A uniform rectangular block of height h whose base is a square of side a rests on a rough horizontal plane. The plane is gradually tilted about a line parallel to two edges of the base then, (where   is the coefficient of friction)
 
 

Let rectangle OPQR be the vertical section of the block through its centre of gravity G, where
OP = a and PQ = h
When the block is just on the point of toppling over, let  be the inclination of the plane to the horizontal and in this case the vertical line through G must pass through O. from G draw GL perpendicular to OP. 
Then in $\Delta OGL$ , we have
  $\tan \alpha =\frac{OL}{GL}=\frac{\frac{1}{2}a}{\frac{1}{2}h}=\frac{a}{h}\mathrm{................}\left(i\right)$
When the block is on the point of sliding, the inclination of the plane to the horizontal must
 be $\lambda$ , given by 
  $\mu =\tan \lambda$
  The block will slide or topple over according as  
i.e.,   $\tan \lambda <or>\tan \alpha$
i.e.,  $\mu <or>a/h$
i.e.,  $\mu h<or>a$
   $a>or<\mu h$