If the plane in question be chosen perpendicular to the direction
of the vector-sum of the given forces, the vector-sum of the
components Q is zero, and these components are therefore
equivalent to a couple (§ 4). Hence any three-dimensional
system can be reduced to a single force R acting in a certain line,
together with a couple G in a plane perpendicular to the line.
This theorem was first given by L. Poinsot, and the line of action
of R was called by him the central axis of the system. The combination
of a force and a couple in a perpendicular plane is termed
by Sir R. S. Ball a wrench. Its type, as distinguished from its
absolute magnitude, may be specified by a screw whose axis is
the line of action of R, and whose pitch is the ratio G/R.
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| Fig. 42. |
The case of two forces may be specially noticed. Let AB be the shortest distance between the lines of action, and let AA′, BB′ (fig. 42) represent the forces. Let α, β be the angles which AA′, BB′ make with the direction of the vector-sum, on opposite sides. Divide AB in O, so that
and draw OC parallel to the vector-sum. Resolving AA′, BB′ each into two components parallel and perpendicular to OC, we see that the former components have a single resultant in OC, of amount
whilst the latter components form a couple of moment
Conversely it is seen that any wrench can be replaced in an infinite number of ways by two forces, and that the line of action of one of these may be chosen quite arbitrarily. Also, we find from (2) and (3) that
The right-hand expression is six times the volume of the tetrahedron of which the lines AA′, BB′ representing the forces are opposite edges; and we infer that, in whatever way the wrench be resolved into two forces, the volume of this tetrahedron is invariable.
To define the moment of a force about an axis HK, we project the force orthogonally on a plane perpendicular to HK and take the moment of the projection about the intersection of HK with the plane (see § 4). Some convention as to sign is necessary; we shall reckon the moment to be positive when the tendency of the force is right-handed as regards the direction from H to K. Since two concurrent forces and their resultant obviously project into two concurrent forces and their resultant, we see that the sum of the moments of two concurrent forces about any axis HK is equal to the moment of their resultant. Parallel forces may be included in this statement as a limiting case. Hence, in whatever way one system of forces is by successive steps replaced by another, no change is made in the sum of the moments about any assigned axis. By means of this theorem we can show that the previous reduction of any system to a wrench is unique.
From the analogy of couples to translations which was pointed out in § 7, we may infer that a couple is sufficiently represented by a “free” (or non-localized) vector perpendicular to its plane. The length of the vector must be proportional to the moment of the couple, and its sense must be such that the sum of the moments of the two forces of the couple about it is positive. In particular, we infer that couples of the same moment in parallel planes are equivalent; and that couples in any two planes may be compounded by geometrical addition of the corresponding vectors. Independent statical proofs are of course easily given. Thus, let the plane of the paper be perpendicular to the planes of two couples, and therefore perpendicular to the line of intersection of these planes. By § 4, each couple can be replaced by two forces ±P (fig. 43) perpendicular to the plane of the paper, and so that one force of each couple is in the line of intersection (B); the arms (AB, BC) will then be proportional to the respective moments. The two forces at B will cancel, and we are left with a couple of moment P·AC in the plane AC. If we draw three vectors to represent these three couples, they will be perpendicular and proportional to the respective sides of the triangle ABC; hence the third vector is the geometric sum of the other two. Since, in this proof the magnitude of P is arbitrary, It follows incidentally that couples of the same moment in parallel planes, e.g. planes parallel to AC, are equivalent.
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| Fig. 43. |
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| Fig. 44. |
Hence a couple of moment G, whose axis has the direction (l, m, n) relative to a right-handed system of rectangular axes, is equivalent to three couples lG, mG, nG in the co-ordinate planes. The analytical reduction of a three-dimensional system can now be conducted as follows. Let (x1, y1, z1) be the co-ordinates of a point P1 on the line of action of one of the forces, whose components are (say) X1, Y1, Z1. Draw P1H normal to the plane zOx, and HK perpendicular to Oz. In KH introduce two equal and opposite forces ±X1. The force X1 at P1 with −X1 in KH forms a couple about Oz, of moment −y1X1. Next, introduce along Ox two equal and opposite forces ±X1. The force X1 in KH with −X1 in Ox forms a couple about Oy, of moment z1X1. Hence the force X1 can be transferred from P1 to O, provided we introduce couples of moments z1X1 about Oy and −y1X1, about Oz. Dealing in the same way with the forces Y1, Z1 at P1, we find that all three components of the force at P1 can be transferred to O, provided we introduce three couples L1, M1, N1 about Ox, Oy, Oz respectively, viz.
It is seen that L1, M1, N1 are the moments of the original force at P1 about the co-ordinate axes. Summing up for all the forces of the given system, we obtain a force R at O, whose components are
and a couple G whose components are
where r = 1, 2, 3 . . . Since R2 = X2 + Y2 + Z2, G2 = L2 + M2 + N2, it is necessary and sufficient for equilibrium that the six quantities X, Y, Z, L, M, N, should all vanish. In words: the sum of the projections of the forces on each of the co-ordinate axes must vanish; and, the sum of the moments of the forces about each of these axes must vanish.
If any other point O′, whose co-ordinates are x, y, z, be chosen in place of O, as the point to which the forces are transferred, we have to write x1 − x, y1 − y, z1 − z for x1, y1, z1, and so on, in the preceding process. The components of the resultant force R are unaltered, but the new components of couple are found to be
| L′ = L − yZ + zY, |
By properly choosing O′ we can make the plane of the couple
perpendicular to the resultant force. The conditions for this
are L′ : M′ : N′ = X : Y : Z, or
| L − yZ + zY | = | M − zX + xZ | = | N − xY + yX |
| X | Y | Z |
