Two small spheres of masses `m_(1)` and `m_(2)` are connected by a light right rod.The system is placed between a rought floor and smooth verticle wall as shown in Fig. The co-efficent of friction between the floor and the sphere of mass `m_(2)` is` mu` .Find the minimum value of `theta` so that the system of masses does not slip

Explanation

For the equation of `m_(1)`

`sum F_(x) = 0 rArr N_(1) - T cos theta = 0`
`rArr N_(1) = T cos theta`
`sum F_(x) = 0 rArr T sin theta - m_(1) g = 0` …(i)
`T sin theta= m_(2) g` …(ii)
For the equilibrium of `m_(2)`
`sum F_(x) = 0 rArr T cos theta - mu N_(2) = 0`
`rArr T cos theta = mu N_(2)` ....(iii)
`sum F_(x) = 0 rArr N_(2) - T sin theta - m_(2) g = 0`
`rArr N_(2) = T sin theta + m_(2)g` ...(iv)
(ii) + (iv) `rArr N_(2) = ( m_(1) + m_(2)) g`
substituting `N_(2)` from (v) in (iii) we obtain
`T cos theta = mu (m_(1) + m_(2)g` ....(vi)
(ii) + (vi) `tan theta = (m_(1))/(mu(m_(1) + m_(2))) rArr theta = cot^(-1) [mu (1 + (m_(1))/(m_(2)) )]`