if the minimum force required to drag a heavy block is n times its own magnitude when the force is applied in an inclination of coefficient of friction with the horizon.find coefficient of friction.

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