`sumF_(x)=-n_(w)=0` …….i
`sumF_(Y)=n_(g)-m_(1)g-m_(2)g=0`……….ii
`sum tau_(A)=i-m_(1)g(L/2)costheta=m_(2)gxcostheta=n_(2)Lsintheta=0`
From the torque equation `n_(2)=[1/2m_(1)g+(x/L)m_(2)g]cottheta`
Then from eqn i
`f=n_(w)=[1/2m_(g)+(x/L)m_(2)g]cottheta`
from eqn ii
`mu=(f_(x=d))/(n_(g))=(((m_(1))/2+(m_(2)d)/L)cottheta)/(m_(1)+m_(2))`