Given a mechanism with the link lengths Lx = 0.62 m, Ly = 0.19 m, L...

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Given a mechanism with the link lengths L₁x = 0.62 m, L₁y = 0.19 m, L₂ = 0. 325 m, L3 = 0.16 m, and L4 = 0. 18 m, L5x = 0.1 m, L5y = 0.16 m, L₁ = 0.35 m, lg2 = 0.17115 m, lg3 = 0.09026, lg4 = 0.09m, lg6 = 0.175m, I₂ = 0.089191665 kg⋅m², I3 = 0.01389689, kg-m² I4 = 0.01435385 kg⋅m², and I6= 0.130437 kg⋅m², find the values for 03, 04, 05, 06, W3, W4, W5, W6, A3, A4, A5, and α6 as shown in Figures 1, 2, and 3 for the open circuit of the linkage assuming 0₂ = 54.12°, W₂ = -25 rad/sec, and a₂ = -15 rad/sec². The mass density is p=30000kg/m³, the link thickness T is 0.02 m, the joint shaft and hole diameter is 0.025 m, and the link radius is 0.02 m. There is an external torque (t N.m) on one of the links, applied at the centre of mass on one of the links, and at the same time an external force of F, acts on link 5, applied at point p and lf [m] away from point A. Please note that depending on your student ID, you will get a different F, and lf values (Fp = 479N) (Lp = 0. 979m) Find F12, F32, F62, F43, F54, F56, and F15 at the joints and the driving torque T₁2 needed to maintain motion with the given angular velocity w₂ = 25 rad/sec and acceleration α₂ = 15 rad/sec² for this instantaneous position of the link (Figures 1 to 3). 1) Find the values of 03, 04, 05, 06, W3, W4, W5, W6, a3, a4, as, and α6 Y'2 Link 2 Link 1 p0.025 m 0.08 m Y'6 9 6 0.04 m 0₂ T12 Origin Position (0, 0) L3 A 7g3 D E в 04 X • Link 3 1g6 03 Link 4 L₁ Link 6 Lix Y'4 fr A P L4 90° 0.08 C Link 5 Lsy 125x 8³y ROO2 m 3 LIV X Figure 1 Dynamic force analysis of a mechanism (front view) Figure 2 Dynamic force analysis of a mechanism (bottom view) Fazy R₁₂ F624 F12x F₂34 Fxzy 32 P₂2 F62x F₂014 F434 F23x R₂6 F43x Ag3 26x F341 Fax R54 R56 A₂6 FSAY Fsort A₂² Fasv FSAK Fosv R45 F56x Fasx FISH R15 F65x Ags T12 Figure 3 Dynamic force analysis of a mechanism (free-body diagrams) FISK Given a mechanism with the link lengths L₁x = 0.62 m, L₁y = 0.19 m, L₂ = 0. 325 m, L3 = 0.16 m, and L4 = 0. 18 m, L5x = 0.1 m, L5y = 0.16 m, L₁ = 0.35 m, lg2 = 0.17115 m, lg3 = 0.09026, lg4 = 0.09m, lg6 = 0.175m, I₂ = 0.089191665 kg⋅m², I3 = 0.01389689, kg-m² I4 = 0.01435385 kg⋅m², and I6= 0.130437 kg⋅m², find the values for 03, 04, 05, 06, W3, W4, W5, W6, A3, A4, A5, and α6 as shown in Figures 1, 2, and 3 for the open circuit of the linkage assuming 0₂ = 54.12°, W₂ = -25 rad/sec, and a₂ = -15 rad/sec². The mass density is p=30000kg/m³, the link thickness T is 0.02 m, the joint shaft and hole diameter is 0.025 m, and the link radius is 0.02 m. There is an external torque (t N.m) on one of the links, applied at the centre of mass on one of the links, and at the same time an external force of F, acts on link 5, applied at point p and lf [m] away from point A. Please note that depending on your student ID, you will get a different F, and lf values (Fp = 479N) (Lp = 0. 979m) Find F12, F32, F62, F43, F54, F56, and F15 at the joints and the driving torque T₁2 needed to maintain motion with the given angular velocity w₂ = 25 rad/sec and acceleration α₂ = 15 rad/sec² for this instantaneous position of the link (Figures 1 to 3). 1) Find the values of 03, 04, 05, 06, W3, W4, W5, W6, a3, a4, as, and α6 Y'2 Link 2 Link 1 p0.025 m 0.08 m Y'6 9 6 0.04 m 0₂ T12 Origin Position (0, 0) L3 A 7g3 D E в 04 X • Link 3 1g6 03 Link 4 L₁ Link 6 Lix Y'4 fr A P L4 90° 0.08 C Link 5 Lsy 125x 8³y ROO2 m 3 LIV X Figure 1 Dynamic force analysis of a mechanism (front view) Figure 2 Dynamic force analysis of a mechanism (bottom view) Fazy R₁₂ F624 F12x F₂34 Fxzy 32 P₂2 F62x F₂014 F434 F23x R₂6 F43x Ag3 26x F341 Fax R54 R56 A₂6 FSAY Fsort A₂² Fasv FSAK Fosv R45 F56x Fasx FISH R15 F65x Ags T12 Figure 3 Dynamic force analysis of a mechanism (free-body diagrams) FISK