Two blocks are in contect moving on an inclined plane of inclination `alpha` with the horizontal. The masses of the blocks are equal to `m_(1)` and `m_(2)` and the coefficients of friction between the inclined plane and the blocks are equal to `mu_(1)` and `mu_(2) (mu_(1) gt mu_(2))` respectively. Find (a) common acceleration of the blocks and (b) the contact force between the blocks.

Explanation

(a) `N_(1) = m_(1) g cos alpha , f_(1) = mu_(1) N_(1) = m_(1) g cos alpha`
`N_(2) = m_(2) g cos alpha , f_(2) = mu_(2) N_(2) = mu_(2) m_(2) g cos alpha`
Along the plane :
`(m_(1) + m_(2))g sin alpha - (f_(1) + f_(2)) = (m_(1) + m_(2)) a`
`a = ((m_(1) + m_(2)) g sin alpha - (mu _(1) m_(1) + mu_(2) m_(2)) g cos alpha)/((m_(1) + m_(2)))`
(b) Consider motion of `m_(1)` : (let the contact force between the blocks in `N` )
`m_(1) g sin alpha + N - f_(1) = m_(1) a`
`m_(1) g sin alpha + N - mu_(1) m_(1) g cos alpha = m_(1) a`
`N = m_(1) g sin alpha - ((mu_(1) m_(1) + mu_(2) m_(2)) g cos alpha m_(1))/((m_(1) + m_(2)))`
`= mu_(1) m_(1) g cos alpha - (m_(1) g cos alpha(mu_(1)m_(1) + mu_(2) m_(2)))/((m_(1) + m_(2)))`
` = m_(1) g cos alpha [(mu_(1) (m_(1) + m_(2) ) - (mu_(1) m_(1) + mu_(2) m_(2)))/((m_(1) + m_(2)))]`
`= ((mu_(1) - mu_(2)) m_(1) m_(1) g cos alpha)/((m_(1) + m_(2)))`
Since `mu_(1) gt mu_(2)` , direction `N` is correct.