1x d b q Mtah k J=J= 0.02 kg.m, F=15 N, q=1.5 rad/s, q = 0...

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1x₂ d b q Mtah k J₂=J= 0.02 kg.m², F=15 N, q=1.5 rad/s, q = 0 rad/s² In the mechanism given in the figure, the mass of the sliding elements is m-20 g. Also all rods It is manufactured from steel with p=7800 kg/m³ and cross-sectional area A=5 mm × 5 mm. The rods are homogeneous and their center of mass is at the midpoints of the rods. a) Obtain the dynamic equilibrium equations for the bond strengths. b) The solution of the system of equations you will obtain by writing the dynamic equilibrium equations the q coordinate and the bond strength at the O2 point in the range q = 0°-360°. to the q principal coordinate (in 1° increments) depending on the draw. a=25 cm, b=35 cm,, c=55 cm, d= 105 cm, e-85 cm 1x₂ d b q Mtah k J₂=J= 0.02 kg.m², F=15 N, q=1.5 rad/s, q = 0 rad/s² In the mechanism given in the figure, the mass of the sliding elements is m-20 g. Also all rods It is manufactured from steel with p=7800 kg/m³ and cross-sectional area A=5 mm × 5 mm. The rods are homogeneous and their center of mass is at the midpoints of the rods. a) Obtain the dynamic equilibrium equations for the bond strengths. b) The solution of the system of equations you will obtain by writing the dynamic equilibrium equations the q coordinate and the bond strength at the O2 point in the range q = 0°-360°. to the q principal coordinate (in 1° increments) depending on the draw. a=25 cm, b=35 cm,, c=55 cm, d= 105 cm, e-85 cm

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Dynamic equilibrium equations for the bond strengths The dynamic equilibrium equations can be derive... View the full answer