and 0 = Wa sin2 0 cos - W'a' sin cos 0, which two equations are sufficient for the determination of e and p. COR. If the rods are equal in all respects, these two and 0 = sin2 0 cos 0 – sin2 & cos ; the last of which gives θ = φ .(4), or 1 = cos2 + cos e cos & + cos3Ó.............. (B). Let us consider these two results separately, and whence the position of the rods is known. This position is symmetrical with regard to a vertical line through C. (2) The equations (B) and (7) shew that and are interchangeable, and consequently there are, besides the symmetrical position just found, two unsymmetrical positions of equilibrium, similarly situated on each side of the first found position. They may be found by means of (7) and (B). 27. A solid of any form is placed with its convex surface upon a horizontal plane; to find the position of equilibrium. Let x = f(x, y) be the equation of the surface, referred to three rectangular axes in the body: and let y≈ be the co-ordinates of the point in contact with the horizontal plane, and xyz those of the centre of gravity referred to the same axes. Then the plane on which the body stands being a tangent plane, if a ßy be the inclinations of the co-ordinate П П axes to the horizon, α, - B, y will be the in 2 2 2 clinations of the vertical through the point of contact to the co-ordinate axes; this vertical line is a normal, and therefore But since the solid rests upon a point, the vertical through that point must pass through the centre of gravity of the solid, i. e. the normal at the point of contact passes through the centre of gravity of the solid; hence the equations of the normal give will enable us to find x, y, x; and thence a, ß, y from (1). Ex. Suppose the solid to be the eighth part of a sphere comprehended between three rectangular planes: to find the position in which it will rest with its convex surface on a horizontal plane. Let its equation be a2 = x2 + y2 + x2; .. d.%= and d≈ – – 2. = hence making substitution in equations (2) we obtain 00 = y =2; these two equations joined with the equation of the surface of the ellipsoid give 28. To determine the nature of the equilibrium when a body of given form rests upon a given curve surface. At the point of contact of the given body with the surface on which it rests in equilibrium, the two surfaces will have a common normal, which will be vertical and pass through the centre of gravity of the body. Let DAd (fig. 104) bẹ this normal, A being the point of contact of the two surfaces BC, bc; and D, d being the centres of curvature of the arcs BC, be corresponding to the point A; and let G be the centre of gravity of the body. Let now the body be made to roll over a very small arc AP, and thereby to come into the position b'c'; A', G', d' being the new positions of A, G, d; and P being the new point of contact. By this movement the point A' will trace out a small portion of an epicycloid, which at the very beginning of the motion is perpendicular to the surface at A; hence A' begins to move along the line Ad. We suppose the displacement of the body so small that A' is in Ad. Draw Pp vertical. Draw Pp vertical. If Pp pass through G the body is still in equilibrium; but if G' lie figure,) the body when left original position; and lastly, to the right of Pp (as in the to itself will roll back into its if Glie to the left of Pp, the body will roll farther from its first position. Hence the first position is one of stable, unstable, or neuter equilibrium according as A'p is >< or = A'G'. To express this result analytically, let p, p' be the radii of curvature DA, dA. Then because the lines Pp, DA' (for A' is in the line Dd) are parallel; A'd p + p ᎠᏢ A'p Hence the equilibrium is stable, unstable, or neuter according as COR. 1. If the surface on which the body rests be concave, we must account p negative in the above result. COR. 2. If the surface be a plane, we must make P infinite, and then since in that case, the equilibrium will be stable, unstable, or neuter according as is >< or = AG. COR. 3. If the lower surface of the body be a plane, we must make p' infinite, and then the result is Ex. Find what segment of a paraboloid will rest in a position of neuter equilibrium upon a spherical surface of given radius. Let be the length of the axis of the paraboloid, 4m its latus rectum; and a the radius of the spherical surface. Then from Ex. 5, p. 101, we have 29. A string is stretched along a smooth curve line of any form by two equal forces, required the unit of pressure exerted by it upon the cylinder at any point. (Fig. 105.) Let AHB be the curve line along which the string is stretched by the two equal forces P, Q. Let HH' be a very small arc, and at H, H' draw tangents meeting in K, and normals HO, H'O. Join KO, and put ▲ HOH' = 80. The portion of string HH' is kept in equilibrium by the tensions at H, H', each of which is equal to P or Q; and by the reactions of the curve line HH', which being smooth, the reaction at every point will be in the direction of a normal. Hence the resultant of all the reactions on HH' will pass through O, and as it must also pass through K, it acts in the line OK. Hence by Art. 28, resultant reaction on HH': P:: sin HKH': sin OKH :: sin HOH': cos KOH :: 80: 1 ultimately; .. resultant reaction on HH' = P.SO. |