Find the acceleration `a_(1) , a_(2) and a_(3)` of the three blocks shown in Fig if a horizontal force of `10 N` is applied on

(a) `2 kg` blocks, (b) `3 kg` blocks and ( c) `7 kg` blocks (Take `g = 10 ms^(-2))`

Explanation

When force of `10N` is applied on `2 kg` blocks . The limiting friction force between `2 kg and 3 kg` blocks
`f_(l_(1)) = 0.2 xx 2 kg = 0.2 xx 2 xx 10= 4 N`
The limiting friction force between`3 kg` and `7 kg`
`f_(l_(2)) = 0.3 xx 5g`
`= 0.3 xx 5 xx 10 = 15 N`
After proper analysis , we find that `2 kg ` will slip over `3 kg` but `3 kg` will not slip over `7 kg`
Thus the we have `a_(1) = (10 - 4)/(2) = 3ms^(-2)`
`a_(2) = a_(3)= (4)/(3 +7) = 0.4ms^(-2)`
Since `a_(1)` is coming out to be greater than `a_(2), 2 kg` will slip over `3 kg`

b. When force of `10 N` is applied on `3 kg` blocks
Let us assume that all these blocks move together with common acceleration a then

`a = (10)/(2 xx 3 + 7) = (5)/(6) ms^(-2)`
Under this assumption
Friction between `2 kg` and `3 kg`
`f_(1) = 2a = (2 xx 5)/(6) = (5)/(3)N lt f_(l_(1))`
Friction between `3 kg` and `7 kg `
`f_(2) = 7a = (7 xx 5)(6) (35)/(3)N lt f_(l_2)`
Hence, no slipping occut any where
c. When force of `10 N` is applied on `7 kg` block. The acceleration of the whole system

`a = (10)/(2 xx 3 + 7) = (5)/(6) ms^(-2)`
The pseudo force on `2 kg` block `= (2 xx 5)/(6) = (5)/(3)N` ,which is less than the friction forces between `2 kg` and `3 kg` blocks so they move together.
Now check whether `2 kg` and `3 kg ` move together over `7 kg` blocks The pseudo force on `( 2+ 3) kg` is `5 xx (5)/(6) = (25)/(6)N` , which is also less than friction force between ` 3kg` and `7 kg` blocks move with a common acceleration of `5//6ms^(-2)` Therefore
`a = a_(1) = a_(2) = a_(3) (10)/(2 + 3 + 7)= (5)/(6)ms^(-2)`