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 `4kg` mass
If `theta = 30^(@), u_(k) = 0.3`
`a_(4) = g(sin theta -u_(k) cos theta) =10 [(1)/(2)-0.3xxsqrt3/(2)]=2.6ms^(-2)`
The acceleration of `2kg` mass
`a_(2)=10[(1)/(2)-0.2xxsqrt3/(2)]=3.27ms^(-2)`
`:. a_(2) lt a_(4)`
Thus there will be contact force between the block and they move to together If 'a' is the common acceleration,
`(m_(1)+m_(2))a=`
`(m_(1)+m_(2)) g sintheta-(mu_(1)m_(1)+mu_(2)m_(2))g cos theta`
`6a = 6xx 10 xx (1)/(2) - (0.3 xx 4+0.2 xx 2) xx 10xxsqrt3/(2)`
`6a =30 -13.856 rArr a =2.7 ms^(-2)`
For `4kg` mass, `mg` in `theta + f_("contact") -f_("friction") = ma`
`4 xx 10 xx (1)/(2) + f_(c) - 0.3 xx 10 xx 10 xx sqrt3/(2) = 4xx 2.7`
`f_(c) = 10.8 + 10.4 - 20 rArr f_(c) =1.2N` .