A pen of mass 'm' is lying on a piece of paper of mass M placed on a rough table. If the coefficient of friction between the pen and paper, and, the paper and table are `mu_(1)` and `mu_(2)`, respectively, then the minimum horizontal force with which the paper has to be pulled for the pen to start slipping is given by-

Options

A. `(m + M) (mu_(1) + mu_(2)) g`
B. `(m mu_(1) + M mu_(2)) g`
C. `{m mu_(1) + (m + M) mu_(2) }g`
D. `m (mu_(1) + mu_(2))g`

Answer

A

Explanation

For pen to start slipping maximum horizontal force on it is `f = mu_(1) mg`

`therefore a = mu_(1) g` is the maximum common acceleration for both pen and paper .
F , B , D for both pen and paper
`f_(1) = mu_(1) N_(1)` and `N_(1) = W_(1)`
`therefore f_(1) = mu_(1) mg`
`therefore N_(2) = N_(1) + W_(2)`
`F - f_(1) - f_(2) =` Ma
also `f_(2) = mu_(2) N_(2) = mu_(2) (m + M) g " " therefore F = f_(1) + f_(2) ` + Ma


`F = mu mg + mu_(2) (m + M)g + M (mu_(1) g)`
`F = (m + M) (mu_(1) + mu_(2)) g`