Given the 40Bar Mechanism Draw accurate acceleration diagram

Given the 40-Bar Mechanism: Draw accurate acceleration diagram. Find the following omega_3, omega_4, alpha_3, alpha_4, A^T_Ao2, A_Ao2^t, A_Ao2^ToT, A_BA^T, A_BA^R, A_BA^TOT, A_Bor^T, A_Bo4^R, A_Bo_9^TOT (Magnitude & direction.) (Same as a above).

Solution

solution:

answer for question first

1)here for drawing velocity polygon i choose ascale of 1 cm/s=1 cm

then from fixed point o2 draw line at angle of -25 degree for 4 cm,from that again draw perpendicular to link Ab at angle of -75 degree for unknown length,then draw line from point o2 at angle of -34 degree and meeting of this two line indicate B point,

measurement are

o2A=4 cm

Vo2a=4 cm/s

AB=1 cm

Vab=1 cm/s

Vab=Rab*w3

w3=.2 rad/s

o4B=4.7 cm

Vo4b=4.7 cm/s

w4=Vo4b/Ro4b=.7833 rad/s

2)for drawing velocity polygon we must know following thing for each link

for O2A

centrifugal accelaration=Fco2a=Vo2a^2/Ro2a=4 cm/s2

tangential accelaration+Fto2a=Ro2a*m=4*2=8 cm/s2

for link AB

centrifugal accelaration=Fcab=Vab^2/Rab=.2 cm/s2

for link O4B

centrifugal accelaration=Fco4b=Vo4b^2/Ro4b=3.6816 cm/s2

start drawing polygon for scale

2 cm/s2= 1 cm

draw centrifugal component along the line and then tangential component perpendicular to it,we get point A,the dra centrifugal component for Ab and for o$B along length and draw perpendicular to both of them for tangential accelartion and point of meet is B

tangential accelaration

for link AB

Ftab=3.8*2=7.6 cm/s2

for link o4B

Fto4b=4.3*2=8.6 cm/s2

total accelaration

Fo2a=9 cm/s2

Fab=7.8 cm/s2

Fo4b=9.6 cm/s2

angular accelartion is obtain from tangential accelaration

m3=Ftab/Rab=7.6/5=1.52 rad/s2

m4=Fto4b/Ro4b=8.6/6=1.43 rad/s2