Homework 4 solutions: Statement: The linkage in Figure P6-5f has the dimensions and coupler angle given below.Find 3,VV and Vc for the position shown for V-10 in/sec in the direction shown.Use the velocity difference graphical method. Given: Link lengths and angles: Coupler point: Link 3(4 to B) b=1.8.in Distance A to C 卫：=1.44n Coupler angle 63:=128.deg Angle BAC 8=49.deg Slider 4 angle 64=59.deg Input slider velocity VA=10-imsec-1 Solution: See Figure P6-5f and Mathcad file P0618a. 1.Draw the linkage to a convenient scale.Indicate the directions of the velocity vectors of interest Direction ofVB 0 05 1in Direction of VBA 3 Direcion ofVCA 49.000 59.000° 128.000° 2 2.The magnitude and sense of the velocity at point A. 7a=10.000 日A=180deg sec 3.Use equation 6.5 to(graphically)determine the magnitude of the velocity at point B,the magnitude of the relative velocity VA,and the angular velocity of link 3.The equation to be solved graphically is VB=VA+VBA 5in/sec a.Choose a convenient velocity scale and 儿L⊥ layout the known vector V. b.From the tip of V draw a construction line VaA with the direction of V magnitude unknown. 4.784 c.From the tail of V,draw a construction line with the direction of Ve,magnitude unknown. d.Complete the vector triangle by drawing 3.436 VBA from the tip of V to the intersection of the Ve construction line and drawing Ve from the 38.000° 59.000° tail of V to the intersection of the VBA construction line V
4.From the velocity triangle we have: Velocity scale factor: h=5-inseeI in VB=3.438.in-ky B=17.190n 0B=59deg sec VBA:=4.784m-k 'BA=23.920n 0BA=38.deg sec 5.Determine the angular velocity of link 3 using equation 6.7. -VBA 03= b o3=-13.289ad sec 6.Determine the magnitude and sense of the vector Vc using equation 6.7. VcA;=p03 'cA=19.136m sec 日c4:=(128-49-90)deg 0c4=-11.000deg 7.Use equation 6.5 to (graphically)determine the magnitude of the velocity at point C.The equation to be solved graphically is VC=VA+VCA a.Choose a convenient velocity scale and layout the known vector VA. b.From the tip of V,layout the (now)known vector Vc4. c.Complete the vector triangle by drawing Vc from the tail of V to the tip of the Vc vector. 0 5 in/sec LLL⊥L」 Y 1.902 N X 11.000° 22.572° 3.827- VCA 8.From the velocity triangle we have: Velocity scale factor: 元=5mecl in Vc=1.902 in-ky 'c=9.510n 0c=-22.572deg sec