Find the accelerations of blocks A and B for the following cases.
(A) `mu_(1)=0 " and "mu_(2)=0.1" (P)" a_(A)=a_(B)=9.5m//s^(2)`
(B) `mu_(2)=0 " and "mu_(1)=0.1 " (Q)"a_(A)=9m//s^(2), a_(B)=10m//s^(2)`
(C) `mu_(1)=0.1 " and "mu_(2)=1.0 " (R)"a_(A)=a_(B)=g=10m//s^(2)`
(D) `mu_(1)=1.0 " and "mu_(2)=0.1 " (S)" a_(A)=1, a_(B)=9m//s^(2)`

Explanation

(a) R, (b) Q, (c) P, (d) S
(i) FBD in (case(i))
`{mu_(1)=0, mu_(2)=0.1}`

While friction’s work is to oppose the relative motion and here if relative motion will start then friction comes and without relative motion there is no friction so both the block move together with same acceleration and friction will not come.
` rArr =a_(A)=a_(B)=10m//s^(2)`
(ii) Friction between wall and block A oppose relative motion since wall is stationary so friction wants to slop block A also and maximum friction will act between wall and block while there is no friction between block.

Note : Friction between wall and block will oppose relative motion between wall and block only it will not do anything for two block motion.

`a_(A)=9m//s^(2),a_(B)=10m//s^(2)`
(iii)
Friction between wall and block will be applied maximum equal to 1N but maximum friction available between block A and B is 10 N but if this will be there then relative motion will increase while friction is to oppose relative motion. So friction will come less than 10 so friction will be f that will be static.

for system (20 - 1) `=2timesa`
rArr a=19/2=9.5m//s^(2)`
(iv) `a_(A)=(11-10)/1=1m//s^(2)`
`a_(B)=(10-1)/1=9m//s^(2)`