Static and Kinetic Friction

Transcript

0:00Introduction

Hello everyone! One small puzzle first. There is a heavy box on the floor. You push it gently. It does not move. You push harder. Still nothing. Then you push really hard, and suddenly it slides, and now it feels easier to keep it moving! So, is friction one fixed number, or does it change? Hmm. Today: static and kinetic friction. In JEE, friction questions come up often, and they build on today's idea. So, let's begin.

0:32What is friction?

So, first: what exactly is friction? When two surfaces touch, they push on each other with a contact force. The part perpendicular to the surface is the normal reaction, capital N, and the part along the surface, small f, is friction. Okay? If the surfaces are sliding, it is kinetic friction, and it opposes the relative motion. If they are not sliding yet, but tend to, it is static friction, and it opposes the tendency of relative motion. Notice the word relative. Right? Let's test it. There is a crate on the floor of a truck, and the truck starts to accelerate forwards. Which way does friction act on the crate? Forwards! If you said backwards, careful, that is the trap. Relative to the floor, the crate tends to slide backwards. Friction opposes that tendency, so it acts forwards, and gives the crate its forward acceleration. Got it?

1:26Static friction adjusts itself, up to a limit

Now look at this graph. We pull a block on a floor horizontally with a force, capital F, on the horizontal axis. Small f, on the vertical axis, is the friction. The block does not move. We pull harder and harder. What does friction do? It grows with the pull, exactly! The block stays at rest, so friction balances the pull: f equals capital F, the blue line. Static friction adjusts itself. With me? But only up to a limit. f sub s, small s for static, is less than or equal to mu sub s times N; that is the sign with a line under it. Mu is the Greek letter, and mu sub s is the coefficient of static friction. This largest value is the limiting friction, the top dashed line. Now the pull goes past it, and the block slips. Does friction rise, stay the same, or drop? It drops, to the green line! Once slipping starts, friction is kinetic: f sub k equals mu sub k times N, with mu sub k the coefficient of kinetic friction; that is the lower dashed line. The normal force N does not change, so however hard we pull, f sub k stays the same: the green line is flat. And it is nearly independent of speed too. Right? And usually, mu k is less than mu s; see the less-than sign. So, our puzzle! Once the box slides, friction drops below the static limit. That is why it felt easier to keep it moving. Clear?

3:00The trap: static friction is not always at its limit

Now, the biggest trap here. There is a block resting on a rough horizontal floor, and nothing is pushing it. What is the friction on it? Mu s times m g? No, no, careful! It is zero. There is no tendency to slide, so no friction is needed. Remember, mu s N is only the largest possible static friction. Got it? So how do we find static friction? First, use Newton's laws to find the friction needed for no slipping. Then compare it with mu s N, okay? If the friction needed is at most mu s N, there is no slipping, and friction is just that needed value. If it is more, the block slips, and friction becomes mu k N, right? And one more law. For the same normal force, friction does not depend on the area of contact. Wide face down or narrow face down, same friction. Clear?

3:53Angle of friction and angle of repose

Now, two angles. N and f together make the total contact force. When the block is just about to slip, the angle between this total force and the normal is the angle of friction: lambda, the Greek letter. Friction is along the surface, N is perpendicular, and f is now at its limit. So, what is tan lambda? Yes, exactly! Tan lambda is f over N, and f is at its limit, mu s N, so we get mu s N over N, which is just mu s. Tan lambda equals mu s. Really simple, okay? Next, the angle of repose, theta sub r: theta is the Greek letter, and small r is for repose. It is the steepest incline on which a block can rest without sliding. With me? And tan theta r also equals mu s. So theta r equals lambda: the angle of repose equals the angle of friction. Same angle, two names! Clear?

4:54Example 1: friction on a 10 kg box at 30, 50 and 60 N

Now, remember our puzzle? Here it is with numbers. There is a ten kilogram box, k g for kilograms, on a floor with mu s zero point five and mu k zero point four. Find the friction when it is pushed horizontally with, part a, thirty newtons; part b, fifty; part c, sixty. The upright N after each number is newtons. And find its acceleration in part c. Pause the video, and try it first. Done? Okay, let's check. Take g as ten metres per second squared, the m s to the power minus two. The push is horizontal, so N equals m g, and the limiting friction, mu s m g, is zero point five times ten times ten: fifty newtons. Right? Part a: thirty newtons. So the friction is the limiting friction, fifty newtons, yes? No, no, careful! That is our trap. Thirty is less than fifty, so the box stays at rest, and static friction only balances the push: thirty newtons. Clear? Part b: fifty newtons, exactly the limit. The box is just on the point of slipping, and friction is fifty newtons, its largest value. Got it? Part c: sixty is more than fifty, so static friction cannot hold it. The box slides! Friction is now kinetic: mu k m g, zero point four times one hundred, forty newtons. Friction dropped from fifty to forty, just like our puzzle. Okay? And the acceleration: net force over mass. Sixty minus forty is twenty, over ten: two metres per second squared. Not sixty minus fifty; once it slides, use the kinetic value. Right?

6:41Example 2: angle of friction and total contact force

Now a slightly harder one. For surfaces with mu s equal to zero point seven five, find the angle of friction, and the total contact force on a four kilogram block that is just about to slide on a horizontal floor. Pause the video, and try it. Done? Okay, let's check. Tan lambda equals mu s: zero point seven five, which is three over four. So lambda is thirty seven degrees, the standard value we take when tan is three over four; the little circle means degrees. Clear? Now the forces. The floor is horizontal and nothing else presses on the block, so N equals m g: four times ten, forty newtons. Okay? Just about to slide, so friction is at its limit, mu s N: zero point seven five times forty, thirty newtons. This time, mu s N is correct. Got it? Now the total contact force. Forty plus thirty, so seventy newtons, right? No, no, wait! N and f are perpendicular, so we combine them like the sides of a right triangle. The tick-shaped sign is the square root: root of forty squared plus thirty squared, fifty newtons. And it makes the angle of friction with the normal: thirty seven degrees. Three, four, five again! With me?

8:04Quick revision: static vs kinetic friction, angle of repose

So, quick revision. Static friction adjusts itself, up to mu s N, only as much as needed. Once the block slides, friction is mu k N, usually smaller. And tan lambda and tan theta r both equal mu s. Now go and try the practice problems on this page. You will find them easy, I promise. All the best!