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_91)g=0 rArr(m_(2)-2m_(1))g=0rArrm_(2)=2m_(1)`
(b) If the upwards acceleration of `m_(1)` is a, then acceleration of `m_(2)` is `a//2` downwards For mass `m_(2): m_(2)g-2T=m_(2)((a)/(2))rArr2m_(2)g-4T=m_(2)a`
For mass `m_(1): T-m_(1)g=m_(1)arArra=(2m_(2)-4m_(1))/(m_(2)+4m_(1))g=(2(8-12))/(8+24)g=-(g)/(4)`
Negative sign shows that acceleration is oppposite to considered direction i.e. it is downwards for `m_(1)` and upwards for `m_(2)`.