These two solids are of equal weight. They are also of equal length and breadth. Hence, length and breadth of the log are not taken into account. But, they are of unequal depths. For, beech (Sp. gr. ·85), being lighter than water (Sp. gr. 1), will obtain sufficient support, that is, will float, by displacing less than its own bulk of water; and length and breadth being the same in both, the depth of the water displaced will be less than the whole thickness of the log of beech. The Sp. gr. of the wood being 85, any bulk of it will weigh ·85 of the weight of the same bulk of water. Hence, the log in question must displace 85 of its own bulk of water; and this it will do by sinking 85 of its whole thickness. .. Depth sunk 85 of 2 ft. 1 in. = 25 in. X 100 = lin. in. = = 1. ft. 9 in. PROOF. Suppose any length and breadth for the piece of timber, say 40 ft. long, and 4 ft. wide. Then, Specific gravity of beech is 85, .. Weight of 1 cu. ft. of beech is 850 oz. avoir. The weight of displaced water being, then, equal to that of the log, the latter floats when immerged to the depth named, 1∙ ft. 9 in. FIFTH EXAMPLE WORKED OUT. said log sink in Vinegar, Sp. gr. 1.02? How far would the 850 oz. avoir. Weight of 1 cu. ft. of vinegar = 1020 oz. avoir. = 850 =1020 or 8 wt. of cu. ft. of vinegar. A cu. ft. or any mass of beech (Sp. gr. 85) will float by dis placing of its own bulk of vinegar (Sp. gr. 1.02). .. Parallelopipedon of such timber, floating evenly, displaces of its own depth of vinegar; or, in present example, of 2 ft. 1 in. 1. ft. 8 .. Depth log will sink = PROOF. 1. ft. 8 in. 10. li. Suppose the timber to be 1 ft. long and 1 ft. wide. And, Wt. of vinegar 1020 oz. X 199 = = .. Block, by sinking in Vinegar to a depth of 1 ft. 8 in. displaces its own weight of the liquid, and is consequently supported by, or floats in it. SIXTH EXAMPLE WORKED OUT. What additional weight, placed on the top of a squared log of beech, 19 ft. by 3 ft. by 15 in., will just sink it to the water's edge; Specific Gravity of the wood being 85? Example completed. The log, when immersed to the water's edge, displaces its own bulk or (19 × 3 × 1) cubic feet = 71 cu. ft. But its own weight is supported by displacing only 85 of its bulk. Hence, Bulk of water supporting} added weight = = 15 of bulk of Log. = 1000 oz. X 10.6875 10687.5 oz. PRIN. VIII. =5′ cwt. 3. gr. 23 lb. 15 oz. 8. dr. Added weight The combined bodies are, together, of the same weight as the water displaced, which therefore supports them. From this Example we derive a simple rule. To find what weight, not immersed, a body lighter than a liquid will support when fully immersed in that liquid. Subtract weight of the supporting body from weight of its own bulk of the liquid. The difference is the weight which may be added unimmersed. SEVENTH EXAMPLE WORKED OUT. What weight of iron, Sp. gr. 7.25, attached to the bottom of above-mentioned log, will just sink it to the water's edge? Example completed. Here, the iron, being also immersed, is partly supported by displacing its own bulk of water. A cubic foot of iron weighs But displaces only 7250 oz. 1000 oz. of water. Hence, of each cubic foot of iron, 6250 oz. derive support from the log of wood. Now, the loaded log displaces 10687.5 oz. of water beyond what is requisite for its own support. (See last Example.) = .. No. of cu. ft. of iron 10687-56250. = = 1.71 cu. ft. Weight of iron = 7250)• oz. × 1·71 — 6· cwt. 3′qr. 18· lb. 13′ oz. 8′dr. = PROOF. Whole bulk of water displaced = United bulk of wood and iron. = 71-25 e. ft. + 1·71 c. ft. = 72·96 cu. ft. = 72960. oz. .. Weight of water displaced = 1000′ oz. × 72·96 Now, this should be equal to joint weight of the wood and iron. And it is so, for = Weight of wood 850 oz. X 71·25 = 60562·5 oz. :. United weights supported = (Wt. of water) =72960- oz. From the examination of this example we may also deduce a useful rule. For, suppose the question to be confined to a single cubic foot of beech, Sp. gr. 85; then: of that Any bulk of beech, Sp. gr. 85, will, then, support bulk of iron, Sp. gr. 7·25. In other words 125 cu. ft. of beech will support 3 cu. ft. of iron, both being wholly immersed. .. 1 cu. ft. of iron requires the support of 125 cu. ft., or 41 cu. ft. of beech. And this latter fraction is the reciprocal of the former. To find what bulks of any two materials, one specifically heavier, and the other specifically lighter than a given liquid, must be combined, so that the united bodies may just sink to the surface of that liquid. First. Find the difference of weight of a cubic foot of the liquid and of the same bulk of the lighter body. Second. Find the difference of weight of a cubic foot of the heavier body and of a cubic foot of the liquid. Third. Divide the first of those differences by the second. The result will express the fraction or multiple of its own bulk which the lighter body can support of the heavier. Fourth. Divide the second of the differences by the first, and the result will express the multiple or fraction of its own bulk which the heavier body will require of the lighter to support it. PROOF OF LAST EXAMPLE. oz. 125 cu. ft. of beech weigh, 850· × 125·· Oz. = 106250. = 21750. = 128000. Arithmetical Symbol. The sign placed between two quantities, shews that their difference is to be taken. Hence, if F represent one quantity, and B represent another, FB means that the less of the two, whether placed first or second, is to be taken from the other. .. If F be the greater, F~ B F-B; = B-F. Let F be weight of a certain bulk of any fluid, and B be weight of the same bulk of any solid, whether specifically heavier or lighter than the fluid. Suppose the solid to be put into a vessel containing a sufficient quantity of the said fluid; Then the fraction FB B will coincide with I. Ratio of unimmersed bulk to immersed. If the solid be specifically lighter than the fluid. II. Ratio to lighter body of additional unimmersed weight which it will sustain. III. Ratio of unsupported weight to that supported by fluid. If the solid be specifically heavier than the fluid. IV. Ratio to heavier body of additional support requisite to sustain it. V. Fraction of heavier body which must be excavated to make it float in the fluid. (See EXERCISE CXVI., Example 7.) Let the fluid be distilled water, Sp. gr. 1; the heavier solid, iron, Sp. gr. 7-25; and the lighter solid, cork, Sp. gr. 24. Then the following instances will illustrate and prove the foregoing statements. I. Since a cu. ft. of cork weighs 240 oz., and a cu. ft. of water weighs 1000 oz.; and the cork will float when it displaces its own weight of water, .. Any mass of cork will float by displacing 240 of its bulk of 1000 1 1000 760 1000 .. Displacing of bulk supports of weight. But there are of bulk of cork not immersed by its own weight. .. To immerse these also,, of its own wt. must be superimposed. .. Add unimmersed weight which cork, fully immersed, 760 F-B of its own wt. (II.) B 1 240 A cu. ft. of iron weighs 7250 oz., but displaces only 1000 oz. of Fraction of weight of iron requiring extra support is also 7250-1000. B-F of its w2 = V. (IV.) B Hence, evidently, the fraction of the weight of iron which must be excavated that the hollow vessel may swim in water, also EIGHTH EXAMPLE WORKED OUT. 1. What weight of flint-glass, Sp. gr. 3, would 1 cwt. of alder-wood, Sp. gr. 8, support in distilled water? II. What weight of alder will support 1 cwt. of glass? |