ltbr. Consider the situation as shown in the figure. Initially blocks are at rest o the floor. A constant upward force F is applied on the pulley Assuing the pulley smooth and light and the string massless, find the acceleration of each blocks and pullay if F is (a) 10N,
(b). 60 N
(c) 120 N
Given `m_(1)=2kg,m_(2)=4kg`

Explanation

`F=2TimpliesT=(F)/(2)`
(a) when `F=10N,T=5N`
No block will move `a_(1)=a_(2)=a_(p)=0`
(b) when `F=60N,T=30N`
`30-20=2a_(1)impliesa_(1)=5m//s^(2)`
The block of mass 4 kg will not move `a_(2)=0`
if the block moves a distance `x_(0)`, the poulley will move `(x_(0))/(2)` hence the acceleration of pulley
`a_(p)=(a_(1))/(2)=2.5m//s^(2)`
(c) when `F=120N,T=60N`
`60-20=2a_(1)impliesa_(1)=20m//s^(2)`
`60-40=4a_(2)impliesa_(2)=5m//s^(2)`
`a_(p)=(a_(1)+a_(2))/(2)=(20+5)/(2)=12.5m//s^(2)`

the acceleration of pulley can be explained in the following manner. Let the blocks be moving with acceleration 'a' with respect to the pulley
`a_(1)=a+a_(p)` ....(i)
`a_(2)=a_(p)-a` ....(ii)
`a_(p)=(a_(1)+a_(2))/(2)`