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.

Explanation

In figure (a), for the motion of mass m, T-mg=ma....... (1)

For motion of mass 2 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`