For the following mechanism Calculate degree of freedom and

For the following mechanism Calculate degree of freedom and number of loops. Show position variables on the figure. Write loop closure equations in complex form. Write real and imaginary parts of LCE, VLCE and ALCE. Show each term in VLCE and ALCE as velocity and acceleration vectors (e.g. bar a_A, bar v_B/A). Determine the angular position, velocity and acceleration of link 4.

Solution

solution:

1)here DOf of mechanism by grublers criteria is

DOF=3n-2p1

n=4

p1=3+1=4

DOF=4

2)here loop closure equation is written as

QB=BoAo+AoA+AQ

in complex form as

35e^im4=-90+35e^i(-45)+iSaqe^im4

as m3=m4+90

hence by taking its complex conjugate and removingSaq we get final eqaution as

Saq^2=R1^2+R4^2-R2^2-2R1R4cosm2

on putting value in equation we get

Saq=60.375 mm

and again putting value in loop closure equation we get for real part as

35cosm4=-90+35cos45-60.375sinm4

35cosm4+60.375sinm4=-65.25

on applying triangle we get

sin(30.1+m4)=-.935

m4=-99.33

m3=-9.33

4)velocity of slider is given as tangential velocity hence

velocity Saq\'=35*2=70 mm/s

accelaration Saq\'\'=35*1=35 mm/s2

on differentiating loop closure equation we get

35w4ie^im4=35w2ie^im2-60.375w4e^im4+i.Saq\'e^im4

on solving for real part we get

w4=4.7909 rad/s

5)on differentiating we get accelaration of link

-35w4^2e^im4+35iA4e^im4=-35w2^2e^im2-60.375w4^2e^im4-2Saq\'w4e^im4+iSaq\'\'e^im4+35iA2e^im2-60.375A4e^im4

on solving for real part we get A4=6.6044 rad/s2