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Two blocks are kept an incline in contac...

Two blocks are kept an incline in contact with each other. Masses of blocks are `m_(1)` and `m_(2)` and coefficients of friction are `mu_(1)` and `mu_(2)` respectively. The angle of inclination is `theta`. Determine
(a) acceleration of blocks, and
(b) force `F` with which the blocks press against each other

Text Solution

Verified by Experts

Consider the force diagrams of `m_(1)` and `m_(2)`. R represents contact force between `m_(1)` and `m_(2)` .
`N_(1) , N_(2)` are normal reactions between blocks and the inclined palne .

Balancing forces along y-axis :
`N_(2) = m_(1) g "cos" alpha ........... (i) `
Forces along x-axis :
`N_(2) = m_(2) g"cos" alpha .........(ii)`
`m_(1)g "sin" alpha + R - mu_(1)N_(1) = m_(1)a ........ (iii)`
`m_(2) g "sin" alpha + R - mu_(2)N_(2) = m_(2)A ............. (iv)`
`(m_(1) + m_(2)) g "sin" alpha - mu_(1)m_(1) g "cos" alpha - mu_(2) m_(2) g "cos" alpha = (m_(1) + m_(2))a`
`a = g[(m_(1) + m_(2)) "sin" alpha - (mu_(1) m_(1) + mu_(2) m_(2)) "cos" alpha]// (m_(1) + m_(2)), `
g = `9.8 m//s^(2)`
Substituting the expression for a , in Eqn. (iii) or (iv) , we get
`R = (mu_(2) - mu_(1))m_(1) m_(2) g "cos" alpha//(m_(1) + m_(2))`
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