Wedges

Transcript

0:00Introduction

Hello everyone! Okay, a puzzle. Put a block on a perfectly smooth slope and let go. It slides down, right? Now I push the slope sideways, hard enough, and the block just sits there. Push harder, and it even climbs up! How can a smooth slope hold a block? That is today's lesson: wedges. Wedge problems come up often in JEE, and one idea, the pseudo force, makes them simple. So, let's begin.

0:30Sit on the wedge: add the pseudo force

Look at the figure, on the right. Ignore the car on the left for now. Here is a wedge with angle theta, the Greek letter, at its base. And all surfaces are smooth: no friction anywhere. Okay? The wedge accelerates to the left: the dark arrow pointing left, labelled a with a little arrow on top. We call it a nought, a with a small zero below. Notice: it moves towards the side the face looks at, so the face pushes into the block. Right? Now sit on the wedge. It is accelerating, so this frame is not inertial. So, what extra force must we add on the block? Yes, exactly! The pseudo force, m times a nought, opposite to the wedge's acceleration. That is the red arrow, m a, pointing right, into the face. Plus the weight m g, the green arrow, and the normal reaction N, the blue arrow. Clear?

1:26Resolve along and across the face

Now resolve along the face. The weight has a part pulling down the slope. And the pseudo force: does its part along the face point up the slope, or down? Up the slope, correct! That part is m a nought cos theta, against the weight's part, m g sine theta. Cancel m, and we get a rel, with small letters r e l below: the block's acceleration relative to the wedge. So, a rel equals g sine theta minus a nought cos theta, down the slope. Got it? Now across the face. The block does not leave it, so the forces across it balance. Both the weight and the pseudo force press into the face. So, N equals m times, bracket, g cos theta plus a nought sine theta. More than m g cos theta! With me? Now, back to our puzzle. When will the block not slide at all? Yes, when a rel is zero! Then g sine theta equals a nought cos theta, so a nought equals g tan theta. That is the figure's label too: at rest if a equals g tan theta. Puzzle solved! Okay? Push harder, so a nought is greater than g tan theta; that is the greater-than sign, opening towards a nought. Then a rel turns negative, and the block moves up the face: our climbing block!

2:45Trap: which way does the wedge accelerate?

Right? Now, the trap. So far the wedge accelerated towards the side its face looks at. What if it accelerates the other way, away from the face? Same formulas? No, no, careful! Now the pseudo force points away from the face. So, change the sign: the arrow on screen means replace a nought with minus a nought. Okay? So, a rel becomes g sine theta plus a nought cos theta: faster down the slope. And N becomes m times, g cos theta minus a nought sine theta: smaller. Right? Since N got smaller, check that N is greater than zero. Then the block stays on the face. Clear?

3:27A free wedge on a smooth floor

One more case: nobody pushes the wedge. Small m is the block, capital M the wedge, and the floor is smooth, so the wedge is free. As the block slides down, will the wedge stay still? No! The block presses on the face and pushes the wedge away, with an acceleration capital A of its own. So the pseudo force on the block points away from the face: N equals m times, g cos theta minus capital A sine theta. And for the wedge, N sine theta equals capital M times capital A, horizontally. With me? Eliminate N, and capital A equals m g sine theta cos theta, over, capital M plus m sine squared theta. The small two up on sine means sine squared. Okay? And relative to the wedge, a rel equals, bracket capital M plus m, times g sine theta, over the same denominator. Both need a smooth face and a smooth floor. Got it?

4:21Example 1: wedge acceleration so the block does not slide

Example one: our puzzle with numbers. So, a wedge of angle thirty seven degrees, the little circle means degrees, is pushed horizontally so that a one kilogram block, the k g, on its smooth face does not slide. Find the wedge's acceleration and the normal force on the block. 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 minus two. Sine thirty seven is three fifths, cos thirty seven four fifths, so tan thirty seven is three quarters. Right? No sliding means a nought equals g tan theta. Ten times three quarters: seven point five metres per second squared. Clear? Now the normal force. Many students write just m g cos thirty seven: eight newtons, and the capital N after the eight is the newton, the unit of force. Is that right? No, no, wait! The pseudo force also presses into the face. So, N equals m times, g cos thirty seven plus a nought sine thirty seven: one times, ten times four fifths, plus seven point five times three fifths. Okay? That is eight plus four point five: twelve point five newtons. Much more than eight! Clear? Topper's check. The block moves with the wedge, so the normal force alone holds up m g and gives m a nought sideways. So, N is the root of ten squared plus seven point five squared: twelve point five newtons again. Perfect!

5:52Example 2: time to slide down a wedge moving away

Now a harder one. So, a smooth wedge of angle thirty degrees moves with a constant two metres per second squared, away from the side its face looks at. Then a block starts from rest at the top of the face, eight point eight metres along the slope from the bottom. How long does it take to reach the bottom, and does it stay on? Pause and try. Hint: which way does the pseudo force point? Done? Okay. Many students jump to our first formula: a rel equals g sine thirty minus two cos thirty. Correct? No, no, careful! That formula is for a wedge accelerating towards the face. Here it moves away, so a nought is minus two metres per second squared. Change the sign! So, a rel equals g sine thirty plus two cos thirty. That is five plus root three, about six point seven three metres per second squared, down the slope; wavy equals means approximately. Okay? Now, the stay-on check. N equals m times, g cos thirty minus two sine thirty, that is m times, five root three minus one. Five root three is much bigger than one, so N is greater than zero: it stays on. Right? Last part: the time. Relative to the wedge, the block starts from rest with constant acceleration. Which equation of motion? Yes, distance equals half times acceleration times t squared. Eight point eight equals half times six point seven three times t squared. The double arrow means which gives: t equals root of two times eight point eight, over six point seven three. About one point six two seconds, the small s. Very good!

7:40Quick revision: pseudo force, when it rests, the sign trap

So, quick revision. Sit on the wedge, add the pseudo force m a nought, and resolve along and across the face. Towards the face: a rel equals g sine theta minus a nought cos theta, and at rest when a nought equals g tan theta. Away from the face: change the sign of a nought, and check N greater than zero. Now try the practice problems on this page. All the best!