In the previous problem, suppose `m_2=2.0kg and m_3=3.0 kg`. What should be the mass m so that it remains at rest?

Explanation

Let the block `m_1` moves upward with accelerastion `a_1 ` and the two block `m_2 and m_3` have relative cceleration `a_2` due to the diffference of weight between them,
So, the actual acceleratin of the blocks `m_1 , m_2 and m_3` will be `a_1(a_1-a_2) and (a_1+a_2)` as shown
From ure 2, `T-1g-1a_2=0`....i.
From ure 3, `
` T/2 -2 g-2(a_1-a_2)=0` ........ii. `
From ure 4,
` T/2 -3g-3(a_1+a_2)=0` ........iii.
From equations I. and ii. elimiN/Ating T, we get,
`1g+1a_2=4g+4(a_1+a_2)`
`rarr 5a_2-4a_1=3g` ...........iv.
From equations ii and iii we get
` 2g+2(a_1-a_2)=3-3(a_1-a_2)`
`rarr 5a_1 +a_2=g` .........v.
Solving equations iv and v we get,
` a_1 = 2g/29`
`and a_2 =g-5a_1`
`= g-(10g)/29=(19g)/29`
So, `a_1 -a_2= (2g)/(17g)/29`
and `a_1+a_2 =(2g)/29+(19g)/29=(21g)/29`
So, acceleratiion of `m_1, m_2 and m_3` are ` `(19g)/29 (up) , (17g)/29` (down) and `(21g)/29 (down), respectively.
Again, from `u=0, S=20 cm= 0.2 cm and
` a_2= (19g)/29 [g=10 m/s^2]`
:. S=ut+1/2 at^2`
`rarr 0.2=1/2xx19/29gt^2`
` rarr t=0.25 sec` .