1Velocity and Acceleration AnalysisProblem 1.4.4The planar mechanism considered is shown in Fig. 3.19. The followingdata are given: AB=0.150 m, BC=0.400 m, CD=0.370 m, CE=0.230 m,EF =CE, La=0.300 m, Lb=0.450 m, and Lc=CD. The constant angularspeed of the driver link 1 is 60 rpm. Find the velocities and the accelerationsof the mechanism for φ = φ1= 30◦.The joints have the the coordinates [m]:xA= yA= 0;xD= 0.3, yD= 0.45;xB= 0.129904, yB= 0.075;xC= −0.0689445, yC= 0.422073;xE= −0.298288, yE= 0.404712;xF= −0.37, yF= 0.186177.Velocity of joint BThe velocity of the point B = B1on the link 1 isvB= vB1= vA+ ω1× rAB= ω1× rB,where vA≡ 0 is the velocity of the origin A ≡ O.The angular velocity of link 1 isω1= ω1k =πn30k =π(60)30k = 6.28319k rad/s.the position vector of point B isrAB= rB− rA= rB= xBı + yB + zBk = 0.129904ı + 0.075 m.The velocity of point B2on the link 2 is vB2= vB1because between the links1 and 2 there is a rotational joint.The velocity of B1= B2isvB= vB1= vB2=ı  k0 0 ωxByB0=ı  k0 0 6.283190.129904 0.075 0= −0.471239ı+0.81621 m/s.2Velocity of joint CThe points B2and C2are on the link 2 andvC2= vB2+ ω2× rBC= vB+ ω2× (rC− rB), (1)where the angular velocity of link 2 is ω2= ω2k (ω2is unknown).The points D3and C3are on the link 3 andvC3= vD3+ ω3× rDC= ω3× (rC− rD), (2)where vD= vD3≡ 0 and the angular velocity of link 3 isω3= ω3k (ω3is unknown).Equations (1) and (2) give (vC2= vC3)vB+ ω2× (rC− rB) = ω3× (rC− rD),orvB+ı  k0 0 ω2xC− xByC− yB0=ı  k0 0 ω3xC− xDyC− yD0(3)Equation (3) represents a vectorial equation with two scalar components onx-axis and y-axis and with two unknowns ω2and ω3vBx− ω2(yC− yB) = −ω3(yC− yD),vBy+ ω2(xC− xB) = ω3(xC− xD),or−0.471239 − ω2(0.422073 − 0.075) = −ω3(0.422073 − 0.45),0.81621 + ω2(−0.0689445 − 0.129904) = ω3(−0.0689445 − 0.3).It resultsω2= −1.1307 rad/s and ω3= −2.82169 rad/s.The velocity of C isvC= ω3× (rC− rD) = −ω3(yC− yD)ı + ω3(xC− xD)= −(−2.82169)(0.422073 − 0.45)ı + (−2.82169)(−0.0689445 − 0.3)= −0.0788027ı + 1.04105 m/s.3Velocity of joint EThe points E3and D3are on the link 3 andvE= vE3= vD3+ ω3× rDE= ω3× (rE− rD)= −ω3(yE− yD)ı + ω3(xE− xD)= −(−2.82169)(0.404712 − 0.45)ı + (−2.82169)(−0.298288 − 0.3)= −0.127788ı + 1.68819 m/s.Velocity of joint FThe points F4and E4are on the link 4 andvF= vF4= vE4+ ω4× rEF= vE+ ω4× (rF− rE), (4)where the angular velocity of link 4 is ω4= ω4k (ω4is unknown).On the other hand the velocity of F is along the vertical axis (y-axis)because slider 5 translates along y-axisvF= vF5= vF4= vF. (5)Equations (4) and (5) givevE+ ω4× (rF− rE) = vF,orvE+ı  k0 0 ω4xF− xEyF− yE0= vF (6)Equation (6) represents a vectorial equation with two scalar components onx-axis and y-axis and with two unknowns ω4and vFvEx− ω4(yF− yE) = 0,vEy+ ω4(xF− xE) = vF,or−0.127788 − ω4(0.186177 − 0.404712) = 0,1.68819 + ω4(−0.37 + 0.298288) = vF.It resultsω4= 0.58475 rad/s and vF= 1.64625 m/s.4Acceleration of joint BThe acceleration of the point B = B1on the link 1 isaB= aB1= aB2= aA+ α1× rB+ ω1× (ω1× rB) = α1× rB− ω21rB= −(6.28319)2(0.129904ı + 0.075) = −5.1284ı − 2.96088 m/s2.The angular acceleration of link 1 is α1=˙ω1= 0.Acceleration of joint CThe points C2and B2are on the link 2 andaC2= aB2+ α2× rBC− ω22rBC= aB+ α2× (rC− rB) − ω22(rC− rB), (7)where the angular acceleration of link 2 is α2= α2k (α2is unknown).The points C3and D3are on the link 3 andaC3= aD3+ α3× rDC− ω23rDC= α3× (rC− rD) − ω23(rC− rD), (8)where aD= aD3≡ 0 and the angular velocity of link 3 isα3= α3k (α3is unknown).Equations (7) and (8) giveaB+ α2× (rC− rB) − ω22(rC− rB) = α3× (rC− rD) − ω23(rC− rD),oraB+ı  k0 0 α2xC− xByC− yB0− ω22[(xC− xB)ı + (yC− yB)] =ı  k0 0 α3xC− xDyC− yD0− ω23[(xC− xD)ı + (yC− yD)] . (9)Equation (9) represents a vectorial equation with two scalar components onx-axis and y-axis and with two unknowns α2and α3aBx− α2(yC− yB) − ω22(xC− xB) = −α3(yC− yD) − ω23(xC− xD),aBy+ α2(xC− xB) − ω22(yC− yB) = α3(xC− xD) − ω23(yC− yD),5or−5.1284 − α2(0.422073 − 0.075) − (−1.1307)2(−0.0689445 − 0.129904)= −α3(0.422073 − 0.45) − (−2.82169)2(−0.0689445 − 0.3),−2.96088 + α2(−0.0689445 − 0.129904) − (−1.1307)2(0.422073 − 0.075)= α3(−0.0689445 − 0.3) − (−2.82169)2(0.422073 − 0.45).It resultsα2= −22.33 rad/s2and α3= −2.20443 rad/s2.The acceleration of C isaC= α3× (rC− rD) − ω23(rC− rD)= [−α3(yC− yD) − ω23(xC− xD)]ı + [α3(xC− xD) − ω23(yC− yD)]= [−(−2.20443)(0.422073 − 0.45) − (−2.82169)2(−0.0689445 − 0.3)]ı+ [(−2.20443)(−0.0689445 − 0.3) − (−2.82169)2(0.422073 − 0.45)]= 2.87595ı + 1.03567 m/s2.Acceleration of joint EThe points E and D are on the link 3 and the acceleration of E isaE= aD+ α3× rDE− ω22rDE= α3× (rE− rD) − ω23(rE− rD)= [−α3(yE− yD) − ω23(xE− xD)]ı + [α3(xE− xD) − ω23(yE− yD)]= [−(−2.20443)(0.404712 − 0.45) − (−2.82169)2(−0.298288 − 0.3)]ı+ [(−2.20443)(−0.298288 − 0.3) − (−2.82169)2(0.404712 − 0.45)]= 4.66371ı + 1.67947 m/s2.Acceleration of joint FThe points F and E are on the link 4 andaF= aE+ α4× rEF− ω24rEF= aE+ α4× (rF− rE) − ω24(rF− rE), (10)where α4= α4k (α4is unknown) is the angular acceleration of link 4.6The slider 5 moves along the y-axis and the acceleration of F isaF= aF5= aF. (11)Equations (10) and (11) yieldaE+ α4× (rF− rE) − ω24(rF− rE) = aF,oraF+ı  k0 0 α2xF− xEyF− yE0−ω24[(xF− xE)ı + (yF− yE)] = aF. (12)Equation (12) represents a vectorial equation with two scalar components onx-axis and y-axis and with two unknowns α4and aFaEx− α4(yF− yE) − ω24(xF− xE) = 0,aEy+ α4(xF− xE) − ω24(yF− yE) = vF,or4.66371 − α4(0.186177 − 0.404712) −