Question

For the figure shown below, link 2 moves at a constant angular v loop closure is shown in the figure below elocity of 2π Rad/sec cew. 2. The following information is also known about the linkages: Linkage l R(inch) ļexdeg 180 16 30 126941 39.06 a. Specify the velocity unknowns for the linkage system.(10 points) Perform velocity analysis using Chace's method to determine the unknowns(numerical value of unknowns)(15 point:s) b.

Answer 1

The angular velocity of link AB is given as $\omega$_AB =  2$\pi$ rad/s (CCW)

linear velocity of link AB (direction is perpendicular to link AB) , V_AB = (AB)* $\omega$_AB = 2$\pi$ * 16 = 100.531 inch/s

here, the unknowns are the angular velocity of link PC and the linear velocity of slider 4

The unknowns can be determined by velocity analysis

The figure of mechanism along with velocity triangles are shown below

Note here, Capital letters denote the mechanism and small letter alphabets represent respective velocities.

The velocity diagram is drawn by dropping perpendicular lines from links AB and CP ( as velocity is perpendicular to links) and completing the triangle by drawing a line from B parallel to link CP( this line represents the velocity of slider 4 which moves along link CP)

0 t 6 1) e-12-6949 39 06 36

AB ao 9. 06 66° Pc 34 06 50-11
from the velocity triangle,

V_PC/sin(80.94) = V_AB/sin(90) = V_slider/sin(9.06)

we got, V_AB = 100.531 inches/s

therefore, V_PC = V_AB*sin(80.94)/sin(90) = 100.531*sin(80.94) = 99.2767 inches/s

V_slider = V_AB*sin(9.06)/sin(90) = 15.8305 inches/s

we know, V_PC = (PC)* $\omega$_PC , where $\omega$_PC is the angular velocity of link PC and (PC) is the length of link PC

from this equation,  $\omega$_PC = V_PC/(PC) = 99.267/12.6944 = 7.82 rad/s

hence we get the unknowns, $\omega$_PC= 7.82 rad/s and V_slider = 15.8305 inches/s