In Fig block 1 is placed on top of block `2` .Both of then have a mass of `1 kg` The coefficient of friction between block 1 and 2 are `mu_(s) = 0.75` and `mu_(k) = 0.60` The table is frictionless A force `p//2` is applied on block `1 ` in the left and force `P` on block `2` to the right .Find the minimum value of `p` such that sliding accore between the two blocks

Explanation

Let both blocks moves together the friction beween the block should be static

For block (2) `P - f = 1a`...(i)
For block (1) ` f - P//2 = 1a`...(ii)
From Eq. (i) and (ii) . `P//2 = 2a` or `a = (P)/(4)` and `f = (3P )/(4)`
As `f` is static `f = (5P)/(4) le mu_(s) mg`
`mu_(s) mg = 0.75 xx 10 = 7.5`
Thus `P le (4)/(5)mu_(s) mg = (4)/(3) xx 0.75 xx 10`
`P = 10 N`