1. no Slip anywhere
whole system moves with acceleration a, which is given by, a = (F-μN)/(20+30+40) = (F-88.2) / 90 m/s2 ................(1)
where μ is the friction coefficient between ground and block-A. N is the normal reaction force between the surfaces of ground and block-A
from the free body diagram given for block-A, we get the no slip condition for block-A is ( mA × a )< f ,
where f is frictional force between the surfaces of A and B.
mA × a < μ×mA ×g .........(2)
using (1) and (2), we can write (F-88.2) < 90×μ×g or ( F-88.2) < 90×0.1×9.8 or F < 176.4 N .................(3)
Hence if the applied force F is less than 176.4 N, there will not be any slip between blocks.
2. Block-B sliding on Block-C
Free body diagram of block-B is given in figure.
we can see that if applied force exceeds sum of firctional forces f1 and f2 , block-B slides on block-C
F >(f1 + f2) = (0.1×20 + 0.2×50)×9.8 = 117.6 N
3. block-A slips on Block-B
As we already calculated for 'no slip anywhere', block-A slips if applied force exceeds 176.4 N