Application of Pascal's law. Part 1 (1943)

[Music] In the preceding film, we observed some of the principles of the behavior of liquids which are summed up in Pascal's law. Pressure exerted on a confined liquid is transmitted unddeinished to all portions of the interior of the containing vessel. Applying Pascal's law, we can see that in a simple hydraulic system, a force applied to the input piston results in pressure which is transmitted unddeinished through the liquid and acts on all portions of the interior of the containing wall. Part of the interior of the system on which the pressure acts is the output piston which moves when the pressure acts on it. We are already familiar with this type of hydraulic system. Let us modify the system slightly by squaring the corners until we now have a simple diagram of an hydraulic jack. We have in this system a small input or a piston, a larger output or B piston, both of which are flush with the oil which is completely confined within the system. The input and output pistons are free to move when a force is applied to one of them. Let us apply a 50 lb weight to our input piston which has an area of bottom surface of 1 square in. If the weight of 50 lb on the input piston acts on an area of 1 square in, then the pressure on the input piston is 50 lb per square in. This pressure of 50 lb per square in is, as Pascal's law states, transmitted unddeinished through the system and acts on all portions of the interior of the containing wall. In actual practice, the input piston with the 50 lb weight on it would move down and displace oil which would move the output piston up. In order that we may study this system in detail, however, we are not showing this motion here. This pressure of 50 lb per square in transmitted through the system acts on the output piston which is free to move. The area of bottom surface of the output piston on which this pressure acts is five square in. And hence the total force acting to push the output piston upward is the pressure 50 lb per square in times the area 5 square in. This gives us a total force of 250 lb. This means that a weight of 250 lbs is necessary to balance the weight on the input piston. Here we are purposely neglecting friction which is small and also the weight of the pistons. In this system as it is shown here, the weights would just balance each other. A very small additional force on the input piston would cause it to move down and to displace liquid which would move the output piston upward. For the sake of working in round numbers, however, let us say that the 50 lb weight on the input piston will just lift the 250 lb weight on the output piston. And so we see that a 50 lb weight on the input piston can move a 250 lb weight on the output piston. In actual motion, the system would look like this with the 50 lb weight moving the input piston through a large distance and the output piston exerting a 250 lb total force through a small distance. We have seen several examples of pistons moving when a force or weight is applied to them. Let us observe this more closely for a moment and consider the fact that if no force is applied to the piston, it will not move. This can be said in another way. The piston offers resistance to being moved and if motion is to occur, a force is necessary to overcome this resistance. Overcoming this resistance in order to move the piston we call doing work. Let us consider for a moment what work in the mechanical sense is. We can explain it best by a simple illustration. A man lifting a box. In the scientific sense of the word, no work is done no matter how hard the man tugs or lifts unless the box is moved. When the box begins to move, work is being done. It is the actual work accomplished that we measure, not the effort expended in doing it. When a 100 lb weight is lifted through a vertical distance of 3 feet, we say that 300 ft-lb of work has been done. If we lift one pound through a vertical distance of 1 ft, we do 1 foot-pound of work. Thus, work is the force exerted times the distance through which it moves. Expressed as a formula, work equals force time distance. In our example of a 100b box being lifted onto a truck, we exerted a force of 100 lb through a vertical distance of 3 ft and did 300 ft-lb of work. Remember the formula for work, force times distance. You will use it often in hydraulics. For smaller distances measured in inches, the work done is expressed in inch pounds. Returning to our example of force being applied to a piston and moving it, we can see that as force is applied to the piston and the piston move through a distance, work is being done. And the work done by the piston is the force applied to it times the distance through which it moves. We shall proceed now to investigate how work is done by means of a hydraulic system. Let us consider a specific example. We shall find first exactly how much work is done by the input piston as it moves to the bottom of its stroke. Let us apply a weight of 50 lb to the input piston and assume that its stroke is 10 in. When a force moves through a distance, work equal to the force times the distance is done. The work done by the input piston then is 50 lb * 10 in 500 inb. This work done by the input piston is called the work input or work put into the system. We have just found the work done by the input piston by multiplying the force on the piston times the distance through which it moves. The formula for work can be applied to find the work done by the output piston. This will be the force on the output piston times the distance the output piston is moved. A moment ago, we found the total force on the output piston. This force was the pressure in the system. 50 lb per square in times the area of the output piston, 5 square in or a total force of 250 lb. The work done by the output piston will equal the force on it, 250 lb, times the distance through which this force moves. We do not know this distance. We must find it in order to calculate the work output. One way to find the distance the output piston moves is to find the volume of liquid displaced by the input piston as it moves down. The input piston, as we have seen, has an area of bottom surface of one square in and moves 10 in to the bottom of its stroke. As the input piston moves down, it displaces a cylindrical column of oil 10 in high and 1 square in in cross-section. The volume of this cylinder is 10 cub in. Since the oil displaced is not compressible, this 10 cub in must be transferred to some other part of the system where there is room for it. The only other movable part of the system is the output piston. And so the oil displaced by the input piston raises the output piston. The column of liquid which moves the output piston has changed shape but its volume is the same. The area of the ends of this column of liquid is the same as the area of the bottom surface of the output piston 5 square in. Since its volume is 10 cubic in, the height of this column must be 2 in. And since this column of displaced liquid is what causes the output piston to move upward, the amount the output piston has moved is 2 in. And now let us view this system in motion schematically. A force of 50 lb having been applied to it, the input piston moves to the bottom of its stroke, a distance of 10 in. As it moves down, it displaces liquid which pushes up the output piston a distance of 2 in. Now that we know the distance it moves, we can find the work done by the output piston. This work will be the force acting on the piston times the distance it moves. A few minutes ago, we found that the total force on the piston was 250 lb and expressed this by saying that this meant the piston could just lift a 250 lb weight. The work done by the output piston then is the force of 250 lb times the distance of 2 in that the piston moves or a total of 500 inb. But the work done by the input piston which we found a moment ago was also 500 in lb. We found it by multiplying the force on the input piston 50 lb times the distance it moves 10 in. Thus we find that the work input and the work output are the same. If we neglect friction and possible leakage and assume the system to be 100% efficient.