Question

kinematics

In this system, rotation of link AB is translated into linear movement of piston P. If the angular velocity of link AB is (o-1rad/sec), find the linear velocity of piston P (point D in the picture). Remember: There are multiple ways to solve this problem, one is to use relative velocity vector equation as we worked out in class. Or another method is to use the center of instantaneous zero velocity for link BD For this part of the homework rewrite the solution from class, and solve for Vp and BD using the vector equation for relative velocity of D to B. -l=81n .. r 3 in 40°

Answer 1

By sine rule: 3/sin beta = 8/sin 40 degree doublerightarrow beta = 13.95 degree w_AB = 1rad/s (clockwise) doublerightarrow w^rightarrow_AB = -1 k^cap rad/s # For AB: v^rightarrow_B = v^rightarrow_A + v^rightarrow_B/A doublerightarrow v^rightarrow_B = 0 + w^rightarrow_AB times r^rightarrow_B/A = (-1k^cap) times (3 cos 40 degree i^cap + 3 sin 40 degree j^cap) doublerightarrow v^rightarrow_B = -2.298 j^cap + 1.925 i^cap # For BD: v^rightarrow_D = v^rightarrow_B + v^rightarrow_D/B doublerightarrow v^rightarrow_D i^bar = v^rightarrow_B + w^rightarrow_BD times r^rightarrow_D/B doublerightarrow v^rightarrow_D i^bar = (-2.298 j^cap + 1.925 i^cap) + w_BD k^cap times (8 cos 13.95 degree i^cap - 8 sin 13.95 degree j^bar) doublerightarrow v^rightarrow_D i^cap = -2.298 j^cap + 1.928 i^cap + 7.764 w_BD j^cap + 1.928 w_BD i^Cap #By comparing j^cap and i^cap: j: 0 = -2.298 + 7.764 w_BD doublerightarrow w_BD = 0.296 rad/s circlearrowright i: v_D = 1.928 + 1.928(0.296) doublerightarrow v_D = 2.5 in/s