In the sitution shown in figure there is no friction between `2 kg` and ground.

a. For what maximum value of force `F` can all three blocks move together?
b. Find the value of force `F` at which sliding starts at other rough surfaces.
c. Find acceleration of all blocks, nature and value of friction force for the following values of force `F` (i) `10N` (ii) `N` and (iii) `25 N`

Explanation

Friction between blocks `A and B`
`f_(l_1) = f_(k_1)= 0.5 xx 1 xx g = 5N`
Friction force between blocks `B` and `C`
`f_(l_2) = f_(k_2)= 0.2 xx (1 + 2) xx g = 6N`
maximum posible accceleration of blocks `A` can be
`a_(A ,max) = 0.5 g = 5 ms^(-2)`
Maximum posible accceleration of blocks `C`:
`a_(C .max) (f_(l2))/(2)= (6)/(2)= 3 ms^(-2)`
For the entire system to move together, we have to take lesser ecceleration
a. Maximum force `F` that can be applied for the entire system to move togather : `F_("max") = m_("total")a`
`rArr F_("max")= (1 + 2+ 2) xx 3 = 15 N`
b. Sliding will start between blocks `B` and blocks `C` if `F gt 15 N`

For sliding to start between blocks `A` and blocks `B` : This will accur when acceleration of combined blocks` A` and blocks `B` will exceed`5 ms^(-2)`
`F - f_(k_2_) = (m_(A) + m_(B))a`
` F - 6 = ( 1+ 2) xx 5 rArr F = 21 N`
c. i `F = 10N lt 15N` , so entire system will move together with common acceleration

`a = (10)/(1 + 2 + 2) = 2 ms^(-2)`
Friction between blocks `B` and blocks `C`
`f_(2) = 2a = 2 xx 2 = 4N`
Friction between blocks `A` and blocks `B`
`f_(1) = 1 xx a = 1 xx 2 = 2N`
ii . `F = 18 N`, here `15 N lt Flt 21 N`, so sliding start between blocks `B` and blocks `C` and not between blocks `A` and `B`

Acceleration of blocks `C`
`a = - f_(k_2)//m = 6//2 = 3 ms^(-2)`
Acceleration of blocks `A` and `B`
`a = (18 - 6)/(1 + 2) = 4 ms^(-2)` Friction between blocks `C` and ground is
`f_(k_(2)) =6N`
Friction between blocks `A` and `B`:
`f_(1) = 1 xx 4 = 4 N`

iii `F = 25 N gt 21 N`, so sliding occure at all the contact surface friction between blocks `A` and `B` is `f_(k_1) = 5 N`
Friction between blocks `B` and `C` is `f_(k_2) = 6 N`
Accelerations of blocks `A` and `B` are their maximum possible acceleration
Acceleration of block `B` :
`a = (25 - 5 - 6)/(2) = 7ms^(-2)`