Time derivative of Eq. 1 AC el +AC bjelo-DC yj elr = 0 We need to determine AC and ė. AC (cos +j sin 9) + AC è j(cos +j sin 6) -DC y j(cos y +j sin y) = 0 AC (cos +j sin 0) + AC (-sin 0 + j cos 0) -DC (-siny+jcos y) = 0 Real=0 AC cos 0-AC è sin 0 + DC y sin y=0 AC cos 30°-17.32 x x sin 30° + 10 x 1 x sin 60° = 0 (Eq. 1) Imag=0 AC sin + AC è cos 0-DC y cosy = 0 AC sin 30° +17.32 è cos 30° -10 x 1 x cos 60° = 0 (Eq. 2) Solve the two equations 1 and 2 with the two unknowns, you get Velocity of sliding of block C: AC = -0.05 m/s Angular velocity of member AB: 0.5 rad/s b) Energy method (No masses) Στο Στο W=0 with =>P.V+M.Y=0 The velocity of B: V=AB. 6 (-sin 30° + cos 30°) VB = 0.4x0.5x (-sin 30° + j cos 30°) The force at P: P= -1000/ and P.V=-1000x (0.4 x 0.5 x cos 30°) = -173.2 then Eq. 4>-173.2+Mx1=0=> M=173.2 N.m F.B.D. of DC 30° 30° D DX 30° M N is perpendiculare to the axes of slip along the axis of transmission ΣM/D = 0 (a = 0, constant angular velocity) M - N x 0.1 x cos 30° = 0 N = 173.2 0.1 x cos 30° = 2000N For the quick-return mechanism shown below, the rod DC rotates with a constant angular velocity of 1 rad/s CCW. a) For the position shown find the angular velocity of member AB and the velocity of sliding of block C within the member AB. b) If a force P = 1 kN is applied vertically at B, determine the couple moment M that must be applied to member DC. Also find the force transmitted through the slider joint at C Neglect the masses of all members. Dimensions are in cm. 30 B 40 10 D 60