between them. In Fig. 53 the siphon is the inverted U-shaped tube open at both ends and entirely full of water. In such a tube the water will continue to flow until it reaches the same level in both vessels. A complete explanation of the theory of the siphon involves principles of flowing liquids which cannot be gone into here. The simplest explanation is one which shows that the liquid in the tube cannot be at rest. If the liquid in the siphon were at rest, then, according to what was learned in section 23, points on the same level in the two arms must have the same pressure. It will be seen that they do not. At the level of the free surface in the lower vessel the pressure both inside and outside of the tube is atmospheric pressure, provided that the liquid is at rest; but at this level in the other vessel the pressure must be much greater than atmospheric. Such inequalities can be easily shown to exist at other levels. It follows that the liquid in this tube cannot be in equilibrium and at rest. It must be in motion. Water will flow through a siphon which is not more than about 30 feet higher than the level of the source; the maximum height for mercury is about 29 inches. Why? FIG. 53 By means of the inverted siphon water in a pipe line may be taken under a river and up on the other side. Water will flow through this pipe without the aid of pumps if the outlet is lower than the place where the water enters the pipe. PROBLEMS 1. Compute the pressure in bars and also in centimeters of mercury in the following cases: (a) 600 cm. under water; (b) at the bottom of an open vessel which has mercury 10 cm. deep, with 20 cm. of water on top of the mercury. (The atmospheric pressure is 74 cm. in both cases.) 2.A tube closed at one end is inverted in a dish of mercury, with the mercury standing 20 cm. high in the tube. Find the pressure in the upper part of the tube. The barometer stands 74 cm. high. Reduce the answer to dynes per square centimeter. 3. What is the total weight in pounds of the air in a column 1 sq. ft. in cross section extending from the ground up, when the barometer is 30 in. high? 4. Compute the pressure in bars inside the inverted cylinder of Fig. 43 if h is 20 cm. and if the barometer is 74 cm. high. (The liquid is water.) 5. What is the approximate weight of air per cubic foot when the pressure is 75 lb./sq. in. greater than the normal pressure of the atmosphere? 6. If the air in an automobile tire is at a pressure of 80 lb./sq. in. more than atmospheric pressure, what is the ratio of the density to that of air at normal pressure? (Assume that the atmospheric pressure is 15 lb./sq. in.) 7. The total buoyant force on one of the United States army balloons is 6300 lb. when in air at normal pressure and density. What is the volume of the balloon in cubic feet? 8. Show that the lifting-power of a balloon filled with hydrogen is about 7 per cent greater than that of one filled with helium. 9. The area of the piston in an apparatus similar to that shown in Fig. 40 is 20 sq. in. The opening in the upper part is sealed, and a weight of 50 lb. is hung from the piston. Neglecting the effects of friction, find the pressure of the air on the inside of the cylinder. (Assume the atmospheric pressure to be 15 lb./sq. in.) 10. It is found that the pressure in the gas mains on the top floor of an office building is greater than in the basement. Explain. 11. If air at normal pressure has a mass of 1.3 kg. per cubic meter, and hydrogen 0.09 kg., how large must the gas bag of a balloon be if the total load has a mass of 300 kg. and if it is to be filled with hydrogen at normal pressure ? 12. From what depth in water has an air bubble risen which doubles its volume when it reaches the surface? (The barometric pressure is 75 cm.) 13. (a) What is the apparent loss of mass on account of the buoyant effect of the atmosphere on 1000 gm. of brass? (b) on 1000 gm. of mercury? (c) If brass weights are used to weigh 1000 gm. of mercury, what will be the magnitude of the error produced by the buoyancy of the air? (The air is at normal pressure and density. Density of brass = 8.4, of mercury = 13.6.) 14. To what height can a liquid of 1.8 specific gravity be raised inside a siphon if the barometric pressure is 76 cm. of mercury? 15. What would be the maximum height over which glycerin, of specific gravity 1.26, can be siphoned when the barometer is 75 cm.? 16. The pressure of air in a vessel as measured by an open-tube gauge (Fig. 45) is 30 cm. of water. Find the ratio of the density of this air to that of the atmosphere if the barometer is 74 cm. high. 17. The closed arm of a gauge such as is illustrated in Fig. 46 contains a vacuum; the liquid is mercury. If h is 2 cm., what fraction of air remains in the vessel to which the gauge is connected, the original pressure being 74 cm. of mercury? 18. When the vessel attached to the gauge of Fig. 46 is at atmospheric pressure, 75 cm., h is 5 cm. Find the pressure in the vessel when h is 30 cm. and the volume in the closed end is half of what it was in the first case. (The liquid in the gauge is mercury.) CHAPTER V UNIFORMLY ACCELERATED MOTION Velocity, 49. Acceleration, 50. Velocity acquired under constant acceleration, 51. Distance traveled in uniformly accelerated motion, 52. Special cases, 53. Geometrical interpretation, 54. Falling bodies, 55. Speed and velocity, 56. Projectiles, 57. Independence of motions, 58. In some of the chapters which follow there will be constant reference to the motion produced by forces and to the effects of such motions. It will be of great help if some of the simpler concepts and laws of motion are kept clearly in mind. For that reason there is given in this chapter an introduction to the subject of accelerated motion. Special attention should be given by the student to the technical term acceleration, for it must be clearly understood in order to avoid confusion later. 49. Velocity. Everyone is familiar with the concept of velocity and with the computations of distance traveled when the time and average velocity are known. The law used for such computations is or Distance = velocity X time, s = vt. (7) This law applies to two cases: (1) when the velocity v is constant during the time t; (2) when the velocity is changing and v is the average velocity during the time t. Equation (7) really states a definition of velocity and may be called the defining equation of velocity. Such an equation cannot be logically verified by experiment, for it is true because v is always given the numerical value which will satisfy the equation. 50. Acceleration. We are frequently more interested in the changes in velocity than in the velocity itself. For example, some types of locomotives are designed to acquire a high speed more quickly than other types. To the designer of such a locomotive the point of interest is how quickly the velocity can be changed, or, in other words, what the rate of change of the velocity can be. This rate is such an important quantity that a special name has been given to it. It is called the acceleration. When the unit of time used is the second, this term is defined as follows: Acceleration is the change per second of the velocity. The meaning of this can be clearly understood by studying the data given in the following table. The first column gives the number of seconds after a smooth, heavy ball has begun falling; the second column gives the approximate velocities, in feet per second, that the ball has acquired at the end of the periods given in the first column; the third column gives approximately the total distances passed over by the ball during the corresponding number of seconds. 1 The table shows that the velocity is increasing at a uniform rate, for during each second the ball gains a velocity of 32 feet per second. Hence the acceleration is constant and equal to 32 feet per second per second.* The rate of change of velocity, and the rate of change in distance s, while related, are not the same thing and should not be confused. If the student will compute the rate of change in the distances in the last column of the table, he will find that that rate is not a constant but increases with time. 51. Velocity acquired under constant acceleration. Experiments with falling bodies show that they tend to move with a constant * The student should see that time enters twice in computing the acceleration, which is a rate of change of a rate. Since it is customary to state after the numerical value of a quantity the units used, the word second appears twice in the above. The student should be able to find a case where the acceleration might be stated as 10 feet per minute per second. Often the units are abbreviated; as, 32 ft./sec.2 |