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 to 2/AD DB, which is equal to 2√x (a - x) or 2/ax x2, if a be put to denote the whole height AB of the vessel, and x = AD the depth of the hole. Hence 2/ax fluxions, ax x, or ax - x2, must be a maximum. In 2xx 0, or a - 2x0, and 2x = a, or xa. So that the hole D must be in the middle between the top and bottom; the same as before found at the end of the scholium above quoted. 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 vessel as before, and be the greatest distance spouted by the fluid, DG, on the plane bc. Here, as before, bG 2 AD Db = 2 √√x (c− x) = 2√cx x2, by putting abc, and AD = x. So that 24 cx x2 or cx x2 must be a max imum. And hence, like as in the former question, D B x = c = Ab. So that the hole D must be made in the middle between the top of the vessel, and the given plane, that the water may spout farthest. 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 r 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. Therefore ax 2x + x = 0; 2x(a x + z) = x2 + xz, as well as z, is a maximum. Hence, by art. 54 of the Fluxions, ax 2xx zx = 0, or a conseq. z = 2x a; and hence bG becomes barely 2x. But as the given angle Gвb is 30°, the sine of which is ; therefore BG = 2Bb or 2z, and bG2 = Bb2 = 3x2 = 3(2 x − a)2, or bG BG2 = ±(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 √ 3 3±√3 x= ~3± 1 = 4 a, the value of AD required. Note. In the Select Exercises, page 269, this answer is brought out 6+/6 a, by taking the velocity proportional to the root of half the altitude only. 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. Let ABC represent the cone of the glass, and DHE the ball, touching the sides in the points D and E, the centre of the ball being at some points F in the axis Gc of the cone. B E H Put AC = VAG2 + GC2 = 6% = c, AD = FE = FH the radius of the ball. The two triangles ACG and DCF are equiangular; theref. сх AG: AC: DF: FC, that is, a ; c :: x: = FC; hence a a and GHGF + FH = b+ x a GF GC FC b the height of the segment immersed in the water. Then (by rule 1 for the spherical segment, page 51), the content of the said immersed segment will be (6DF 2G11) × GH2 b+ -) × (x + b − (7)2 × 1·0472, CX α - a * ·5236 = (2x which must be a maximum by the question; the fluxion of this made 0, and divided by 2x and the common factors, × (6- -x) 2a + c -x - b) x -x2=0; a a a abc 2, the ra this reduced gives x = (ca) x (c + 2a) dius of the ball. Consequently its diameter is 41 inches, as required. 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 following 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 , s. Then, because the spaces are as the times and velocities conjointly, and the velocities as the forces and times; we shall have, 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 v2s 32%, which call 2g, namely, g 16 feet, or 193 inches. Then the above four theorems, in this case, become as here below: And from these are deduced the following four theorems, In these last four theorems, the force f, though variable, is supposed to be constant for the indefinitely small time t, and they are to be used in all cases of variable forces, as the former ones in constant forces; namely from the circumstances of the problem under consideration, an expression is deduced for the value of the force f, which being substituted in one of these theorems, that may be proper to the case in hand; the equation thence resulting will determine the corresponding values of the other quantities, required in the problem. When a motive force happens to be concerned in the question, it may be proper to observe, that the motive force m, of a body, is equal to fq, the product of the accelerative force, and the quantity of matter in it q; and the relation between these three quantities being universally expressed by this equation m = qf, it follows that, by means of it, any one of the three may be expelled out of the calculation, or else brought into it. Also, the momentum, or quantity of motion in a moving body, is qu, the product of the velocity and matter. It is also to be observed, that the theorems equally hold good for the destruction of motion and velocity, by means of retarding forces, as for the generation of the same, by means of accelerating forces. And to the following problems, which are all resolved by the application of these theorems, it has been thought proper to subjoin their solutions, for the better information and convenience of the student. PROBLEM I. To determine the time and velocity of a body descending, by the force of gravity, dorun an inclined plane; the length of the plane being 20 feet, and its height 1 foot. Here, by Mechanics, the force of gravity being to the force down the plane, as the length of the plane is to its height, therefore as 20:1:: 1 (the force of gravity): =f the force on the plane. Therefore, by theor. 6, vor /4gfs is/4 x 162 × 20 × 20√4 x 16 = 2 × 4% or 8 feet nearly, the last velocity per second. And, 48 By theor. 7, t or √ √ is √ 42 seconds, the time of descending. PROBLEM |