If x, y be the co-ordinates of P, and x, y those of Q, both measured from B as origin, the virtual velocities of P and Q will be respectively da' and da; consequently . Pdx' + Qdx = 0. To this we must join the equations of the two curves y' = F" (x'), and y = F(x), and the equation b = √ x22 + y2 + √ x2 + y2. Ex. Let AD be a circle whose centre is in BA pro duced; and AC a parabola, the directrix of which passes through B. from which equation, BQ being known in terms of x, x may be found. 23. Two given weights P, Q are connected by a string PAQ which is laid across a horizontal cylinder; to find the position and nature of the equilibrium. (Fig. 100.) It is evident the string will lie in a vertical plane perpendicular to the axis of the cylinder. Let C be the centre of the circular section of the cylinder by this plane. Draw CA vertical, and BCD, PM, QN horizontal: join CP, CQ. Then since the length of the string and the radius of the cylinder are given, the angle PCQ is known; let it be denoted by 2a and let a +0, a represent the angles QCA, PCA; and a = CA. Then if be the altitude of the common centre of gravity of P and Q above BD, we have (P+Q) ≈ = P. CM + Q. CN = Pa cos (a-0) + Qa cos (a + 0); .. (P+Q) d。≈ = Pa sin (a-0) Pa cos (a -0) - Qa cos (a + 0) = − (P + Q)ž.............................................. (1). = Now in the position of equilibrium dez 0, and therefore Pa sin (a) = Qa sin (a + 0), from which we find P-Q tan = tan a, P+Q which gives the position of equilibrium, which is unstable because equation (1) shews that is then a maximum. 24. A hollow paraboloid is placed with its vertex downwards and axis vertical; a given rod rests within it, leaning against a pin at the focus, and having its lower end upon the parabolic surface. Find the position of equilibrium. (Fig. 101.) Let PQ be the rod, G its centre of gravity, S the focus of the paraboloid, AS its axis, BAC a section of it by a vertical plane passing through the rod; a = AS, b = PG, r = PS, 0 = ASP; through S draw LS horizontal, and draw MG vertical; let = MG. Then by the nature of the parabola In the position of equilibrium d,z = 0, and therefore r = √2ab; from which the position of the rod is known. Equation (1) shews that the altitude of G is then a minimum; and therefore the position is one of stable equilibrium. 25. A paraboloid, formed by the revolution of a given parabolic area about its axis, is placed with its convex surface upon a horizontal plane; to find the position in which it will rest. (Fig. 102.) Let AC be the axis of the parabola, inclined at an 40 to the horizontal plane: P the point on which it rests; draw PN vertical: then since there is equilibrium the centre of gravity must be in the line PN (Art. 132), but it is also in AC, the axis of the parabola, consequently it is at N. Draw PM perpendicular to AC; let a = AC, b = BC; then b2 the latus rectum = - " But because N is the centre of gravity, AN=2a. 3 П COR. The least value of cot is when =: hence when 2 or, when a is = or < the solid can only rest in equilibrium with its axis vertical. 26. Two heavy rods AC, CB connected by a hinge at C rest on two smooth points D, E, situated in a horizontal line: find the position of equilibrium. (Fig. 103.) Let G, g be the centres of gravity; and W, W' the weights of the rods AC, BC; R, R' the reactions of the points D, E which will be at right angles to the rods, because the points on which they rest are smooth. Join DE, and let 0, denote the angles CDE, CED; and put CG = a, Cg=a', DE=b. The rod AC is kept in equilibrium by three forces, viz. its own weight at G, the reaction Rat D, and the tension of the hinge C; to avoid the last, (the magnitude and direction of which are unknown, and are not required,) let us take the moment of these forces about C; find ...R.DC-W. a cos 00............(1). Proceeding in a similar manner with the beam CB, we R'.EC - W'.a' cos p = 0... (2). Again, when the equilibrium is once established, we may suppose the hinge C to become rigid; under this hypothesis the rigid body ACB is kept in equilibrium by four forces, viz. R, R', W and W'. Hence resolving them vertically and horizontally, we find R cos +R' cos and R sin – W W' = 0......(3), - R' sin = 0......(4). These four are all the independent equations which can be derived from the mechanical properties of the machine; there are however two others, which express its geometrical properties, derived from the triangle DCE, viz. which being substituted in (3) and (4) give W + W' = sin (0 + 4) ( Wa cos2 0 b sin o + W'a' cos p b sin |