Dynamic Force Analysis: R-RTR mechanism 11 Dynamic Force Analysis1.1 Joint Forces AnalysisThe R-RTR mechanism shown in Fig. 1.1 has the dimensions: AB = 0.14 m,AC = 0.06 m, and CF = 0.2 m. The driver link 1 makes an angle φ =φ1=π3rad with the horizontal axis and rotates with a constant speed ofn = n1= 30 π rpm. The position vectors of the points A, B, C, and F arerA= 0ı + 0 m,rB= rC2= xBı + yB = 0.07ı + 0.121 m,rC= xCı + yC = 0ı + 0.06 m,rF= xFı + yF = 0.150ı + 0.191 m.The position vectors of the mass centers of links 1 and 3 arerC1= xC1ı + yC1 =xB2ı +yB2 = 0.035ı + 0.06 m,rC3= xC3ı + yC3 =xC+ xF2ı +yC+ yF2 = 0.075ı + 0.125 m.The mass center of the slider 2 is at B (B = C2): rC2= rB.The height of the links 1 and 3 is h = 0.01 m. The width of the slider2 is wSlider= 0.05 m, and the height is hSlider= 0.020 m. All three movinglinks are rectangular prisms with the depth d = 0.01 m.The density of the material is ρSteel= ρ = 8000 kg/m3. The gravitationalacceleration is g = 9.807 m/s2.Inertia forces and momentsLink 1The mass of the link 1 ism1= ρ AB h d = 0.112 kg.The acceleration of C1isaC1= −3.40932ı − 5.90511 m/s2.The inertia force on link 1 at C1isFin 1= −m1aC1= 0.381844ı + 0.661373 N.Dynamic Force Analysis: R-RTR mechanism 2The gravitational force isG1= −m1g  = −1.09838  N.The total force on link 1 at the mass center C1isF1= Fin 1+ G1= 0.381844ı − 0.437011 N.The mass moment of inertia isIC1= m1(AB2+ h2)/12 = 0.000183867 kg · m2.The moment of inertia isM1= Min 1= −IC1α1= 0.Link 2The mass of the link 2 ism2= ρ2hSliderwSliderd = 0.08 kg.The acceleration of C2= B isaC2= −6.81864ı − 11.8102 m/s2.The inertia force on link 2 at C2= B isFin 2= −m2aC2= 0.545491ı + 0.944818 N.The gravitational force isG2= −m2g  = −0.78456  N.The total force on link 2 at the mass center C2isF2= Fin 2+ G2= 0.545491ı + 0.160258 N.The mass moment of inertia isIC2= m2(h2Slider+ w2Slider)/12 = 0.0000193333 kg · m2.The moment of inertia isM2= Min 2= −IC2α2= −0.00169109k N · m.Dynamic Force Analysis: R-RTR mechanism 3where α2= 87.47 rad/s2.Link 3The mass of the link 3 ism3= ρ CF h d = 0.112 kg.The acceleration of C3isaC3= −20.6416ı − 6.4373 m/s2.The inertia force on link 3 at C3isFin 3= −m3aC3= 3.30266ı + 1.02997 N.The gravitational force isG3= −m3g  = −1.56912  N.The total force on link 3 at the mass center C3isF3= Fin 3+ G3= 3.30266ı − 0.539153 N.The mass moment of inertia isIC3= m3(CF2+ h2)/12 = 0.000534667 kg · m2.The moment of inertia isM3= Min 3= −IC3α3= −0.0467673k N · m.where α3= α2= 87.47 rad/s2.The external moment on link 3 isM3ext= −Sign(ω3)|Mext|k = −1000k N · m,where ω3= 14.0619 rad/s and Mext= |Mext| = 1000 NDetermine the moment M required for dynamic equilibrium and the jointforces for the mechanism.Dynamic Force Analysis: R-RTR mechanism 4Newton-Euler equations of motionFor each moving link the Newton-Euler equations of motion are writtenmiaCi=XF(i)and ICiαi=XM(i)Ci,where Ciis the center of mass of the link i.The force analysis will start with link 3 because the moment M3extisknown.Link 3For the link 3, Fig. 1.2(a), the equations of motion for planar motion givem3aC3=XF(3)= F03+ G3+ F23,IC3α3=XM(3)C3= rC3C× F03+ rC3Q× F23+ M3ext,orm3aC3x= F03x+ F23x,m3aC3y= F03y− m3g + F23y,IC3α3k =ı  kxC− xC3yC− yC30F03xF03y0+ı  kxQ− xC3yQ− yC30F23xF23y0+M3extk,orm3aC3x= F03x+ F23x,m3aC3y= F03y− m3g + F23y,IC3α3= (xC− xC3)F03y− (yC− yC3)F03x+(xQ− xC3)F23y− (yQ− yC3)F23x+ M3ext,where the unknowns areF03= F03xı + F03y, F23= F23xı + F23y,and the position vector rQ= xQı + yQ of the application point of the jointforce F23.Dynamic Force Analysis: R-RTR mechanism 5Numerically the previous system of equations becomes−3.30266 = F03x+ F23x, (1.1)−1.02997 = −1.56912 + F03y+ F23y, (1.2)0 = −0.065F03x− 0.075F03y+ 0.125F23x− 0.075F23y+F23yxQ− F23xyQ− 1000 . (1.3)The application point Q of the joint force F23is on the line BC:yB− yCxB− xC=yQ− yCxQ− xCor 0.874 −yQ− 0.06xQ= 0. (1.4)The joint force F23is perpendicular to the sliding direction BC:F23· rBC= 0 or − 0.07F23x− 0.061F23y= 0. (1.5)There are five scalar equations, Eqs. (1.1) throught (1.5), and six unknowns,F03x, F03y, F23x, F23y, xQ, yQ. The force analysis will continue with link 2.Link 2The Newton-Euler equations for slider 2, Fig. 1.2(b), givem2aC2=XF(2)= F12+ G2+ F32,IC2α2=XM(2)C2= rBQ× F32,orm2aC2x= F12x− F23x,m2aC2y= F12y− m2g − F23y,IC2α2k =ı  kxQ− xByQ− yB0−F23x−F23y0,orm2aC2x= F12x− F23x,m2aC2y= F12y− m2g − F23y,IC2α2= (xQ− xB)(−F23y) − (yQ− yB)(−F23x),where the new unknown is introduced (the reaction of link 1 on link 2):F12= F12xı + F12y.Dynamic Force Analysis: R-RTR mechanism 6Numerically, the previous equations becomes−0.545491 = F12x− F23x, (1.6)−0.944818 = −0.78456 + F12y− F23y, (1.7)0.00169109 = −0.121244 F23x+ 0.07 F23y− xQF23y+ yQF23x= 0. (1.8)Now there is a system of eight scalar equations, Eqs. (1.1) through (1.8),eight unknowns, and the solution isF03= F03xı + F03y = 7078.41 ı − 8093.7  N,F23= F23xı + F23y = −7081.72 ı + 8094.24  N,F12= F12xı + F12y = −7082.26 ı + 8094.08  N,rQ= xQı + yQ = 0.069 ı + 0.121  m.Link 1The force equation for the driver link 1, Fig. 1.2(c), givesm1aC1=XF(1)= F21+ G1+ F01.The reaction of the ground 0 on the link 1 isF01= m1aC1− F12− G1= −7082.64 ı + 8094.52  N.The sum of the moments about the mass center C1for link 1 gives theequilibrium momentIC1α1k =XM(1)C1= rC1B× F21+ rC1A× F01+ M,orM = rC1B× F12− rC1A× F01= 712.632 k + 712.671 k = 1425.3 k N · m.Another way of calculating the moment M required for dynamic equilib-rium is to write the moment equation of motion for link 1 about the fixedpoint AIAα1k =XM(1)A= rAC1× G1+ rAB× F21+ M =⇒ M = rB× F12− rC1× G1,where IA= IC1+ m1(AB/2)2.Dynamic Force Analysis: R-RTR mechanism 7D’Alembert’s principleFor each link two vectorial equations are written:XFj+ Fin j= 0 andXMCj+ Min j= 0, (1.9)wherePFjis the vector sum of all external forces (resultant of externalforce) on link j, andPMCjis the sum of all external moments on link jabout the mass center Cj.The force analysis will start with