1 in the perpendicular AH. Now, if a denote AH, C=BD, X=AG, and y=EF any line parallel to the base BD; then as a ¦ c¦ ¦ x ¦ y fluent yxx fluent yx fluent x2xx3 fluent xx x = AH, when x be AH. comes AH: consequently CH = In like manner, the centre of gravity of any other plane triangle, will be found to be at of the altitude of the triangle; the same as it was found in prop. 43, Mechanics. EXAM. 2. In a parabola; the distance from the vertex is 3x, or 3 of the axis. EXAM. 3. In a circular arc; the distance from the centre of the circle, is r the radius. cr where a denotes the arc, c its chord, and EXAM. 4. In a circular sector; the distance from the centre : where a, c,r, are the same as in exam.3. EXAM. 5. In a circular segment; the distance from the cen C3 tre of the circle is ; were c is the chord, and a the area, 12α of the segment. EXAM. 6. In a cone, or any other pyramid; the distance from the vertex is 3x, or 3 of the altitude. EXAM. 7. In the semisphere, or semispheroid; the distance from the centre is 3r, or of the radius: and the distance from the vertex § of the radius. EXAM. 8. In the parabolic conoid; the distance from the base is x, or of the axis. And the distance from the vertex 2 of the axis. EXAM. 9. In the segment of a sphere, or of a spheroid; the distance from the base is the segment, and a the whole axis, or diameter of the sphere. EXAM. 10. In the hyperbolic conoid; the distance from the 2α + x báse is 6a+4x; where x is the height of the conoid, and « the whole axis or diameter. a PRACTICAL 1 123, PRACTICAL QUESTIONS. QUESTION I. A LARGE Vessel, of 10 feet, or any other given depth, and of any shape, being kept constantly full of water, by means of a supplying cock, at the top; it is proposed to assign the place where a small hole must be made in the side of it, so that the water may spout through it to the greatest distance on the plane of the base. Let AB denote the height or side of the vessel; D the required hole in the side, from which the water spouts, in the parabolic curve DG, to the greatest distance BG, on the horizontal plane. By the scholium to prop. 68, Hydraulics, the distance BG is always equal ax-x2, to 2 AD. DB, which is equal to 2x(x-a) or 2 124. QUESTION II. If the same vessel, as in Quest. 1, stand on high, with its bottom a given height above a horizontal plane below; it is proposed to determine where the small hole must be made so as to spout farthest on the said plane. Let the annexed figure represent the Here, as before, ba=2AD. Db =2 or cx r2 must be a maximum. And hence, like as in the former question, x = 1c = lab. So that the hole D must be made in the i middle middle between the top of the vessel, and the given plane that the water may spout farthest. 125. QUESTION III. But if the same vessel, as before, stand on the top of an inclined plane, making a given angle, as suppose of 30 degrees, with the horizon; it is proposed to determine the place of the small hole, so as the water may spout the farthest on the said inclined plane. Here again (D being the place of the hole, and BG the given inclined plane), bG 2/ad. db = 2 √x(a x ± z), putting z = Bb, and, as before, a = AB, and x AD. Then be must still be a maximum, as also Bb, being in a given ratio to the maximum BG, on account of the given angle B. x2± xz, as well as z, the Fluxions, ax conseq. 2 = 2x becomes barely 2x. the sine of which is BG2 · Bb2 = 322 = 3 Therefore ax is a maximum. Hence, by art. 54 of a; ba 2 α z and hence bo = 2 √√x(α 'x x ± z} But as the given angle Geb is ≈ 30o, ; therefore BG = 2Bb or 2z, and bg2 — (2x-a)2, or be = ±(2x a) 3. Putting now these two values of be equal to each other, gives the equation 2x=(2x-a) 3, from which is found a√3 3±√3 √3±1 4. a, the value of AD required. Note. In the Select Exercises, page 252, this answer is 6+/6 a, by taking the velocity proportional to brought out 1 126. QUESTION IV. It is required to determine the size of a ball, which, being let fall into a conical glass full of water, shall expel the most water possible from the glass; its depth being 6, and diameter 5 inches. AC = √AG2+Gc2 = 61 AD = FE = FH x the radius of the ball. The two triangles ACG and DCF are equiangular; theref. AG : AC :: DF: Fc, that is, a : c :: x : сх CX FC; hence α GF GC – FC = b. and GH = GF + FH = b + x GF α the height of the segment immersed in the water. Then (by rule 1 for the spherical segment, p. 427 vol. 1.), the content of the said immersed segment will be (6DF ×·5236 = (2x− b + =) × (x + b a сх α 2GH) X GH2 )2 X 1.0472, which must be a maximum by the question; the fluxion of divided by 2x and the common factors, this made = 0, and gives ·× (b 2a+c (c-a)X(c+2a) α =21, the radius Consequently its diameter is 411 inches, as re PRACTICAL EXERCISES CONCERNING FORCES ; WITH THE RELATION BETWEEN THEM AND THE TIME, VELOCITY, AND SPACE DESCRIBED. BEFORE entering on the following problems, it will be convenient here, to lay down a synopsis of the theorems which express the several relations between any forces, and their corresponding times, velocities, and spaces, described; which are all comprehended in the followig 12 theorems, as collected from the principles in the foregoing parts of this work. LET f, F, be any two constant accelerative forces, acting on any body, during the respective times t, T, at the end of which are generated the velocities v, v, and described the spaces, 8, Then, because the spaces are as the times and velocities conjointly, and the velocities as the forces and times; we shall have, ន. t And if one of the forces, as F, be the force of gravity at the surface of the earth, and be called 1, and its time T be= 1"; then it is known by experiment that the corresponding space s is 16 feet, and consequently its velocity v=2s= 321, which call 2g, namely, g16 feet, or 193 inches. Then the above four theorems, in this case, become as here below: 12 And from these are deduced the following four theorems, for variable forces, viz. |