The lengths ds, ds′ are proportional to the velocities of the points to whose paths they belong, and the proportions of those velocities to each other are deducible from the construction of the machine by the principles of pure mechanism explained in Chapter I.
§ 90. Static Equilibrium of Mechanisms.—The principle stated in the preceding section, namely, that the energy exerted is equal to the work performed, enables the ratio of the components of the forces acting in the respective directions of motion at two points of a mechanism, one being the point of application of the effort, and the other the point of application of the resistance, to be readily found. Removing the summation signs in equation (52) in order to restrict its application to two points and dividing by the common time interval during which the respective small displacements ds and ds′ were made, it becomes P ds/dt = R ds′/dt, that is, Pv = Rv′, which shows that the force ratio is the inverse of the velocity ratio. It follows at once that any method which may be available for the determination of the velocity ratio is equally available for the determination of the force ratio, it being clearly understood that the forces involved are the components of the actual forces resolved in the direction of motion of the points. The relation between the effort and the resistance may be found by means of this principle for all kinds of mechanisms, when the friction produced by the components of the forces across the direction of motion of the two points is neglected. Consider the following example:—
| Fig. 126. |
A four-bar chain having the configuration shown in fig. 126 supports a load P at the point x. What load is required at the point y to maintain the configuration shown, both loads being supposed to act vertically? Find the instantaneous centre Obd, and resolve each load in the respective directions of motion of the points x and y; thus there are obtained the components P cos θ and R cos φ. Let the mechanism have a small motion; then, for the instant, the link b is turning about its instantaneous centre Obd, and, if ω is its instantaneous angular velocity, the velocity of the point x is ωr, and the velocity of the point y is ωs. Hence, by the principle just stated, P cos θ × ωr = R cos φ × ωs. But, p and q being respectively the perpendiculars to the lines of action of the forces, this equation reduces to Pp = Rq, which shows that the ratio of the two forces may be found by taking moments about the instantaneous centre of the link on which they act.
The forces P and R may, however, act on different links. The general problem may then be thus stated: Given a mechanism of which r is the fixed link, and s and t any other two links, given also a force ƒs, acting on the link s, to find the force ƒt acting in a given direction on the link t, which will keep the mechanism in static equilibrium. The graphic solution of this problem may be effected thus:—
(1) Find the three virtual centres Ors, Ort, Ost, which must be three points in a line.
(2) Resolve ƒs into two components, one of which, namely, ƒq, passes through Ors and may be neglected, and the other ƒp passes through Ost.
(3) Find the point M, where ƒp joins the given direction of ƒt, and resolve ƒp into two components, of which one is in the direction MOrt, and may be neglected because it passes through Ort, and the other is in the given direction of ƒt and is therefore the force required.
| Fig. 127. |
This statement of the problem and the solution is due to Sir A. B. W. Kennedy, and is given in ch. 8 of his Mechanics of Machinery. Another general solution of the problem is given in the Proc. Lond. Math. Soc. (1878-1879), by the same author. An example of the method of solution stated above, and taken from the Mechanics of Machinery, is illustrated by the mechanism fig. 127, which is an epicyclic train of three wheels with the first wheel r fixed. Let it be required to find the vertical force which must act at the pitch radius of the last wheel t to balance exactly a force ƒs acting vertically downwards on the arm at the point indicated in the figure. The two links concerned are the last wheel t and the arm s, the wheel r being the fixed link of the mechanism. The virtual centres Ors, Ost are at the respective axes of the wheels r and t, and the centre Ort divides the line through these two points externally in the ratio of the train of wheels. The figure sufficiently indicates the various steps of the solution.