The Sin long crank i.e. link 2 shown weighs 2 lb. Its centre of Gravity (CG)...

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The Sin long crank i.e. link 2 shown weighs 2 lb. Its centre of Gravity (CG) is at 3in and 30 from the line of centers. Its mass moment of inertia about its CG is 0.05 lb-in-s. Its acceleration is defined in its Local nonrotating coordinate system (LNCS) x, y. Its kinematic data are shown below. The coupler i.e. link 3 is 15 in long and weighs 4 lb. Its CG is at 9 in and 45 from the line of centers. Its mass moment of inertia about its CG is 0.10 lb-in-s. Its acceleration is defined in its LNCS, x, y. Its kinematic data are shown below. The sliding joint on link 3 has a velocity of 96.95 in/s in the +Y direction. There is an external force of 50 lb applied at point P which is located at 2.7 in and 201 from x-axis of link 3 Local Coordinate system. The coefficient of friction is 0.2. i. ii. iii. 0 deg 60 03 deg 99.59 00 rad/sec 30 @3 rad/sec -8.78 2 CG 3" 2 30 rad/sec -10 3 rad/sec -136.16 B 15" aG in/sec 2700.17 @ 89.4 aG3 in/sec 3453.35 @ 254.4 Draw a Free Body Diagram of each moving link of the mechanism. Derive equations that are solvable for the input torque and reaction forces at every joint. Write the equations derived from part (ii) in matrix form. The Sin long crank i.e. link 2 shown weighs 2 lb. Its centre of Gravity (CG) is at 3in and 30 from the line of centers. Its mass moment of inertia about its CG is 0.05 lb-in-s. Its acceleration is defined in its Local nonrotating coordinate system (LNCS) x, y. Its kinematic data are shown below. The coupler i.e. link 3 is 15 in long and weighs 4 lb. Its CG is at 9 in and 45 from the line of centers. Its mass moment of inertia about its CG is 0.10 lb-in-s. Its acceleration is defined in its LNCS, x, y. Its kinematic data are shown below. The sliding joint on link 3 has a velocity of 96.95 in/s in the +Y direction. There is an external force of 50 lb applied at point P which is located at 2.7 in and 201 from x-axis of link 3 Local Coordinate system. The coefficient of friction is 0.2. i. ii. iii. 0 deg 60 03 deg 99.59 00 rad/sec 30 @3 rad/sec -8.78 2 CG 3" 2 30 rad/sec -10 3 rad/sec -136.16 B 15" aG in/sec 2700.17 @ 89.4 aG3 in/sec 3453.35 @ 254.4 Draw a Free Body Diagram of each moving link of the mechanism. Derive equations that are solvable for the input torque and reaction forces at every joint. Write the equations derived from part (ii) in matrix form.

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Lets go through each point stepbystep i Draw a Free Body Diagram FBD of each moving link of the mechanism We cant physically draw the diagrams here bu... View the full answer