Force Analysis 1Dynamic Force AnalysisIn this section the motor moment Mmrequired for the dynamic equilib-rium of the considered mechanism, shown in Fig. 1(a), is calculated. Thejoint reaction forces are also calculated. The widths of the links 1, 3, and 5are AB = 0.140 m, F D = 0.400 m, and respectively, EG = 0.500 m. Theheight of the links 1, 3, and 5 is h = 0.010 m. The width of the links 2 and4 is wSlider= 0.050 m, and the height is hSlider= 0.020 m. All five movinglinks are rectangular prisms with the depth d = 0.001 m. The angle of thedriver link is φ =π6rad and the angular velocity is n = 50 rpm. The exter-nal moment applied on link 5 is opposed to the motion of the link. Becauseω5= 0.917 k rad/s, the external moment vector will be Me= −100 k N·m.The density of the material is ρSteel= ρ = 8000 kg/m3. The gravitationalacceleration is g = 9.807 m/s2. The center of mass locations of the linksi = 1, 2, ..., 5 are designated by Ci(xCi, yCi, 0).Inertia forces and momentsTo calculate the inertia moment Miand the total force Fifor the linki = 1, 2, ..., 5, the mass mi, the acceleration of the center of mass aCi, thegravity force Gi, and the mass moment of inertia ICiare needed.Link 1The mass of the link ism1= ρ AB h d.The position, velocity, and acceleration for the center of mass C1arerC1= rB/2, vC1= vB/2, and aC1= aB/2.The inertia force isFin1= −m1aC1.The gravitational force isG1= −m1g .The total force on link 1 at the mass center C1isF1= Fin1+ G1.Force Analysis 2The mass moment of inertia isIC1= m1(AB2+ h2)/12.The moment of inertia isM1= Min1= −IC1α1.Link 2The mass of the link ism2= ρ hSliderwSliderd.The position, velocity, and acceleration for the center of mass C2arerC2= rB, vC2= vB, and aC2= aB.The inertia force isFin2= −m2aC2.The gravitational force isG2= −m2g .The total force on slider 2 at B isF2= Fin2+ G2.The mass moment of inertia isIC2= m2(h2Slider+ w2Slider)/12.The moment of inertia isM2= Min2= −IC2α2.Link 3The mass of the link ism3= ρ F D h d.The position, velocity, and acceleration for the center of mass C3arexC3= xC+ (F D/2 − CD) cos φ3, yC3= yC+ (F D/2 − CD) sin φ3,rC3= xC3ı + yC3, vC3= ˙xC3ı + ˙yC3, and aC3= ¨xC3ı + ¨yC3.Force Analysis 3The inertia force isFin3= −m3aC3.The gravitational force isG3= −m3g .The total force at C3isF3= Fin3+ G3.The mass moment of inertia isIC3= m3(F D2+ h2)/12.The total moment on link 3 isM3= Min3= −IC3α3.Link 4The mass of the link ism4= ρ hSliderwSliderd.The position, velocity, and acceleration for the center of mass C4arerC4= rD, vC4= vD, and aC4= aD.The inertia force isFin4= −m4aC4.The gravitational force isG4= −m4g .The total force on slider 4 at D isF4= Fin4+ G4.The mass moment of inertia isIC4= m4(h2Slider+ w2Slider)/12.The moment of inertia isM4= Min4= −IC4α4.Force Analysis 4Link 5The mass of the link ism5= ρ EG h d.The position, velocity, and acceleration for the center of mass C5arexC5= (EG/2) cos φ5, yC5= (EG/2) sin φ5,rC5= xC5ı + yC5, vC5= ˙xC5ı + ˙yC5, and aC5= ¨xC5ı + ¨yC5.The inertia force isFin5= −m5aC5.The gravitational force isG5= −m5g .The total force on link 5 at C5isF5= Fin5+ G5.The mass moment of inertia isIC5= m5(EG2+ h2)/12.The moment of inertia isM5= Min5= −IC5α5.The numerical values areF1= 0.018ı − 0.099 N, M1= 0 N · m,F2= 0.026ı − 0.063 N, M2= −0.00002k N · m,F3= 0.049ı − 0.333 N, M3= −0.00621k N · m,F4= −0.036ı − 0.063 N, M4= 0.00001k N · m,F5= −0.055ı − 0.410 N, M5= 0.00481k N · m.Force Analysis 5Dyad MethodEDD DyadThe last dyad with the links 4 and 5 has the following joints: E rotation,D translation, and D rotation (ER, DT, DR). The links 4 and 5 form a RTRdyad.The application points of the unknown joint reaction forces F05and F34are at E and D (Fig. 3). The unknown joint reaction forces F05and F34arecalculated from the relations• sum of all the forces on links 4 and 5 is zero:XF(4&5)= F05+ F5+ F4+ F34= 0,orXF(4&5)· ı = F05x+ F5x+ F4x+ F34x= 0, (1)XF(4&5)·  = F05y+ F5y+ F4y+ F34y= 0. (2)• sum of the moments of all the forces and moments on links 4 and 5about D is zero:XM(4&5)B= rDE× F05+ rDC5× F5+ M5+ M4+ Me= 0(rE− rD) × F05+ (rC5− rD) × F5+ M5+ M4+ Me= 0. (3)• sum of all the forces on link 5 projected onto the sliding direction DEis zero:XF(5)· rDE= (F05+ F5) · (rE− rD) = 0. (4)The components F12x, F12y, F43x, and F43yare calculated from Eqs. (1), (2), (3),and (4).The force components F54xand F54yare computed from the sum of all theforces on link 4:XF(4)= F54+ F4+ F34= 0,orXF(4)· ı = F54x+ F4x+ F34x= 0 =⇒ F34x= −F54x− F4x,XF(4)·  = F54y+ F4y+ F34y= 0 =⇒ F34y= −F54y− F4y.Force Analysis 6The coordinates xPand yPof the application point P of the force F54are calculated from the equations(rE− rD) × (rP− rD) = 0,XM(4)D= rDP× F54+ M4= (rP− rD) × F54+ M4= 0. (5)CBB DyadThe next dyad with the links 2 and 3 has the following joints: C rotation,B translation, and B rotation (CR, BT, BR). The links 2 and 3 form a RTRdyad.The unknown joint reaction forces F03and F12have the application pointsat C and B (Fig. 3). The joint forces are calculated with the equations• sum of all the forces on links 2 and 3 is zero:XF(2&3)= F12+ F2+ F3+ F03− F34= 0,orXF(2&3)· ı = F12x+ F2x+ F3x+ F03x− F34x= 0, (6)XF(2&3)·  = F12y+ F2y+ F3y+ F03y− F34y= 0. (7)• sum of the moments of all the forces and moments on links 2 and 3about B is zero:XM(2&3)B= (rD− rB) × (−F34) + (rC− rB) × F03+(rC3− rB) × F3+ M3+ M2= 0. (8)• sum of all the forces on link 2 projec ted onto the sliding direction BCis zero:XF(2)· (rC− rB) = (F12+ F2) · (rC− rB) = 0. (9)The components F12x, F12y, F03x, and F03yare calculated from Eqs. (6), (7), (8),and (9).Force Analysis 7The force components F32xand F32yare computed from the sum of all theforces on link 2:XF(2)= F12+ F2+ F32= 0 =⇒ F32= −F12− F2,where F32= F32xı + F32y.The coordinates of the application point Q(xQ, yQ) of the force F32arecalculated from the equations(rB− rC) × (rQ− rC) = 0,XM(2)B= rBQ× F32+ M2= (rQ− rB) × F32+ M2= 0. (10)Driver linkThe motor moment MmThe motor moment needed for the dynamic equilibrium of the mechanismis Mm= Mmk (Fig. 4). A moment equation is written for the link 1 aboutthe rotation joint AXM(1)A= rAB× F12+ rAC1× F1+ M1+ Mm= 0 =⇒Mm= −rB× F12− rC1× F1− M1.Reaction force F01The rotation joint A between the links 0 and 1 is