Friction in Accelerating Frames

Transcript

0:00Introduction

Hello everyone! Okay, one small puzzle before we start. Hold a block against the front face of a cart, and let go just as the cart speeds up, hard. No hand, no glue, nothing. And the block stays there; it does not fall! So who is holding it up? Keep that in mind. Today: friction when the surface itself is accelerating, on a cart and on a wedge. Questions like these do come up in JEE, and one simple step handles them. So, let's begin.

0:33The method: pseudo force first, then friction

Alright, here is the whole method. Work in the frame of the accelerating body, the cart or the wedge. But an accelerating frame is not inertial, so Newton's laws do not work there as they are. So what must we add? Yes, exactly! The pseudo force: minus m a nought vector. Here a nought, the a with the little arrow on top and the small zero below, is the frame's acceleration, and m is the block's mass. The minus sign means opposite to that acceleration. Right? And then, the nice part. After adding it, treat friction exactly as for a stationary surface. Nothing new to learn! Got it? Friction can go up to its limit, mu N, okay? Mu, this Greek letter, is the coefficient of friction, and capital N is the normal reaction. But the pseudo force can press the block into the surface, so find N with it included. Most mistakes happen right here!

1:32Block on the front face of a cart

Now, the first situation: a block on the vertical front face of a cart, and the cart accelerates forwards at a. In the cart's frame, which way does the pseudo force on the block point? Backwards, correct! Opposite to the cart's acceleration, with size m times a. So it presses the block straight into the front face. With me? In the cart's frame the block is at rest, so the face pushes back just as hard: N equals m a. Okay? Now, gravity pulls the block down along the face, so friction acts up the face, against that. At most mu N, that is mu m a. Right? The block does not fall if this friction can hold the weight m g, with g the acceleration due to gravity. So mu m a greater than or equal to m g; that sign is greater than, with a line under it. Cancel m: a is at least g over mu. Heavy or light block, same answer. Clear?

2:28Block on a rough wedge

Now the harder one: a wedge. Theta, this Greek letter, is its angle, and its face is rough. We push the wedge horizontally at a, towards the side its face looks at. With me? In the wedge's frame, the pseudo force m a acts on the block, opposite to the push. Along the face, does it push the block up, or down? Up the face, yes! Gravity's part along the face, m g sine theta, pulls the block down. The pseudo force's part, m a cos theta, pushes it up. Okay? Now the normal reaction. On a fixed incline, N is m g cos theta. So here also, N is just m g cos theta, isn't it? No, no, wait! The pseudo force also has a part into the face, m a sine theta. So N equals m g cos theta plus m a sine theta: more pressing, so more friction available. Right? Now, two cases. Small a: gravity wins, and the block tends to slide down. Large a: the pseudo force wins, and it tends to slide up. Clear?

3:33The range of accelerations on the wedge

So, take the lowest acceleration first. Here the block is just about to slide down. Which way does friction act on it? Up the face, correct! It is at its limit, mu N, okay? Balance along the face, and m cancels. So a min, with the small word min below a, is g times, sine theta minus mu cos theta, over, cos theta plus mu sine theta. Got it? Now the highest acceleration. The block is about to slide up, so friction flips: down the face. So a max is g times, sine theta plus mu cos theta, over, cos theta minus mu sine theta. Only the signs of the mu terms flip. One small condition: this needs mu tan theta less than one. If mu tan theta is one or more, the denominator is not positive, and there is no upper limit at all. Right? Between these two limits, the block stays at rest on the wedge: a min, less than or equal to a, less than or equal to a max. With me? Now a topper's check. Put mu equal to zero, a smooth wedge. Both limits become g sine theta over cos theta: g tan theta, the smooth-wedge result from pseudo forces. Friction just widens it into a range. Nice, isn't it?

4:48Example 1: least cart acceleration so the block won't fall

Okay, example one, our puzzle! Read it with me: a block is held against the vertical front face of a cart only by friction, and mu is zero point five. Find the least acceleration of the cart that keeps it from falling. Pause the video, and try it first. Done? Okay, let's check. First, the friction. Limiting friction is mu m g, as usual, right? No, no, careful! The weight m g acts along the face, not into it. Into the face, in the cart's frame, is the pseudo force m a. So N equals m a, not m g. Got it? Friction acts up the face, at most mu m a, and it must hold the weight m g. So mu m a greater than or equal to m g, and a is at least g over mu. Right? Take g as ten metres per second squared. So a min is ten over zero point five: twenty metres per second squared; on screen, m s to the power minus two. And our puzzle is solved: the cart's face pushes the block, and that push lets friction hold it up. Very good!

5:58Example 2: no-slip acceleration range on a 37 degree wedge

Now a harder one. Read it with me: a block rests on the rough face of a thirty seven degree wedge, the little circle meaning degrees, and mu is zero point five. The wedge is pushed horizontally, towards the side its face looks at. Find the range of accelerations for which the block does not slip. Pause and try it. Hint: friction points a different way at each limit. Done? Okay, let's check. Take sine thirty seven degrees as zero point six, and cos thirty seven degrees as zero point eight. So mu cos theta is zero point four, and mu sine theta is zero point three. With me? Lower limit: about to slide down, so friction acts up the face. Ten times, zero point six minus zero point four, over, zero point eight plus zero point three. Two over one point one: twenty over eleven, about one point eight metres per second squared. The wavy equals sign means approximately. Okay? Upper limit: about to slide up, so friction acts down the face. Ten times, zero point six plus zero point four, over, zero point eight minus zero point three. Ten over zero point five: twenty metres per second squared. Clear? So, any push below twenty keeps the block safe, right? Oh no, careful! Below twenty over eleven, gravity wins and the block slides down the face. So the block does not slip from twenty over eleven, about one point eight, up to twenty metres per second squared. Perfect!

7:39Quick revision: pseudo force first, then friction

So, quick revision. In an accelerating frame, add the pseudo force, minus m a nought. Then friction as usual, but find N with the pseudo force in. Cart: the acceleration must be at least g over mu. Wedge: it lies between a min and a max. Now go and try the practice problems on this page. All the best!