Consider the system of masses and pulleys shown in fig. with massless string and fricitionless pulleys.
(a) Give the necessary relation between masses `m_(1)` and `m_(2)` such that system is in equilibrium and does not move.
(b) If `m_(1)=6kg` and `m_(2)=8kg`, calculate the magnitude and direction of the acceleration of `m_(1)`.

Explanation

(a) Applying newton's law `m_(2)g-2T=0` (because there is no acceleration) and T - `m_(1)g=0`
`implies(m_(2)-2m_(1))g=oimpliesm_(2)=2m_(1)`
(b) If the upwards acceleration of `m_(1)` is a, then accleration of `m_(2)` is `a/2` downwards for mass `m_(2):m_(2)g-2T=m_(2)((a)/(2))implies2m_(2)g-4T=m_(2)a` for mass `m_(1),T-m_(1)g=m_(1)a`
`impliesa=((2m_(2)-4m_(1))/(m_(2)+4m_(1)))g=(2(8-12))/(8+24)g= -g/4`

Negative sing shows that acceleration is opposite to considered direction i.e. it is downwards for `m_(1)` and upward for `m_(2).`