screw, to the circumference of the circle described by the power in one revolution.

4

To find the power that should be applied to raise a given weight.

RULE. - As the distance between the threads of the screw is to the circumference of the circle described by the power, so is the power to the weight to be raised.

NOTE.

One third of the power is lost in overcoming friction.

21. If the threads of a screw be 1 inch apart, and a power of 100 pounds be applied to the end of a lever 10 feet long, what force will be exerted at the end of the screw?

22. If the threads of a screw be er must be applied to the end of a to raise 100,000 pounds?

Ans. 75,398.20+lbs. an inch apart, what powlever 100 inches in length Ans. 79.5774+-lbs.

23. If the threads of a screw be an inch apart, and a power of 79.5774+ pounds be applied to the end of a lever 100 inches in length, what weight will be raised?

Ans. 100,000lbs. 24. If a power of 79.5774+ pounds be applied to the end of a lever 100 inches long, and by this force a weight of 100,000 pounds be raised, what is the distance between the threads of the screw? Ans. an inch.

25. If a power of 79.5774+ pounds be applied to the end of a lever, raising by this force a weight of 100,000 pounds, what must be the length of the lever, if the threads of the screw be an inch apart? Ans. 100 inches.

WEDGE.

The wedge is composed of two inclined planes, whose bases are joined.

When the resisting forces and the power which acts on the wedge are in equilibrio, the weight will be to the power as the height of the wedge to a line drawn from the middle of the base to one side, and parallel to the direction in which the resisting force acts on that side.

To find the force of the wedge.

RULE. As half the breadth or thickness of the head of the wedge is to one of its slanting sides, so is the power which acts against its head to the force produced at its side.

26. Suppose 100 pounds to be applied to the head of a wedge that is 2 inches broad, and whose slant is 20 inches long, what force would be affected on each side?

Ans. 2000lbs.

27. If the slant side of a wedge be 12 inches long, and its head 1 inches broad, and a screw whose threads are of an inch asunder be applied to the head of this wedge, with a power of 200 pounds at the end of the lever, 16 feet long, what would be the force exerted on the sides of the wedge? Ans. 5147184.3+lbs.

SECTION LXXXI.

SPECIFIC GRAVITY.*

To find the specific gravity of a body.

RULE. Weigh the body both in water and out of water, and note the difference, which will be the weight lost in water; then, as the weight lost in water is to the whole weight, so is the specific gravity of water to the specific gravity of the body. But if the body whose specific gravity is required is lighter than water, affix to it another body heavier than water, so that the mass compounded of the two may sink together. Weigh the dense body and the compound mass separately, both in water and out of it; then find how much each loses in water by subtracting its weight in water from its weight in air; and subtract the less of these remainders from the greater; then say, as the last remainder is to the

*The specific gravity of a body is its weight compared with water; the water being considered 1000.

weight of the body in air, so is the specific gravity of water to the specific gravity of the body.

NOTE.

A cubic foot of water weighs 1000 ounces.

1. A stone weighed 10 pounds, but in water only 6 pounds. Required the specific gravity. Ans. 2608.6+. 2. Suppose a piece of elm weigh 15 pounds in air, and that a piece of copper, which weighs 18 pounds in air and 16 pounds in water, is affixed to it, and that the compound weighs 6 pounds in water. Required the specific gravity of the elm. Ans. 600.

SECTION LXXXII.

STRENGTH OF MATERIALS.

THE force with which a solid body resists an effort to separate its particles or destroy their aggregation can only become known by experiment.

There are four different ways in which the strength of a solid body may be exerted; first, by resisting a longitudinal tension; secondly by its resisting a force tending to break the body by a transverse strain; thirdly, in resisting compression, or a force tending to crush the body; and, fourthly, in resisting a force tending to wrench it asunder by torsion. We shall, however, only consider the strength of materials as affected by a transverse strain.

When a body suffers a transverse strain, the mechanical action which takes place among the particles is of a complicated nature. The resistance of a beam to a transverse strain is in a compound ratio of the strength of the individual fibres, the area of the cross section, the distance of the centre of gravity of the cross section from the points round which the beam turns in breaking.

The following are the facts and principles on which mechanics make their calculations.

1. A stick of oak one inch square and twelve inches long, when both ends are supported in a horizontal position, will sustain a weight of 600 pounds; and a bar of iron of the same dimensions will sustain 2190 pounds.

2. The strength of similar beams varies inversely as their

lengths; that is, if a beam 10 feet long will support 1000 pounds, a similar beam 20 feet long would support only 500 pounds.

3. The strength of beams of the same length and depth is directly as their width; that is, if there be two beams, each 20 feet long and 6 inches deep, and one of them is 6 inches wide and the other but 3 inches, the former will support twice the weight of the latter.

4. The strength of beams of the same length and width is as the squares of their depths; that is, if there be two beams, each of which is 20 feet long and 4 inches wide, but one is 6 inches deep and the other is 3 inches deep, their strength is as the squares of these numbers. Thus, 6 x6 = 36; 3 × 3 = 9; that is, the strength of the former is to the latter as 36 to 9. It will, therefore, sustain four times the weight of the latter. Thus, 369 = 4.

5. To compare the strength of two beams of the same length, but of different breadth and depth, we multiply their widths by the squares of their depths, and their products show their comparative strength. Thus, if we wish to ascertain how much stronger is a joist that is 2 inches wide and 8 inches deep, than one of the same length that is 4 inches square, we multiply 2 by the square of 8, and 4 by the square of 4; thus, 2 × 8 × 8= 128; 4 x 4 x 4 = 64; 128 642. Thus, we see that although the quantity of material in one joist is the same as in the other, yet the former will sustain twice the weight of the latter. Hence "deep joists" are much stronger than square ones, which have the same area of a transverse section.

6. To compare the strength of two beams of different lengths, widths, and depths, we multiply their widths by the squares of their depths, and divide their products by their lengths, and their quotients will show their comparative strength. Therefore, if we wish to ascertain how much stronger is a beam that is 20 feet long, 8 inches wide, and 10 inches deep, than one 10 feet long, 6 inches wide, and 5 inches deep, we adopt the following formulas :

The strength of the former, therefore, is to the latter as 40 to 15; that is, if the first beam would sustain a weight of 40cwt., the latter would sustain only 15cwt.

7. Having all the dimensions of one beam given, to find

another, part of whose dimensions are known, that will sustain the same weight. We multiply the width of the given beam by the square of its depth, and divide this product by the length, and the result we call the reserved quotient; then, if we have the length and breadth of the required beam given to find the depth, we multiply the reserved quotient by the length of the required beam, and divide the product by its width, and the quotient is the square of the depth of the required beam. If the length and depth of the required beam were given to find the width, we multiply the reserved quotient by the length of the required beam, and divide this product by the square of the depth of the required beam, and the quotient is the breadth. But if the width and depth of the required beam were given to find the length, we multiply the width of the required beam by the square of the depth, and divide this product by the reserved quotient, and the result is the length of the required-beam.

8. A triangular beam will sustain twice the weight with its edge up that it will with its edge down. Hence split-rails have twice the strength with the narrow part upward, which they have with the narrow part downward.

9. In making the above calculations, we have not noticed the weight of the beam itself, and in short distances it is of but little consequence; but where a long beam is required, its weight is of importance in the calculation.

10. A beam supported at one end will sustain only one fourth part the weight which it would if supported at both ends.

11. The tendency to produce fracture in a beam by the application of a weight is greatest in the centre, and decreases towards the points of support; and this ratio varies as the square of half the length of the beam to the product of any two parts where the weight may be applied. Hence the tendency of a weight to break a bar 8 feet long, when applied to the centre, to that of the same weight, when applied 3 feet from one end, is as 4 x 4 16 to 3 x 5 = 15.

QUESTIONS TO BE PERFORMED BY THE PRECEDING RULES.

1. If a stick of oak 1 inch square and 12 inches long, when both ends are supported in a horizontal position, will sustain a weight of 600 pounds, how many pounds would a similar stick sustain, that was 36 inches long? Ans. 200 pounds.

2. If a beam 4 inches square and 12 feet long would support a weight of 1000 pounds, how many pounds would a similar beam support, that was 3 feet long? Ans. 4000 pounds.