I.6 Dynamic Force Analysis 0Contents6 Dynamic Force Analysis 16.1 Equation of Motion for the Mass Center . . . . . . . . . . . . 16.2 Angular Momentum Principle for a System of Particles . . . . 46.3 Equations of Motion for General Plane Motion . . . . . . . . . 66.4 D’Alembert’s principle . . . . . . . . . . . . . . . . . . . . . . 96.5 Free-Body Diagrams . . . . . . . . . . . . . . . . . . . . . . . 116.6 Joint Forces Analysis Using Individual Links . . . . . . . . . . 116.7 Joint Force Analysis Using Contour Method . . . . . . . . . . 136.8 Joint Force Analysis Using Dyads . . . . . . . . . . . . . . . . 196.9 Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 236.10 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34I.6 Dynamic Force Analysis 16 Dynamic Force AnalysisFor a kinematic chain it is important to know how forces and moments aretransmitted from the input to the output, so that the links can be properlydesignated. The friction effects are assumed to b e negligible in the forceanalysis presented here.6.1 Equation of Motion for the Mass CenterConsider a system of N particles. A particle is an object whose shape andgeometrical dimensions are not significant to the investigation of its motion.An arbitrary collection of matter with total mass m can be divided into Nparticles, the ith particle having mass, mi, Fig. 6.1m =NXi=1mi.A rigid body can be considered as a collection of particles in which thenumber of particles approaches infinity and in which the distance betweenany two points remains constant. As N approaches infinity, each particleis treated as a differential mass element, mi→ dm, and the summation isreplaced by integration over the bodym =Zbodydm.The position of the mass center of a collection of particles is defined byrC=1mNXi=1miri, (6.1)where riis the position vector from the origin O to the ith particle. AsN → ∞ the summation is replaced by integration over the bodyrC=1mZbodyr dm, (6.2)where r is the vector from the origin O to differential element dm.I.6 Dynamic Force Analysis 2The time derivatives of Eq. (6.1) givesNXi=1mid2ridt2= md2rCdt2= maC, (6.3)where aCis the acceleration of the mass center. The acceleration of themass center can be related to the external forces acting on the system. Thisrelationship is obtained by applying Newton’s laws to each of the individualparticles in the system. Any such particle is acted on by two types of forces.One type is exerted by other particles that are also part of the system. Suchforces are called internal forces (internal to the system). Additionally aparticle can be acted on by a force that is exerted by a particle or object notincluded in the system. Such a force is known as an external force (externalto the system). Let fijbe the internal force exerted on the jth particle bythe ith particle. Newton’s third law (action and reaction) states that the jthparticle exerts a force on the ith particle of equal magnitude, and oppositedirection, and collinear with the force exerted by the ith particle on the jthparticle, Fig. 6.1fji= −fij, j 6= i.Newton’s second law for the ith particle must include all of the internal forcesexerted by all of the other particles in the system on the ith particle, plusthe sum of any external forces exerted by particles, objects outside of thesystem on the ith particleXjfji+ Fexti= mid2ridt2, j 6= i, (6.4)where Fextiis the external force on the ith particle. Equation (6.4) is writ-ten for each particle in the collection of particles. Summing the resultingequations over all of the particles from i = 1 to N the following relation isobtainedXiXjfji+XiFexti= maC, j 6= i. (6.5)The sum of the internal forces includes pairs of equal and opposite forces.The sum of any such pair must be zero. The sum of all of the internal forceson the collection of particles is zero (Newton’s third law)XiXjfji= 0, j 6= i.I.6 Dynamic Force Analysis 3The termPiFextiis the sum of the external forces on the collection of particlesXiFexti= F.One can conclude that the sum of the external forces acting on a closedsystem equals the product of the mass and the acceleration of the masscenterm aC= F. (6.6)Considering Fig. 6.2 for a rigid body and introducing the distance q inEq. (6.2) givesrC=1mZbodyr dm =1mZbody(rC+ q) dm = rC+1mZbodyq dm. (6.7)It results1mZbodyq dm = 0, (6.8)that is the weighed average of the displacement vector about the mass centeris zero. The equation of motion for the differential element dm isa dm = dF,where dF is the total force acting on the differential element. For the entirebodyZbodya dm =ZbodydF = F, (6.9)where F is the resultant of all forces. This resultant contains contributionsonly from the external forces, as the internal forces cancel each other. Intro-ducing Eq (6.7) into Eq. (6.9) the Newton’s second law for a rigid body isobtainedm aC= FThe derivation of the equations of motion is valid for the general motion ofa rigid body. These equations are equally applicable to planar and three-dimensional motions.I.6 Dynamic Force Analysis 4Resolving the sum of the external forces into cartesian rectangular com-ponentsF = Fxı + Fy + Fzk,and the position vector of the mass c enterrC= xC(t) ı + yC(t)  + zC(t) k,the Newton’s second law for the rigid body ism¨rC= F, (6.10)orm¨xC= Fx, m¨yC= Fy, m¨zC= Fz. (6.11)6.2 Angular Momentum Principle for a System of Par-ticlesAn arbitrary system w ith the mass m can be divided into N particlesP1, P2, ..., PN. The position vector of the ith particle relative to an originpoint, O, is ri= rOPiand the mass of the ith particle is mi, Fig. 6.3. Theposition of the mass center, C, of the system is rC=NXi=1miri/m. The positionof the the particle Piof the system relative to O isri= rC+ rCPi. (6.12)Multiplying Eq. (6.12) by mi, summing from 1 to N, the following relationis obtainedNXi=1mirCPi= 0. (6.13)The total angular momentum of the system about its mass center C, is thesum of the angular momenta of the particles about CHC=NXi=1rCPi× mivi, (6.14)I.6 Dynamic Force Analysis 5where vi=dridtis the velocity of the particle Pi.The total angular momentum of the system about O is the sum of the angularmomenta of the particlesHO=NXi=1ri× mivi=NXi=1(rC+ rCPi) × mivi= rC× mvC+ HC, (6.15)or the total angular momentum about O is the sum of the angular momentumabout O due to the velocity vCof the mass center of the system and the totalangular momentum about the mass center, Fig. 6.4.Newton’s second law