Two blocks of masses 4 kg and 2 kg are in contact with each other on an inclined plane of inclination `30^(@)` as shown in the figure . The coefficient of friction between 4 kg mass and the inclined plane is `0.3` , whereas between 2 kg mass and the plane is `0.2` . find the contact force between the blocks .

Explanation

The acceleration of 4 kg mass
If `" " theta = 30^(@) , mu_(k) = 0.3`
`a_(4) = g ("sin" theta - mu_(k) "cos" theta)`
`= 10 [(1)/(2) - 0.3 xx (sqrt3)/(2) ] = 2.6 m s^(-2)`
The acceleration of 2 kg mass
`a_(2) = 10 [(1)/(2) - 0.2 xx (sqrt3)/(2) ] = 3.27 m s ^(-2)`
`therefore " " a_(2) gt a_(4)`
Thus , there will be contact force between the blocks and they move together . if 'a' is the common acceleration ,
`(m_(1) - m_(2)) a = (m_(1) + m_(2)) g "sin" theta - (mu_(1) m_(1) + mu_(2) m_(2)) g "cos" theta `
`6a = 6 xx 10 xx (1)/(2) - (0.3 xx 4 + 0.2 xx 2 ) xx 10 xx (sqrt3)/(2)`
`6a = 30 - 13.856`
`implies " " a = 2.7 m s^(-2)`
For , 4 kg mass , mg sin `theta + f_("constant") - f_("friction") = ma`
`4 xx 10 xx (1)/(2) + f_(c) - 0.3 xx 10 xx 10 xx (sqrt3)/(2) = 4 xx 2.7 `
`f_(c) = 10.8 + 10.4 -20`
` therefore " " f_(c) = 1.2 N`