and supposing the specific gravity of the fluid, or the value of s' to be expressed by unity; then, these equations become x2 — b2 (1 − s), and x2-1.732b cos. 4 x —— b ̊ (1 − s). Resolving these equations by the rules which the writers on algebra have laid down for that purpose, we shall have for the pure quadratic, In the equations, (251), it is obvious that the values of x and y are assignable, whatever may be the value of s, provided that it is less than unity; and since x and y are each expressed by the same quantity, it follows that they are equal to one another, and consequently the body will float in equilibrio, when the immersed side or base of the section is parallel to the surface of the fluid. 410. EXAMPLE. If the floating prism be of fir from the forest of Mar, of which the specific gravity is 0.686, that of water being unity; then we have xy=0.566, and if the value of b, or the side of the equilateral triangle be 28 inches, we get x=y=0.56×2815.69 inches; and the position of equilibrium corresponding to this common value of x and y, is exhibited in the annexed diagram, where IK is the horizontal surface of the fluid, DCE being the extant portion of the floating body, and ABED the I part immersed below the plane of floatation; CD and CE being respectively equal to 15.69 inches. Bisect AB in F, and draw the straight lines FD and FE to inter B K sect the surface of the fluid in the points D and E; then, because the triangle ABC is equilateral, and CD equal to CE by the construction, it follows, that FD and FE are equal to one another; this satisfies one of the conditions of equilibrium, and we have now to inquire if the area of the immersed portion A BED, is to the area of the whole section ABC, as the fraction 0.686 is to unity. Now, by the principles of mensuration, we know that the area of a plane triangle, of which the three sides are equal, is expressed by one fourth of the square of the side, drawn into the square root of the number 3; consequently, the area of the whole section ABC, is therefore, the area of the immersed part ABED, is (a' — a") = {b2 √ 3 — 4x2 √/ 3 = 0.433 (b3 — x2) ; 0.433 (b2x2): 0.4336 0.686 1, and by equating the products of the extremes and means, it is But b is 28 and ≈ 15.69 inches; therefore, if these values of b and r be substituted instead of them in the preceding equation, we shall have In this case also, one of the conditions of equilibrium is satisfied; hence we conclude, that the position which we have represented above is the true one, since both the conditions upon which the equilibrium depends, have been fulfilled by the results as obtained from the reduction of the formula. The value of x and y, as exhibited in equations (252), will indicate two other positions of equilibrium, subcontrary to each other; but in order that those positions may be consistent with the conditions of the problem, it becomes necessary to assign the limits of s, or the specific gravity of the floating body; for it is manifest, that beyond certain limits, the conditions specified in the problem cannot obtain. 411. Now, in the case of the isosceles triangle, it has been shown, that the greater limit of the specific gravity, is s= a2 262' and consequently, when the triangle is equilateral, $===0.5; and moreover, it has also been shown, that when the triangle is isosceles, the lesser limit of the specific gravity, is (8b2 — a2) Xa2 which, when the triangle is equilateral, becomes 7 s= =0.4375, 16 and the arithmetical mean of these, from equation (247), is Let therefore this value of s be substituted instead of it in the expressions, class (252), and we shall obtain. 412. The positions of equilibrium, as indicated by these values of x and y, are as represented in the annexed diagrains, where IK is the horizontal surface of the fluid, ABED, a bed the immersed, and DEC, dec the extant portions of the section corresponding to the positions ABC and abc, in which CD and ce are each equal to 25.95 inches, and CE, cd equal to 16.05 inches, being the respective values of x and y, as determined from equation (252). Bisect AB and ab in the points F and ƒ, and draw the straight lines FD, FE and fd, fe intersecting the horizontal surface of the fluid in the points D, E and d, e; then, when the body floats in a state of equilibrium, the lines FD, FE, fd and fe are equal among themselves. This is very easily proved, for since the triangle ABC is equilateral, the angle ACB is equal to sixty degrees, and consequently its half, or the angles ACF and BCF are each of them equal to thirty degrees; therefore, by the principles of Plane Trigonometry, we have DFCD+CF-2C D.CF cos.30°, and similarly, by the same principles, we get but according to the conditions of equilibrium, these are equal, hence we have CD2 2C D.C F cos.30° C E2-2CE.CF cos.30°; = therefore, by substituting the analytical expressions, and transposing, we get x2-y-2d cos.30°(x-y), and dividing both sides by (xy), we shall have 2d cos.30°x+y. By Plane Trigonometry cos.30° sin.60°, and by the property of the equilateral triangle, we have d=b sin.60°; consequently, by substitution, we get 2b sin2.60° x + y ; or numerically, we obtain 2×28×25.95 +16.0542. 413. Hence it appears, that in so far as the equilibrium of floatation depends upon the equality of the lines FD and FE, the condition is completely satisfied, and the same may be said respecting the lines fd and fe; but it is manifest, that another condition must be fulfilled before the body attains a state of perfect quiescence, and that is, that the area of the immersed part ABED, is to the area of the whole section A BC, as the specific gravity of the solid body, is to that of the fluid on which it floats, or as 0.46875 to unity: now, this condition is evidently satisfied, when xyb(1-0.46875), therefore, numerically we obtain 25.95 X 16.05282 × 0.5312541.65. Here then, both the conditions of equilibrium are satisfied, and from this we infer, that the positions exhibited in the diagram are the true ones, the downward pressure of the body in that state, being perfectly equipoised by the upward pressure of the fluid. 414. What we have hitherto done respecting the positions of equilibrium, has reference only to a solid homogeneous triangular prism, floating on the surface of a fluid with its axis of motion* horizontal; * When a solid homogeneous body, in a state of equilibrium on the surface of a fluid is disturbed by the application of an external force, it will endeavour to restore itself by turning round a horizontal line passing through its centre of gravity, and this line on which the body revolves, is called the axis of motion. but there are various other forms, which are not less frequent in the practice of naval architecture, nor less important as subjects of theoretical inquiry some of these we now proceed to investigate. PROBLEM LVIII. 415. Suppose that a solid homogeneous body in the form of a rectangular prism, floats upon the surface of a fluid of greater specific gravity than itself, in such a manner, that only one of its edges falls below the plane of floatation :— It is required to determine what position the body assumes, when it has attained a state of perfect quiescence. Let ABCD be a vertical section, at right angles to the horizontal axis passing through the centre of gravity of the rectangular prism, and let IK be the surface of the fluid, on which the body floats in a state of equilibrium, ABCnm being the extant portion, and mDn the part which falls below the plane of floatation. Bisect mn in F and Dn in н, and draw the straight lines DF and mн, intersecting each other in I C K 急便 points A, C and B, D by the diagonals AC and BD, intersecting in G the centre of gravity of the rectangular section ABCD, and draw Gg. Then, because the body floats upon the surface of the fluid in a state of equilibrium according to the conditions of the problem; it follows from the laws of floatation, that the straight line Gg is perpendicular to IK. Through F the point of bisection of mn the base of the immersed triangle, and parallel to go, draw FP meeting the diagonal BD in the point P, and join Pm, Pn; therefore, because the straight line go is perpendicular to mn the plane of floatation, it is evident that FP is also perpendicular to mn, and consequently, pm and Pn are equal to one another. This is a condition of equilibrium which holds universally, and another is, that the area of the immersed triangle mDn, is to the area of the whole section ABCD, as the specific gravity of the solid, is to |
