In the figure shown , the coefficient of static friction between B and the wall is `(2)/(3)` and the coefficient of kinetic friction between B and the wall is `(1)/(3)`. Other contacts are smooth. Find the minimum force `F` required to lift B, up. Now if the force applied on A is slightly increased than the calculated value of minimum force, then find the acceleration of B. mass of A is 2m and the mass of B is m. Take `tantheta=(3)/(4)`

Explanation

The FBDs of A and B are
For A to be in equilibrium:
`F=Nsintheta` .(1)
For B to just lift off,
`Nsintheta=mg+mu_SN^(`)` .(2)
For horizontal equilibrium of B,
`N^(`)=Nsintheta` .(3)
From (2) and (3)
`N(costheta-mu_Ssintheta)=mg` or `N((4)/(5)-(2)/(3)xx(3)/(5))mg`
or `N=(5)/(2)mg` .(4)
From equation (1) `F=Nxx(3)/(5)`
`because` `F=(3)/(2)mg`
(ii) The acceleration of the block A be a and B be b
`F-Nsintheta=2ma` .(1)
`Ncostheta-mg-mu_SN^(`)=mb` .(2)
`N^(`)=Nsintheta` .(3)
From constraint `=`
`asintheta=bcostheta`
Solving (1),(2),(3) and (4) we get
`impliesb=(3g)/(7)`