Two blocks `m_(1) = 4 kg` and `m_(2) = 2` kg connected by a weightless rod slide down a plane having an inclination of `37^(@)` . The coefficient of dynamic friction of `m_(1)` and `m_(2)` with the inclined plane are `mu_(1) = 0 .75` and `mu_(2) = 0 . 25` respectively Find the common acceleration of the two blocks and tension in the rod Take is `sin37^(@) = 0.6` and `cos 37^(@) = 0 .8` .

Explanation

As shown in let T be the tension in the rod connecting the two blocks and a be common acceleration of the two blocks . As is know
` R_(1) = m_(1) g cos 37^(@) , R_(2) = m_(2) g cos 37^(@)`
The components of weights of two blocks down the incline ` = m_(1) g sin 37^(@) + m_(2) g sin 37^(@)`
`= (m_(1) + m_(2) ) g sin 37^(@)`
Total frictional force up the plane ` = F_(1) + F_(2) `
` = (mu_(1) m_(1) + mu_(2) m_(2)) g cos 37^(@)`
The equation of motion of the combination of two blocks can be written as
`(m_(1) + m_(2)) a = (m_1 + m_(2) ) g sin 37^(@)`
` - mu_(1) m_(1) + mu_(2) m_(2) ) g cos 37^(@)`
`(4 + 2 ) a = (4+2 ) 9.8 xx 0.6`
` - (0 .75 xx 4 + 0.25 xx 2 ) xx 9.8 xx 0.8 `
`6 a = 35.28 - 27.44 = 7.84 `
`a = (7.84)/(6) = 1.30 m//s^(2)`
The equation of motion of block `m_(1)` can be written as
` T + m_(1) g sin 37^(@) - mu_(1) m_(1) g cos 37^(@) = m_(1) xx a`
`T + 4 4 xx 9.8 xx 0.6 - 0.75 xx 4 xx 9.8 xx 0.8 `
`4 xx 1.30`
`T + 23 .52 - 23 . 52 = 5 .20`
` T = 5 .20 N `