Consider the situation shwon in Fig. 7.49 . The horizontal surface below the bigger block is smooth . The coefficient of friction between the blocks is `mu` . Find the minimum and maximum accelerations with which the system should move in order to keep the smaller blocks at rest with respect to the bigger block .

Explanation

If the system is not accelerated , the block B will pull the block A . Let a be the minimum acceleration , so that A and B are at rest relative to C . As A has a tendency to move towards the pulley ,while B downwards ,so force of friction on A will act away from the pulley while on B upwards .
Now as A is at rest relative to C , so its equation of motion is
`T - f_(L) = ma ` or `T = ma + mu mg " " ..... `(i)
[as `f_(L) = muR_(1) = mu g`]
and as block B is at rest relative to C , so for its vertical and horizontal equilibrium , we have
`T + f_(L) = "mg" ` and `" " R_(2) = "ma" `
or ` " " T = "mg" - mu "ma" " " ["as f_(L) = mu R_(2) = mu "ma" ] ........ `(ii)
Eliminating T between Eqns. (i) and (ii) ,
ma ` + mu "mg" = "mg" - mu "ma" `
i .e ., `" " a_("min") = (1- mu)/(1 + mu) g " " ........`(iii)
Now , if the acceleration of the system is continously increased , there will come a situation when A will have a tendency of slipping away from the pulley P and B upwards . In direction opposite to intended motion , i.e., on A towards pulley while on B downwards so that equations of motion for A and B will become
`T + f_(L) = "ma" "and" T = "mg" + f'_(L)`
or `" " T = "ma" - mu "mg" "and" T = "mg" + mu "ma" `
Eliminating T between these,
ma `- mu "mg" = "mg" + mu "ma" `
or `" " a_("max") = (1 + mu)/(1- mu) g`