A block of mass m placed on a rough inclined plane of inclination theta = 30^@ can just be prevented from sliding down by applying a force Fi up the plane and it can be just made to slide up the plane by applying a force F_2 up the plane. If the coefficient of friction between the block and the inclined plane is (1)/( 2 sqrt(2sqrt(3)) the relation between F_1and F_2 is

A block of mass m placed on a rough inclined plane of inclination thet

1.	A block of mass m placed on a rough inclined plane of inclination theta = 30^@ can just be prevented from sliding down by applying a force Fi up the plane and it can be just made to slide up the plane by applying a force F_2 up the plane. If the coefficient of friction between the block and the inclined plane is (1)/( 2 sqrt(2sqrt(3)) the relation between F_1and F_2 is
Answer» `F_2 =4F_1` `f_2 - 3F_1` `F_2-2F_1` `F_2 -F_1` Solution :The FORCE`F_1`requiedto preventto blockfromslidingdownis `F_1= mgsin theta- mumg cos theta ` The force` F_2`requiedto maketheblock moveupthe planeis ` F_2 = mgsintheta+ mu MG cos theta` adding(i) and (II) , we get ` F_2 -F_1= 2 mu mgcostheta` ` therefore(F_2+F_1)/(F_2 -F_1) = ( 2 mgsin theta)/( 2 mumgcos theta) = ( tan theta)/( mu) = (tan 30^@)/(1 // 2 sqrt(3)) =2` ` orF_2 + F_1 = 2F_2 - 2F_1 ` ` or 3F_1= F_2orF_2=3F_1`