Find the minimum value of `F` to move the lower block between the upper block and the surface . Also, find acceleratuion of each block.

Explanation

`N_(1) = 20 N`
`(f_(1))_(max) = mu_(s) N_(1) = 0.3 xx 20 = 6 N`
`uarr : N_(2) - N_(1) - 30 = 0 rArr N_(2) = 50 N`
`(f_(2))_(max) = mu_(s) N_(2) = 0.5 xx 50 = 25 N`
`F_(min) = (f_(1))_(max) + (f_(2))_(max) = 6 + 25 = 31 N`
When `F = F_(min)` , the blocks are in motion.
`f_(1) = (f_(1))_(k) = mu_(k) N_(1) = 0.2 xx 20 = 4 N`
Acceleration of the upper block , `4 = 2a_(1) rArr a_(1) = 2 m//s^(2)`
`f_(2) = (f_(2))_(k) = mu_(k) N_(2) = 0.4 xx 50 = 20 N`
Acceleration of the lower block
` F - [(f_(1))_(k) + (f_(2))_(k)] = 3a_(2)`
`31 - [ 4 + 20] = 3a_(2) rArr a_(2) = (7)/(3) m//s^(2)`
Note : For the lower block , we can also solve as there is no acceleration in the vertical direction .
` N_(2) = 50`
`(f_(2))_(max = mu_(s) N_(2) = 0.5 xx 50 = 25 N`