LECTURE XVIII. CONTENTS. Useful Data regarding Fresh and Salt Water-Examples I. II. III. IV.--Centre of Pressure-Immersion of Solids-Law of Archimedes-Floating Bodies-Example V.-Atmospheric Pressure-The Mercurial Barometer-Example VI.-Low Pressure and Vacuur Water Gauges Example VII.-The Siphon-Questions. Useful Data regarding Fresh and Salt Water.-We will commence this Lecture by giving some useful data regarding the weights, &c., of fresh and salt water, and then work out a few more examples for the pressures on immersed surfaces, finishing with the immersion of solids in fluids, &c. FRESH WATER SALT WATER 4 ft. 3 ft. -4 ft = I. (Specific gravity* 1. I cubic foot weighs 62.5 lbs., or 1000 oz. I ton occupies 35.84 cubic feet. I atmosphere = 14.7 lbs. per sq. in. = 29.92 in. mercury = 33.9 (say 34) ft. head of water. I foot of head='43 lb. on sq. in. I lb. on the sq. in. = 2.308 ft. head. H.P. in a waterfall cubic ft. per minute x (Specific gravity=1026. I ton occupies 35 cubic ft., or 2181 gallons. 1 ft. EXAMPLE I.-A cubical box or tank with a closed lid, the length of a side of which is 4 feet, rests with its base horizontal, and an open vertical pipe enters one of its sides by an elbow. The tank is full of water, and the pipe contains water to the height of foot above the top of the tank. What are the pressures of water on the top, bottom, and sides of the tank? (Given the weight of a cubic foot of water=62 lbs.) (S. and A. Exam. 1887.) * Specific gravity is the ratio of the weight of a given bulk of a substance, to the weight of the same bulk of pure water. ANSWER. (1) For the pressure on the top The depth of c.g. of the top from free surface = H = 1'. Total pressure on top=HAW=1′ × (4′ × 4′) × 62·5 lbs. — 1000 lbs. (2) For the pressure on the bottom The depth of c.g. of the bottom from the free surface = H=5′. .. Total pressure on bottom = HAW 5000 lbs. = · 5′ × (4′ × 4′) × 62.5 lbs. (3) For the pressure on each of the sides— The depth of c.g. of each side from the free surface = ... Total pressure on each side = HAW=3′ × (4′ × 4′) × 62 5 lbs. = 3000 lbs. = EXAMPLE II.—A cylindrical vessel, 30 inches long and 6 inches in diameter, is sunk vertically in water, so that the base, which is horizontal, is at a depth of 25 inches below the surface of the water. Find the upward pressure in pounds on the base of the vessel. The weight of a cubic foot of water is 62 lbs., and π= 3*1416. (S. and A. Exam. 1889.) -H=25 feet ANSWER. The depth of c.g. of the base from the free surface = The total pressure on base = HAW=2·083 × •196 × 62·5 = 25.5 lbs. EXAMPLE III.—A water tank, 8 feet long and 8 feet wide, with an inclined base, is 12 feet deep at the front and 6 feet deep at the back, and is filled with water. Find the pressure in lbs. on each of the four sides and on the base; water weighing 62 lbs. per cubic foot. ANSWER. In answering a question of this kind the student will find it best to draw a figure representing the water tank and the positions of the centres of gravity of each side and of the base in the manner shown by the accompanying illustration. The only point that presents any difficulty is the c.g. of the side DEFC, and of the correspondingly opposite one. This might be done by first finding the c.g. of the DEFH, viz., G.; second, of the A HFC, viz., G.; third, by joining these two points with a line G,G,, and taking a distance along it from G, towards G, inversely proportional to the areas of the DEFH and the ▲ HFC; this would give a point G, the c.g. of the whole side = 4·6′ from surface. But it will evidently be easier to treat the pressures on the ☐ and separately, and then to add them together in order to obtain the total pressure on the whole side DEFC. SHOWING POSITIONS OF THE CENTRES OF GRAVITY. G, represents centre of gravity of area ABCD G1 DEFH HFC ENMF BCFM. Let H1, H., &c., represent depths of G1, G2, &c. Then H1 = DC=6′; H,=EF=3'. G, is of HC below the line HF (see Lecture III., re position of c.g. of certain areas). .. H=6+=8'; H1 = EF?'. G, is at a depth below the surface=the mean between the edges BC and FM of the base BCFM. ... H ̧= {(DC+EF) = (12+6)=9'. 5 Total pressure on area— ABCD=H,A,W 6' x (12' x 8') × 62.5 = 36,000 lbs. DEFH = HA2W = 3' x (6' x 8') x 62.5 = 9,000 lbs. HFC=H ̧A ̧W = 8′ × (—, × 8') × 62.5 =12000 lbs. X DEFC-DEFH + HFC = 9000 + 12000 = = 21000 EXAMPLE IV.A sluice gate is 4 feet broad and 6 feet deep, and the water rises to a height of 5 feet on one side and 2 feet on the other side. Find the pressure in pounds on the gate. ANSWER. The net pressure on the sluice gate is evidently equal to the difference of the pressures on the two sides. Total pressure on— Free Surface NETT PRESSURE ON SLUICE GATE. Back side = H,A,W = 2·5′ × (4 × 5') × 62.5 = 3125 lbs. = Subtracting the front from the back] =2625 lbs. Centre of Pressure. In the case of a plane area immersed in a liquid, the "centre of pressure" is the point at which the resultant of all the pressures of the fluid acts. If the plane be horizontal, the resultant naturally acts at the centre of the figure, and therefore the centre of pressure agrees with the centre of gravity of the figure. In the case of a vertical rectangle, having one of its edges in the surface of liquid, like a dock-gate or a sluice, the centre of pressure will be at a point of the depth from the free surface and at the middle of the breadth of the immersed portion. We will have to prove this in our Advanced Course, and perhaps refer to the position of the centre of pressure in other cases. Immersion of Solids.-Archimedes' Discovery.-If a solid be immersed in any fluid (whether liquid or gas), it displaces a quantity of that fluid equal to its own volume. This is evident from the principle of impenetrability-viz., “two bodies cannot occupy the same space at the same time." Hence we have a simple method of determining the volume of any irregular body by plunging it into a liquid, and noting the cubic contents of the liquid displaced, by letting it run into a measure of known capacity, such as a graduated jar. This principle was first discovered by Archimedes, a philosopher of Syracuse, in the year 250 B.C. The story of this discovery is related by Vitruvius, who states that Hero, a king, sent a certain weight of gold to a goldsmith to be made into a crown. Suspecting that the workman had kept back part of the gold, he weighed the crown, but found that it was the same as the weight of the gold previously sent by him to the goldsmith. He was, however, not satisfied with this test, so he consulted Archimedes, and asked him whether he could find out if the crown was adulterated. Not long afterwards the philosopher, on going into his bath (which happened to be full of water), observed that a quantity of the water was displaced. He immediately conjectured that the water which ran over must be equal to the volume of the immersed part of his body. He was so overjoyed at the discovery that he jumped out of the bath and ran naked to the king, exclaiming, Evpnka! evpnka! (I have discovered! I have found out!) He then began to experiment with the crown by taking a quantity of pure gold of the same weight, and observed its displacement in water. Next he ascertained by the same process the volume of the same weight of silver, and finally the volume of the crown, which actually displaced more water than its equivalent weight of pure gold. In this interesting manner the fraud of the artificer was detected, to his great astonishment and chagrin, and a Law of Nature was discovered. Floating Bodies.-A body is entirely supported by the fluid. said to float in a fluid when it is In order that a body may float, the forces acting upon it must be in equilibrium. Now, as may be seen from the case illustrated by the accompanying figure, there are only two forces to be considered-viz., the weight of the body acting vertically downwards through its centre of gravity G1, and the pressure of the fluid acting vertically upwards through the centre of gravity G, of the displaced fluid. The horizontal pressures of the fluid on the body are in equilibrium by themselves, and simply tend to compress it so that they do not affect the question. The upward CONDITIONS OF EQUILIBRIUM IN THE |