Aα − Bβ = AB − αβ = AB (1 − cos φ) = 1⁄2AB·φ2,
ultimately. Since this is of the second order, the products F·Aα and F·Bβ are ultimately equal.
| Fig. 46. | Fig. 47. |
The total work done by two concurrent forces acting on a particle, or on a rigid body, in any infinitely small displacement, is equal to the work of their resultant. Let AB, AC (fig. 46) represent the forces, AD their resultant, and let AH be the direction of the displacement δs of the point A. The proposition follows at once from the fact that the sum of orthogonal projections of AB>, AC> on AH is equal to the projection of AD>. It is to be noticed that AH need not be in the same plane with AB, AC.
It follows from the preceding statements that any two systems of forces which are statically equivalent, according to the principles of §§ 4, 8, will (to the first order of small quantities) do the same amount of work in any infinitely small displacement of a rigid body to which they may be applied. It is also evident that the total work done in two or more successive infinitely small displacements is equal to the work done in the resultant displacement.
The work of a couple in any infinitely small rotation of a rigid body about an axis perpendicular to the plane of the couple is equal to the product of the moment of the couple into the angle of rotation, proper conventions as to sign being observed. Let the couple consist of two forces P, P (fig. 47) in the plane of the paper, and let J be the point where this plane is met by the axis of rotation. Draw JBA perpendicular to the lines of action, and let ε be the angle of rotation. The work of the couple is
P·JA·ε − P·JB·ε = P·AB·ε = Gε,
if G be the moment of the couple.
The analytical calculation of the work done by a system of forces in any infinitesimal displacement is as follows. For a two-dimensional system we have, in the notation of §§ 3, 4,
| Σ(Xδx + Yδy) | = Σ{X(λ − yε) + Y(μ + xε)} |
| = Σ(X)·λ + Σ(Y)·μ + Σ(xY − yX) ε | |
| = Xλ + Yμ + Nε. |