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 force F = 2 mg. Calculate the accelerations in the two cases.

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

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 force F = 2 mg. Calculate the accelerations in the two cases.
Answer» Solution :In figure (a), for the motion of mass m, T-mg=ma....... (1) For motion of MASS2 m figure (b) 2mg -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`