The following synopsis shows the scheme on which the treatment
is based:-
Part I.-Statics.
I. Statics of a particle.
2. Statics of a system of particles.
3. Plane kinematics of a rigid body.,
4. Plane statics.
5. Graphical statics.
6. Theory of frames.
7. Three-dimensional kinematics of a rigid body.
8. Three-dimensional Statics.
9. Work.
10. Statics of inextensible chains.
11. Theory of mass-systems.
Part 2.-Kinetics.
12. Rectilinear motion.
13. General motion of a particle.
14. Central forces. Hodograph.
15. Kinetics of a system of discrete particles.
16. Kinetics of a rigid body. Fundamental principles.
17. Two-dimensional problems.
18. Equations of motion in three dimensions.
19. Free motion of a solid.
20. Motion of a solid of revolution.
21. Moving axes of reference.
22. Equations of motion in generalized co-ordinates.
23. Stability of equilibrium. Theory of vibrations.
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B '-v ~ '~ 5. FIG. I. R represented by the diagonal OC of the parallelogram determined by OA, OB. This is of course a physical assumption whose propriety is justified solely by experienced We shall see later that it is implied in Newton's statement of his Second Law of motion. In modern language, forces are compounded b¥“ gztonaddition ”; thus, if we draw in succession vectors HK, KL to represent P, Q, the force R is represented by the -9 -9 -> vector HL which is the “ geometric sum” of HK, KL. By successive applications of the above rule any number of forces acting on a particle may be replaced by a single force which is the vector-sum of the given forces; this single force 6 9 -> 9 is called the resultant. Thus if AB, BC, CD . ., HK be vectors representing the given forces, the resultant will be given % - by AK. It will be understood that the figure ABCD . . K need not be confined to one plane. If, in particular, the point K coincides with A, so that the resultant vanishes, the given system of forces is said to be Q n Q P R FIG. 2. in equilibrium-t.e. the particle could remain permanently at rest under its action. This is the proposition known as the polygon of forces. In the particular case of three forces it reduces to the triangle of forces, viz. “If three forces acting on a particle are represented as to magnitude and direction by the sides of a triangle taken in order, they are in equilibrium.” A sort of converse proposition is frequently useful, viz. if three forces acting on a. particle be in equilibrium, and any triangle be constructed whose sides are respectively parallel to the forces, the magnitudes of the forces will be to one another as the corresponding sides of the triangle. This follows from the fact that all such triangles are necessarily similar. As a simple example of the geometrical method of treating statical problems we may consider the equilibrium of a particle on a “ rough " inclined plane. The usual empirical law of sliding friction is that the mutual action between two plane surfaces in contact, or between a particle and a curve or surface, cannot make with the normal an angle exceeding a certain limit 7 called the angle of friction. If the conditions of equilibrium require an obliquity greater than this, sliding will take place. The precise value of A will vary with the nature and condition of the surfaces in contact. In the case of a body simply resting on an inclined plane, the reaction must of course be vertical, for equilibrium, and the slope a of the plane must therefore not exceed K. For this reason A is also known as the angle of repose. If a > }, a force P must be applied in order to maintain equilibrium; let 6 be the inclination of P to the plane, as shown in the left-hand diagram. The relations between this force P, the gravity W of the body, and the reaction S of the plane are then determined by a triangle of forces HKL. Since the inclination of S H P - . 51

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~ '. w s, 10. W L2 K FIG. 3., to the normal cannot exceed K on either side, the value of P must lie between two limits which are represented by LIH, L2H, in the right-hand diagram. Denoting these limits by Pl, P2, we have P1/W=L1H/HK=sin (a-X)/cos (0-l-K), P2/W=L2H/HK=sin (a-1-A)/cos (0-A). It appears, moreover, that if 0 be varied P will be least when LIH is at right angles to KL1, in which case P1=W sin (a-X), corresponding to 6 = -}. . Just as two or more forces can beg combined into a single resultant, so a~ single force., may be resolved into components