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.

Explanation

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`