The object is released from rest at the position shown The w

The object is released from rest at the position shown. The weight per unit length is 10 lb/ft. Determine the acceleration of point A. G is 6.6364\' from A I_G = 65.97 slug-ft^2 alpha = 4.3354 md/s^2

Solution

solution;

1)here mass of component is givenas

W=w1*(L)

L=6+4*4=22 ft

W=10*22=220 lb or 978.608 N

2)center of gravity of system is found out by varignon\'s thereom

220*k=6*10*3+4*10*6+2*4*10*8+4*10*10=1460

k=6.636 ft or 2.022 m

3)moment of inertia of system is given as about pointA

I=mk^2=220*6.636^2=9688.02 lb ft2 or 301.11 slug ft2 or 408.25 kg m2

4)where at point A forces acting are

F1=mgcos60=489.30 N at angle of 30 to horizontal

5)here torque acting on point A is

T=I*m

m=angular accelartion

T=mgsin60*2.022=1713.64 Nm

1713.64=408.2542*m

m=4.1975 rad/s2

here accelaration at point A is

a=k*m=2.022*4.1975=8.4873 m/s2

5)accelaration along X direction atA is equal in opposite direction is

ax1=acos60=4.2436 m/s2

or another approach is

6)accelaration due to force F1 is

F1h=F1cos30=423.74 N=m*ax1

ax1=4.2478 m/s2

7)resultant accelaration in x direction is

ax1=4.2478 m/s2=13.93 ft/s2