For the fourbar link mechanism shown below find the velociti

For the four-bar link mechanism shown below, find the velocities and accelerations of Points A and B. Then, find the acceleration of Point P.

Use the following data for the link mechanism:

Solution

solution:

1) here given 4 bar chain mechanism when draw on paper it gives angle a3 =90 degree and a4=119 degree for given condition graphically hence loop closure equation is given by

R1\'+R2\'+R3\'+R4\'=3.368+15.87i

2)on differentiating we get by removing constant term as

R2w2(cosa2+sina2i)+R3w3(cosa3+sina3i)+R4w4(cosa4+sina4i)=0

2*10(cos30+sin30i)+7w2(cos90+sin90i)+9w4(cos119+sin119i)=0

for solving for real and imaginary part we get

w4=-3.9695 rad/s

w3=3.035 rad/s

3) again differentiating we get accelaration of links

(R2t2-R2w2^2)(cosa2+sina2i)+(R3t3-R3w3^2)(cosa3+sina3i)+(R4t4-R4w4^2)(cosa4+sina4i)=0

t=angular accelaration of link

on solving for real and imaginary part we get

t4=-23.93 rad/s2

t3=68.12 rad/s2

5) velocity and accelaration of point A is given by

va=R2*w2=2*10=20 m/s

T=R2*t2=2*0=0 m/s2

6)velocity and accelaration of point B is given by

vba=R3*w3=7*3.035=21.245 m/s

Tba=R3*t3=7*68.12=476.84 m/s2

where absolute is given by

vb=vba+va=21.245+20=41.245 m/s

Tb=Tba+Ta=476.84+0=476.84 m/s2

7)velocity and accelaration of point P is given by

vpa=Rp*w3=6*3.035=18.21 m/s

Tpa=Rp*t3=6*68.12=408.72 m/s2

where absolute is given by

vp=vpa+va=18.21+20=38.21 m/s

Tp=Tpa+Ta=408.72+0=408.72 m/s2