730 MECHANICS now necessarily perpendicular to its plane) we have a system of two forces acting in non-intersecting lines. Then we have the following curious proposition, which may easily be proved from the formulas already given : When a system of forces is reduced in any manner ivhat- ever to tivo, the volume of the tetrahedron of which these are opposite edges is constant. Symmet- 227. The most symmetrical pair of resultant forces is rical re- found thus. Take any point P (fig. 61) in the central axis, duction. and draw through it a line APA , , T ^ perpendicular to it, and bisected Aft r at P. Substitute for the single force at P its halves acting at A and A respectively. Combine these respectively with the forces AQ, A Q of the couple when A A is made its arm. The system is thus reduced to two equal forces AR, A R , whose direc tions are interchangeable by a rotation of two right angles about the central axis. Let P with direction cosines A, /j., v be one of the forces, and let x, y, z be its point of application. Then X = 2(PA), Y = 2(P/0, Z = 2(PiO are the components of the single force at the origin. Also L Y Fig. 61. General reduc tion. are the components of the resultant couple. If we shift the origin to the point a, b, c the first three quantities are unaltered, but the couples become L = L + cY-bZ M =M + aZ-cX The point a, b, c is on the central axis if the axis of the resultant couple be parallel to the single force, i.e., if k = M. = -_ = e, suppose ; A- i Zj or L= eX-cY + JZ, M= cX+eY-aZ, Either of these sets gives the equations of the central axis. The resultant force and couple are in one plane, and therefore the resultant is a single force in the central axis, when By the values of L , M , N , above, we see that this is equivalent LX + MY + NZ-0. "When this last condition is not satisfied, we see that the value of the left hand member which, from the way in which it occurs, must obviously be an invariant, is where e has the same value as in the three .equations above. Minding s 228. One of the most remarkable of the many curious theorem, theorems connected with the single resultant of a system of forces is that of Minding. We have seen that, in general, the resultant may be put in the form of a single force and a couple in a plane perpendicular to it. Tf we now suppose the system of forces to be shifted into a new position such that their points of application, their magnitudes, and the angles between their directions two and two, all remain unchanged, the resultant force will be of the same magni tude as before, but the couple will in general be different. Of the infinitely infinite number of possible positions which the forces may assume, an infinite number correspond to a zero couple. Minding has shown that the lines of action of these single resultants consist of all lines passing through each of two curves, fixed in the body, an ellipse and an hyperbola, in planes perpendicular to each other. The proof of the proposition gives an interesting example of the use of Rodrigues s coordinates ( 83). The most obvious mode of attacking this question would be to resolve the applied forces into three groups, parallel respectively to three rectangular axes which revolve with them, and to choose those axes so that the sum of the resolved parts does not vanish parallel to any one of the three. Each of these systems of parallel forces has its own "centre" ( 222), so that the final resolution gives three forces, each of a given magnitude, acting in any mutually perpendicular directions at three definite points in the body. This, however, is not analytically so simple as the following. We refer the body to fixed axes Ox, Oy, Oz, to be afterwards speci fied. As the origin and the directions of these axes nre at our disposal, we may impose six conditions. Now suppose the forces to be resolved parallel to a set of rectangular axes Ox , Oy , Oz which will be considered afterwards to rotate with them. Such a system of axes may, at starting, have any assigned position. This gives us three conditions more. Let then A, B, C be the components, parallel to the second set of axes, of the force applied at the point whose coordinates referred to the first system are a, b, c. Let the direction cosines of the second system in any of its future positions, referred to the first system, be l lt m^, n-; l z , w 2 > 2> ^s> m & r h rc ~ spectively. Then the force at a, b, c has the following components : + C1 3 parallel to Ox ,, ,, Oy The expressions for the resultant force and couple at the origin will evidently depend upon the following twelve quantities, besides the direction cosines, viz : 2A, 2B, 20, 2(Aa), 2(A6), 2(Ac) , 2(Ba), 2(B6), 2(Bc) , 2(Ca), 2(06), 2(Cc) . Assume 2B = 0, 20 = 0, i.e., let Ox be always parallel to the direc tion of the resultant force. Next, let 2(Aa) = 0, 2(A&) = 0, 2(Ac) = 0, i.e., let the origin be chosen as the "centre" ( 222) of the forces parallel to the resultant force. As we have still four conditions to impose, we select the following: 2(Ba) = 0, 2(Bc) = 0, 2(Ca) = 0, 2(CZ>) = 0. These express that the plane of the couple due to the forces C passes through Oy, while that of the forces B passes through Oz. Write now The force and couple at the origin are These are equivalent to a single force if ( 227) (/^ - ?i 1 7 2 )0 - (^ma - mj^y = , or m 3 f3-n 2 y=Q ....... (1). This is the required condition. When it is satisfied, the equations of the line in which the single force jt acts are any two of the condition that these three agree being (1). Eliminate l : between the last two, and we get SM-ntf^lM-lzflC ...... (3). Now introduce in (1), in the first of (2), and in (3), Rodrigues s values of the cosines ( 83), and they become respectively (yz - wx}p - (yz + wx}y = , (xz - wy)-n - (wz + xij}= (yz + wx)$ - (yz - wx)y , Q(wz + xy) - fr(xz - wy) = (xz + wy}y-rj - (xy - wz)$. Rearranging according to y, z, and yz, - y}y - (xq - w)z + (0 + y}wx = , the second of which may be put, by means of the first, in the form (wil + xQy - (XT) - wC)z + 2(p + y)wx=Q. These three equations involve z/ , x, y, z in the form of the ratios only of the last three to the first. The last two are linear in ]L . Solving them, and substituting in the first we find, w w
finally, a biquadratic in .