The diagram in Figure 1 shows a 4bar linkage system The angu

The diagram in Figure 1 shows a 4-bar linkage system. The angular velocity of bar 1 (link AB) is a constant ?(dot)=10 rad/s.

a. Determine the velocity and acceleration of points A, B, C, and D when ?1= 90?.

B L20.2m -0 2 L3=0.2m A 0.1m D//0; 0.2m Figure 1: Diagram of four-bar linkage

Solution

solution:

1) here above problem is solved by vector algebra method,where loop closure equation is given by

ad\'=ab\'+bc\'+cd\'

in vector form as

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

z2=90

wher R1=.2236 m and and angle z1=153.43 degree

2) here by chace solution we get

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

so we get

R3\'-R4\'=R\'=-.1788i-.01054j

where its unit vector is given by

R^=-.9983i-.05884j

3) here by chace solution we get

x2+y2=R3\'^2

x2+(R-y)2=R4^2

here value of x,y as pint of intersection as

y=R^2+R3^2-R4^2/2*R=.08955 m

on putting value we get

x=.1788 m

here R3 and R4 is given as

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

R3\'=-.2677i+.01578j

from this we get angle z3=-3.373

R4\'=R\'-R3\'=.0889i-.0263j

angle z4=-16.49 degree

5)here by by differenting this equation we get

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

on differenting removing zero term we get

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

on solving for i and j part we get

.1996w2+.1917w3=0

-.0117w2-.0567w3+1=0

on solving we get

w2=20.81 rad/s

w3=-21.67ad r/s

6) on again differntiating we get accelartion equation as follows

R2a2(R2^*k^)-R2w2^2(R2^)+R3a3(R3^*k^)-R3w3^2(R3^)+R4a4(R4^*k^)-R4w4^2(R4^)=0

on equating we get finall answer as follows

.2cos3.373*a3+.2cos16.49*a4-176.51=0

-.2sin3.373*a3-.2*sin16.49*a4+21.754=0

on solving we get

a3=544.28 rad/s2

a4=249.65 rad/s2

8) in the same way quaestion b can be solved for angle 30 degree