there being no correction, since when x = 0, u= 0, and t = 0. Hence, the time of the descent of the balls is Again, when the iron is disengaged from the wood, the wooden ball will ascend with the force 47 93. (S – s) 3 and, if u' be its velocity at the end of the time t', having then described the space x' from the bottom, the resistance of the water is Х x #r* X s. 2g Hence, the actual force by which the ball is accelerated upwards is ar's v'dv' 93. (S – s) x 0" 3 49 d.x' 129 v'du' Х 16gr.(S-8) 3sve 47 F= S logr. 15 2g x {C – 1. (16gr. S-s — 3svée)} S = 2g 16gr. S B' B' · 0'2 by supposition. Hence, as before, v=7B'XV (I-e-My A' + Х 2 2 a's) + ✓ B Now, since by the question the whole time elapsed during the descent and ascent is given, let it = T; then T=t+ť = x x GB + JB) + A/B 1. {1 + V(1—-^)} + x l. {1 + V(1—-^=)} A/B ✓ {) X X + Syr (S + S' – 2s) (2 x S2 / 38 {ris x l.(1 + vi-e-4-) + AJ Sgr + S – 2s) ✓ 1. (1 + 1 (1-07-)} whence x may be found by approximation. 1 Х + 1 Х 742. In Newton's construction (Schol. Prop. XXXIV, Vol. II.) since the < CQO = 2 2 CSO = 2 S at the vertex of the cone; therefore in the limit of the altitude OD, or when CRO = 90°, CSQ must be 45°; :: OCS = 45°, and its supplement = 135o. Now in the solid of least resistance, whatever the nature of that solid may be, the resistance upon the nascent end of it, corresponding to FG, will evidently be the same as upon the nascent truncated cone generated by the tangent at that end. Hence, when this nascent cone is of minimum resistance, the nascent solid will be also of least resistence; and the surface of the truncated solid of least resistance makes therefore with its smaller end an angle of 135o. 743. Let AC = a, CD = B, and let the co-ordinates at the required point be x, y. Also, let R be the resistance upon the base of* ; then the resistances upon wy?, and #AQ? = y*, are 4a 4.3 + P y R X respectively. Hence the resistance upon the surface of the frustum of the cone generated by PQA is (Vince's Fluxions; prop. 133) R. yo y? 3Rx р 3pR Х Х 22 489 ) 4x2 + yo 4a p+4r. Moreover, if R' be the resistance upon the whole paraboloid, it will easily be shewn from the above cited prop. of Vince that the resistance upon the paraboloidal frustum is 2 R' ydy Х 32 dx2 dy? ydy d.rs 1+ dy Hence, by the question, we have 3pR y + R + R' ydy e minimum dy + It :: by the rule, we get R- 288 2R ydy . = 0. d.rs dye But since y' = pr, we have by substitution, pdor dr + R = 0 (v + 4x) – 4pp + 48) = - 3p and this being resolved, gives r=; which will determine the required position of P. 744. If R and R' denote the resistances upon the circle and hexagon, A and A' their areas, and 0, 8the inclinations of the directions to their planes; then (Vince prop. 28) R: R' :: sin. 8 x cos. 8 X A: sin.° 6' x cos. 6' X A But by the question, sin. 8 = sin, 30 = 1 and sin. O'=sin. 60°= N 3 Also, A = ar (r being the radius of the circle) and 2 A' = 6 x 13 373 X ተ = Х 2 2 745. Let a be the length of either of the equal sides of the trapezium, 6 their inclination to the other sides. Also, let m be the length of that parallel side which is resisted by the fluid ; and suppose A, the resistance on a when moving parallel to m, A' the resistance when moving perpendicularly to m, and M the resistance upon m. Moreover, let R be the resistance when moving I ly to the fluid. Then A : R:: sin.3 0 : 1 R: M:: a : m A: M:: a sin. 3 0 : m. A' : M:: a x cos. 3 0 : m :: 2A' + M: M:: 2a cos.3 0 + m : m But M : A:: m : a sin.3 8' :: 2A' + M : A :: 2a cos.3 0 + m ; a sin. 3 0 the analogy required. 746. The reason is because of the oscillatory motion of the waves. They rise and fall in the manner of a pendulum. (See Newton, Vol. II., prop. 44). 747. 1 The faces are inclined to the diagonal of a cube at the angle 0 = sin. N 3 :: If R denote the resistance upon the face when its direction is I to its surface, the resistance when the direction is parallel to the diagonal is R X sin.3 0 = 373 But there are three faces thus resisted; therefore the whole resistance is R Q. E. D. 3 R |