85. "In a quadrilateral there are given, the length and position of the lower base, the lengths of the two sides, the length of the upper base and the position of a point through which it passes: required to construct the quadrilateral." [No construction of this prob. has been received.-Prof. J. Scheffer writes: "If we describe a circle about one of the extrmities of the lower base of the quadrilateral as a centre, with a radius equal to one of the sides, and another circle about the other extremity of the lower base as a centre, with a radius equal the other side, we reduce the problem to the following: If two circles are given as to magnitude and position, to lay a line of given length between the two circumferences, which passes throngh a given point. The line of given length is of course the upper base. As the line may be laid between the outer circumferences, or one outer and one inner circumference, or two inner circumferences, the problem admits in general of four solutions. This problem belongs to those which cannot be solved by means of the straight line and circle only, and, therefore, exceeds the power of Plane Geometry." Marcus Baker says that in the solution of 85 the equation is involved to the 8th degree.] 86. "Prove that the attraction of a sphere of uniform density upon an external point is the same as if all the matter of the sphere were concentrated at its centre." SOLUTION BY A. B. NELSON M. D., DANVILLE, KY. Let A be the point, and suppose the sphere to be generated by the revolution of a semicircle about its diameter EF. Take the origin of coordinates at the centre C, and let CD and PD be the coordinates of any point P of the semicircle. Then the elemental ring of matter generated by the particle P, 2луdуdx. = 1; and of y, 0 and of an elemental ring upon A, measured along CA is (1-2). The attraction AP [(a + x)2 + y2] Hence the sum of the attractions of all the particles of the sphere is which is equal to the attraction of the sphere at the distance a, if all its particles were concentrated at the centre. [For solution of this problem Mr. Siverly refers to the following works: Simpson's Fluxions p. 450, Emerson's Fluxions p. 370, Vince's Fluxions p. 116, Dealtry's Fluxions p. 146, Poisson's Mechanics Art. 100, Mecanique Celeste Book II, Chap. 2, and Earnshaw's Dynamics p. 333. To which may be added, Newton's Principia Book I, Prop. LXXI.] 87. "There are n tickets in a bag numbered 1, 2, 3, n. A man draws three tickets together at random and is to receive a number of shillings equal to the product of the numbers he draws. Find the value of his expectation." SOLUTION BY J. M. GREENWOOD AND W. H. BAKER, KANSAS CITY, MO. Let & be the sum of the products taken three in a set, p the number of products, and the value of the chance. and p Ρ To find 8, (see Todhunter's Algebra, Art. 227), we have ... 16) (1 + 2 + 3 + . . . n)3 = 13 + 23 + 33 + . . . n3 + 3. 12 (2 + 3 + . . . π) +3.22(1+3+4+ But 3.n2(1+2+3+ n ... n) + = n(n + 1) .... 2) + ... ... n2(n+1)2(2n+1) n2 (n + 1)2 4 2 n2(n + 1)2(n —— '1) (n — 2); 48 = (5) .(6) 88. "An ellipse revolves about its latus rectum; show that the volumes of the solids generated by the larger and smaller segments are respectively equal to" &c. SOLUTION BY E. B. SEITZ, GREENVILLE, OHIO. The equation to the ellipse referred to the latus rectum and major axis, is Hence the volume of the solid generated by the larger segment is and the volume of the solid generated by the smaller segment is 89. "A sphere, radius r, rolls down the surface of another sphere of the same material, radius R, placed on a horizontal plane. The surfaces of both spheres and plane are rough enough to secure perfect rolling. Determine the motion of the spheres, the point of separation and the equation of the curve described by the center of the upper sphere." SOLUTION BY WALTER SIVERLY, OIL CITY, PA. It is evident from the principle of the motion of the center of gravity that the spheres will roll in opposite directions. Let A and B be the centers of the spheres at any time t after the beginning of the motion, O being the initial position of A; D the point of contact of the spheres; C, C' the points that were in contact at the beginning of the motion; m, m', the masses of A and B respectively; O the origin of horizontal and vertical coordinates; (x, 0), (x, y) the coordinates of A and B; F the friction between the spheres, F" the friction between the lower sphere and the plane; A, 0' the angles through which the spheres have respectively revolved, y the inclination of AB to the vertical, ß being the initial value of 4. Since F" is the only force acting on the system horizontally, ዋ (1) = 2m'g(R+ r) (cos ẞcos q). Eliminating F, F" from (1), (2), (3) and integrating once, 49m + 45m′ — 25m' cos2 + 20m'cos y) dy2 70g φ ) at 2 = R + r (cosß (9) R+r(m+m2) (cosp―0084) (10) At the point of separation each sphere moves uniformly horizontally, hence Differentiating (6) once, Substituting de d2x d'x' dt2 dt2 =-2 cos ẞ (49m + 45m′), from (10), differentiating and putting + = dt obtain 25m' cos3y-40m' cos3y-[3(49m+45m′)—20m'cos ẞ] cos which determines the point of separation. Differentiating (6), (7) once and substituting for x' its value from (5), Eliminating do by (11), then dy by (12) and substituting for sin y, cos y, 7(m + m') dx=(—7m + 2m') y R + r' ydy 2m' (R+r) dy √[(R + r)3 — y2] _v[(R+r)2 — y3]' Integrating and observing that initially x=(R+r) sin ẞ, y=(R+r) cos ß, 7(m+m')x = (7m+2m')v ́[(R+r)2—y3]+5m′(R+r) sin ß+2m′(R+r) × cos-1[y÷(R+r)]—ß }, the equation to the required path. Mr. Adcock finds for the equation of the curve, (in which a=Siverly's B), [2(r+R)y—y2]1⁄2\ — 2m2(r+B) [sin-1 ([2(r+y=-*) — a] + [2(r+ R)y — y2 ] *. x= 90. 7m+2m' "Let an oblate ellipsoid of revolution of homogeneous density rotate about one of its greatest diameters." &c. (See page 160). SOLUTION BY WALTER SIVERLY. The attraction of the ellipsoid at any point in the shorter diameter at a distance p from the center, as shown in works on attraction, is and the centrifugal force = pa. Also the attraction at any point on the longer diameter is and the centrifugal force = pa2. Let u the pressure at the point distant p from the center on the shorter diameter then a2)pdp. u = C — p2(P — a2). At the surface the pressure = 0. ... C C = b2(P— a2), u = (b2 — p2) (P — a2), center where p=0. Similarly the pressure at the center on the longer diameter= a2(Qa2). Hence for equilibrium, b2(P — a2) = a2(Q a2), or (1 (1 — e2) (P — a2) = Q — a3. Substituting values of P and Q and reducing, |