Block `B` rests on surface . If the coefficient of static friction between `A` and `B` is `mu = 0.4`. Determine the acceleration of each , if
(a) `F = 30 N` and
(b) `F = 250 N (g = 10 m//s^(2))`

Explanation

Limiting friction between `A` and `B`
`f_(L) = mu N = 0.4 xx 100 = 40 N`
(a) Both the blocks will have a tendency to move together with same acceleration (say a).
So, the diagram is as shown.

Equation of motion are,
`30 - f = 10 xx a` ...(i)
`f = 25 xx a` ...(ii)
Solving these two equations, we get
`a = 0.857 m//s^(2)`
and `f = 2142 N`
As this force is less than `f_(L)` both the block will move together with same acceleration ,
`a_(A) = a_(B) =0857 m//s^(2)`
(b) `250 - f = 10a` ...(iii)
`f= 25 N` ...(iv)
Solving eqs, (ii) and (iv), we get
`f = 178.6 N`

As `f gt f_(L)` , slipping will take place between two block and
`f = f_(L) = 40 N`
`a_(A) = (250 - 40)/(10)`
`= 21.0 m//s^(2)`

`a_(A) = (40)/(25) = 1.6 m//s^(2)`