A cubical block rests on an inclined plane of coefficient of friction mu =(1)/( sqrt(3)) . What should be the angle of inclination so that the block just slides down the inclined plane?

A cubical block rests on an inclined plane of coefficient of friction

1.	A cubical block rests on an inclined plane of coefficient of friction mu =(1)/( sqrt(3)) . What should be the angle of inclination so that the block just slides down the inclined plane?
Answer» `30^@` `60^@` `45^@` `90^@` SOLUTION :`As TANTHETA = mu` ` theta= (1)/( sqrt(3))`or ` theta= tan^(-1)((1)/( sqrt(3))) = 30^@`