The pulley arrangements of figures (a) and (b) are identical . The mass of the rope is negligible . In figure (a), the mass m is lifted up by attaching a mass 2m to the other end of the rope . In figure (b) , m is lifted up by pulling the other end of the rope with a constant downward forceF=2mg . Calculate the accelerations in the two cases.

The pulley arrangements of figures (a) and (b) are identical . The mas

1.	The pulley arrangements of figures (a) and (b) are identical . The mass of the rope is negligible . In figure (a), the mass m is lifted up by attaching a mass 2m to the other end of the rope . In figure (b) , m is lifted up by pulling the other end of the rope with a constant downward forceF=2mg . Calculate the accelerations in the two cases.
Answer» SOLUTION :In figure (a), for the MOTION of mass m , T-mg=ma For motion of mass 2M figure (B) 2 mg -T (2m)a…(2) Adding equation (1) and (2) , we get `2mg-mg= 2ma+ma, mg=3ma, a = g//3` (b)In figure `T^(1)-mg=ma^(1)` But `T^(1)=2mg , :. 2mg-mg =a^(1), ` Therefore `a^(1)=g`