The pulley arrangements of figures (I) and (II) are identical. The mass of the rope is negligible. In figure (I), the mass m is lifted up by attaching a mass 2m to the other end of the rope. In figure (II), 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 (I) and (II) are identical. The mas

1.	The pulley arrangements of figures (I) and (II) are identical. The mass of the rope is negligible. In figure (I), the mass m is lifted up by attaching a mass 2m to the other end of the rope. In figure (II), 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 mass 2m figure (b) `2mg-T=(2m)a ""` …..(2) Adding EQUATION (1) and (2), we get `2mg-mg=2ma+ma` `mg=3ma RARR a=(g)/(3)` Case II : In figure (C ) `T^(1)-mg = ma^(1)` But `T^(1)=2mg` `therefore 2mg-mg=a^(1)` `therefore a^(1)=g`