Consider two blocks A and B of mass 3 kg and 9 kg respectively. Now, block A is placed on block B which is at rest on a table where the coefficient of friction between A and B is 0.4 and between B and the surface of the table is also 0.4. A force F is applied on B horizontally such that the block A does not slide over B. The maximum value of force F is ________N . (Take `g = 10 m//s ^(2))`

Answer

`96.00`

Explanation

Taking (A+ B) as system ,
`F 0 mu ( m _(A) + m _(B)) g = (m _(A) + m _(B))`
a `therefore a = ( F - mu ( m_(A) + m _(B)) g )/(( m _(A) + m _(B)))`
`therefore a = ( F - 0.4 xx 12 xx 10)/(12) = ( F - 48)/(12)`
But, `a _("max") = mu g = 0.4 xx 10 = 4`
` therefore F - 48 = 48`
` therefore F = 96 N`