of the girth, taken in inches ; divide this product by 144, and the quotient is the contents in cubic feet. Note. – The girth is usually taken about f of the distance from the larger to the smaller end. 5. How many cubic feet in a stick of timber, which is 30 feet long, and whose girth is 40 inches ? Ans. 205 feet. 6. If a stick of timber is 50 feet long, and its girth is 56 inches, what number of cubic feet does it contain ? Ans. 68 feet. 7. What are the contents of a log 90 feet long, and whose circumference is 120 inches ? Ans. 5621 feet. SECTION LXXIX. PROBLEM 1. To find the time in which pendulums of different lengths would vibrate ; that which vibrates seconds being 39.2 inches. The time of the vibrations of pendulums are to each other as the square roots of their lengths; or their lengths are as the squares of their times of vibrations. RULE. — As the square of one second is to the square of the time in seconds in which a pendulum would vibrate, so is 39.2 inches to the length of the required pendulum. EXAMPLES. 1. Required the length of a pendulum that vibrates once in 8 seconds. 12: 82 :: 39.2in. : 2508.8in. = 2091's feet, Ans. 2. Required the length of a pendulum that shall vibrate 4 times a second. Ans. 20 inches. 3. Required the length of a pendulum that shall vibrate once a minute. Ans. 3920 yards. 4. How often will a pendulum vibrate whose length is 100 Ans. once in 5.53+ seconds. PROBLEM II. To find the weight of any body, at any assignable distance above the earth's surface. * For demonstrations of the following problems, the student is referred to Enfield's Philosophy, or to the Cambridge Mathematics. The gravity of any body above the earth's surface decreases, as the squares of its distance in semidiameters of the earth from its centre increases. Rule. — As the square of the distance from the earth's centre is to the square of the earth's semidiameter, so is the weight of the body on the earth's surface to its weight at any assignable distance above the surface of the earth, and vice versa. 5. If a body weigh 900 pounds at the earth's surface, what would it weigh 2000 miles above its surface ? Ans. 400lbs. 6. Admitting the semidiameter of the earth to be 4000 miles, what would be the weight of a body 20,000 miles above its surface, that on its surface weighed 144 pounds ? Ans. 4lbs. 7. How far must a body be raised to lose half its weight ? Ans. 1656.85+ miles. 8. If a man at the earth's surface could carry 150 pounds, how much would that burden weigh at the earth, which he could sustain at the distance of the moon, whose centre is 240,000 miles from the earth's centre ? Ans. 540,000lbs. 9. If a body at the surface of the earth weigh 900 pounds, but being carried to a certain height weighs only 400 pounds, what is that height ? Ans. 2000 miles. PROBLEM III. By having the height of a tide on the earth given, to find the height of one at the moon. Rule. - As the cube of the moon's diameter, multiplied by its density, is to the cube of the earth's diameter, multiplied by its density, so is the height of a tide on the earth to the height of one at the moon. 10. The moon's diameter is 2180 miles, and its density 494 ; the earth's diameter is 7964 miles, and its density 400. If, then, by the attraction of the moon, a tide of 6 feet is raised at the earth, what will be the height of a tide raised by the attraction of the earth at the moon ? Ans. 236.8+ feet. Note. — The above question is on the supposition that the moon has seas and oceans similar to those on the earth, but astronomers at the pres. ent time doubt their existence at this secondary planet. PROBLEM IV. To find the weight of a body at the sun and planets, having its weight given at the earth. If the diameters of two globes be equal, and their densities different, the weight of a body on their surfaces will be as their densities. If their densities be equal, and their diameters different, the weight of a body will be as ļof their circumference. If their diameters and densities be both different, the weight of a body will be as of their semidiameters, multiplied by their densities. Therefore, having the weight of a body on the surface of the earth given, to find its weight at the surface of the sun and the several planets, we adopt the following Rule. — As of the earth's semidiameter, multiplied by its density, is to f of the sun's or planet's semidiameter, multiplied by its density, so is the weight of a body at the surface of the earth to the weight of a body at the surface of the sun or planet. 11. If the weight of a man at the surface of the earth be 170 pounds, what will be his weight at the surface of the sun, and the several planets, whose densities, &c., are as in the following table? Density. Diameter. 1 Semidiameter. 1 Semidiameter. 100 883246 441623 294415 Jupiter, 94.5 89170 44585 29723 Saturn, 79042 39521 26347 Earth, 400 7964 3982 2654 494 2180 1090 726 Sun, Moon, PROBLEM V. To find how far a heavy body will fall in a given time, near the surface of the earth. Heavy bodies near the surface of the earth fall 16 feet in one second of time ; and the velocities they acquire in falling are as the squares of the times; therefore, to find the distance any body will fall in a given time, we adopt the following Rule. - As the square of 1 second is to the square of the time in seconds that the body is falling, so is 16 feet to the distance in feet that the body will fall in the given time. 12. How far will a leaden bullet fall in 8 seconds ? 12 : 82 : : 16ft. : 1024ft. = Answer. 13. How far would a body fall in 1 minute ? Ans. 10 miles 1600 yards. 14. How far would a body fall in 1 hour ? Ans. 39,272 miles 1280 yards. 15. How far would a body fall in 9 days? Ans. 1,832,308,363 miles 1120 yards. PROBLEM VI. The velocity given, to find the space fallen through to acquire that velocity. RULE. — Divide the velocity by 8, and the square of the quotient will be the distance fallen through to acquire that velocity. 16. The velocity of a cannon-ball is 660 feet per second. From what height must it fall to acquire that velocity? Ans. 68061 feet. 17. At what distance must a body have fallen to acquire the velocity of 1000 feet per second ? Ans. 2 miles 5065 feet. PROBLEM VII. The velocity given per second, to find the time. RULE. — Divide the velocity by 8, and a fourth part of the quotient will be the time in seconds. 18. How long must a body be falling to acquire a velocity of 200 feet per second ? Ans. 64 seconds. 19. How long must a body be falling to acquire a velocity of 320 feet per second ? Ans. 10 seconds. PROBLEM VIII. The space through which a body has fallen given, to find the time it has been falling. Rule. — Divide the square root of the space in feet fallen through by 4, and the quotient will be the time in seconds in which it was falling. 20. How long would a body be falling through the space of 40,000 feet ? Ans. 50 seconds. 21. How long would a ball be falling from the top of a tower, that was 400 feet high, to the earth? Ans. 5 seconds. PROBLEM IX. The weight of a body and the space fallen through given, to find the force with which it will strike. RULE. — Multiply the space fallen through by 64, then multiply the square root of this product by the weight, and the product is the momentum, or force with which it will strike. 22. If the rammer for driving the piles of Warren Bridge weighed 1000 pounds, and fell through a space of 16 feet, with what force did it strike the pile ? „16 x 64 = 32 32 X 1000 = 32,000lbs. Answer. 23. Bunker Hill Monument is 220 feet in height; what would be the momentum of a stone, weighing 4 tons, falling from the top to the ground ? Ans. 1,063,184.6+ lbs. SECTION LXXX. MECHANICAL POWERS. That body which communicates motion to another is called the power. The body which receives motion from another is called the weight. The mechanical powers are six, the Lever, the Wheel and Axle, the Pulley, the Inclined Plane, the Screw, and the Wedge. THE LEVER. The lever is a bar, movable about a fixed point, called its fulcrum or prop. It is in theory considered as an inflexible line, without weight. It is of three kinds; the first, when the prop is between the weight and the power; the second, when the weight is between the prop and the power; the third, when the power is between the prop and the weight. A power and weight acting upon the arms of a lever will balance each other, when the distance of the point at which the power is applied to the lever from the prop is to the distance of the point at which the weight is applied as the weight is to the power. Therefore, to find what weight may be raised by a given power, we adopt the following RULE. — As the distance between the body to be raised, or balanced, and the fulcrum or prop, is to the distance between the prop and the point where the power is applied, so is the power to the weight which it will balance. |