Pascal's Law and Hydraulic Machines

Transcript

0:00Introduction

Hello everyone! Okay, a small puzzle first. At a service centre, a car of fifteen hundred kilograms sits on a hydraulic lift. And the force holding it up, on the other side, is just one hundred and forty seven newtons. That is the weight of a fifteen kilogram bag! Small force, huge load. So, is this free energy? Hmm, then where is the catch? Today we learn Pascal's law, and by the end, we will find that catch ourselves. This idea does appear in JEE questions, so let's begin.

0:36Pascal's law: pressure travels undiminished

First, a quick recap. Pressure, capital P, equals F over A: the normal force, at right angles to the surface, divided by the area. Its unit is the pascal, capital P small a: one newton per square metre. The minus two on the metre means per square metre, okay? Now, take a closed container full of liquid, with a piston. The fluid is enclosed and incompressible: its volume does not change when squeezed. Push the piston, and the pressure below it rises by delta P, where delta, the little triangle, means change in. Does this extra pressure arrive weaker at the far end? No, not at all! Pascal's law says: a change of pressure applied to an enclosed, incompressible fluid is transmitted undiminished, without any loss, to every part of the fluid and the walls. Every single point gets that full extra pressure. Clear? Now, a very common mistake. Every point gets the same delta P, so every part feels the same force too, right? No, no, careful! Pascal's law transmits pressure, not force. Force equals pressure times area; the cross means times. So the same delta P on a bigger area gives a bigger force. That is the whole secret of hydraulic machines. Got it? And in a car's hydraulic brakes, the pedal raises the pressure in the brake fluid. The same delta P reaches every wheel cylinder. Their pistons are larger than the one at the pedal, so they press the brake pads with a larger force. With me?

2:14The hydraulic lift: small force, large load

So, the hydraulic lift: two cylinders joined at the bottom, full of liquid, each with a piston. The small piston has area A one, and we push it with force F one. The small one written below is a subscript. The large piston has area A two, and carries the load, F two. Okay? Now tell me. Under the two pistons, what is the same: the force, or the pressure? The pressure, exactly! So F one over A one equals F two over A two. Write this down. This is the heart of today's lesson. Clear? Rearrange it: F two equals F one, times A two over A one. So the force is multiplied by the area ratio. Give the large piston a hundred times the area, and your push becomes a hundred times the force. With me? Now a trap. Many questions give diameters, not areas. So if you double the diameter of the large piston, the force doubles too, right? No, no, wait! For a circular piston, the area is pi d squared over four, with d the diameter and pi the Greek letter on screen. The double arrow means therefore. So the area ratio is the diameter ratio squared: the small two outside the bracket. Double the diameter, and the force becomes four times. Got it?

3:32No free energy: the small piston moves further

Alright, back to our puzzle. Small force, huge load. So do we get out more energy than we put in? Of course not! Say the small piston moves down a distance d one, and the large one moves up d two. The work we put in is F one times d one; the work that comes out is F two times d two. Okay? And energy is conserved, so F one d one equals F two d two. Nothing comes for free. Right? So when F one is small, d one must be large. The small piston moves a long way, and the large one rises only a little. That is the catch! With me? Another way to see it. The fluid is incompressible, so the volume leaving the small cylinder equals the volume entering the large one. Volume is area times distance, so A one d one equals A two d two. Clear? So here is a topper's trick. The machine multiplies the force by A two over A one, and divides the distance by the same ratio. The line with two dots means divided by. Okay?

4:40Example 1: force on a 5 cm2 piston that holds up a car

Now, let's solve our car puzzle. The car of mass fifteen hundred kilograms rests on a lift piston of area zero point zero five square metres; the small two above the m means square metres. What force on a piston of area five square centimetres holds it up? Take g as nine point eight. Pause the video, and try it first. Done? Okay, let's check. The load is the weight of the car: F two equals m g, fifteen hundred times nine point eight. Fourteen thousand seven hundred newtons, the capital N on screen. Okay? Now the small piston: five square centimetres. One centimetre is ten to the power minus two metres, so five square centimetres is five times ten to the power minus two square metres, right? No, no, careful! Many students lose marks here. The unit is squared, so the conversion is squared too: one square centimetre is ten to the power minus four square metres. So A one is five times ten to the power minus four square metres. Clear? Now Pascal's law: F one equals F two, times A one over A two. That is fourteen thousand seven hundred, times five times ten to the power minus four, over zero point zero five. Okay? The area ratio is zero point zero one, just one hundredth. So F one is fourteen thousand seven hundred times zero point zero one: one hundred and forty seven newtons! That is the small force from our puzzle. And with the wrong conversion, your answer would be a hundred times too big. Right?

6:16Example 2: how far the car rises for a 20 cm push

Now, one step harder. In the same lift, the small piston is pushed down twenty centimetres. How far does the car rise? Pause and try. Hint: think about volume. Alright. Both pistons sit on the same fluid. So if one goes down twenty centimetres, the car goes up twenty centimetres, right? Oh no, careful! The fluid is incompressible, so the same volume moves across: A one d one equals A two d two. So d two equals d one, times A one over A two. With me? And A one over A two is zero point zero one. So d two is twenty times zero point zero one: zero point two centimetres, which is two millimetres, the m m on screen. Push twenty centimetres, and the car rises just two millimetres! Got it? Let us check with energy. Twenty centimetres is zero point two zero metres. So F one d one is one hundred and forty seven times zero point two zero: twenty nine point four joules, the capital J on screen. Okay? And two millimetres is zero point zero zero two metres. So F two d two is fourteen thousand seven hundred times zero point zero zero two: again twenty nine point four joules. Equal! You get a hundred times the force, but push a hundred times further. No free energy, just a fair trade.

7:39Quick revision: Pascal's law, F1/A1 = F2/A2, F1d1 = F2d2

Clear? So, quick revision. Pascal's law: a change of pressure on an enclosed, incompressible fluid reaches everywhere, undiminished. Hydraulic lift: F one over A one equals F two over A two; with diameters, square the ratio. Energy: F one d one equals F two d two, so the small piston moves much further. Now try the practice problems on this page. Watch out for the diameter twist! All the best!