RULE.-Multiply the space fallen through by 64, and the square root of the product will be the answer. Reason. Because the velocity acquired at the end of any number of seconds, is equal to twice the mean velocity with which the body falls during that time. NOTE.-If a question requires the reverse of a statement, as by the rule, it will be when the velocity with which a falling body strikes an obstacle, is given, to find the perpendicular space fallen through. Example. There is a flume 20 feet in length, and one end is 24 feet higher than the other. What would be the velocity of a stream of water through it per second ? 2.5X64 160, and 160—12.649 feet, Ans. The reverse, thus, 12.649×12.649,-64-2 feet. NOTE.-If the velocity per second, 12.649, be multiplied by 62.5 (lb. per cubic foot,) for clear water, by 63 lb. for dirty water, and by 64 lb. for sea water, the product would be the momentum or force, a fluid running with such velocity, would strike against a fixed obstacle. 7. Knowing the weight of a body, and the space fallen through, to find the force with which it will strike. RULE. Find the velocity by case 2, and the product of the weight and velocity will be the answer. NOTE.-If the weight and striking force is given, to find the space fallen through. RULE. Divide the force by the weight, and the quotient is the velocity. Then divide the square of the velocity by 64, and the quotient is the space fallen through. Example. If the hammer used in driving the piles of Charlestown bridge weighed 24 tons, and fell through a space of 10 ft., with what force did it strike the pile? 2 tons 4500 lb., and 10×64-25.3, velocity; then, 4500 multiplied by 25.3-113850 lb., Ans. The reverse; thus, 113850 lb., force, 4500-25.3, and 25.3 multiplied by 25.3÷64-10 feet, space, Ans. 8. Given, the number of seconds from the time a body projected directly upwards returns to the earth, to find how high it ascended. RULE.-Multiply half the time, in seconds, by 4, and the square of the product is the answer. Reason See Principle 4. Example.-If a ball discharged from a gun, directly upwards, returns to the earth in 14 seconds, how high did it ascend? 14÷2-7, and 7×4-28; then, 28x28-784 ft., Ans. 9. To find the depth of a well by dropping a stone into it, also the time of the stone's descent and of the sound's ascent. RULE. Take a line of any length, and by case 2, Pendulums, find the time from the dropping of the stone until you hear it strike the bottom. Then multiply this time by 73088, and to this product add 1304164, and from the square root of the sum take 1142. Divide the square of the remainder by 64, the quotient will be the depth of the well, in feet. Divide the depth by 1142, the quotient will be the time of the sound's ascent, which, being taken from the whole time, will be the time of the stone's descent, in seconds. Reason-7308816×4×1142; 1142 ft. being the distance which sound moves in a second; and 1304164-1142 squared; and 64-16X4; for the reason of 16X4, see Principle 2. Example.-I dropped a stone into a well, and a string with a plummet, which measured 25 inches from one end of the string to the middle of the plummet, made 5 vibrations before I heard the stone strike the bottom. Required the depth of the well, time of the stone's descent, and of the sound's ascent. 25 39.2.7985,5-4 seconds, from the dropping of the stone to the hearing of it strike the bottom. Then,/73088×4,+1304164,-1142-121.53. And 121.53X121.53,64-230.77 feet, depth of the well. Then, 230.77-1142.2 of a second, time of the sound's ascent. And, 4-2-3.8 seconds, time of the stone's descent. RIVERS AND FLUIDS. PRINCIPLES. 1. If the breadth of a river is uniform, the motion of the water is accelerated in the same manner with any body moving down an inclined plane. And the force with which a body descends down an inclined plane is to that with which it would descend freely in air, as the elevation of the plane is to its length, or as the sine of the angle of inclination is to radius. 2. The water at the bottom of a river runs faster than the water at the surface. And the velocity of each drop of water in a river of uniform depth, is as the square root of its distance from the SUS level of the surface of the fountain, and that distance is greater at the bottom than at the surface of the river. 3. The depth of a river continually decreases as it runs, if not fed by tributary streams. Because, the farther a river runs, the swifter it runs, and the more water passes in the same time. 1. Given, the height of a head of water above the sluice, to find what the height must be to discharge any given proportion more. RULE. To the square root of the first height, add the given proportion of the root, and the square of the sum will be the answer. Example. If a head of water be 64 feet above the sluice, how high must it be to discharge one fifth as much more in the same time? 6.25 2.5, which plus a fifth of itself is 3. Then, 3×3 9 feet, height, Ans. NOTE 1.-For other problems, their rules and examples, see case 6, its Rule, Reason, Notes, and Example of Falling Bodies. NOTE 2.-Water being a yielding substance, it loses two thirds of its power in producing effects. P. S. To find the perpendicular pressure of fluids on the bottom of vessels. RULE.-Multiply the area of the bottom by the altitude of the fluid, and that product by its specific gravity, gives the pressure required. Example. Suppose a vessel is 3 feet wide, 5 feet long, and 4 feet high. What is the pressure on the bottom, it being filled with water to the brim? 3X5 15 square ft., area of the bottom. 15X4=60 cubic ft. 60X62.5 3750 lb. 1 T. 13 cwt. 1 qr. 26 lb., Ans. 2. To find the quantity of pressure against the sluice or bank, which pens a body of water. RULE.-Multiply the area of the sluice, under water, by the depth of the centre of gravity, (which is equal to half the depth of water,) in feet, and that product again by 624 for clear water, or 64 for sea-water, the product is the answer, in pounds. Example.-If the length of a sluice be 30 feet, and the depth of the water 4 feet, what is the pressure against the side of the sluice? 30X4=120 feet, area of the side, and 120×2, (depth of the centre of gravity,) gives 240 cubic feet. Then, 240X62.3-15000 lb., Ans. PENDULUMS. PRINCIPLES. 1. The time of an oscillation, in the cycloid, is to the time of a heavy body's descent through half its length, as the circumference of a circle to its diameter, which is as 3.1416 to 1; and a heavy body descends freely, by the attraction of gravitation, * 193.5 inches in the first second, nearly, in our latitude. 2. The time of the vibration of a pendulum vibrating in a chord of a circle, is equal to the time in which a body falling freely, would descend through eight times the length of the pendulum. 3. The times of the vibrations of pendulums of different lengths, are as the square roots of their lengths. 4. The vibrations of a pendulum of the same length, are all performed in nearly the same time, whether the vibrations are a large or small, but as the time of the latter is rather the least, a great number of them together would make a sensible difference of time; and as in cold weather the wheels of a clock are obstructed considerably more than in warm, so as not to communicate so much force to the pendulum, a clock will gain time in winter; another cause of its gaining is, the cold contracts the length of the pendulum rod a little, which makes it vibrate rather faster. 5. The power of gravity is greatest at the poles of the earth, and least at the equator, hence, the greater the latitude of a place, the longer must be the pendulum, to vibrate in any given time. A pendulum vibrating seconds, in lat. 51° 31', is of an inch longer than one vibrating seconds, at the equator. NOTE.-39.2 inches is the length of a pendulum vibrating seconds in lat. 51° 31'. 1. To find the length of a pendulum vibrating in any given time. RULE.-Multiply the square of the time in seconds, by 39.2, and the product will be the length of the pendulum in inches. Example-Required the lengths of several pendulums, which will respectively swing fourth seconds, half seconds, seconds, and minutes. 25X.25×39.2-2.45 inches, to swing 4th seconds. .5x.5X39.29.8 in. to swings. 1X1X39.2 39.2 inches; or, thus, as 3.1416×3.1416:1:: 193.5: 19.6 inches, half the length, Ans. 39.2 in. for sec.; 60×60×39.22 miles and 1200 feet, to swing minutes. 2-To find the time which a pendulum of any given length will ving. RULE.-Divide the given length by 39.2, and the quotient will be the square of the time in seconds. Example.-How often will a pendulum of 9.8 inches vibrate in a second? Thus, 9.8-39.2.5 of a second, that is, it vibrates ha.f seconds. AIR BALLOON. 1.—Given, the weight to be raised by a balloon, to find its diameter. RULE. As the specific difference between common and inflammable air is to one cubic foot, so is any weight to be raised, to the cubic feet contained in the balloon. Divide the cubic feet by .5236, and the cube root of the quotient will be the diameter required; but, to raise it, the diameter must be something greater, or the weight something less. Example.-I would construct a spherical balloon, of sufficient capacity to ascend with 4 persons, weighing one with another, 160 lb., and the balloon and a bag of sand weighing 60 lb. Required the diameter of the balloon. By the Table of Specific Gravities, I find a cubic foot of common air weighs 1.25 oz., and a cubic foot of inflammable air, 12 oz., each Avoirdupois weight. Then, 1.25-12-1.13 oz., difference. And, 160×40,+60-700 lb. 11200 oz. Then, as 1.13: 1ft.:: 11200 oz.: 9911.5044, the cube root of which is 26.65 feet, required diameter, Ans. 2.-Given, the diameter of a balloon, to find the weight it is capable of raising. RULE.-Multiply the cube of the diameter by .5236, and the product will be the content in cubic feet. Then, as a cubic foot is to the specific difference between common and inflammable air, so is the content of the balloon to the weight it will raise. Example. The diameter of a balloon is 26.65 feet. What weight is it capable of raising? 26.65X26.65×26.65.52369911.5044 cubic feet. Then, as 1 cubic foot: 1.13:: 9911.5044: ounces in 700 lb., nearly, Ann. |