now let's start this video with a problem a downward force of 100 newtons is applied to the small piston with a diameter of 50 centimeters in the hydraulic lift system shown below what is the upward force exerted by the large piston with a diameter of 2 meters so if we push down the small piston with a force let's call it f1 the large piston is going to be pushed up with a force of f2 now according to pascal's law the pressure applied to an enclosed fluid is transmitted undiminished to every portion of that fluid so what does it mean with regard to the hydraulic lift system this means that the pressure exerted on a small piston is equal to the pressure exerted by the large piston so we could say that p1 is therefore equal to p2 and pressure is defined as force divided by area so since we're looking for f2 let's solve for it let's multiply both sides by a2 so here's the equation that you need the output force or the upward force exerted by the large piston is going to be f1 multiplied by the ratio of the areas of the large piston with respect to the small piston so if you exert a downward force on a small piston you're going to get a larger upward force on the large piston so the piston with the larger area is going to be associated with the greater force and that's what you want to take from this hydroglyph system it's a force multiplying device you can apply a small force on a small piston and get a larger force on the large piston so let's go ahead and answer part a now the area is pi r squared because the pistons are circular nature now we're given the diameter the radius is half of the diameter so if you have a circle here's the diameter and this is the radius so f2 is going to be f1 which is 100 newtons times a2 which is pi r squared and the diameter of the large piston is 2 meters which means the radius is one meter now the radius of the small piston is 25 centimeters since the diameter is 50. it's half of that but we need to convert 25 centimeters into meters so if we divide that by 100 that's going to be point 25 meters now we could cancel pi so one squared divided by .25 squared is 16 and 16 times 100 is 1600 so f2 the output force is 1600 newtons so that's the answer to part a so let's move on to part b what is the mechanical advantage of this hydraulic lift system mechanical advantage is the ratio between the output force and the input force so the output force the force exerted by the machine is f2 which is 1600 newtons the input force is the force that we applied to the machine that's 100 newtons so the mechanical advantage of this particular hydraulic lift system is 16. so if we were to apply an input force of a thousand this particular machine would multiply that input force by 16. so we would get an output force of 16 000 and so that's what the mechanical advantage of a machine tells you it tells you what factor the machine will multiply your input force so if we apply a force of 10 newtons the machine is going to multiply that by 16 and exert an output force of 160 newtons now let's move on to part c if the input force of a hundred newtons pushes the small piston down by two meters how high will the large piston rise so we're going to push this down by a distance of two meters how high will this piston rise let's call it d2 now there's two ways to get this answer the first way is to realize that the volume of the fluid inside the hydraulic lift system is the same so typically there's some type of fluid inside this hydroglyph system and the change in volume in the small piston has to equal the change in volume of the large piston because the volume of the fluid can't change so therefore we need to say that v1 is equal to v2 the volume is the product of the area and the height so v1 is a1 times h1 but we use d to represent the height so you can say d is vertical displacement so a1 times d1 and v2 is going to be a2 times d2 so a1 that's pi r1 squared since we still have a circular cylinder and a2 is going to be pi times r2 squared so we don't need pi we could cancel that in this problem so r1 that's the radius of the small piston is 25 centimeters which is point 25 meters and then d1 we're pushing down the small piston by two meters so d1 is two r2 that's half of two so that's one and so our goal is to calculate d2 so it's going to be .25 squared times two so d2 is 0.125 meters so what does this mean it means that there's no such thing as a free lunch the machine didn't multiply our force by a factor of 16 from nothing there was a cost that was a price to pay for notice that the small force is associated with a longer distance and the large force is associated with a smaller distance so in order to multiply the force by a factor of 16 we had to pay the price by exerting the small force for a longer distance and so the machine exerted a large force for a shorter distance and so that's why the force was multiplied the work has to be the same the work done by the small force has to equal the work done by the large force because energy has to be conserved and so we could get this answer as well by using the conservation of energy principle so the input work has to equal the output work so the work that we apply has to equal the work that the machine will do if there's no friction so this is the ideal situation if this machine is 100 efficient the energy that we apply to the machine has to equal the energy that the machine puts out as well so the work done by the input force is just force times displacement f1 times d1 the work done by the output force is going to be the same f2 times d2 so we applied a small force of 100 newtons for a distance of 2 meters and so the large force of 1600 newtons will be applied for a shorter distance of 0.125 so it's going to be 100 times 2 divided by 1600 and thus that will give you the same answer of 0.125 meters so before we start this problem you may want to pause the video to see if you could do it yourself so we have a 1500 kilogram car and it's placed on a large piston which has a radius of four meters what is the minimum upward force that the large piston must exert to lift the car now gravity exerts a downward weight force on the car which is mg and the piston exerts an upward force f2 on the car so the minimum upward force that the piston has to exert in order to lift the car at constant speed is mg if f2 exceeds mg then the car is accelerating upward but to lift the car at constant speed those two forces have to be equal so the net force is zero so we can say that f2 the minimum upward force has to be at least mg so that's going to be 1500 kilograms times the gravitational acceleration of 9.8 so in this example f2 is 14 700 newtons so that's the answer to part a now let's move on to part b if the mechanical advantage of the hydraulic lift system is 20 what is the minimum downward force that should be applied to lift the car so we know that a small force is associated with the small piston and a large force is associated with the large piston so if the mechanical advantage is 20 that means that the small force has to be 20 times less than a large force so it's going to be 14 700 divided by 20. so therefore f1 is 735 newtons for those of you who want to use an equation to get the answer here's what you need to do the mechanical advantage is the ratio between the output force and the input force so we're looking for f1 so first let's multiply both sides by f1 so f1 times mechanical advantage is equal to f2 so now we need to divide both sides by the mechanical advantage so what we have left over is that f1 is going to be f2 divided by the mechanical advantage which is what we did and that's 14 700 divided by 20 and that gave us an input force of 735 newtons now let's move on to part c what is the radius of the small piston so how can we get the answer to that question well according to pascal's law we know that the pressure exerted on these two pistons have to be the same so that means that f1 over a1 has to equal f2 over a2 now i'm going to rearrange the equations so first let's cross multiply so first we have f1 times a2 that's equal to f2 times a1 now i'm going to divide both sides by f1 and simultaneously by a1 so on the right side a1 will cancel and on the left side f1 will cancel so what this tells us is that the ratio of the forces f2 over f1 is equal to the ratio of the areas a2 over a1 and we know that mechanical advantage is basically equal to the output force over the input force the output force being f2 and the input force being f1 so we could say that the mechanical advantage is basically the ratio between the areas of the large piston and the small piston and in this example the area is pi r squared so a2 is going to be pi r2 squared and a1 is pi r1 squared so we could cancel pi in this problem the mechanical advantage is 20 r2 that's the radius of the large piston that's 4 meters so that's going to be 4 squared and we're looking for r1 so let's cross multiply so 1 times 4 squared is 16 and that's going to equal 20 times r1 squared so we need to divide both sides by 20. so 16 divided by 20 as a decimal that's 0.8 so 0.8 is equal to r1 squared so now we need to take the square root of both sides so r1 is point eight nine four meters as you could see it's much less than four meters so that's how you could find the radius of the small cylinder or the small piston if you know the radius of the large piston and if you know the mechanical advantage now let's move on to part d what is the pressure exerted by the large piston so pressure is force divided by area so the force exerted by the large piston we know it's 14 700 newtons and the area is pi r squared and the radius of the large piston is 4 meters so it's fourteen thousand seven hundred divided by pi times four squared or sixteen so the pressure is two hundred ninety two point pascals keep in mind one pascal is one newton per square meter so that's the pressure in the fluid and that pressure is exerted throughout the fluid so the pressure on the small piston is the same as the pressure on the large piston keep that in mind and that's based on pascal's principle part e how far should the small piston descend so looking for d1 if we want to lift the car by two meters so d2 is two meters the easiest way to get the answer is to use conservation of energy so the work done by the small piston is equal to the work done by the large piston or rather the work done on the small piston is equal to the work done by the large piston so f1 d1 is equal to f2 d2 so f1 we said was 735 newtons and d1 we're gonna push down the small piston by some unknown amount we need to calculate d1 and f2 that's the large force that's 14700 and we know that d2 is 2 meters that's how high the car is being lifted up now keep this in mind the large force is always associated with the shorter distance the small force has to be exerted over a longer distance so d1 should be much larger than two meters in fact since the mechanical advantage is 20 we know that the distance that we have to push the small cylinder down has to be 20 times the distance that the car is going to lift up so the mechanical advantage is the ratio between d1 and d2 so d1 is going to be 20 times 2 so it's going to be 40 meters so you can also use this equation to get the answer or you could say 14 700 times 2 divided by 735 and that will give you the same answer 40 meters so there's multiple ways in which you can get the right answer so now let's review some formulas that you want to keep in mind when dealing with the hydraulic lift system the first thing is the mechanical advantage is the ratio between the output force and the input force and it's also the ratio between the area of the large piston and the small piston and it's the ratio of the distance that you have to push down the small piston by the large piston so if you increase the area of the piston the output force will increase but the distance that the large piston has to move will decrease so make sure you understand that relationship and also if you need to find the output force it's equal to the input force multiplied by the ratio of the areas and if the cylinder is circular the area is pi r squared and keep in mind that the diameter is twice the value of the radius this is not the distance that you push down the piston by so just make sure you understand that and then the input work f1 d1 has to equal the output work f2 d2 and the volume has to be constant so v1 equals v2 and the pressure is constant p1 is equal to p2 and for volume you can say that a1 times d1 is equal to a2 times d2 so there's a lot of formulas that you could use for the hydroglyph system and so if you want to write them down here it is thanks for watching you