Using the analytical method find the angular velocities of l

Using the analytical method, find the angular velocities of links 3 and 4 and the linear velocities of points A, B and P_1 In the XY (global) coordinate system for the linkage shown in the figure below in the position shown. Assume theta_2 (Global) = 45 degree in the Xy coordinate system and Omega_2 = 10 rad/s, The coordinates of point P_1 on link 4 are (114.68, 33.19) with respect to the xy (local) coordinate system.

Solution

1)here loop closure equation to given problem is given as follows

we solve our problem in global coordiante only

R1\'=R2\'+R3\'+R4\'

R1=79.7 in

m1=angle=126.58

R2=14

m2=45(for analysis m2 input angle and uniform velocity w2=10rard/s)

a2=0 rad/s2

R3=80 in

R4=51.26 in

R1\'-R2\'=R3\'+R4\'

R3\'+R4\'=-57.39i+54.1j=R\'

R^=-.727i+.6859j

R=78.87

hence by chace solution method for calculating angle of link 3 and 4

x^2+y^2=R3^2

x^2+(S-y)^2=R4^2

here on putting and

y=R^2+R3^2-R4^2/2R

from equation solving we get

x=48.8541 in,y=63.35 in

R3\'=x(R^*K^)+yR^

R3\'=-81.57i+76.96j

R3^=-.7273i+.6862j

m3=-43.33 or 136.66

R4\'=R\'-R3\'

R4\'=24.18i-22.86j

R4^=.7266i-.6871j

m4=-43.39 or 136.607

here on differntiating loop closure equation we get

0=R2w2(R2^*K^)+R3w3(R3^*K^)+0=R4w4(R4^*K^)

on putting and solving we get that

as we want to know w4,hence

R4w4=-R2w2(R2^k^).(R3^k^)*k^/(R4.k^)(R3.k^)k^

on solving we get

w4=.07894 rad/s

for w3

R3w3=-R2w2(R2^k^).(R4^k^)*k^/(R3.k^)(R4.k^)*k^

w3=.04894 rad/s

4)where linear velocity point A and B and P1 are given as

Va=R2*w2=14*10=140 in/s

Vb=R3*w3=80*.04894=3.9152 in/s

Rp1\'=33.19/cos10.027=33.7048 in

Rp1=R4+Rp1\'=33.7048+51.26=84.9648 in

Vp1=Rp1*w4=84.9648*.07894=6.707in/s