2. A labourer was hired for 60 days upon this condition, that for every day he wrought, he should receive 75c.; and for every day he was idle, should forfeit 371c.; at the expiration of the time he received $18: How many days did he work, and how many was he idle? Ans. He was employed 36 days, and was idle 24. 3. There is a fish, whose head is 10 feet long; his tail is as long as his head and half the length of his body, and his body as long as the head and tail: What is the whole length of the fish ? Ans. 80 feet. 4. A farmer, having driven his cattle to market, received for them all $320, being paid at the rate of $24 per ox, $16 per cow, and $6 per calf: there were as many oxen as cows, and 4 times as many calves as cows: How many were there of each sort ? Ans. 5 oxen, 5 cows, and 20 calves. PERMUTATION. PERMUTATION is the method of finding how many different ways the order or position of any given number of things may be changed or varied. To find the number of permutations or changes that can be made of any given number of things, all different from each other. RULE. Multiply all the terms of the natural series of numbers, from one up to the given number, continually together, and the last product will be the answer required.* *The reason of this rule may be shown thus, any one thing a is capable of one position only as, a. Any two things a and b are capable of two variations only; as ab, ba; whose number is expressed by 1X2 If there be three things a, b, and c; then any two of them, leaving out the third, will have 1×2 variations; and consequently, when the third is taken in, there will be 1×2×3 variations; and so on as far as you please. 309. What is Permutation ?—310. What is the rule for finding the number of changes that can be made of any given number of things, all different from each other? EXAMPLES. 1. How many changes or variations can be made of the three first letters of the alphabet ? 2. Christ church in Boston, has 8 bells: how many changes may be rung on them? Ans. 40320. 3. Nine gentlemen met at an inn, and were so well pleased with their host, and with each other, that in a frolick, they agreed to tarry so long as they, together with their host, could sit every day in a different position at dinner: Pray how long, had they kept their agreement, would their frolick have lasted? Ans. 3628800 days, 9935 years. GAUGING. GAUGING is the art of measuring all kinds of casks or vessels used for liquor, and of determining the quantity they will contain. The instruments used in gauging are the gauging rod, callipers, the sliding rule, and Gunter's scale. RULE. Take the dimensions of the cask in inches, viz. the diameter at the bung and head, and the length of the cask; subtract the head diameter from the bung diameter, and note the difference. If the staves of the cask be much curved or bulging between the bung and head, multiply the difference between the bung and head diameter by 7; if not quite so much curved, by 65; if they curve yet less, by 6; and if they are almost or quite straight, by 55, and add the product to the head diameter, the sum will be a mean diameter, by which the cask is reduced to a cylinder. 311. What is Gauging?— -312, What are the rules for measuring casks, and finding their contents in gallons? Square the mean diameter, thus found, then multiply it by the length; divide the product by 294 for wine, or by 359 for ale or beer, and the quotient will be the content in gallons. NOTE 1.-These divisors are found by dividing 231 and 282 by .7854-If the square of the mean diameter be multiplied by .7854 and the product multiplied by the length, the last product will be the content in cubic inches, which being divided by 231 for wine, or by 282 for ale or beer, the quotient will be the content in gallons. NOTE 2.-The length and head diameter are usually taken by callipers, allowing for the thickness of both heads, 1 inch, 11 inch, or 2 inches, according to the size of the cask. The head diameter must be taken close to its outside, and for small casks, add 3 tenths of an inch; for casks of 30, 40, or 50 gallons, 4 tenths; and for larger casks 5 or 6 tenths; and the sum will be very nearly the head diameter within. The bung diameter is usually taken by the gauging rod, and in taking it, observe by moving the rod backward and forward, whether the stave, opposite the bung, be thicker or thinner than the rest, and if it be, make allowance accordingly. EXAMPLE. What is the content in wine, and ale or beer gallons, of a cask, whose bung diameter is 35 inches, head diameter 27 inches, and length 45 inches? 313. Squared: 1062.76 How do you obtain the divisors named in the rule?314. How are the differ ent diameters of the cask to be ascertained? MECHANICAL POWERS. OF THE LEVER OR STEELYARD. It is a principle in mechanics, that the power is to the weight, as the velocity of the weight, is to the velocity of the power. Therefore, to find what weight may be raised or balanced by any given power, say; 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. If a man weighing 160 lbs. rest on the end of a lever 10 feet long, what weight will he balance on the other end; supposing the prop one foot from the weight. The distance between the weight and the prop being 1 foot, the distance from the prop to the power is 10-19 feet; therefore, as 1 foot 9 feet :: 160 lbs. : 1440 lbs. Ans. In giving directions for making a chaise, the length of the shafts between the axletree and backband, being settled at 9 feet, a dispute arose whereabout on the shafts the centre of the body should be fixed. The chaisemaker advised to place it 30 inches before the axletree; others supposed 20 inches would be a sufficient incumbrance for the horse: Now supposing two passengers to weigh 336 pounds, and the body of the chaise 84 pounds more; what will the beast in both these cases bear more than his harness? Weight of the passengers and chaise=420 lbs. and 9 feet=108 inches : The proportion for the wheel and axle (in which the power is applied to the circumference of the wheel, and the weight is raised by a rope, which coils about the axle as the wheel turns round) is as the diameter of the axle is to the diameter of the wheel, so is the power applied to the wheel, to the weight suspended by the axle. 315. What is a general principle in mechanics? 316. What is the power of the lever or steelyard, or how do you find what weight may be raised or balanced by any given power? 317. What is the power of the wheel and axle. or how do you find what weight suspended from the axle, will be balanced by any given power applied to the wheel? If the diameter of the axle be 6 inches, and the diameter of the wheel, 60 inches; what weight suspended from the axle will be balanced by a power of 1 lb. applied to the wheel? The power is to the weight which is to be raised, as the distance between two threads of the screw, is to the circumference of a circle described by the power applied to the end of the lever. RULE. Find the circumference of the circle described by the end of the lever; then, as that circumference is to the distance between the spiral threads of the screw; so is the weight to be raised, to the power which will raise it, abating the friction which is not proportional to the quantity of surface, but to the weight of the incumbent part; and, at a medium, part of the effect of the machine is destroyed by it, sometimes more and sometimes less. There is a screw whose threads are an inch asunder; the lever by which it is turned 30 inches long, and the weight to be raised a ton, or 2240 lbs. What power or force must be applied to the end of the lever, sufficient to turn the screw-that is, to raise the weight? : The lever being the semi-diameter of the circle, the diameter is 60 inches; then, as 113: 355 60 188.496 inches nearly, the circumference. Therefore, 188.496 : 1 : 2240: 11.88+ Ans. Let the weight be 2240lb. the power 11.88lb. and the lever 30 inches; Required the distance between the threads? lbs. As 2240 Ibs. in. in. 11.88: : 288.496: 1 nearly, Ans. Let the power be 11.88lb., the weight 2240lb., and the threads an inch asunder, to find the length of the lever. lbs. lbs. in. in. As 11.88 2240 :: 1: 188.5; then, as 355: 113: 188.5: 60 inches nearly, the diameter, and 60÷2-30 inches, Ans. 318 What is the power of the screw?-319. What is the rule for finding the power which must be applied to the end of the lever, sufficient to turn the screw, that is, to raise ny given weight?. -320. When required. how do you find the distance between the ir al threads of the screw; or the length of the lever? |