These are the three necessary and sufficient equations of equilibrium, when any forces act on a free body in one plane. 25. If there were a fixed point in the plane of the forces, we might take it for the origin of co-ordinates O, and its resistance would destroy the effect of the resultant force R, and we should have the condition of equilibrium only G=0; or there must be no tendency to rotation around the fixed point. 26. PROP. To prove the principle of virtual velocities for forces acting in one plane on a point, and on a rigid body at different points. 2 a P2 A P2 DEFINITION. If any forces as P1 and P2 act at a point as A in the figure, and this point is displaced through an indefinitely small space Aa, and we draw perpendiculars Pia and from a upon the directions of the forces, then the distances Ap1 and Apa are called the virtual velocities of the forces P1 and P2; and Ap1 being measured in the direction of force P1 is called positive, Ap, being measured in the direction of force P2 produced is called negative. The principle of virtual velocities is thus enunciated: If any number of forces be in equilibrium at one or more points of a rigid body, then if this body receive an indefinitely small disturbance, the algebraic sum of the products of each force into its virtual velocity is equal to zero. This principle is true when the forces in equilibrium act at any points and in any planes on a rigid body; but we shall in this treatise only prove the case when the forces act either at one point or at different points, in one plane, because the general case requires a knowledge of analytical geometry of three dimensions. First. To prove the principle when the forces act all at one point. Let A be the point at which the forces P1, P2, &c. . . . P, act. D = Aa cos. (α1-0) =Aa (cos. a1 cos. O + sin. a1 sin. 0) and P1.v1=P1Aa(cos. a1 cos. @ +sin. a1 sin. Ø) =Aa (cos. 0. P1 cos. a1+ sin. 0. P1 sin. a1) 2 P2. v=Aa (cos. 0. P2 cos. a2+ sin. 0. P2 sin. a) and so for the other forces, therefore, we have 2 P1. v1+P2. v2+P3.v3+&c. ... Pn.vn=Σ(P.v); say, 1 2 + Pn cos.an) +sin. 0 (P1 sin. a1 + P2 sin. a+ &c. . . . + P2 sin. a)} 2 P1 sin. a+P sin. a+ &c. .+P sin. a=0 ... we have Σ (P.v)=0; or, the principle is true when the forces act all at one point. Second. Let the forces act at different points or particles of the body in one plane. We have now to consider these points connected together by rigid lines or rods without weight, which transmit the reactions of the particles upon each other. These reactions must be considered together with the other forces. be the reaction of the particle A1 upon the particle A2 Am be the particles. Let vaja, Vaa Vaa Vaa &c. &c. be the corresponding virtual velocities; then ra12=ra, rajaz=raza,, &c. &c. from the nature of reactions. -Vaga1, &c., which we must shew. Also Vaja2 Draw the perpendiculars ap, bq. Then if the line ab is parallel to AB, the point to be proved is evidently true. When ab is not parallel to AB, let them meet when produced, if necessary, in some point C. Since the displacements are indefinitely small, the perpendiculars ap, bq coincide with circular arcs having C for center, and but Ap=Cp-CA=Ca- CA Bq=Cb― CB=(Ca+ab)− (CA+AB) = Ca-CA ... since AB=ab = Ap, but measured in the opposite direction to the reaction of B upon A, and is therefore negative. Let the sum of the products of all the external forces into their virtual velocities, acting on particle A, be (Pa1• Va1) Since each particle is in equilibrium from the 0=Σ(Pa1• Va1)+ra ̧a2• Và ̧a2+ˆã‚a ̧•Va1az+&c. &c. (Pam Vam) action of the In taking the sum of the products for all the particles, the products of the reactions into their virtual velocities will disappear, being in pairs equal in magnitude with contrary signs; therefore we have, Σ(Pa1• Va1)+(Pa2•Va2) +Σ(Paz•Vaz)+&c. +Σ(Pam•Vam)=0 or generally, when there is equilibrium, (P.v)=0. 27. CONVERSELY. If the sum of the products of the forces into their virtual velocities be equal to zero, or Σ(P.v)=0, then there will be equilibrium. For if the forces be not in equilibrium, they will be equivalent either to a single force or a single couple. (Art. 24.) In the first case, let R be the single resultant force, then a force equal and opposite to R will reduce the system to equilibrium; let u be its virtual velocity for any displacement. Since there is now equilibrium, we have, by the preceding article, Σ(P. v)+R'.u=0 But by hypothesis Σ(P.v)=0 .. R.u=0; which being true for all small displacements of the body, we must have R=0, or the body was in equilibrium from the action of the original forces. In the second case, if the forces were equivalent to a resultant couple, it would be balanced by an equal and opposite couple. Let the forces of this opposite couple be Q and Q', and their virtual velocities for any displacement be q and q' respectively. Since they will reduce the system to equilibrium, we have by the preceding article E(P. v)+Q.q+Q'. q' =0 but Σ(P.v)=0 .. Q.q+Q'. q' =0 for all displacements, which is impossible unless Q and Q' each=0, since they are equal and parallel forces, and act at different points. CHAPTER V. ON THE CENTER OF GRAVITY. 28. The center of gravity of a body is that point at which the whole weight of a body may be considered to act, and would produce the same mechanical effect as the weight of the body actually does. The weights of all the particles of a body, acting vertically downwards, are parallel forces, so that the center of gravity coincides with the center of parallel forces for such weights. From the definition it arises, that if the center of gravity of a body be a fixed point, the body will balance about that point in all positions. This property of the center of gravity often. furnishes the means of determining its position practically. In regular and symmetrical figures, as cubes, spheres, cylinders, thin plates which are circular, elliptic, or regular polygons, &c. it is evidently the center of the body, or point about which it is symmetrical. 29. PROP. If a body be in equilibrium, suspended from any point, or resting with one point of contact upon another body, then the center of gravity lies in the vertical line through that point of suspension or contact respectively. Let A in the figures be the points of suspension and contact respectively; draw the vertical lines Aw. If the whole weight of the body act in these vertical lines, it will be supported by the reactions of the fixed points A, or when the centers of gra A m m A |