A force `F` is applied on block `A` as shown in the figure. The contact force between `A` and `B` and between the blocks `B` and `C` respectively are (Assume frictionless surface)

Options

A. `(F)/(7),(2F)/(7)`
B. `(6F)/(7),(4F)/(7)`
C. `F,(F)/(7)`
D. `(4F)/(7),(6F)/(7)`

Answer

B

Explanation

`a=(F)/(m+2m+4m)=(F)/(7m)`
Let normal force between `A` and `B` is `N_(1)` , Then
`N_(1)=(2m+4m)a=(6f)/(7)`
and between `B` and `C` is `N_(2)` , then
`N_(2)=4ma=(4F)/(7)` .