Mixed Problems: Set 2 (Friction and Accelerated Frames)

Transcript

0:00Introduction

Hello everyone! Okay, one small puzzle before we start. You push a five kilogram box on a floor with fifteen newtons. Your friend says: mu for this floor is point four, friction is mu times N, so it is twenty newtons. Twenty backwards, against your fifteen forwards? Then the box should move towards you! That cannot be right, can it? Today is set two of the mixed JEE Main problems: friction and accelerated frames. Questions like these come up often. Three quick ideas, then two full problems. So, let's begin.

0:38Idea 1: static friction is only as much as needed

So, idea number one, and it cracks our puzzle. Here is the box: five k g, five kilograms, on a floor with mu s equal to point four. Mu is this Greek letter, the coefficient of friction, and the small s means static. The push is fifteen newtons; the upright N after fifteen means newtons. Okay? First, the limit: mu s times capital N, where capital N is the normal reaction. The floor is level and nothing else presses down, so N equals m g, the weight. With g equal to ten, as this chapter takes it: point four times five times ten, twenty newtons. Clear? So your friend was right? The friction on the box is twenty newtons? No, no, careful! Twenty newtons is only the largest friction this floor can give. Look at the graph: capital F is the pull, small f is the friction. At rest, f equals F, up to the dashed level mu s N, okay? Beyond that, the box slips, and friction drops to mu k N, with small k for kinetic. Our push is fifteen, less than twenty, this less than sign. So the box stays at rest. At rest means no net force. So friction balances the push: fifteen newtons, not twenty. Puzzle solved! Static friction adjusts itself: only as big as it needs to be, up to its limit. Got it?

2:04Idea 2: constant speed down a rough incline

Alright, idea number two. Here, a block slides down a thirty seven degree incline at constant speed; the little circle means degrees. Find mu k, the coefficient of kinetic friction. In the figure, theta, this Greek letter, is the angle of the incline. On the left, the block tends to slide down, so friction f acts up the slope. On the right, it moves up with velocity v, so friction acts down. Friction opposes the slipping, okay? Our block slides down, with nothing else pushing it. So, a equals g times, sine theta minus mu k cos theta. The g sine theta part pulls it down the slope; the g times mu k cos theta part is friction holding it back. Right? Now, the block moves at constant speed, in a straight line. So what is its acceleration? Zero, exactly! No change in velocity, so no acceleration. Put a equal to zero, and g drops out: sine theta equals mu k cos theta. With me? Divide by cos theta. Mu k equals tan theta: tan thirty seven degrees, taken as three over four, point seven five. Topper's trick: a block sliding down at constant speed means mu k is just tan theta. No mass, no g, one line! Clear?

3:24Idea 3: a pseudo force in an accelerating car

Now, idea number three: accelerated frames. Here, a bob hangs in a car accelerating at ten m s to the power minus two: metres per second squared. Find the angle of the string, and the tension. In the left half of the figure, theta is the angle of the string from the vertical, and the arrow a, with a small arrow on top for a vector, is the car's acceleration. The right half is the same trick on a wedge; you will meet it in this set. Sit inside the car. The car accelerates, so its frame is not inertial, and Newton's laws need a fix there. Give every body a pseudo force, m times a, opposite to the car's acceleration: the red arrow, m a, pointing backwards. But careful: the pseudo force exists only in the car's frame. From the ground, there is none, so never mix the two frames in one equation. Okay? In the car's frame, the bob hangs at rest. So the tension in the string must balance which forces? Both together, m g, the weight, and m a, yes! Capital T is the tension, and the string lines up along their resultant. Right? So tan theta is m a over m g, that is, a over g. Ten over ten, one. So theta is forty five degrees, with the string leaning backwards, as in the figure. And the tension is the size of that resultant: m times the square root of, g squared plus a squared. That is m root two hundred, ten root two times m, where m is the mass of the bob. One pseudo force, and the car becomes a simple equilibrium problem. Got it?

5:02Example 1: block pressed against a wall, does it slip

Now, Example one. There is a one kilogram block, pressed against a wall by a horizontal force of forty newtons, and mu is point three. Does it slip? And what is the friction? Pause the video, and try it first. Done? Okay, let's check. Look at what presses the block into the wall: the forty newton push. So the wall pushes back with N equal to forty newtons. Not m g! The weight acts along the wall, not into it. Okay? So the limit of friction is mu times N: point three times forty, twelve newtons. Clear? So the friction on the block is twelve newtons, right? Oh no, careful! Remember idea one. First ask what friction has to hold up. Only the weight: m g, one times ten, ten newtons. And ten is less than twelve, so the wall can hold it. So the block does not slip. Friction is ten newtons, upwards, just enough to balance the weight. Twelve was only the limit. Same trap as our puzzle, isn't it?

6:07Example 2: crate on an accelerating truck, does it slip

Now a harder one, friction and an accelerating frame together. There is a truck accelerating at four metres per second squared, with a crate on its floor; mu s and mu k are both point three. Does the crate slip? And what is its acceleration relative to the truck? Pause and try. Hint: find the most that friction can give. Done? Let's check. Friction is the only horizontal force on the crate, at most mu s m g. Divide by the mass m: the most acceleration it can give is mu s g, point three times ten, three metres per second squared. The mass cancels. Okay? But the truck speeds up at four. So, can the crate keep up with the truck? No, it cannot! Four is more than three; this sign is greater than. So the crate slips, and friction becomes kinetic. Right? Now, careful with the direction. The crate slips backwards on the floor, so kinetic friction on it acts forwards. So from the ground, the crate still speeds up forwards, at mu k g, three metres per second squared. The small word crate written under the a just means the crate's acceleration. With me? Relative to the truck: three minus four, minus one. So the crate slips at one metre per second squared, backwards, towards the rear of the truck. Clear? Quick check with idea three. In the truck's frame: pseudo force, m times four, backwards; friction, m times three, forwards. Net, m times one, backwards: the same answer. Two frames, one answer, just never mixed. Got it?

7:49Quick revision: static friction, incline, pseudo force

So, quick revision. Static friction is only what is needed, at most mu s N, so always compare first. Constant speed down an incline means mu k equals tan theta. In an accelerating frame, add the pseudo force m a, opposite to the acceleration, and never mix frames. Now go and try the practice problems on this page. You will enjoy them, I promise. All the best!