hi,

in the situation u mention in ur qn ,

f up   nmg sin(a)               tan(a)=mu (given in qn)

f down   mg
  
f towards rt  nmg cos(a)
 f towards lft   mu*mg(1-nsin(a))   {total dwnward f*mu}

if this is the min. then,  f  lft = f rt =>  mu*mg(1-nsin(a))= nmg cos(a)

                                            => mu = ncos(a)/{1-nsin(a)}



                                                  put sin(a)= mu/sqr root{1-mu^2}
                                          
                                                   &  cos(a)= 1/sqr root{1-mu^2
                                           then cross multiply u wud get

                                          u^2{1+u^2}=n^2*{(1+u^2)}^2
                                      =>   u^2 =n^2* {1+u^2}
                                      =>    1/n^2 = 1/u^2 +1
                                       =>   u = n/sqr root{1-n^2}.
                 








						

							
I have solved the answer in one of your previous posts..plzz follow through this link:
http://askiitians.com/forums/Mechanics/10/3639/friction.htm