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Examples for Practice.
1. A. has 200 yards of linen at 25 cents per yard, ready money, which he barters with B. at 30 cents per yard, taking buttons at 11 cents per gross, which are worth but 8 cents; how many gross of buttons will pay for the linen ; who gets the best bargain ; and by how much ?
Ans. 515,45 gross, B. gains $6,364. 2. How much corn at $1 per bushel, must I give in exchange for 8 cwt. of sugar, at $9,45 per cwt. ?
Ans. 75 bu 2 pe. 3} qts. 3. How much gold of 15, 17, and 22 carats fine, must be mixed with 5 oz. of 18 carats fine, so that the composition may be 20 carats fine ?
Ans. 5 oz. of 15 carats fine, 5 of 17, and 25 of 22. 4. A. merchant would mix 4 sorts of wine, of several prices, viz. at 75 cents, $1,25, $1,50, and $1,625 per. gallon ; of these, he would have a mixture of 60 gallons, worth $1,167 per gallon ; what quantity of each sort must he take ?
Ans. 241145 gal. at 75 cts. 1118 gal. of each of the others.
I will now give a few practical rules for Gauging, and the Mechanical Powers, with several interesting Mathematical Problems.
I have omitted the demonstrations to Gauging, as it would require a more extensive knowledge of mathematics to understand them, than the pupil can, at present, be supposed to possess, or the limits of this work will allow me to give. The practical part is so easy, that the pupil will have no difficulty in obtaining a practical knowledge of the rule without the demonstrations.
Gauging, is taking the dimensions of a cask in inches, to find its contents in gallons, by the following
Method. To the head diameter add two thirds of the difference between the head and budg diameters, and the sum will
MECHANICAL POWERS OF THE LEVER.
be the mean diameter; but if the staves be but little curying, add only six tenths of this difference. Square the mean diameter, which, multiplied by the length of the cask, and the product divided by 294, for wine, or by 359 for ale, the quotient will be the answer in gallons.
Nore. There is a difference in the thickness of the heads of different casks, for which proper allowances must be made in taking the length The head diameter must be taken close to the chimes ; and for small casks, add 3 tenths of an inch; for casks of 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.
Examples. How many ale or beer gallons will a cask hold, whose bung diameter is 28 inches, head diameter 22 inches, and whose length is 38 inches ? 28 bung diam.
22 head diam. 22 head diam.
4 two thirds diff.
359)25688(71 % gallons. How many wine gallons will a cask hold, whose bung diameter is 20 inches, head diameter 17 inches, and whose length is 24 inches ?
Ans. 29 gal. 135. qts.
MECHANICAL POWERS OF THE LEVER.
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 lb. rest on the end of a lever, 12 feet long, what weight will he balance on the other end, if the prop be 1 foot from the weight.
1 : 11 :: 160 : 1760 lb. Ans. At what distance from a weight of 1760 lb. must a prop be placed, so that a weight of 160 lb. applied, 11 ft. from the prop, will raise it? 1760 : 160 ::11 : 1 ft. Ans.
In giving directions for making a chaise, the length of the shafts between the axletree and back board, 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 encumbrance for the horse: Now supposing two passengers to weigh 3 cwt. and the body of the chaise cwt. more; what will the beast, in both these cases, bear more than his harness?
Weight of the chaise and passengers 3 cwt.=420 lbs. and 9 feet=108 inches.
As the weight applied to the axle is to the weight or power applied to the wheel, so is the diameter of the axle to the diameter of the wheel;
OR, As the diameter of the wheel is to the diameter of the axle, so is the weight applied to the axle to the weight applied to the wheel.
If the diameter of the axle be 4 inches, and the weight applied to it be 20 lb. what must be the diameter of the wheel, so that 2 lb, at the wheel will just balance the 20 lb. at the axle ?
Ib. Ib. in.
If the diameter of the axle be 6 inches, and that of the wheel be 48 inches, what weight applied to the wheel will balance 1268 lb. at the axle ?
If the diameter of the wheel be 50 inches, and that of the axle 5 inches, what weight applied to the axle will balance 2 lb. at the wheel ?
The power is to the weight to be raised, as the distance between two threads of the screw is to the circumference of a circle described by the power applied at the end of the lever.
The diameter of a circle is to its circumference as 7 is to 22, or more correctly, as 113 is to 355: Hence to find the circumference of the circle, described by the power at the end of the lever, say; As 7 is to double the lever, so is 22 to the circumference; or as 113 is to double the lever, so is 355 to the circumference; 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 of the effect of the machine for friction.
There is a screw whose threads are one inch asunder; the lever by which it is turned, 72 inches long, and the weight to be raised, 3560 Ib; what power must be applied to the end of the lever, sufficient to turn the screw, that is, to raise the weight?
7 : 144 :: 22 : 4524 circumference.
in. in. lb. Then, as 4524 : 1 :: 3560 : 71141 lb Ans.
If the lever be 72 inches, the circumference 4524 inches, the threads i inch asunder, and the power 73762 1b. 7,8651b; what is the weight to be raised ?
lb. 1 : 452,571 :: 7, 865 : 3559,360915 Ans.
There is a loss of a few ounces, by the fractions.
The diameter of a circle being given, to find the circumference.
As 7 is to the diameter, so is 22 to the required circumference; or as 113 is to the diameter, so is 355 to the required circumference.
There is a circle whose diameter is 12 feet, what is its circumference ?
7 : 12 :: 22 : 37,7142.
or, 113 : 12 :: 355 : 37,6991.
The circumference of a circle being given, to find the diameter.
As 22 is to the circumference, so is to the required diameter; or as 355 is to the circumference, so is 113 to the required diameter.
There is a circle whose circumference is 48 feet; what is its diameter ?
22 : 48 :: 7 : 15,272
To find the area of a circle ;
Multiply half the diameter by half the circumference, and the product will be the area.
There is a circle whose diameter is 15,278 feet; and its circumference 48 feet; what is its area?
7,639 half the diameter.
24 half the circumference.
183,336 feet = area.