, which is just mu g. The mass cancels completely. Clear? Now, if it starts with speed u, how far does it go? Use the equations of motion with a final speed of zero, and you get the stopping distance: s equals u squared over two mu g. The little two up on u means squared. Okay? And the stopping time: t equals u over mu g. Speed lost, over the retardation. Really simple, right? Now a quick test. Double the starting speed. Does the stopping distance also double? Oh no, careful! The distance goes with u squared. This sign, which looks like an open alpha, means is proportional to. So double the speed, and the stopping distance becomes four times. Classic trap. Got it?

2:14Force at an angle: pull or push?

Now, apply a force, capital F, at an angle theta, this Greek letter, above the horizontal: a pull, like a rope. Its upward part, F sine theta, takes some of the weight off the floor, so the floor has to push less. So N equals m g minus F sine theta. With me? Now the same force, at the same angle, but pushing from behind, below the horizontal. What happens to N now? It goes up! Now F sine theta presses the block into the floor. So N equals m g plus F sine theta. Okay? And friction is mu times the normal reaction. Pull, and N falls, so friction falls. Push, and N rises, so friction rises. So, pulling is easier than pushing. Remember this one; it comes up often. Clear?

3:06The least force that moves the block

Now, back to our rope, pulling at angle theta. How much force just gets the block moving? The forward part, F cos theta, must equal the largest friction: mu times the normal reaction. And that normal reaction is m g minus F sine theta, the part inside the bracket. Right? Collect the F terms on one side. F equals mu m g, over cos theta plus mu sine theta. Got it? Now, which angle makes this force the least? Hint: look at the bottom of the fraction. Yes, exactly! The force is least when the bottom is largest. And that happens when tan theta equals mu. That is the topper's trick: the best angle has its tan equal to mu. Okay? At that angle, the bottom becomes the root of one plus mu squared; the long tick is the root sign. So F min, where the small word below F is short for minimum, equals mu m g over root of one plus mu squared. It acts at tan inverse mu above the horizontal; the minus one up on tan means inverse tan. With me?

4:15Block pressed against a wall

One more situation. Press a block against a vertical wall, using a horizontal force F, pointing into the wall. The wall pushes back just as hard, so here N equals F, okay? But F is horizontal, and the weight pulls straight down. So what holds the block up? Friction, correct! The block tends to slide down, so friction acts up, along the wall. And it can be at most mu times N, that is, mu times F, right? So the block does not slip if mu F is greater than or equal to m g; that is this sign, with the line under it. Press harder, and more friction is available. Clear?

4:57Example 1: stopping distance and time of a braking car

Example one. Read it with me: a car moving at twenty metres per second, written m s to the power minus one, brakes with locked wheels on a road with mu equal to zero point five. Find the stopping distance and the time. Pause the video, and try it first. Done? Okay, first: the mass of the car is not given. Can we still solve it? Yes, we can! Locked wheels means the tyres slide on the road, so the car is just like our block sliding freely, and its mass cancels. And we take g as ten metres per second squared, written m s to the power minus two, okay? So the retardation is mu g: zero point five times ten, that is five metres per second squared. Right? Now the stopping distance: u squared over two mu g. Twenty squared is four hundred, over two times five, which is ten. Forty metres. With me? And the time: u over mu g, so twenty over five. Four seconds. That's it. Perfect!

6:03Example 2: least force to drag a 10 kg block

Now, remember our puzzle? Here it is. Read it with me: find the least force that can drag a ten kilogram block along a floor with mu equal to zero point seven five, and its direction. Pause the video, and try it. Done? Okay, let's check. First, the obvious way: pull straight along the floor. Then N is just m g, one hundred newtons, and you need mu m g: zero point seven five times one hundred, which is seventy five newtons. Careful: the upright N after a number is the newton, not the normal reaction. Okay? So, is seventy five newtons the least? Most students say yes. What do you say? No! Tilt the rope. The force is least when tan theta equals mu, zero point seven five. So theta is about thirty seven degrees: the wavy equals sign means approximately, and the small circle means degrees. Right? Now the bottom: root of one plus mu squared. Mu squared is zero point five six two five, so we get root of one point five six two five, which is one point two five. Clear? So F min is seventy five over one point two five: sixty newtons, at thirty seven degrees above the horizontal. Less than seventy five! Puzzle settled: tilting the rope helps. Perfect! Quick check. With tan theta exactly zero point seven five, sine theta is zero point six and cos theta is zero point eight. N is one hundred minus sixty times zero point six: sixty four newtons. Friction, mu N, is forty eight newtons, and the forward pull, sixty times zero point eight, is also forty eight. They match! Got it?

7:55Quick revision: stopping distance, pull vs push, least force

So, quick revision. Sliding freely, the retardation is mu g, the stopping distance is u squared over two mu g, and the stopping time is u over mu g. Pull at an angle, and N is m g minus F sine theta; push, and it is m g plus F sine theta. So pulling is easier. The least force is mu m g over root of one plus mu squared, at tan inverse mu above the horizontal. And on a wall, no slipping if mu F is at least m g. Now go and try the practice problems on this page. All the best!