PLease help with steps Find the sliding velocity of link 6 a

PLease help with steps

Find the sliding velocity of link 6 and the angular velocities of links 3 and 5. V_6 ~ 44 in/sec, omega_3 ~ 1.2 rad/sec, omega_5 ~ 0.85 rad/sec Determine the velocity of link 6.

Solution

velocity of B w.r.t A will be perpendicular to the direction of AB, because B can\'t move along AB(rigid rod)

so V_ba (which is V_b as A,E etc are the frame of ref.) will be 40^o from horiontal

now, from the picture 29.5 is the sum of cos components of AB and BC, ie 29.5=12.5cos(50)+22.4cos(X), solving this will give X as 16.6^o, from this we can find the angles of CD,BD etc

velocity of C wrt B will be perpendicular to BC, ie. 16.6^o from vertical or 73.4^o from horizontal,so draw a line from B in that angle, and draw a horizontal line from A as C moves horizontally as CE is vertical. the point where these lines meet is C in velocity diagram

vel of D wrt C and wrt B can be drawn from C and B in the angles from the picture and the line meets at D

now, to find angles of FD, find height of D by solving 10.4+ABcos50=2+FDsinX+BDsin73.4, gives X as1.4^o , BD can be found from pythagoras theorem.

draw line at 1.4^o from D and horizontal from A,E as it slides on the surface. they meet at F

distance of F from A,E is the absolute velocity of F ie 6

from angular velocity of AB, find velocity of B and scale accordingly in the velocity diagram, V_b=37.5 in/sec

V_f / V_b=1.16, V_f =43.6

Angular velocity of a link can be found by dividing the velocity of a point wrt another point by distance between the points (it is the defenition of angular velocity)

so for 5 V_cb/CB =1.1rad/sec

other ang.vel can be found like this

NB accuracy of the answer will depend on the drawing precision